Quantum Mechanics (QM) from Special Relativity (SR)
The RoadMap to the SRQM Interpretation of Quantum Mechanics

A physical derivation of Quantum Mechanics (QM) using only the assumptions of Special Relativity (SR) as a starting point...
Quantum Mechanics is not only compatible with Special Relativity, QM is derivable from SR!
This is the SRQM Interpretation of Quantum Mechanics, a thesis by John B. Wilson.

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Quantum Mechanics is derivable from Special Relativity
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Site last modified: 2015-Nov


The following is a derivation of Quantum Mechanics (QM) from Special Relativity (SR).
It basically highlights the few extra physical assumptions necessary to generate QM given SR as the base assumption.
The Axioms of QM are not required, they emerge instead as Principles of QM based on SR derivations.
This is the SRQM Interpretation of Quantum Mechanics, a thesis by John B. Wilson.

See also presentation as PDF: SRQM.pdf
or as OpenOffice Presentation SRQM.odp
There is also an alternate discussion at SRQM.html
Also, see 4-Vectors & Lorentz Scalars Reference for lots more info on Four-Vectors (4-Vectors) in general

Where do we start?
I will begin with some basic assumptions and background stuff on Special Relativity (SR).
I will introduce the SR Physical 4-Vectors.
I will show how these 4-Vectors are related to one another in purely relativistic theory.
Then we leap...
I will show a couple of relations that have previously been considered quantum assumptions,
  but which are in fact fundamentally no different from the other SR relations.
I will derive a relativistic quantum wave equation based just on these few SR relations.
I will derive the principles of Quantum Mechanics (QM), assuming no axioms of QM.

Basics

There are a couple of paradigm assumptions that we need to get clear from the start.
Relativistic stuff is *IS NOT* the generalization of Classical stuff.
Classical stuff *IS* the low-velocity limiting-case approximation of Relativistic stuff.
This includes Classical Mechanics, Classical EM, and Classical QM (meaning the non-relativistic Schrödinger QM Equation).

Also, we will not be considering the commonly used Newtonian limits via {c-->Infinity} or {ћ-->0}.
Neither of these is a valid assumption, for the following reasons:
(1:)
Both (c) and (ћ) are unchanging physical constants and Lorentz Invariants. Taking a limit where these change is unphysical.
(2:)
Consider E = ћω.  If ћ-->0 then E-->0.  Then Classical EM light rays/waves have zero energy.
Consider E = pc.  If c-->Infinty then E-->Infinity.  Then Classical EM light rays/waves have infinite energy.
Obviously neither of these is true in the Newtonian limit.
The correct way to take the limits is:
The low-velocity non-relativistic limit {|v|<<c}, which is a physically occuring situation.
The Hamilton-Jacobi non-quantum limit {ћ|∇·p|<<(p·p)}, which is a physically occuring situation.


Given that, assume that Einstein's General Relativity (GR) is essentially correct and our starting point.

Consider the [Low Mass = {Curvature ~ 0}] limiting case.

This gives Special Relativity < = > Minkowski Spacetime, which has the following properties:
===================================================================
The Principle of Relativity: The requirement that the equations describing the Laws of Physics have the same form in all admissible Frames of Reference
In other words, they have the same form for all Inertial Observers
Mathematically this is Invariant Interval Measure ΔR·ΔR = (cΔt)2-Δr·Δr = (cΔt)2-| Δr| 2 = (cΔτ)2 = Invariant
This is known as:
Lorentz Invariance (Covariance) (for rotations and boosts {Aμ' = Λμ'ν Aν})
Poincaré Invariance (Covariance) (for rotations, boosts, and translations {Aμ' = Λμ'ν Aν + ΔAν})
"Flat" Spacetime = Minkowski Metric ημν = ημν = DiagnolMatrix[+1,-1,-1,-1], with ηαν ηνβ = δαβ
which uses my preferred Metric Sign Convention ,see 4-Vectors & Lorentz Scalars Reference for reasoning behind this
Elements of Minkowski Spacetime are Events (a time, a spatial location)
These elements are represented by 4-Vectors , which are actually Tensors {technically (1,0)-Tensors}
4-Vector notation: A =     ...   = (a0,a) = (a0,a1,a2,a3) = > (at,ax,ay,az)
Tensor notation:    Aμ = (aμ) = (a0,ai) = (a0,a1,a2,a3) = > (at,ax,ay,az)
4-Vectors can be used to describe Physical Laws
Scalar Products of 4-Vectors give Invariant Lorentz Scalars {(0,0)-Tensors} ex. A·B = Aμ ημν Bν = A'·B'
The Isometry group (the set of all "distance-preserving" maps) of Minkowski Spacetime is the Poincaré Group
Poincaré Group Symmetry: (a non-Abelian Lie Group with 10 Generators)
  = {1 time translation P0 + 3 space translations Pi + 3 rotations Ji + 3 boosts Ki}
SL(2,C) x R1,3 is the Symmetry group of Minkowski Spacetime: i.e. the double cover of the Poincaré Group, as this includes particle symmetries
Poincaré Algebra has 2 Casimir Invariants = Operators that commute with all of the Poincaré Generators
These are { PμPμ = (m)2, WμWμ = -(m)2j(j+1) }, with Wμ = (-1/2)εμνρσJνρPσ is the Pauli-Lubanski pseudovector
Casimir Invariant Eigenvalues  = { mass m , spin j }, hence mass *and* spin are purely SR phenomena, no QM axioms required
This Representation of the Poincaré Group is known as Wigner's Classification in Particle_Physics_and_Representation_Theory
Speed of Light c = Invariant Lorentz Scalar = constant
Infinitesimal Invariant Interval Measure dR·dR = (cdτ)2 = (cdt)2-dr·dr = (cdt)2-| dr| 2
see also The Wightman Axioms
===================================================================

Light Cone

       | time-like interval(+)

       |             / light-like interval(0)
worldline
       |         c           --- space-like interval(-)
\  future /
  \    |     /
    \  |  /
      \| /now
      /| \
    /  |   \        elsewhere
  /    |     \
/   past   \
       |        -c

SR 4-Vectors - Conventions and Properties


*Note* Numeric subscripts and superscripts on variables inside the vector parentheses typically represent tensor indices, not exponents
In the following, I use the Time-0th-Positive (t0+) SR metric sign convention ημν = ημν = DiagnolMatrix[+1,-1,-1,-1] in Cartesian coords.
I use this convention primarily because it Starts on a Positive note :)
Actually, it reduces the number of minus signs in Lorentz Scalar Magnitudes,
since there seem to be many more time-like physical 4-vectors than space-like.
More importantly, this sign convention is the one matched by the QM Schrödinger Relations later on...
Alternate conventions are (t0-), (t4+), and (t4-).  Always check which convention is being used when reading other SR and GR texts.

I always choose to have the 4-Vector refer to the upper index tensor of the same name. {eg. A = Aμ}
In addition, I like the convention of having the (c) factor in the temporal part for correct dimensional units. {eg. 4-Position R = (ct,r)}
This allows the SR 4-Vector name to match the classical 3-vector name, which is useful when considering Newtonian limiting cases.
I will use UPPER case bold for non-indexed 4-vectors (A = Aμ), and lower case bold for non-indexed 3-vectors (a = ai).

All SR 4-Vectors have the following properties:
==================================

A = Aμ = (a0,ai) = (a0, a) = (a0,a1,a2,a3)     = > (at,ax,ay,az): A typical 4-vector

       Aμ = (a0,ai) = (a0,-a) = (a0 ,a1, a2, a3)  = > (at, ax, ay, az)
                           = (a0,-a) = (a0,-a1,-a2,-a3) = > (at,-ax,-ay,-az): A typical 4-covector

with Aμ = ημνAν  and  Aμ = ημνAν: Tensor index lowering and raising with the Minkowski Metric

The main idea that makes a generic 4-Vector into an SR 4-Vector is that it must transform properly according to a Lorentz Transformation.
(Aμ' = Λμ'ν Aν) with Λμ'ν as the Lorentz Transformation tensor.  {β = v/c = βx + βy + βz} and {γ = 1/√[1-β2]}

Typical Lorentz Transformation, for a frame shift in the -direction:
Λμ'ν = {for -boost}

γ

xγ

0

0

xγ

γ

0

0

0

0

1

0

0

0

0

1

So (at, ax, ay, az)' = (γat - γβxax ,-γβxat + γax, ay, az){for -boost Lorentz Transform}

An Inverse Lorentz Transformation is equivalent to just reversing the direction of the boost, i.e. change the signs on the β.
So (at, ax, ay, az)' = (γat + γβxax , γβxat + γax, ay, az){for -boost Inverse Lorentz Transform}


General Lorentz Transformation, for a frame shift in any direction :
Λμ'ν = {for -boost}

γ

xγ

yγ

zγ

xγ

1+(γ-1)(βx/β)2

( γ-1)(βxβy)/(β)2

( γ-1)(βxβz)/(β)2

yγ

( γ-1)(βyβx)/(β)2

1+( γ-1)(βy/β)2

( γ-1)(βyβz)/(β)2

zγ

( γ-1)(βzβx)/(β)2

( γ-1)(βzβy)/(β)2

1+( γ-1)(βz/β)2


An SR 4-Vector will still represent the same physical object after a Lorentz Transformation is applied.


A·B = Aμ ημν Bν = Aν Bν = Aμ Bμ = +a0b0-a·b = +a0b0-a1b1-a2b2-a3b3 The Scalar Product relation, used to make Invariant Lorentz Scalars
If the scalar product is between tensors with multiple indices, then one should use tensor indices for clarity, otherwise the equation remains ambiguous.
{eg. U·Fμν = Uα·Fμν = ? = > UαηαμFμν =   UμFμν  or  UαηανFμν =   UνFμν }
Importantly, A·B = (a0ob0o) = Ao·Bo and  A·A = (a0o)2 = Ao·Ao, the Lorentz Scalar Product can quite often be set to the "rest values" of the temporal component.
This occurs when the 4-Vector A is Lorentz-Boosted (Aμ' = Λμ'ν Aν) to a frame in which the spatial component is zero:  A = (a0, a) == > Ao = (a0o, 0)
[A·B > 0] --> Time-Like
[A·B = 0] --> Light-Like / Photonic / Null
[A·B < 0] --> Space-Like


The Invariant Rest Value of the Temporal Component Rule:
==========================================
β = v/c = u/c
4-UnitTemporal T = γ(1,β) = U/c
4-Velocity U = γ(c,u) = cT
Generic 4-Vector A = (a0,a)
A·T = (a0, a)·γ(1,β) = γ(a0*1 - a·β) = γ(a0 - a·β) = (1)(a0o - a·0) = a0o
A·T = a0o
The Lorentz Scalar product of any 4-Vector with the 4-UnitTemporal gives the Invariant Rest Value of the Temporal Component.
This makes sense from a vector viewpoint - you are taking the projection of the generic vector along a unit-length vector in the time direction.
A·U = c*a0o
The Lorentz Scalar product of any 4-Vector with the 4-Velocity gives c*Invariant Rest Value of the Temporal Component.
It's the same thing, just multiplied by (c).
I will call these ( A·T = a0o or A·U = c*a0o ) the "Invariant Rest Value of the Temporal Component Rule".
This will get used extensively later on...
There is an analogous relation with the 4-UnitSpatial.


The Scalar Product-Gradient-Position Relation:
=================================
4-Position R = (ct,r)
4-Gradient = X = (t/c,-)
Generic 4-Vector A = (a0,a), which is not a function of X

A·X = (a0,a)·(ct,x) = (a0*ct - a·x) = Θ is equivalent to [Θ] = [A·X] = A

{ A·X = Θ } <==> { [Θ] = A }

Proof:
Let A·X = Θ
[Θ] = [A·X] = [A]·X + A·∂[X] = (0) + Aκ·∂μ[Xν] = Aκ·ημν = Aκηκμημν = Aμημν = Aν = A

Let [Θ] = A
A·X = (a0,a)·(ct,x) = (a0*ct - a·x) = (t[Θ]/c*ct + [Θ]·x) = (t[Θ]*t + [Θ]·x) = Θ

*Note*
f = f(t,x) ==> df = (∂tf) dt + (∂xf) dx
f = ∫df = ∫(∂tf) dt + ∫(∂xf) dx
f ==> (∂tf) ∫dt + (∂xf) ∫dx = ∂tf *t + ∂xf *x   {if the partials are constants wrt. t and x, which was the condition from A not a function of X}
This comes up in the SR Phase and SR Analytic Mechanics.


Basis Representation & Independence (Manifest Covariance):
===========================================
When the components of the 4-vector { A }are in (time scalar,space 3-vector) form { (a0,a) }then the 4-vector is in spatial basis invariant form.
Once you specify the spatial components individually, you have picked a basis or representation.  I indicate this by using { = > }.
e.g. 4-Position X = (ct,x) = {Space Basis independent representation}
= > (ct,x,y,z) = {Cartesian/rectangular representation}
= > (ct,r,θ,z) = {Cylindrical representation}
= > (ct,r,θ,φ) = {Spherical representation}
These can all indicate the same 4-Vector, but the components of the 4-vector will vary in the different bases.

Likewise, the Metric changes a bit in different bases:
Cartesian: ημν = Diag[+1,-1,-1,-1]; ημν = Diag[+1,-1,-1,-1]; Det[ημν] = -1
Cylindrical: ημν = Diag[+1,-1,-r2,-1]; ημν = Diag[+1,-1,-1/r2,-1]
Spherical: ημν = Diag[+1,-1,-r2,-r2sin(θ)]; ημν = Diag[+1,-1,-1/r2,-1/r2sin(θ)]
with ηαν ηνβ = δαβ
μμ) = 1/(ημμ)

Now, once you are in a space basis invariant form, e.g. X = (ct,x), you can still do a Lorentz boost and still have the same 4-Vector X.
It is only when using 4-Vectors directly, (eg. X·Y, X+Y), that you have full Spacetime Basis Independence.
Knowing this, we try to find as many relations as possible in 4-Vector and Tensor format, as these are applicable to all observers.

Standard Physical SR 4-Vectors

Consider the following Physical SR 4-Vectors:
=================================
4-Position R = (ct,r) = > (ct,r,θ,z); X = (ct,x) = > (ct,x,y,z)
4-Velocity U = γ(c,u)
4-Acceleration A = γ(cγ̇,γ̇u) = γ(cγ̇,γ̇u + γa) = (cγ4uu̇,γ4uu̇u + γ2a)
4-Differential dR = (cdt,dr); dX = (cdt,dx)
4-Displacement ΔR = (cΔt,Δr); ΔX = (cΔt,Δx)
4-Momentum P = (E/c,p) = (mc,p)
4-MomentumDensity G = (Eden/c,pden) = (ue/c,g)
4-Force F = γ(Ė/c,) = γ(Ė/c,f) = γ(ṁc,f)
4-ForcePure Fp = γ(f·u/c,f)
4-ForceEM FEM = γq( (u·E)/c , (E) + (u x B) )
4-ForceHeat Fh = γṁ(c,u) = γ2o(c,u)
4-ForceScalar Fs = k(∂t[Φ],-[Φ])
4-ForceDensity Fden = γ(Ėden/c,fden)
4-NumberFlux N = (cn,nu) = Σa [ ∫
δ4(X-Xa(τ))dxa/dτ ]
4-CurrentDensity J = (cρ,ρu) = (cρ,j)
4-VectorPotential A = (φ/c,a) = A[X] = A[(ct,x)]  = (φ[(ct,x)]/c, a[(ct,x)]), often used as AEM
4-PotentialMomentum Q = (U/c,q) = q(φ/c,a)
4-TotalMomentum PT = (ET,pT) = (H,pT) = (E/c+U/c,p+q) = (E/c+qφ/c,p+qa)
4-WaveVector K = (ω/c,k) = (ω/c,ω/vphase) = (ω/c,ωu/c2) = (ω/c)(1,β) = (1/cT,/λ)
4-Gradient = X = (t/c,-) = (t/c,-del) = > (t/c,-∂x,-∂y,-z) = (∂/c∂t,-∂/∂x,-∂/∂y,-∂/∂z)
4-Polarization Ε = (ε0,ε) = > (ε·β,ε), with (ε·β = 0) *Note* this can have complex coefficients, and comes from standard EM theory (non-QM)
I choose this particular set because each of these is considered a basic SR 4-Vector,
meaning that no QM axioms are required for the definition of the 4-Vector.


To this list I will add some Mathematical 4-Vectors:
====================================
4-Zero = Z = (0,0)
4-UnitTemporal T = γ(1,β)
4-UnitSpatial S = γ[β] (n̂·β,) with ( any direction)  or  S = (n̂·β,) with (n̂·β = 0)
4-Null N = k(1,)
These are handy for purely mathematical operations,
as each of these are physically dimensionless.


Several if not all of these 4-Vectors may found in the texts on Relativity and Special Relativity.
I particularly like the texts by physicist Wolfgang Rindler, noted for his work and contributions to the field of Relativity.
*Note* The 4-Polarization can be found in Rindler's "Introduction to Special Relativity" in Section 43-Electromagnetic Waves,
as well as some other texts on Classical Electrodynamics.

Standard Physical SR 4-Vector Properties

The 4-Position R = (ct,r); X = (ct,x) gives the location of an event in spacetime.
Typically this would be the location of an SR particle.
t is the temporal location and r the spatial location.
The Lorentz Scalar Product R·R = (ct)2-r·r = (ct)2-|r| 2
R·R = (cto)2 = (cτ)2 {for time-like separation (+)interval}
R·R = 0 {for light-like separation (0)interval = Null/Photonic}
R·R = -(ro)2  {for space-like separation (-)interval}
gives the invariant distance or interval or measure between events.
When the spatial part r is 0, this gives a Lorentz invariant rest time to = τ.
Relativistic time t = γto = γτ.
=====================


The 4-Velocity U = (γc,γu) = γ(c,u) gives the motion of an event in spacetime.
Typically this would be the motion of an SR particle.
γ is the SR gamma factor ( γ = 1/√[1-β2]; β = u/c ), c is the temporal velocity, u is the spatial velocity.
The Lorentz Scalar Product U·U = γ(c,u)·γ(c,u) = γ2(c,u)·(c,u) = γ2(c2 - u·u) = c2γ2(1 - β·β) = (c)2
says that all SR events have invariant 4-Velocity magnitude of (c) in the temporal direction.
When the spatial part u is 0, this gives a Lorentz invariant rest temporal velocity c.
see the "Invariant Rest Value of the Temporal Component Rule"
Interestingly, the 4-Velocity has only 3 independent components, unlike the other SR 4-Vectors here, which typically have 4 independent components.
The reason for this is the constraint imposed by the Lorentz Scalar Product.
Since U·U = (c)2 is a constant, this lowers the number of degrees of freedom by 1, leaving only 3 independent components.
This is actually quite handy, as multiplication by certain other Lorentz Scalars allows the result to have 4 independent components.
Thus, several 4-vectors are actually just scalar multiples of the 4-Velocity. (e.g P = moU and J = ρoU)
In a rest frame, U = > Uo = (c,0)
If we then apply an Inverse Lorentz Transform: Uβ' = Λβ'αUα = γ[[1 β],[β 1]](c,0) = γ([1*c + β·0],[ β*c +1*0]) = γ([c + 0],[ u + 0]) = γ(c,u) = U
=====================


The 4-Acceleration A = γ(cγ̇,γ̇u) = γ(cγ̇,γ̇u + γa) = (cγ4uu̇,γ4uu̇u + γ2a) gives the acceleration of an event in spacetime.
Typically this would be the acceleration of an SR particle.
γ is the SR gamma factor ( γ = 1/√[1-β2]; γ̇ = dγ/dt; β = u/c ), c is the temporal velocity, u is the spatial velocity, a = u̇ is the spatial acceleration.
** Take care not to confuse with the 4-VectorPotential A **
The Lorentz Scalar Product A·A = γ(cγ̇,γ̇u + γ)·γ(cγ̇,γ̇u + γ) = γ2[( cγ̇ )2 - (γ̇u + γ)2] = -(ao)2 = -(α)2
In an instantaneously co-moving reference frame, we get Ao = (0,) = (0,ao) = (0,α)
The Lorentz Scalar Product Ao·Ao = (0,α)·(0,α) = (02 - α·α) = -(α)2
says that the 4-Acceleration is always spatial.
So, events move temporally along their worldlines, with accelerations acting normally to bend/curve the worldlines spatially.
=====================


The 4-Differential dR = (cdt,dr); dX = (cdt,dx) gives the differential form of the 4-Position.
dt is the temporal differential and dr the spatial differential.
The Lorentz Scalar Product dX·dX = (cdt)2-dx·dx = (cdt)2-| dx| 2
gives the invariant differential distance or interval or measure between infinitesimally close events.
When the spatial part dx is 0, this gives a Lorentz invariant rest time displacement dto = dτ.
Relativistic temporal displacement dt = γdto = γdτ.
Basically, the rules of differential calculus still apply to Minkowski Spacetime.
=====================


The 4-Displacement ΔR = (cΔt,Δr); ΔX = (cΔt,Δx) gives the displacement from one event to another in spacetime.
Δt is the temporal displacement and Δr the spatial displacement.
This is just the interval displacement between 4-Position's X1 and X2, with ΔX = X2 - X1
The Lorentz Scalar Product ΔX·ΔX = (cΔt)2-Δx·Δx = (cΔt)2-| Δx| 2
gives the invariant distance or interval or measure between these two events.
When the spatial part Δx is 0, this gives a Lorentz invariant rest time displacement Δto = Δτ.
Relativistic temporal displacement Δt = γΔto = γΔτ.
The 4-Displacement is invariant under both general rotations and translations, unlike the 4-Position, which is invariant only under rotations about the origin.
The reason for this is that the 4-Position is a 4-Displacement that has one of its events "pinned" to the origin (i.e. one of them is the 4-Zero).
=====================


The 4-Momentum P = (E/c,p) = (mc,p) gives the Energy-Momentum content of an SR particle at a particular event.
E is the energy ( = temporal momentum) of the event and p is the spatial momentum.
The Lorentz Scalar Product P·P = (E/c,p)·(E/c,p) = (E/c)2 - p·p = (Eo/c)2 = (moc)2
gives Einstein's famous Mass-Energy Equation, which is equivalent to [ E = mc2 = γmoc2 ].
When the spatial part p is 0, this gives a Lorentz invariant rest energy Eo.
Relativistic energy E = γEo = γmoc2
The Conservation of 4-Momentum unites the Conservation of Energy and Conservation of 3-Momentum into a single law,
and plays a major role in just about all physical effects.
=====================


The 4-Force F = γ(Ė/c,) = γ(Ė/c,f) = γ(ṁc,f) gives the Power-Force acting on an SR particle at a particular event.
γ is the SR gamma factor ( γ = 1/Sqrt[1-β2]; β = u/c ), Ė = dE/dt is the power ( = temporal force) and f = = dp/dt is the spatial force.
The Lorentz Scalar Product F·F = γ(Ė/c,f)·γ(Ė/c,f) = γ2[ (Ė/c)2 - f·f ]

If the force is applied to a particle of stable invariant rest mass mo, (i.e. d/dτ[mo] = 0 <==> F·U = 0),
then the 4-Force is considered to be Pure (space-like), and can be written as:
4-ForcePure Fp = γ(f·u/c,f), with Ė = f·u
The Lorentz Scalar Product Fp·Fp = γ(f·u/c,f)·γ(f·u/c,f) = γ2[ (f·u/c)2 - f·f ] = (- fo·fo) = -(fo)2
An example of a pure force is the EM field:
4-ForceEM FEM = γq( (u·E)/c , (E) + (u x B) ), the force on a charged particle due to an EM field.

If the force applied does not change the particle's 4-Velocity, ie. (A = 0),
then the force is considered to be Heat-like (time-like), and can be written as:
4-ForceHeat Fh = γṁ(c,u) = γ2o(c,u)
The Lorentz Scalar Product Fh·Fh = γṁ(c,u)·γṁ(c,u) = γ22(c2 - u·u) = ṁ2c2 = (ṁc)2 = (Ė/c)2 = (γĖo/c)2
The action-reaction forces that occur in collisions are heat-like.
Another example of an impure force is one derived from a scalar potential:
4-ForceScalar Fs = k(∂t[Φ],-[Φ]), which has U·Fs = k[Φ] = k(d/dτ)[Φ] = Ė = (d/dτ)[Eo]
=====================


The 4-NumberFlux N = (cn,nu) = Σa [ ∫ δ4(X-Xa(τ))dxa/dτ ] gives the Number density - Number flux content of an SR fluid.
n is the number density and nu is the number flux.
The Lorentz Scalar Product N·N = (nc,nu)·(nc,nu) = (nc)2 - (n)2u·u = (noc)2
gives the analogous NumberDensity-NumberFlux equation.
When the spatial part nu is 0, this gives a Lorentz invariant rest charge density no.
Relativistic charge density n = γno
The total count of "items" (N) in a given rest volume (Vo) is a scalar invariant.
N = noVo = γnoVo/γ = nV
The 4-NumberFlux is used in generating the Stress-Energy Tensor for SR Dust
Tdustμν = PμNν = moUμnoUν = monoUμUν = ρmoUμUν
Effectively, any "particle-type" 4-Vector can be transformed into a "density-type" 4-Vector by multiplying by no.
The bit with the integral is the 4D version of the Dirac Delta function.
f(Yμ) = ∫d4Xμ δ4(Xμ-Yμ) f(Xμ), where the Dirac Delta function is a Lorentz Invariant in Minkowski Space.
=====================


The 4-CurrentDensity J = (cρ,ρu) = (cρ,j) gives the EM charge-current density content of an SR EM particle, for example an electron.
ρ is the charge density and j is the current density.
The Lorentz Scalar Product J·J = (ρc,j)·(ρc,j) = (ρc)2 - j·j = (ρoc)2
gives the analogous ChargeDensity-CurrentDensity equation.
When the spatial part j is 0, this gives a Lorentz invariant rest charge density ρo.
Relativistic charge density ρ = γρo
An example of it's use:
(∂·∂)AEM = μoJ: The Non-Homogeneous Maxwell EM Equation (if ∂·AEM = 0 : Lorenz Gauge)
This can be used to derive classical EM.
=====================


The 4-VectorPotential A = (φ/c,a) = A[X] = A[(ct,x)]  = (φ[(ct,x)]/c, a[(ct,x)]),
gives the potential-vector potential content of an SR potential/field which is spread out over spacetime, so technically A = A[X]
φ is the potential and a is the vector potential.
** Take care not to confuse with the 4-Acceleration A **
It is used as the relativistically correct way to specify fields that are "the gradient of a potential".
The most common form is the 4-EM-VectorPotential AEM = (φ/c,a) = AEM[X] = AEM[(ct,x)]  = (φ[(ct,x)]/c, a[(ct,x)]),
The Lorentz Scalar Product A·A = (φ/c,a)·(φ/c,a) = (φ/c)2 - a·a = (φo/c)2
gives the analogous Potential-Vector Potential equation.
When the spatial part a is 0, this gives a Lorentz invariant rest scalar potential φo.
Relativistic scalar potential φ = γφo
Examples of it's use: is
(∂·∂)AEM = μoJ: The Non-Homogeneous Maxwell EM Equation (if ∂·AEM = 0 : Lorenz Gauge)
Fμ = dPμ/dτ = qUυ(∂μAEMν - ∂νAEMμ): The Lorentz Force Equation, the force imposed on a test charge by and EM field
These can be used to derive classical EM.
=====================


The 4-PotentialMomentum Q = (U/c,q) = (qφ/c,qa) gives the Potential Energy-Momentum content the SR field it specifies.
U is the potential energy ( = Potential temporal momentum) of the field and q is the Potential spatial momentum.
The Lorentz Scalar Product Q·Q = (U/c,q)·(U/c,q) = (U/c)2 - q·q = (Uo/c)2
When the spatial part q is 0, this gives a Lorentz invariant rest potential energy Uo.
Relativistic Potential Energy U = γUo
Q = qA
The main idea here is that external fields affecting a test particle contain both energy and momentum.
Both must be accounted for so that the correct dynamics is obtained.
Typically this with be a charged particle (q) interacting with a 4-VectorPotential A
=====================


The 4-TotalMomentum PT = (ET,pT) = (H,pT) = (E/c+qφ/c,p+qa) gives the Total Energy-Momentum content of an SR particle plus the SR field it is in.
H is the Hamiltonian Total energy ( = Total temporal momentum) of the event and pT is the Total spatial momentum.
The Lorentz Scalar Product PT·PT = (H/c,pT)·(H/c,pT) = (H/c)2 - pT·pT = (Ho/c)2
When the spatial part pT is 0, this gives a Lorentz invariant rest Hamiltonian Ho.
Relativistic Hamiltonian H = γHo
PT = P + qA
The 4-TotalMomentum of a system can be split into the 4-Momentum of a Particle + a charge (q)* a 4-VectorPotential.
Essentially, we are examining a lone particle running around in a big field which can affect the particle.
Both the particle and the field have 4-Momentum, and it is the sum of all 4-Momentuṁs that is conserved.
This is used to derive the Classical EM laws.
The Conservation of 4-Momentum unites the Conservation of Energy and Conservation of 3-Momentum into a single law,
and plays a major role in just about all physical effects.
S = -(PT·R) and [S] = - PT.
Getting into Relativistic Analytic Mechanics, we can define an Action (S) as the negative Lorentz Scalar product of the 4-TotalMomentum with the 4-Position,
or the 4-TotalMomentum is the negative 4-Gradient of the Action.
=====================


The 4-WaveVector K = (ω/c,k) = (ω/c,ω/vphase) = (ω/c,ωu/c2) = (ω/c)(1,β) = (1/cT,/λ) gives the description of an SR wave, for example, a photon.
ω is the temporal angular frequency and k is the spatial wave number, which is like an spatial angular frequency.
It can also be described using:
= the unit direction,  vphase = the wave (phase) velocity, u = vgroup = the group velocity, β = u/c = the Relativistic Beta Factor,
T = T/2π = reduced period, λ = λ/2π = reduced wavelength
The Lorentz Scalar Product K·K = (ω/c,k)·(ω/c,k) = (ω/c)2 - k·k = (ωo/c)2
gives the analogous AngularFrequency-WaveVector equation.
When the spatial part k is 0, this gives a Lorentz invariant rest angular frequency ωo.
Relativistic angular frequency ω = γωo
vphase = ω/k = (ω/kx,ω/ky,ω/kz)
vgroup = dω/dk = (∂ω/∂kx,∂ω/∂ky,∂ω/∂kz) = u = event velocity for SR waves
from K·K: ω2 = k2c2 + ωo2
ω2/k2 = c2 + ωo2/k2
vphase2 = c2 + ωo2/k2
vphase = √[c2 + ωo2/k2]
from K·K: ω2 = k2c2 + ωo2
ω = √[k2c2 + ωo2]
vgroup = dω/dk = (1/2)(1/√[k2 c2 + ωo2])*2kc2 = (1/ω)*kc2 =  kc2/ω =  c2/vphase = u = vevent

The 4-WaveVector K is used in the SR Doppler Equations, which give the Doppler red-shift, blue-shift, and transverse Doppler effects.
I need to emphasize here that the 4-WaveVector can exist as an entirely SR object (non-QM).
It can be derived in terms of periodic motion, for which families of surfaces move through space as time increases,
or alternately, as families of hypersurfaces in spacetime, formed by all events passed by the wave surface.
The 4-WaveVector is everywhere in the direction of propagation of the wave surfaces.
More specifically, the 4-WaveVector K is orthogonal to a hypersurface of constant phase Φ {an ensemble of events E(t,x,y,z) that all share the same value of Φ}
i.e. Kμ = -∂Φ/∂Xμ., or in 4-Vector notation: K = -[Φ] The 4-WaveVector is the negative 4-Gradient of the SR Phase.
The phase is chosen so that it decreases by unity as we pass from one wave crest to the next, in the direction of propagation of the waves.
The contravariant vector Kμ = dXμ/dλ is a tangent vector to the particle trajectory.
From this structure, one obtains relativistic/wave optics, without ever mentioning QM.
An SR wave can be written in the form ei(K·X).
Notice an interesting contrast:
4-Position       X = (   ct  , x ), with dimension of [length]
4-WaveVector K = (1/cT, / λ), with dimension of [length-1]
The 4-Position and 4-WaveVectors are inverses, which is why K·X = (ω/c,k)·(ct,x) = (ωt - k·x) = (t/T - n̂·x/λ) = -Φ gives a dimensionless phase.
=====================


The 4-Gradient = X = (t/c,-) = (t/c,-del) = > (t/c,-∂x,-∂y,-z) = (∂/c∂t,-∂/∂x,-∂/∂y,-∂/∂z) is an SR functional that gives the structure of Minkowski Spacetime.
t is the partial wrt. time and is the gradient, the partial vector wrt. spatial dimensions (e.g. {x,y,z} or {r,θ,z} or {r,θ,φ}, etc.).
The Lorentz Scalar Product ∂·∂ = (t/c,-)·(t/c,-) = (t/c)2 - ∇·∇ = (τ/c)2
gives the d'Alembertian equation (a wave equation).
When the spatial part is 0, this gives a Lorentz invariant rest partial to = τ., which would presumably be the partial as measured along a worldline.
The d'Alembert operator (∂·∂) is the Laplace operator of Minkowski Space.  Despite being a functional, the d'Alembertian is still a Lorentz Scalar Invariant.
The Green's function G[X-X'] for the d'Alembertian is defined as (∂·∂)G[X-X'] = δ4[X-X'], with (δ4) as the 4-D Dirac Delta
=====================


The 4-Polarization Ε = (ε0,ε) = > (ε·β,ε), with (ε·β = 0) as a complex-valued 4-vector which can describe the polarization states of light rays.
ε0 is the temporal polarization and ε is the spatial polarization.
The Lorentz Scalar Product Ε·Ε = (ε0,ε)·(ε0,ε) = (ε0)2 - ε·ε ==> -1.  The 4-Polarization is normalized to have a spatial magnitude of -1.
Since the 4-Polarization is orthogonal to the 4-WaveVector, (Ε·K = 0),
0,ε)·(ω/c)(1,β) = (ω/c)(ε0*1 - ε·β) = 0, therefore (ε0 = ε·β),
one can write it as: 4-Polarization Ε = (ε·β,ε).
Normally this would give the 4-Polarization 3 independent components, but the spatial magnitude places an interesting restriction.
Since the 4-vector can have complex components, we make a slight modification:
Ε*·Ε = (ε·β,ε)*·(ε·β,ε)= (ε·β)* (ε·β) - ε*·ε = (ε·β)2 - 1 = -1.
Therefore, (ε·β) = 0
For a massive particle, there is always a rest frame with (β = 0), so there is no further restriction and there are 3 independent polarizations.
For a massless particle, there is never a rest frame, (β = ), so (ε·) = 0 is an additional restriction on the degrees of freedom,
which limits photonic particles/light-rays to 2 independent polarizations, both of which are orthogonal to direction of motion ().
Note that both (Ε*·Ε = -1) and (Ε·K = 0) are Lorentz Invariant equations, and thus true for all observers.
More on this in the 4-UnitSpatial...
=====================


And now the mathematical 4-Vectors...



The 4-Zero Z = (0,0) is a special 4-Vector that is the same for all observers.
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The 4-UnitTemporal T = γ(1,β) gives the dimensionless measure of an event in spacetime.
γ is the SR gamma factor ( γ = 1/√[1-β2]; β = u/c ), 1 is the dimensionless temporal measure, and β is the dimensionless spatial measure, with {β = > 0..1}.
The Lorentz Scalar Product T·T = γ(1,β)·γ(1,β) = γ2(1,β)·(1,β) = γ2(12 - β·β) = γ2(1 - β·β) = (1)
says that all SR events have invariant 4-UnitTemporal magnitude of 1 in the temporal direction, hence the name 4-UnitTemporal.
When the spatial part β is 0, this gives a Lorentz invariant rest temporal measure 1.
In a rest frame, T = > To = (1,0), thus it has 0 independent components.
4-UnitTemporal T = γ(1,β), and has 3 independent components, with 3 degrees of freedom from the motion β.
Generic 4-Vector A = (a0,a)
A·T = (a0, a)·γ(1,β) = γ(a0*1 - a·β) = γ(a0 - a·β) = > (1)(a0o - a·0) = a0o
A·T = a0o
The Lorentz Scalar product of any 4-Vector with the 4-UnitTemporal gives the Invariant Rest Value of the Temporal Component.
This makes sense from a vector viewpoint - you are taking the projection of the generic vector along a unit-length vector in the time direction.
The 4-UnitTemporal has only 3 independent components, given by the 3 degrees of freedom from the motion β.

Part of the reason for introducing this 4-Vector is to appease those physicists who insist on setting (c-->dimensionless 1).
My own opinion is that setting (c-->1) is a poor decision, as c is a fundamental constant with dimensional units of [velocity].
Instead, one can simply divide the 4-Velocity by (c) to get a dimensionless 4-Vector T = U/c,
which accomplishes much of the same math simplification without losing the value of dimensional analysis.
=====================


The 4-UnitSpatial S = γ[β] (n̂·β,) with ( any direction)  or  S = (n̂·β,) with (n̂·β = 0) gives the dimensionless measure of an event in spacetime.
γ is the SR gamma factor ( γ = 1/√[1-β2]; β = u/c ), n̂·β is the dimensionless temporal measure, and is the dimensionless spatial directional measure, with {β = > 0..1}.
The Lorentz Scalar Product S·S = γ[β](n̂·β,)·γ[β](n̂·β,) = γ[β]2(n̂·β,)·(n̂·β,) = γ[β]2((n̂·β)2 - n̂·n̂) = γ[β]2((n̂·β)2 -1 ) = (-1)γ[β]2( 1 - (n̂·β)2 ) = (-1)
says that all SR events have invariant 4-UnitSpatial magnitude of 1 in the spatial direction, hence the name 4-UnitSpatial.
When the spatial motion part β is 0, this gives a Lorentz invariant rest spatial measure -1.
In a rest frame, S = > So = (0,), thus it has 2 independent components, with 2 degrees of freedom due to .
4-UnitSpatial S = γ[β](n̂·β,), and has 3 independent components, with 2 degrees of freedom from , and 1 from the direction of motion β.
Why only 1 and not 3 from the β?  Because the components of β orthogonal to () play no role in the 4-Vector.
Generic 4-Vector A = (a0,a)
A·S = (a0, a)· γ[β](n̂·β,) = γ[β](a0*n̂·β - a·n̂) = > (1)(a0*0 - a·n̂) = - a·n̂
A·S = - a·n̂
If we align S in the direction of A, then A·SA = - a
The Lorentz Scalar product of any 4-Vector with the aligned 4-UnitSpatial gives the Invariant Rest Value of the Spatial Component.
This makes sense from a vector viewpoint - you are taking the projection of the generic vector along a unit-length vector in the same spatial direction.

This 4-Vector goes along with the 4-UnitTemporal, and is orthogonal to it.
T·S = γ(1,β)·γ[β](n̂·β,) = γ[β]γ[β](1,β)·(n̂·β,) = γ[β]γ[β](n̂·β - n̂·β) = 0

Let me back up for a moment:  There are actually 2 classes of  4-UnitSpatial.
Consider the following:
Start with the 4-UnitTemporal as given -> T = γ(1,β)
Let's construct the 4-UnitSpatial just based on the Lorentz Invariants.
Start with a generic 4-vector -> S = (s0,s)
We want the 4-UnitSpatial to be orthogonal to the 4-UnitTemporal.
T·S = 0 = γ(1,β)·(s0,s) = γ(1*s0 - β·s), Hence s0 = β·s
S = (β·s,s)
We also want the 4-UnitSpatial Magnitude to be a negative number (-k), since space is the negative interval in our preferred convention.
S·S = -k = (β·s,s)·(β·s,s) = [(β·s)2 - s·s]
We can write s = (s)
S·S = -k = [(β·s)2 - s·s] = (s)2[(β·)2 - ·] = -(s)2[· - (β·)2] = -(s)2[1 - (β·)2] = -k
Ok, here is where the choice comes in. By splitting s = (s), we have two ways to solve this.
======
(Class #1) Let s = ±√[k]γ[β] and = any direction
S·S = -(s)2[1 - (β·)2] = -(±√[k]γ[β])2[1 - (β·)2] = -(±√[k])2(γ[β])2[1 - (β·)2] =-(±√[k])2 =-(√[k])2 = -k
S·S = -k
S = (β·s,s) with s = ±√[k]γ[β] and = any direction
For (k = 1) we get 4-UnitSpatial S = γ[β](n̂·β,)
======
(Class #2) Let s = ±√[k] and = direction restricted by (β·) = 0
S·S = -(s)2[1 - (β·)2] =-(±√[k])2[1 - (0)] =-(±√[k])2[1] =-(±√[k])2 =-(√[k])2 = -k
S·S = -k
S = (β·s,s) with s = ±√[k] and = direction restricted by (β·) = 0
For (k = 1) we get 4-UnitSpatial S = (·β,), with(·β) = 0

This is actually an interesting exception to the 4-Vector Zero Component Lemma
S' = ΛS = γ'(1*(β·s)+β'·s,β'(β·s) + s) = γ'(β'·s,s) = (β'·s',s'), with s' = γ's
The Lorentz-boosted 4-Vector is in the same format, so we get 4-UnitSpatial S' = (sβ',s'), with(s'·β') = 0,
which is as it should be since S·S = -1 is a Lorentz Invariant equation.
Even though the temporal component is always 0, the overall 4-Vector is not the 4-Zero.
The exception is due to the scalar product. There are 3 ways it can equal 0 = (a·b) = |a| |b| cos[θ].
The Lemma holds if either |a| = 0 or |b| = 0.  The exception occurs when (a) is orthogonal to (b), when (θ = π/2)

Now, to another interesting side effect:
4-UnitSpatial S = (s·β,s), with (s·β) = 0
For a massive particle, there is always a rest frame with (β = 0), so (s) has 3 degrees of freedom.
For a massless particle, there is never a rest frame, (β = ), so (s) has only 2 degrees of freedom due to extra restriction(s·) = 0,
  with the spatial component (s) orthogonal to the direction of motion ().
This exactly how the 4-Polarization works: 4-Polarization Ε = (ε0,ε) = > (ε·β,ε), with (ε·β = 0)
======
For { k = 1} we get 4-Polarization Ε = (ε·β,ε), with (ε·β) = 0
For { k = (Eo/c)2 } we get (Eo/c)S = (Eo/c)(β·s,s) = ((Eo/c)β·s,(Eo/c)s) = (p·s,(Eo/c)s) ==> The 4-Pauli-Lubanski-Vector W
Remember that these were generated by T·S = 0 = U·S = 0.
======

Hmm.... now does this mean there are 2 classes for the 4-UnitTemporal?
Start with the 4-UnitSpatial as given -> S = γ[β](n̂·β,)
Let's construct the 4-UnitTemporal just based on the Lorentz Invariants.
Start with a generic 4-vector -> T = (t0,t)
We want the 4-UnitTemporal to be orthogonal to the 4-UnitSpatial.
S·T = 0 = γ[β](n̂·β,)·(t0,t) = γ[β]((n̂·β)*t0 - n̂·t), Hence t = t0β for any direction .
T = (t0,t0β) = t0(1,β), and we still have 4 independent components.
We also want the 4-UnitTemporal Magnitude to be a positive number (k), since time is the positive interval in our preferred convention.
T·T = k = t0(1,β)·t0(1,β) = (t0)2[(1)2 - β·β] = k
There is really only one choice.
Choose t0 = ±√[k]γ[β], then T =  ±√[k]γ[β](1,β), and with k = 1 we get T = γ[β](1,β) = γ(1,β), with 3 degrees of freedom, which is what we expect.
If we had chosen t0 = ±√[k] with the condition (β·β = 0), then you end up with the static 4-Vector (1,0), which has no degrees of freedom.
=====================


The 4-Null N = k(1,) is the limit of either the 4-UnitTemporal or the aligned 4-UnitSpatial as ( β->1, γ->Infinity ).  It represents photonic/light-like 4-Vectors.
The factor of (k) is somewhat arbitrary, it can take any value and still represent the same light-ray.
This is due to fact that a Lorentz-Boosted photon is still the same photon, but with an altered value of (k) and ().
The Lorentz Scalar Product N·N = k(1,)·k(1,) = k2(1,)·(1,) = k2(12 - n̂·n̂) = k2(1 - n̂·n̂) = (0)
says that all SR events have invariant 4-UnitSpatial magnitude of 0, hence the name Null.
The 4-Null has only 3 degrees of freedom, 1 from the (k), and 2 from the direction of motion ()
=====================

Let me emphasize that all these 4-Vectors are purely from SR - no Quantum Axioms are required.

Standard Physical SR 4-Vector Relations


These SR 4-Vectors can be linked to one another by Lorentz Scalar Invariants.
These Invariants are fundamental constants which can be empirically measured by physical experiment.

As already noted, time and space are linked via the Lorentz Scalar c = √[U·U] = U·T.
The invariant constant (c) may be measured using a light pulse generator/receiver, a mirror, a meter-stick, and a timer.
Set the light source at A, the mirror at B.
Measure the distance AB with the meter-stick.
Fire a light pulse from (A) to the mirror (B) while starting the timer.
Stop the timer when the returning pulse is detected at (A).
c = Distance[A->B->A]/Time[A->B->A].

Note that this factor (c) appears in all Physical SR 4-Vectors, as it links the temporal components to the spatial components,
giving each component the correct physical dimensional units.

R = (ct,r)
U = γ(c,u)
A = γ(cγ̇,γ̇ua)
P = (E/c,p) = (mc,p)
F = γ(Ė/c,f) = γ(ṁc,f)
N = n(c,u)
J = (cρ,j)
A = (φ/c,a)
Q = (U/c,q)
PT = (H/c,pT)
K = (ω/c,k) = (1/cT,/λ)
= (t/c,-)

For some of the dimensionless 4-Vectors, the factor of (c) is hidden in the Beta factor (β = u/c) so that the overall 4-vector is dimensionless.
Ε = (ε0,ε) = > (ε·β,ε), with (ε·β = 0)
T = γ(1,β)
S = γ[β] (n̂·β,) with ( any direction)  or  S = (n̂·β,) with (n̂·β = 0)

========================================
So, to reiterate the invariant scalar products of individual 4-Vectors...
R·R = (cτ)2 or (0) or -(ro2), depending on event interval
U·U = (c)2
A·A = -(ao)2
dR·dR = (cdτ)2
ΔR·ΔR = (cΔτ)2
P·P = (moc)2 = (Eo/c)2
F·F = γ2[ (Ė/c)2 - f·f ]
Fp·Fp = -(fo)2
Fh·Fh = (γĖo/c)2
N·N = (noc)2
J·J = (ρoc)2
A·A = (φo/c)2
Q·Q = (Uo/c)2
PT·PT = (Ho/c)2
K·K = (ωo/c)2 = (1/cTo)2
∂·∂ = (τ/c)2

T·T = (1) Unit Temporal
T·S=(0) = N·N --> They are orthogonal
S·S = (-1) Unit Spatial
========================================

Scalar Products and other SR 4-Vector Relations

Now consider the links between different 4-Vectors.
Some of these are simply mathematical consequences of the 4-Vector definitions, others are found by empirical experiment.
Use of the Lorentz Scalar Product provides many fundamental relations.
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∂·X = 4
∂·X = (∂t/c,-)·(ct,x) = (∂t/c[ct] - (-∇·x)) = (∂t[t] + (∇·x)) = (∂t[t]) + (∂x[x]+∂y[y]+∂z[z]) = (1)+(3) = 4
The divergence or dimensionality of Spacetime is 4 (i.e., that's why they are called 4-Vectors).
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[X] = ημν = ( )·( ) = The Scalar Product Dot
[X] = ∂μXν = (∂t/c,-)(ct,x) =

[(∂t/c)ct,(∂t/c)x,(∂t/c)y,(∂t/c)z]
[-∂xct,     -∂xx,   - ∂xy,    -∂xz ]
[-∂yct,     -∂yx,    -∂yy,    -∂yz ]
[-∂zct,    - ∂zx,    -∂zy,    -∂zz  ] =

 [[ ∂t/[t] , -[x] ]] =

[∂tt,   0  ,  0   ,  0    ]
[0  ,-∂xx,  0   ,  0    ]
[0  ,   0  ,-∂yy,  0     ]
[0  ,   0  ,  0   ,-∂zz  ] =

[1,  0, 0, 0]
[0,-1,  0, 0]
[0,  0,-1, 0]
[0,  0, 0,-1] =

= Diag[1,-1] = ημν
The 4-Gradient of the 4-Position gives the Minkowski Metric ημν.
This is a standard function of the Jacobian Matrix.
The Minkowski Metric ημν essentially is the "dot (·)" of the Scalar Product Relation.
A·B = Aμ ημν Bν = A'·B'
=====================


U·∂ = d/dτ = γ(d/dt)
U·∂ = γ(c,u)·(∂t/c,-) = γ(∂t+u·∇) = γ(d/dt) = d/dτ
The Scalar Product of the 4-Velocity with the 4-Gradient gives the Derivative wrt. Proper Time.
This is a purely mathematical consequence of the definition of 4-Velocity and 4-Gradient.
The function (d/dτ) is a Lorentz Scalar.
see the "Invariant Rest Value of the Temporal Component Rule"
=====================


τ = √[ R·R / U·U ] = √[(cτ)2/(c)2] = √[τ2] = τ
The Proper Time (τ) is a Lorentz Scalar
It is the Invariant Interval between events connected by a worldline that is at rest,
i.e. the temporal distance between event A and event A', for which the event doesn't move spatially.
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U·ΔX = c2Δτ
U·ΔX = γ(c,u)·(cΔt,Δx) = γ(c2Δt - u·Δx) = c2Δto = c2Δτ
This is a relation that can used to determine whether events (at X1 and X2 ) are simultaneous in a given reference frame.
If (U·ΔX  = 0 = c2Δτ ), then the Proper Time displacement is zero, and the event separation ( ΔX = X2 - X1 ) is orthogonal to the worldline at event U.
X1 and X2 are therefore simultaneous for the observer at this worldline event U.
Examining the equation we get γ(c2Δt - u·Δx) = 0.
The condition for simultaneity in an alternate frame (moving at 3-velocity u wrt. the worldline) is Δt = 0, which implies (u·Δx) = 0.
This can be met by:
( | u| = 0 ), the alternate observer is not moving wrt. the events, i.e. is at the same worldline event U.
( | Δx| = 0 ), the events are at the same spatial location (co-local).
( u·Δx = 0 ), the alternate observer's motion is perpendicular to the spatial separation Δx of the events in that frame.
Another case, for orthogonality, but not strict simultaneity, is (c2Δt = u·Δx )
If ( | u| = c ), the worldline for a photon, then | Δx/Δt| = c, the events must be separated by light paths.
========
Note that we can get the a similar result by taking the Lorentz Transform of the separation 4-vector:
ΔX = (cΔt,Δx)
ΔX' = (cΔt',Δx') = ΔXβ' = Λβ'αΔXα = γ[[1 -β],[-β 1]](cΔt,Δx) = γ([1*cΔt - β·Δx],[-β*cΔt + 1*Δx]) = γ(cΔt - β·Δx, -β*cΔt + Δx])
Taking the temporal component:
cΔt' = γ(cΔt - β·Δx)
c2Δt' = γ(c2Δt - u·Δx) = Temporal Component of c*ΔX'
Compare with:
c2Δτ = γ(c2Δt - u·Δx) = U·ΔX
see the "Invariant Rest Value of the Temporal Component Rule"
=====================


U = dR/dτ = (U·∂)[R]
U = dR/dτ = γdR/dt = γ(d/dt)[(ct,r)] = γ(cdt/dt,dr/dt) = γ(c,u) = U
U = (U·∂)[R] = (U·∂[R]) = Uαηαββ[Rμ] = Uαηαβηβμ = Uαδαμ = Uμ = U
4-Velocity is the derivative of 4-Position wrt. Proper Time τ.
=====================


A = dU/dτ = (U·∂)[U]
A = dU/dτ = γdU/dt = γ(d/dt)[γ(c,u)] = γ(d/dt)[(γc,γu)] = γ((d/dt)[γc],(d/dt)[γu]) = γ(cγ', γ'u + γ) = A
A = (U·∂)[U] = (U·∂[U]) = Uαηαββ[Uμ]
= Uαηαβ[[∂t/c (γc),  ∂t/c (γu)],[-(γc),  -u)]]
= Uα[[∂t/c (γc), 0],[0, u)]]
= γ(c∂t/c (γc),  u·∇u))
= γ(c∂t(γ), d/dt (γu))
= γ(cγ', γ'u + γ) = Aμ = A
4-Acceleration is the derivative of 4-Velocity wrt. Proper Time τ.
=====================


dU/dτ = U·A = 0
U·U = (c)2
d/dτ[U·U] = d/dτ[(c)2] = 0
d/dτ[U·U] = d/dτ[U]·U+ d/dτ[U] = A·U+ U·A = 2*(U·A) = 0, hence (U·A) = 0
The 4-Acceleration of an Event is always orthogonal to the 4-Velocity of that Event.
see the "Invariant Rest Value of the Temporal Component Rule"
=====================


N = noU
(nc,nu) = noγ(c,u) = γno(c,u) = n(c,u) = (nc,nu)
4-NumberFlux is the Rest NumberDensity no times the 4-Velocity.
This can be measured in SR fluid experiments, with number density and number flux related in a manner similar to time and space.
The Invariant Count of Particle #, the Scalar N = noVo.
=====================


P = moU = (Eo/c2)U
(E/c,p) = moγ(c,u) = γmo(c,u) = m(c,u) = (mc,mu)
4-Momentum is the Rest Mass mo times the 4-Velocity.
This can be measured in SR collision experiments.
=====================


Pden = G = noP = nomoU = (noEo/c2)U = ρmoU
The 4-MomentumDensity can be obtained by multiplying by the scalar invariant rest number density (no).
When the scalar invariant # of particles N = 1,  this is the equivalent of dividing by the scalar invariant rest volume (Vo)
N = noVo
Start with Energy-Momentum Stress Tensor:
Tμν = (po_m+p/c2)UμUν - pημν
Contract with the 4-Velocity
TμνUν = (po_m+p/c2)UμUνUν - pημνUν
TμνUν = (po_m+p/c2)Uμc2 - pUμ
TμνUν = (c2po_m+p)Uμ - pUμ
TμνUν = c2po_mUμ
TμνUν = c2G
u = EnergyDen, s = EnergyFlux = PoyntingVector = uu = c2g
ue = (εoE·E+B·Bo)/2 = (E·D+B·H)/2 = EM energy density
=====================


J = ρoU = qnoU = qN
(ρc,j) = ρoγ(c,u) = γρo(c,u) = ρ(c,u) = (ρc,ρu)
(ρc,j) = qnoγ(c,u) = qγno(c,u) = qn(c,u) = q(nc,nu)
4-CurrentDensity is the Rest ChargeDensity (qno = ρo) times the 4-Velocity.
This can be measured in SR-EM experiments, with charges and currents related in a manner similar to time and space.
=====================


J = (ρo/mo)P or P = (moo)J
It is derivable from the SR relations P = moU = (Eo/c2)U and J = ρoU
Since both P and J are Lorentz Scalar proportional to U, then by the mathematics of tensors J must be Lorentz Scalar proportional to P.
(i.e. Tensors obey mathematical transitivity, or in this case they are also Right Euclidean   {if a~c and b~c, then a~b} )
J = ρoU = (ρo)/(mo)P = (ρo/mo)P = (γρo/γmo)P = (ρ/m)P
=====================


Σ*[ Pn ] = Z = (0,0)
The Conservation of 4-Momentum.
Possibly one of the most important concepts in all of physics.
The summation Σ* counts all pre-collision terms positively and post-collision terms negatively.
The sum of the 4-momenta of all particles going into a point collision is equal to the sum of the 4-momenta of all particles going out.
Since P = (E/c,p) = (mc,p) = (mc,mu) we get the temporal eqn. {Σ*[mnc] = cΣ*[mn] = 0} and the spatial eqns. {Σ*[mnun] = 0}
=====================


F = dP/dτ = (U·∂)[P] = (d/dτ)[P] = (d/dτ)[moU] = (d/dτ)[mo]U + mo(d/dτ)[U] = (d/dτ)[mo]U + moA
F = dP/dτ = γdP/dt = γ(d/dt)[(E/c,p)] = γ(dE/cdt,dp/dt) = γ(Ė/c,) = γ(Ė/c,f)
4-Force is the derivative of 4-Momentum wrt. Proper Time τ.
A pure 4-Force is one that has ((d/dτ)[mo] = 0), in which case Fp = moA {pure = space-like} and 4-ForcePure Fp = γ(f·u/c,f)
An example of a pure 4-Force is the ElectroMagnetic field, which has 4-ForceEM FEM = γq( (u·E)/c , (E) + (u x B) )
A heat-like 4-Force is one that has (A = 0), in which case Fh = (d/dτ)[mo]U {heat-like = time-like} and 4-ForceHeat Fh = γ2o(c,u) = γṁ(c,u)
A scalar-based 4-Force is one that has (Ė = k(d/dτ)[Φ]), in which case Fs = k[Φ]) {scalar} and 4-ForceScalar Fs = k(∂t[Φ],-[Φ])
=====================


Fden = noF
The 4-ForceDensity can be obtained by multiplying by the scalar invariant rest number density (no).
When the scalar invariant # of particles N = 1,  this is the equivalent of dividing by the scalar invariant rest volume (Vo)
N = noVo
=====================


F·U = (d/dτ)[mo]U·U + moA·U = (d/dτ)[mo]c2 + mo*0 = (d/dτ)[mo]c2 = (d/dτ)[Eo] = γd/dt[Eo] = d/dt[E] = Ė
F·U = γ(Ė/c,f)·γ(c,u) = γ2(Ė - f·u) = γĖo = Ė
If F·U = 0 then (Ė - f·u) = 0 then Ė = f·u then 4-ForcePure Fp = γ(f·u/c,f)
This is true when there is no change in rest mass (rest energy) of a test particle, or again, when the force is conservative or pure.
(i.e. The force is applied to a particle of stable invariant rest mass mo)
We will find that the EM field is a pure force:
Ė = dE/dt = q(u·E)
f = dp/dt = q{(E) + (u x B)}
f·u = q{(E·u) + (u x B)·u} = q{(E·u) + (0)} = q(u·E)
4-ForceEM FEM = γq( (u·E)/c , (E) + (u x B) )
and hence, FEM·U = 0 {for an EM field, it is conservative}
=====================


P·U = (E/c,p)·γ(c,u) = γ(E - p·u) = Eo
The Scalar Product of the 4-Momentum with the 4-Velocity gives the Invariant Rest Energy.
see the "Invariant Rest Value of the Temporal Component Rule"
=====================


AEM·U = (φ/c,a)·γ(c,u) = γ(φ - a·u) = φo
The Scalar Product of the 4-EM-VectorPotential with the 4-Velocity gives the Invariant Rest Scalar Potential.
see the "Invariant Rest Value of the Temporal Component Rule"
=====================


J·U = (ρc,j)·γ(c,u) = γ(ρc2 - j·u) = ρoc2
The Scalar Product of the 4-CurrentDensity with the 4-Velocity gives the Invariant Rest Charge Density times speed-of-light squared..
see the "Invariant Rest Value of the Temporal Component Rule"
=====================


K·U = (ω/c,k)·γ(c,u) = γ(ω - k·u) = ωo
The Scalar Product of the 4-WaveVector with the 4-Velocity gives the Invariant Rest Angular Frequency.
see the "Invariant Rest Value of the Temporal Component Rule"
=====================


K = (ωo/c2)U
4-WaveVector is the Rest AngularFrequency over lightspeed squared (ωo/c2) times the 4-Velocity.
This can be seen by looking at the previous Scalar Product relations.
U·U = (c)2
K·U = (ωo)
K·K = (ωo/c)2
hence:
K = (ωo/c2)U = (ω/c,k) = (ωo/c2)γ(c,u) = γ(ωo/c2)(c,u) = (γωo/c, γωou/c2) = (ω/c, ωu/c2) = (ω/c, ω/vphase)
The wave can move, with the relation:
(| u * vphase| = c2)
(| vgroup * vphase| = c2)
The 4-WaveVector can be measured in SR wave experiments, especially those involving the Doppler Effect.
------
Relativistic Doppler Effect (derivation using 4-Vectors)
K = (k0, k), a generic SR 4-vector under observation, relative to observer (*note*, for the derivation this is not necessarily the 4-WaveVector)
K·U = a Lorentz invariant, upon which all observers agree, see the "Invariant Rest Value of the Temporal Component Rule"
take K·U --> K·Uo = (k0, k)·(c,0) = ck0 = the value of the temporal component of K as seen by observer U
now, let there be an observer Uobs at rest and an emitter Uemit moving with respect to Uobs
Uobs = (c,0): 4-Velocity of observer at rest
Uemit = γ(c,v): 4-Velocity of emitter relative to observer
---------
K·Uobs = (k0, k)·(c,0) = c k0 = ck0_obs = invariant
K·Uemit = (k0, k)·γ(c,v) = γ(ck0 - k·v) = ck0_emit = invariant
---------
K·Uobs / K·Uemit = ck0_obs / ck0_emit = k0_obs / k0_emit = invariant
K·Uobs / K·Uemit = ck0 / γ(ck0 - k·v) = 1 / γ(1 - k·v/k0c) = 1 / γ[1 - (| k| /k0)*(n̂·v/c)] = invariant
---------
k0_obs =
= k0_emit / γ(1 - (| k| /k0)*(n·v/c))
= k0_emit / γ(1 - (| k| /k0)*(n·β))
= k0_emit / γ(1 - (| k| /k0)*(β cos[θ]))
---------
Up to this point, the derivation is general and could be for any 4-Vector K
Now, I am going to assume that K actually is the 4-WaveVector for the rest of the derivation.
K = (ω/c,k)
k0_obs = k0_emit / γ(1 - (| k| /k0)*(n̂·β))
ω_obs /c = ω_emit / cγ(1 - (c| k| /ω)*(n̂·β))
ω_obs = ω_emit / γ(1 - (c| k| /ω)*(n̂·β))
ω_obs = ω_emit / γ(1 - (c/vphase)*(n̂·β))
ω_obs = ω_emit / γ(1 - (n̂·v)/vphase) {for photons or massive particles, with v as the relative velocity between emitter and observer, and vphase is the emission wave velocity}
---------
If K is photonic/light-like/null, then (vphase = c), which gives the Relativistic Doppler Formula for photonic particles
ω_obs = ω_emit / γ(1 - (n̂·v)/c) {for photonic}
ω_obs = ω_emit / γ(1 - (n̂·β)) {for photonic}
ω_obs = ω_emit  √[1+| β| ]√[1-| β| ] / (1 - (n̂·β)) {for photonic}
------------
In the way things are defined,  is unit direction of the photon emission; it always points from the emitter to the observer.
For motion of source away from observer, (n·β) = -β, so ωobs = ωemit / γ(1 + β) = ωemit *Sqrt[1-β]/Sqrt[1+β] = RedShift
For motion of source toward the observer, (n·β) = +β, so ωobs = ωemit / γ(1 - β) = ωemit *Sqrt[1+β]/Sqrt[1-β] = BlueShift
For motion of source tangent to the observer, (n·β) = 0, so ωobs = ωemit / γ(1 - 0) = ωemit / γ = Transverse Doppler Shift
=====================


K = (ωo/Eo)P or P = (Eoo)K
It is derivable from the SR relations K = (ωo/c2)U and P = moU = (Eo/c2)U
Since both P and K are Lorentz Scalar proportional to U, then by the mathematics of tensors K must be Lorentz Scalar proportional to P.
(i.e. Tensors obey mathematical transitivity, or in this case they are also Right Euclidean   {if a~c and b~c, then a~b} )
K = (ωo/c2)U = (ωo/c2)/(Eo/c2)P = (ωo)/(Eo)P = (ωo/Eo)P =  (γωo/γEo)P = (ω/E)P
=====================


K·X = (ω/c,k)·(ct,x) = (ωt - k·x) = -Φ
The Phase of an SR Wave is a Lorentz Scalar.
=====================


K = -[Φ]
The 4-WaveVector can be derived in terms of periodic motion, for which families of surfaces move through space as time increases,
or alternately,as families of hypersurfaces in spacetime, formed by all events passed by the wave surface.
The 4-WaveVector is everywhere in the direction of propagation of the wave surfaces.
More specifically, any particular 4-WaveVector K is the vector orthogonal to a hypersurface of constant phase Φ at that spacetime point/event.
{Each hypersurface is an ensemble of events E(t,x,y,z) that all share the same value of Φ}
The function that provides a vector orthogonal to a surface at each point is the gradient.
i.e. K = -∂[Φ].
The phase is chosen so that it decreases by unity as we pass from one wave crest to the next, in the direction of propagation of the waves.
=====================


[K] = ∂uKv = (∂t/c,-)(ω/c,k) = [ ∂t/c[ω/c],-[k] ] = [[0]]
The 4-Gradient of the 4-WaveVector is the zero matrix assuming the 4-WaveVector is not a function of 4-Position.
=====================


[K·X] = [-Φ] = ∂wuvKuXv] = ∂w[KvXv] = Kvw[Xv] + Xvw[Kv] = Kvηwv + [0] = Kw = K·∂[X]+[K]·X = K
Just more SR Mathematics - The 4-WaveVector K is the 4-Gradient of the SR Phase Φ (a Lorentz Invariant itself).
=====================


∂·J = (∂t/c,-)·(ρc,j) =   (∂tρ + ∇·j) = ? = 0
The 4-Gradient Lorentz Product with the 4-CurrentDensity is zero, or that the Divergence of the 4-CurrentDensity is zero, for a conserved field.
There are no sources or sinks of Current Density when this is true.  This is related to the Conservation of Charge/Current and Noether's Theorem.
The divergence formula may actually be applied in several other circumstances regarding Conserved quantities.
=====================


∂·U = (∂t/c,-)·γ(c,u) = ( ∂tγ + ∇·u] ) = ( ∂tγ + γ( ∇·u ) + ( u·∇[γ] ) ) = ? = 0
The 4-Gradient Lorentz Product with the 4-Velocity is zero, or that the Divergence of the 4-Velocity is zero, for a conserved field.
There are no sources or sinks of 4-Velocity when this is true.
This is related to the Conservation of Charge and Noether's Theorem.
If  ( ∂tγ ) = 0 then the flow is constant or inertial.
If  ( ∇·u ) = 0 then the flow is incompressible, in which case (u) is said to be solenoidal.
If  ( u·∇[γ] ) = 0 then the flow is steady, in which case (γ) is constant along a streamline.
If an incompressible flow also has a curl of zero, so that it is also irrotational, then the flow velocity field is Laplacian.
The divergence formula may actually be applied in several other circumstances regarding Conserved quantities.
Note the following: ∂·J = 0 = ∂·ρoU = ρo∂·U = 0
Typically the charge density (ρo) in question is non-zero, hence it is the (∂·U) = 0.
If the 4-Velocity U is a conservative field, then any of the other 4-Vectors which are just a Lorentz Scalar times the 4-Velocity will be conserved.
∂·U = (∂t/c,-)·γ(c,u) = ( ∂tγ + ∇·u] ) = ( ∂tγ + γ( ∇·u ) + ( u·∇[γ] ) ) = 0
ρo(∂·U)  = ρo(∂t/c,-)·γ(c,u) =  (∂t/c,-)·γρo(c,u) =  (∂t/c,-)·ρ(c,u) =  ( ∂tρ + ∇·u] ) = ( ∂tρ + ρ( ∇·u ) + ( u·∇[ρ] ) ) = 0

This is related to the Advection Equation and Incompressible Flow - See Incompressible Euler Equations, General Continuity Equations, Conservation Form Equations.
Essentially this is the conservation of worldlines...
=====================


AEM = (φo/c2)U
AEM = (φ/c,a) = (φo/c2) γ(c,u) = (γφo/c,γφo/c2u)
4-EM_VectorPotential is the Rest EM Scalar Potential φo over c squared times the 4-Velocity.
This can be measured in SR-EM experiments, with charges and currents related in a manner similar to time and space.
giving temporal component { φ = γφo } and spatial component { a = (γφo/c2)u }
This formula is for a generic 4-VectorPotential
For a point charge one can define:
The Rest EM Scalar Potential (φo) = (qc/4πεo) / [R·U]ret = (q/4πεo) / [R·T]ret is a Lorentz Scalar Invariant.
with [...]ret impling retarded (R·R = 0, the definition of a light signal)
=====================


AEM = (q/4πεoc) U / [R·U]ret = (qμoc/4π) U / [R·U]ret  {for a point charge q}
AEM = (q/4πεoc2) U / [R·T]ret = (qμo/4π) U / [R·T]ret  {for a point charge q}
with [...]ret impling retarded (R·R = 0, the definition of a light signal)
The 4-EM_VectorPotential of a moving point charge (Lienard-Wiechert Potential)
If we use the AEM = (φo/c2)U definition and compare terms with above:
o/c2) = (q/4πεoc) / [R·U]ret
o) = (qc/4πεo) / [R·U]ret
o) = (qc/4πεo) / [c2τ]ret
o) = (q/4πεo) / [cτ]ret
o) = (q/4πεo)/ r   {which is the correct potential for a point charge in its rest frame}
since R·R = (ct)2-| r| 2 = 0 --> ct = r

And likewise, the scalar and vector potential of a moving point charge
φ = (γφo) = (γq/4πεo) / r
a = (φo/c2u = (γφo/c2)u = (γφo/c2)u = (φ/c2)u = ((γq/4πc2εo) / r)u = ((γqμo/4π) / r)u

We can actually simplify the expression a bit by using the 4-UnitTemporal T = U/c
o/c2) = (q/4πεoc2) / [R·T]ret
o) = (q/4πεo) / [R·T]ret
o) = (q/4πεo) / [cτ]ret
o) = (q/4πεo)/ r  {which is the correct potential for a point charge in its rest frame}
since R·R = (ct)2-| r| 2 = 0 --> ct = r
=====================


Q = qA.
The 4-PotentialMomentum Q is a charge (q) time the 4-VectorPotential A.
Essentially, Potentials and Fields carry energy and momentum.
=====================


PT = P + Q = P + qAEM
The 4-TotalMomentum of a system can be split into the 4-Momentum of a Particle + a charge (q)* a 4-VectorPotential.
Essentially, we are examining a lone particle running around in a big field which can affect the particle.
Typically, this is for an 4-EM_VectorPotential, however it could be for any type of SR 4-VectorPotential PT = P + qA
=====================


ν[PTμ] = q∂μ[AEMν]
The 4-Gradient of the 4-TotalMomentum is related to the 4-Gradient of the 4-EMVectorPotential.
Roughly, the change in TotalMomentum, or the TotalForce, is the Gradient of a VectorPotential.
This is one case for which tensor notation is required, since the order of the indices is important when dealing with the 4-Gradient.
This little bit of magic combined with Conservation of 4-TotalMomentum is enough to derive the Faraday EM Tensor.
We will see how this works in the section on the Lorentz Force Equation.
=====================


Fμ = dPμ/dτ = qUυ(∂μAEMν - ∂νAEMμ) = qUνFμν
F = dP/dτ = {qU·Fμν}
F = (U·∂)P = {qU·Fμν}
The Lorentz Force Equation for a charged particle in a EM field.
The 1st version is in single index tensor format.
The 2nd and 3rd versions are in a slightly ambiguous mix of 4-Vector and 2-index Tensor Format,
for which we see the "Invariant Rest Value of the Temporal Component Rule".

We can build this up from the idea that the 4-TotalMomentum is just the sum of the individual charged particle 4-Momentum and q*the spacetime field 4-EM_VectorPotential.
PT = P + qA
P = PT - qA
Pμ = PTμ - qAμ
ν[Pμ] = ∂ν[PTμ] - ∂ν[qAμ]
ν[Pμ] = q∂μ[Aν] - q∂ν[Aμ]: This from the rule about the 4-Gradient on the 4-TotalMomentum ν[PTμ] = q∂μ[AEMν]
ν[Pμ] = q(∂μ[Aν] - ∂ν[Aμ])
Uνν[Pμ] = qUν(∂μ[Aν] - ∂ν[Aμ])
(U·∂)[Pμ] = qUν(∂μ[Aν] - ∂ν[Aμ])
d/dτ[Pμ] = qUν(∂μ[Aν] - ∂ν[Aμ])
Fμ = d/dτ[Pμ] = qUν(∂μ[Aν] - ∂ν[Aμ]) = qUν Fμν
*Note* Here (Fμ) is the 4-Force and (Fμν) is the EM Faraday Tensor.

One way to simplify this is to take one component at a time:
γf1 = γfx = qUν(∂1[Aν] - ∂ν[A1])
= qU0{∂1[A0] - ∂0[A1]} +  qU1{∂1[A1] - ∂1[A1]} + qU2{∂1[A2] - ∂2[A1]} + qU3{∂1[A3] - ∂3[A1]}
= q(γc){(-∂x)[φ/c] - (∂t/c)[ax]} +  q(-γux){0} + q(-γuy){(-∂x)[ay] - (-∂y)[ax]} + q(-γuz){(-∂x)[az] - (-∂z)[ax]}
= qγ{(-∂x)[φ] - (∂t)[ax]) +  (0) + (-uy){(-∂x)[ay] - (-∂y)[ax]} + (-uz){(-∂x)[az] - (-∂z)[ax]}
= qγ{(-∂x[φ] - ∂t[ax]) +  (0) + (-uy)(-∂x[ay] + ∂y[ax]) + (-uz)(-∂x[az] + ∂z[ax])}
= qγ{(Ex) + (0) + (-uy)(-Bz) + (-uz)(By)}
= qγ{(Ex) + (u x B)x}
Thus:
γf = qγ{(Ex) + (u x B)x} + qγ{(Ey) + (u x B)y} + qγ{(Ez) + (u x B)z}

or, we can use the already worked out cells from the EM Faraday Tensor, in another section...
Fμ = d/dτ[Pμ] = qUν(∂μ[Aν] - ∂ν[Aμ]) = qUν Fμν
Fμ = qUν Fμν
Fi = qUν Fiν = q[(U0 Fi0)+(U1 Fi1)+(U2 Fi2)+(U3 Fi3)]
= q[(γc Ei/c)+(-γux * -εi1k Bk)+(-γuy * -εi2k Bk)+(-γuz * -εi3k Bk)]
= γq[(Ei)+(ux * εi1k Bk)+(uy * εi2k Bk)+(uz * εi3k Bk)]
= γq[(Ei)+(u x B)i]

γf = γq{(E) + (u x B)}: The Lorentz Force Equation for f using E and B fields

4-Force F = γ(Ė/c,f) = γ(ṁc,f)
F = dP/dτ = {qU·Fμν}
If we look at the components we get:   **Note** E = Energy, E = Electric Field
γ [Ė/c] = d/dτ [E/c] = qγ [c] · [0   -Ei/c   ] = qγ [c*0 + u·E/c]

f  ]
[ p ]
[u]
[Ei/c ijk Bk ]
[E + u x B]

γĖ/c = dE/cdτ = γq(u·E/c)
γf = dp/dτ = γq{(E) + (u x B)}

Ė = dE/dt = q(u·E)
f = dp/dt = q{(E) + (u x B)}

FEM = γq( (u·E)/c , (E) + (u x B) )
=====================


Fμν = (∂μAEMν - ∂νAEMμ) =

[  0  ,-Ex/c,-Ey/c,-Ez/c ]
[Ex/c,   0   , -Bz  ,  By  ]
[Ey/c,  Bz  ,   0   , -Bx  ]
[Ez/c, -By ,  Bx  ,    0   ] =

[    0   ,  -Ei/c  ]
[+Ei/c ,-εijk Bk]

The Electromagnetic Tensor, also known as the Faraday Tensor, or as the EM Tensor,
can be constructed from the 4-Gradient and 4-EM_VectorPotential 4-Vectors.
Note that the EM Tensor is a 2nd rank, anti-symmetric tensor ( Fμν = -Fνμ ), which gives it some really useful properties.
At this point in the presentation, I am simply showing the results of throwing the 4-Gradient at the 4-EM-VectorPotential.
We can see how this is actually derived (in the section on the Lorentz Force Law).

Note that writing out the complete 4x4 matrix using Fμν = (∂μAEMν - ∂νAEMμ) for each cell gets tedious.
Using slick tensor calculus index gymnastics, let's instead define some meaningful labels for the individual cells:

Fμμ = (∂μAEMμ - ∂μAEMμ) = 0 down the diagonal

F0i = (∂0AEMi - ∂iAEM0) = (∂t/c ai + iφ/c) = - Ei/c
Fi0 = (∂iAEM0 - ∂0AEMi) = (-iφ/c - ∂t/c ai) = +Ei/c
{F0i = -Fi0 as expected...}

Fij = (∂iAEMj - ∂jAEMi)
= ( δilδjm - δjlδim )∂lAEMm
= ( εkij εklm )∂lAEMm
= ( εijk )[- x a]k
= -( εijk )Bk
= -εijk Bk
ijk = -εjik , giving Fij = Fji as expected...}

So, E = (-φ - ∂ta) and B = [ x a]
or,  Ei = cFi0 and Bk = -(1/2)εijk Fij

*Note* For many years many textbooks have said that the electromagnetic (E) and (B) fields were the "real" objects of reality, and that the potential (φ) and vector-potential (a) were mathematical artifacts. This is now known to be incorrect.  (E) and (B) were discovered during classical Euclidean space physics, before it was understood that the 4-EM-VectorPotential A was one of the standard SR 4-Vectors in 4D Minkowski Spacetime.
(E) and (B) are actually derived from the correctly relativistic 4-Gradiant and 4-EM-VectorPotential, and, in fact, may be regarded as a convenience, simply a handy label for each of the 16 cells in the EM Faraday Tensor matrix.  One can write the manifestly covariant Maxwell and Lorentz Force Equations without any reference to (E) and (B) if desired.  However, it turns out that the labels are very useful for writing down the pre-relativistic classical EM equations.  The other reason is that this is the case is because once QM comes into play via the Aharanov-Bohm effect, the 4-EM-VectorPotentialA is required to describe the phenomenon.

One reason for much of the confusion is that there seems to be a discrepancy over the number of degrees of freedom via each choice.
The 4-EM-Potential and the 4-Gradient have 4 independent components each, for a total of 8 degrees of freedom.
The (E) and (B) fields have 3 independent components each, for a total of 6 degrees of freedom.
Which is correct? Scalar Lorentz invariants come to the rescue.

There are some Lorentz Invariants associated with any rank-2 tensor Tμν.
These principle invariants are defined by eigenvalues (λi) of T.

Invariant #1: The Trace: Tr[Tμν] = ημν Tμν = Tνν
Tr[Fμν] = Fνν = F00 + F11 + F22 + F33 = 0+0+0+0 = 0
Hence: Tr[Fμν] = 0

Invariant #2: Half The Inner Product: (1/2)Tμν Tμν = (1/2){Tr[T]2-Tr[T2]}
Fμν Fμν =
F00 F00 + F0i F0i + Fi0 Fi0 + Fij Fij
(0*0) + (- Ei/c)(- Ei/c) + (+Ei/c)(+Ei/c) + (-εijk Bk)(-εijk Bk)
(0*0) + (- Ei/c)(- - Ei/c) + (+Ei/c)(- +Ei/c) + (-εijk Bk)(- - -εijk Bk)
(0) + (- Ei/c)(+ Ei/c) + (+Ei/c)(- +Ei/c) + (εijk εijk Bk * Bk)
(0) + -(E/c)2 + -(E/c)2 + (εij εij )(B)2
(0) + -(2)(E/c)2  + (2)(B)2
2{B2-(E/c)2}
Fμν Fμν = 2{B·B-E·E/c2}
Hence: (1/2)Fμν Fμν = {B·B-E·E/c2}

Invariant #3: The Determinant: Det[Tμν] =?= εμνρσTμ0Tν1Tρ2Tσ3
Ok, this is gonna be ugly... or maybe not... thank you skew-symmetric, dimension 4...
The determinant of a skew-symmetric matrix can always be written as the square of a polynomial in the matrix entries. This polynomial is called the Pfaffian.
Det[Fμν] = Pfaffian[Fμν]2
pf[[ 0, a, b, c]
    [-a, 0, d, e]
    [-b,-d,0, f]
    [-c,-e,-f, 0]] = af-be+cd
Pfaffian[Fμν] = (-Ex/c)(-Bx)-(-Ey/c)(By)+(-Ez/c)(-Bz) = (ExBx/c)+(EyBy/c)+(EzBz/c) = (E·B)/c
Det[Fμν] = {(E·B)/c}2

{#1} doesn't really do anything for us, since 0 = 0 doesn't place any constraints.
But, {#2} and {#3} provide 2 constraints on the total degrees of freedom.

There were originally 8 degrees of freedom from (4-Gradient + 4-VectorPotential), but now there are only 6 due to the 2 constraints.
This matches the total of 6 independent components from the (E) and (B) fields.
This seems correct, since I can think of situations/experiments for which I can set the (E) and (B) fields to whatever I want, and vary each of the components individually to get independent results.
Looking at it a different way:
A 4x4 matrix can have up to 16 independent components.
But, we can always break a matrix into a sum of a symmetric and anti-symmetric matrix.
Tμν = (Tμν+Tνμ)/2 {symmetric} + (Tμν-Tνμ)/2 {anti-symmetric}
For the 4x4 = 16 case, this breaks down into symmetric (at most 10 independent components) + anti-symmetric (at most 6 independent components)
The fact that the Faraday Tensor is 4x4 anti-symmetric (or skew-symmetric) again takes the total of independent components down to 6.
=====================


U·FEM = UμFEMμ = Uμ( qUνFμν ) = qUμUυ(∂μAEMν - ∂νAEMμ) = 0
Beautiful result showing the the EM field is a conservative field.
The contraction of the two 4-Velocities (symmetric) with the EM-Tensor (anti-symmetric) guarantees a zero result.
Hence, the 4-EM-Force is always spatial, and orthogonal to the timelike 4-Velocity.
Proof:
Fμν = -Fμν :Def. of Anti-Symmetric Tensor
Let Vμ be a generic 4-Vector
(Fμν)(Vμ)(Vν) =
(Fνμ)(Vν)(Vμ) :switching dummy indices
(-Fμν)(Vν)(Vμ) :anti-symmetry
(-Fμν)(Vμ)(Vν) :commuting 4-vectors
Hence: (Fμν)(Vμ)(Vν) = -(Fμν)(Vμ)(Vν) = 0
=====================


α(∂αAEMν - ∂νAEMα) = ∂αFαν = μoJν
(∂·∂)AEM - (∂·AEM) = μoJ
∂·Fαν  = μoJ
The full Non-Homogeneous Classical Maxwell EM Equation.
It can be written in a few different informative ways.
The 1st is in single index Tensor format.
The 2nd is in 4-Vector format.
The 3rd is a slightly ambiguous mix of 4-Vector and 2 index Tensor that emphasizes the fact that the Maxwell EM equation is actually just another type of 4-Divergence equation.
The 4-CurrentDensity is the source/sink of the Electromagnetic Tensor.
=====================


να(∂αAEMν - ∂νAEMα) = ∂ναFαν = μoν Jν = 0
(∂·J) = 0 {for an EM field}
The conservation of charge, or continuity equation for an EM field.
Another beautiful result showing the power of tensors.
The contraction of the two 4-Gradients (symmetric) with the EM-Tensor (anti-symmetric) guarantees a zero result. 
=====================


∂·AEM = (∂t/c,-)·(φ/c,a) = (∂tφ/c2+∇·a) = 0
The Lorenz Gauge for Classical EM - Yes, Lorenz, not Lorentz  :)
It simplifies the expression for the Maxwell EM Equations.
Again, it is showing that the 4-Divergence is zero and hence the EM field is conservative, no sources or sinks.
From pure math: ∂·V] = [φ]·V + φ(∂·V), with φ a Lorentz scalar and V a 4-Vector.
AEM = aΕe^(-iK·X) = aΕe^(-iΦ), The 4-EM-VectorPotential has plane wave solutions.
∂·[AEM] = 0 = ∂·[aΕe^(-iΦ)] = a[e^(-iΦ)]·Ε + ae^(-iΦ)(∂·Ε) = aiK·Εe^(-iΦ) + (0) = 0
Hence, K·Ε = 0, the 4-Polarization is orthogonal to the 4-WaveVector.
=====================


(∂·∂)AEM = μoJ {for ∂·AEM = 0 : Lorenz Gauge, if the EM field is conservative}
The "reduced" Non-Homogeneous Classical Maxwell EM Equation, considering a conservative field.
(∂·∂)(φ/c,a) = (∂t2/c2-∇·∇)(φ/c,a) = μo(cρ, j)
Temporal Component: (∂·∂)[φ] = μo(c2ρ) = ρ/εo
Spatial Component: (∂·∂)[a] = μo j
with { εoμo = 1/c2 }
(∂·∂)(AEM·U) = μo(J·U)
(∂·∂o = μoρoc2
(∂·∂o  = ρoo
This is from classical EM, which was actually the starting point for Einstein's theories of Relativity.
Let's examine the 4-VectorPotential of a point charge
φ = (γΦo) = (γq/4πεo) / r
(∂·∂
= (∂·∂) (γq/4πεo)[1/ r]
= (γq/4πεo)(∂·∂)[1/ r]
= (γq/4πεo)4πδ3(r)
= (γq/εo3(r)
= (γq)δ3(r)/εo
= ρ/εo
with ρ = (γq)δ3(r) and thus ρo = (q)δ3(r)
Just a note: The classical Maxwell EM equations do not have Spin included
(∂·∂)AEM = μoJ = μo ρoU = μo(q/Vo)U  = μo q(c/Vo)T
Once spin is included, the equations for QED emerge:
(∂·∂)AEM = μo qψΓψ
with the 4-CurrentDensity J = qψΓψ , 4-DiracGammaMatrix Γ = (γ0,γ)

Finally, just an interesting EM comparison, we can get the Maxwell and Lorentz Force Equations in almost identical format:
Maxwell Eqn α(∂αAEMν - ∂νAEMα) = μoJν ∂·Fαν  = (μo)J
Lorentz Force Eqn Uα(∂νAEMα - ∂αAEMν) = (1/q)Fν U·Fαν = (-1/q)F

The two physical constants obtained from ElectroMagnetic SR are (q) and (μo).
=====================


Ε·Ε = (ε0,ε)·(ε0,ε) = (ε0)2 - ε·ε ==> -1.
The 4-Polarization has a spatial magnitude of -1.
=====================


Ε·K = (ε0,ε)·(ω/c)(1,β) = (ω/c)(ε0*1 - ε·β) = 0
The 4-Polarization is orthogonal to the 4-WaveVector.
From pure math: ∂·V] = [φ]·V + φ(∂·V), with φ a Lorentz scalar and V a 4-Vector
AEM = aΕe^(-iK·X) = aΕe^(-iΦ), the 4-EMVectorPotential has plane wave solutions.
If we choose the Lorenz Gauge, which is appropriate for a conservative field:
∂·[AEM] = 0 = ∂·[aΕe^(-iΦ)] = a[e^(-iΦ)]·Ε + ae^(-iΦ)(∂·Ε) = aiK·Εe^(-iΦ) + (0) = 0
Hence, K·Ε = 0, the 4-Polarization is orthogonal to the 4-WaveVector.
=====================


S = -(PT·R) and [S] = - PT.
Getting into Relativistic Analytic Mechanics, we can define an Action (S) as the negative Lorentz Scalar product of the 4-TotalMomentum with the 4-Position,
or the 4-TotalMomentum is the negative 4-Gradient of the Action.
Using some results from before:
PT = P + qAEM
P·P = (E/c)2 - p·p = (moc)2
P = (E/c,p) = PT - qAEM = -[S] - qAEM = (-∂t[S]/c - qφ/c, [S] - qa)
And putting it all together...
(-[S] - qAEM)·(-[S] - qAEM) = (moc)2
(-∂t[S]/c - qφ/c)2 - ([S] - qa)·([S] - qa) = (moc)2
(∂t[S]/c + qφ/c)2 - ([S] - qa)2 = (moc)2
This is the Relativistic Hamilton-Jacobi Equation including the effects of a 4-VectorPotential.

We can also define a Relativistic Hamiltonian H = γ(PT·U) and a Relativistic Lagrangian L = -(PT·U)/γ
with H + L = pT·u
This comes from the relativistic identity:
γ -1/γ = γβ2
Just multiply all the terms by the Lorentz Scalar (PT·U)
γ(PT·U) -(PT·U)/γ = γ(PT·U2
     H     +     L        =     pT·u
=====================


Tperfectfluidμν = (ρeo+po)UμUν/c2 - poημν
Tdustμν = PμNν = moUμnoUν = monoUμUν = ρmoUμUν = ρeoUμUν/c2
Tvacuumμν = - poημν
Tradiationμν = po(4UμUν/c2 - ημν)
The energy-momentum of a single particle can be specified by a single 4-Vector.
However, the energy-momentum of a fluid requires a 2-index tensor, as neighbor interactions must be included.
In addition, one uses particle densities instead of discrete particles.
One can define a "Stress-Energy" Tensor Tμν as the flux of 4-Momentum Pμ across a surface of constant Xν.
There are 4 main types:
T00 is the flux of 4-Momentum P0 across a surface of constant X0, or flux of Energy across time, which is the energy density.
T0i is the flux of 4-Momentum P0 across a surface of constant Xi, or flux of Energy across space, which is an energy flux (heat conduction).
Ti0 is the flux of 4-Momentum Pi across a surface of constant X0, or flux of i-Momentum across time, which is a momentum density (heat conduction).
Tij is the flux of 4-Momentum Pi across a surface of constant Xj, or flux of i-Momentum across space, which is a momentum flux (stress).
Tij can be broken into 2 subtypes:
Tii is the flux of 4-Momentum Pi across a surface of constant Xi, or flux of i-Momentum across the {i=j}same direction in space, which is the isotropic pressure p = σii.
Tij is the flux of 4-Momentum Pi across a surface of constant Xj, or flux of i-Momentum across an {i<>j}orthogonal direction in space, which is a shear σij.
Tμν is a symmetric tensor, so T0i = Ti0 {The energy flux = momentum density}, Tij = Tji, and the diagonal terms can be non-zero.
All cases below use : ημν = Diag[+1,-1,-1,-1]

Tμν =
T00
ρmc2 = ρe
Energy Density
T0j
cg
Energy Flux
Ti0
cg
Momentum Density
Tij
σij
Momentum Flux

Tμν =
T00
Energy Density
T01 | T02 | T03
Energy Flux
T10
---
T20
---
T30
Momentum Density
T11 | T12 | T13
---   ---    ---
T21 | T22 | T23
---   ---    ---
T31 | T32 | T33
Momentum Flux

with Momentum Flux = Spatial Stress
Tij =
T11
Pressure
xx
T12
Shear
xy
T13
Shear
xz
T21
Shear
yx
T22
Pressure
yy
T23
Shear
yz
T31
Shear
zx
T32
Shear
zy
T33
Pressure
zz



[The Stress-Energy Tensor - Generally {including shear stresses}]
Tμν = ρmo Uμ Uν + p Hμν + (Uμ Qν + Qμ Uν) + Πμν
Tμν =
ρeo Sx/c Sy/c Sz/c
Sx/c xx xy xz
Sy/c yx yy yz
Sz/c zx zy zz

============================================
[For a Perfect Fluid {no spatial heat conduction, no viscosity}]
where ρeo = rest energy density and p = po = pressure, (both of these are Lorentz Scalars)
T
μν = (ρeo+po)UμUν/c2 - poημν
Tμν = (ρmo+po/c2)UμUν - poημν
Tμν =
γ2eo+p) - p γ2eo+p)ux/c γ2eo+p)uy/c γ2eo+p)uz/c
γ2eo+p)ux/c γ2eo+p)uxux/c2 + p γ2eo+p)uxuy/c2 γ2eo+p)uxuz/c2
γ2eo+p)uy/c γ2eo+p)uxuy/c2 γ2eo+p)uyuy/c2 + p γ2eo+p)uyuz/c2
γ2eo+p)uz/c γ2eo+p)uxuz/c2 γ2eo+p)uyuz/c2 γ2eo+p)uzuz/c2 + p


Toμν {restframe}=
ρeo



  p 



  p 



  p 

Toμν {restframe}=
T00 = ρeo T0j = 0
Ti0 = 0 Tij = pδij

The Lorentz Invariant Condition for a perfect fluid: Tr[Tμν] = Tνν = ημν Tμν = (T00) - (T11) - (T22) - (T33) = (ρeo)-(p)-(p)-(p) = ρeo - 3p
This is essentially the scalar magnitude of the Tensor.
Notice the similarity with 4-Vectors:
Scalar Magnitude squared of Tensor Tμν:   ημν Tμν = Tνν
Scalar Magnitude squared of 4-Vector Aμ: ημν AμAν = Aν Aν
Generally for Perfect Fluid: Tμν = (ρeo+po)UμUν/c2 - poημν
Also: ημν ημν = ηνν = δνν = 4
There are at least 3 interesting Invariant scenarios:
Tr[Tμν] = ρeo - 3p = ρeo   { p = 0, Tμν = (ρeo)UμUν/c2}, then all magnitude comes from the energy/mass density of matter particles --> Dust Solution.
Tr[Tμν] = ρeo - 3p = 0      { p = ρeo/3, Tμν = (4ρeo/3)UμUν/c2 - (ρeo/3)ημν = p(4UμUν/c2 - ημν) = p(4Uμ/c2 - ∂μ)Uν}, then the magnitude is 0, and represents light-like/null/photonic particles--> Radiation Solution.
Tr[Tμν] = ρeo - 3p = 4ρeo { p = -ρeo, Tμν = - pημν= + ρeoημν}, then all magnitude comes from the energy density*the Minkowski Metric --> Vacuum Solution.
apparently, another somewhat interesting one is:
If Tr[Tμν] = ρeo - 3p = 2ρeo  { p = -ρeo/3},  --> Curvature Solution, which gives an Einstein Static Universe, but which we now know is not the case.

========================
[For Dust {no particle interaction}]
Tdustμν = (ρeo)UμUν/c2 = (ρmo)UμUν = nomoUμUν = noUμmoUν = NμPν
Todustμν {restframe}=
ρeo



    



    



    
{p = po = 0} for dust {matter particles do not interact}
This is just a perfect fluid with the Lorentz Invariant Condition: Tr[Tμν] = Tνν = ημν Tμν = ρeo = (T00) - (T11) - (T22) - (T33) = (ρeo)-(p)-(p)-(p) = ρeo - 3p = ρeo
Thus {p = 0}
*Note* We have simply gone from momentum of discrete particles Pν = moUν to momentum of a fluid Tμν = Nμ(moUν)
by using the 4-NumberDensity 4-Vector Nμ, where typically ∂μNμ = 0.

===============================================
[For Random Isotropic Radiation/Photons {no particle interaction??}]
Tμν = (4ρeo/3)UμUν/c2 - (ρeo/3)ημν
Tμν = po(4UμUν/c2 - ημν)
Toμν {restframe??, where it is ?random?}=
ρeo



p = ρeo/3



p = ρeo/3



p = ρeo/3
{p = (ρeo/3) = (ρmoc2/3)} for isotropic radiation {particles do not interact}
This is just a perfect fluid with the Lorentz Invariant Condition: Tr[Tμν] = Tνν = ημν Tμν = 0 = (T00) - (T11) - (T22) - (T33) = (ρeo)-(p)-(p)-(p) = ρeo - 3p = 0
Thus {p = (ρeo/3)}




============================
[For Vacuum {?no particle interaction?}]
Tμν =  - poημν
Tμν =  + ρeoημν
Toμν {restframe??}=
ρeo



p = -ρeo



p = -ρeo



p = -ρeo
{p = -ρeo} for vacuum
This is just a perfect fluid with the Lorentz Invariant Condition: Tr[Tμν] = Tνν = ημν Tμν = 4ρeo = (T00) - (T11) - (T22) - (T33) = (ρeo)-(p)-(p)-(p) = ρeo - 3p = 4ρeo
Thus {p = -ρeo}



===============================================
[For ElectroMagnetic Field]

Tμν {no restframe / null}=
c2ρmo = ρeo = (1/2)εo(e2+c2b2
cg =  cεo(e x b)
cg =  cεo(e x b) σij = -εo[eiej + c2bibj -(1/2)δij(e2+c2b2)]
This is just a perfect fluid with the Lorentz Invariant Condition:
Tr[Tμν] = Tνν = ημν Tμν = +(T00) - (T11) - (T22) - (T33)
= +((1/2)εo(e2+c2b2)) - (-εo[exex + c2bxbx -(1/2)δxx(e2+c2b2)]) - (-εo[eyey + c2byby -(1/2)δyy(e2+c2b2)]) - (-εo[ezez + c2bzbz -(1/2)δzz(e2+c2b2)])
= (1/2)εo(e2+c2b2) + (εo[exex + c2bxbx + eyey + c2byby + ezez + c2bzbz]-3(1/2)εo(e2+c2b2)
= -(2/2)εo(e2+c2b2) + (εo[exex + eyey + ezez + c2bxbx + c2byby + c2bzbz]
= -εo(e2+c2b2) + εo(e2+c2b2)
= 0
Tr[Tμν] = 0, as expected for a null EM field
=====================


In all cases, if the Stress-Energy Tensor is conservative, we have the usual 4-Divergence = 4-Zero.
μTμν = Z = (0,0), which is actually 4 equations, Conservation of Energy + Conservation of Momentum for each direction.



 

Ok, let's recap a bit...

R·R = (cτ)2 or (0) or -(ro2), depending on event interval
U·U = (c)2
A·A = -(ao)2
dR·dR = (cdτ)2
ΔR·ΔR = (cΔτ)2
P·P = (moc)2 = (Eo/c)2
F·F = γ2[ (Ė/c)2 - f·f ]
Fp·Fp = -(fo)2
Fh·Fh = (γĖo/c)2
N·N = (noc)2
J·J = (ρoc)2
A·A = (φo/c)2
Q·Q = (Uo/c)2
PT·PT = (Ho/c)2
K·K = (ωo/c)2 = (1/cTo)2
∂·∂ = (τ/c)2

T·T = (1) Unit Temporal
T·S=(0) = N·N --> They are orthogonal
S·S = (-1) Unit Spatial
Every Physical 4-Vector has invariant scalar products.


[X] = ημν = ( )·( ) = The Scalar Product Dot
The Minkowski metric is at the heart of SR.


Σ*[ Pn ] = 0
PT = Σ[Pm] + Σ[qnAn]
Conservation of 4-Momentum and/or 4-TotalMomentum plays a key role.


U·U = (c)2
U·∂ = d/dτ = γ(d/dt)
U·ΔX = c2Δτ
dU/dτ = U·A = 0
U·F = Ė
U·P = Eo
U·AEM = φo
U·J = ρoc2
U·K = ωo
U·FEMμν = 0
Many of the relations are just the "Invariant Rest Value of the Temporal Component Rule".


A = dU/dτ = (U·∂)U
P = moU = (Eo/c2)U
N = noU
J = ρoU = qnoU = qN
K = (ωo/c2)U
AEM = (φo/c2)U
AEM = (q/4πεoc[R·U]ret)U {for a point charge}
Many of the relations are just a Lorentz Scalar (1 independent component) times the 4-Velocity U (3 independent components),
giving a resultant 4-Vector which has (4 independent components).


J = (ρo/mo)P or P = (moo)J
J = qN
K
= (ωo/Eo)P
or P = (Eoo)K
Since many of the 4-Vectors are Lorentz Scalar proportional to 4-Velocity U, they will be proportional to one another via Lorentz Scalars.


∂·X = 4
∂·J = (∂t/c,-)·(ρc,j) =   (∂tρ + ∇·j) = ? = 0
∂·U = (∂t/c,-)·γ(c,u) = ( ∂tγ + ∇·u] ) = ( ∂tγ + γ( ∇·u ) + ( u·∇[γ] ) ) = ? = 0
∂·AEM = (∂t/c,-)·(φ/c,a) = (∂tφ/c2+∇·a) = 0 {In the Lorenz Gauge}
∂·J = 0 {for an EM field}
∂·Fαν  = μoJ
The 4-Divergence is used quite often, and anytime the result is 0 indicates a conservative system (no sinks or sources).


K = -[Φ] and Φ = -K·R: 4-WaveVector is negative 4-Gradient of SR Phase (Φ)
KT = -T] and ΦT = -KT·R: 4-TotalWaveVector is negative 4-Gradient of SR TotalPhase (ΦT)
PT = -[S] and S = -(PT·R): 4-TotalMomentum is negative 4-Gradient of SR Action (S)
ν[PTμ] = q∂μ[AEMν]: The magic behind the EM curtain...
Some very important 4-Vector relations can be defined via 4-Gradient


Ε·Ε = -1.
Ε·K = 0
Polarization States can be understood as simple Lorentz Invariant conditions.


Tperfectfluidμν = (ρeo+po)UμUν/c2 - poημν
Tdustμν = ρeoUμUν/c2, because Tr[Tdustμν] = ρeo, which gives {po = 0}
Tvacuumμν = - poημν, because Tr[Tvacuumμν] = 4ρeo, which gives {po = -ρeo}
Tradiationμν = po(4UμUν/c2 - ημν), because Tr[Tradiationμν] = 0, which gives {po = ρeo/3}
Special subcases of the Stress-Energy Tensor can be understood as simple Lorentz Invariant conditions.


U·U = (c)2
∂·Fαν  = (μo)J
U·Fαν = (-1/q)F
or J = (q)N
P = (Eoo)K
Important Physical constants can be found in the 4-Vector Relations.


================================================

Ok, up to this point, all of the relations are pretty much straight-forward, no doubts, pure SR.
The versatility of the SR Lorentz Scalar Product, and of the SR 4-Gradient are very apparent.
The use of the 4-Gradient as a 4-Divergence is powerful
The connection to Classical EM Theory is apparent.

The next couple of relations will be somewhat controversial if one is stuck in the old paradigm of QM as a separate system from SR.
However, I will present evidence that these relations are also purely from SR, and supported by empirical observation.
A few of the empirical effects are: The PhotoElectric Effect, the Matter-Wave Effect, and the Compton Effect.

================================================

P = ћK
This is the 4D SR version of the de Broglie matter-wave relation. One could argue that this is a quantum axiom.  However, in this case, it is not.
It is derivable from the SR relations P = moU = (Eo/c2)U and K = (ωo/c2)U
Since both P and K are Lorentz Scalar proportional to U, then by the mathematics of tensors P must be Lorentz Scalar proportional to K.
(i.e. Tensors obey mathematical transitivity, or in this case they are also Right Euclidean   {if a~c and b~c, then a~b} )
P = moU = (Eo/c2)U = (Eo/c2)/(ωo/c2)K = (Eoo)K = (γEo/γωo)K = (E/ω)K = (ћ)K
Now, technically, all SR tells us is that there is a invariant Lorentz Scalar with dimensional unit of [action] between these 4-vectors.  It does not give a value. That comes from experiment.  It is an empirical observation that (E/ω) = (ћ) for all known particle types.
However, that is also the case with the Lorentz Scalar speed-of-light (c).  SR says there is a invariant Lorentz Scalar with dimensional unit of [velocity], but the value must be empirically measured.
The constant with dimensional unit [action] turns out to be (ћ = h/2π), with (h) as Planck's constant, can be empirically measured via classical (non-QM) physics knowledge.
This relation is fundamentally no different from the P = (moo)J relation which we derived earlier; the method is the same.
Experiments include: Threshold Voltages of LED's, ring radii of Electron Crystal Diffraction, the momentum of Photo-Electrons, the Watt Balance, etc.

For the LED experiment, one uses several different LED's, each with it's own characteristic wavelength.
One then makes a chart of wavelength (λ) vs threshold voltage (V) needed to make each LED emit,
noting that they are proportional via an unknown constant.
One finds that { λ = h*c/(eV) }, with e = ElectronCharge and c = LightSpeed.
h is found by measuring the slope. Consider this as a blackbox for which no assumption about QM is made.
However, we know that E = eV, and λf = c for photons.
The data force one to conclude that E = hf = ћω.
Applying our 4-Vector knowledge, this means that P = (E/c,p) = ћK = ћ(ω/c,k),
because the temporal equation forces the spatial equation to be true as well due to SR mathematics. Thus
P = ћK

Historically:
Planck discovered h based on statistical-mechanics/thermodynamic considerations of the black-body problem.
Einstein applied Planck's idea to photons in the Photoelectric Effect to give E = ћ ω.
de Broglie realized that every particle, massive or massless, has 3-vector momentum p = ћ k.
Putting it all together naturally produces the SR 4-Vector relation P = ћ K = (E/c,p) = ћ(ω/c,k).
Thus the Matter-Wave Effect was established.

This also gives Action S = ћΦT with S = -(PT·R) = - ћ(KT·R)
=====================


K·K = (ωo/c)2 = (1/cTo)2 = (moc/ћ)2 = (1/λC)2
Now that there is a relation between the 4-Momentum and the 4-WaveVector, we can now notice a few constant relations based on the 4-WaveVector.
*Note* All values with strike-thru̇s are the reduced version of the regular values:
i.e. Reduced Compton Wavelength (λC = λC/2π =  ћ/moc), Reduced Planck's Constant (ћ = h/2π), etc., and, of course, (ωo = 2πνo)
The rest angular frequency (ωo) is related to: the rest reduced period (To), the rest mass (mo), and the Reduced Compton Wavelength (λC), which is *not* a rest wavelength.
The rest value of a massive particle would be when | k| = 0, which is the same as when the particle | λ| = Infinity.
The Compton Wavelength (λC = h/moc) is instead defined to be the wavelength of a photon with the equivalent energy of a massive particle of rest mass (mo).
This can be seen in [ K = (ω/c,k) = (1/cT,/λ) ]. The wavelength is set to the same value as c*period, which would make the 4-WaveVector a null or light-like 4-Vector.
The Compton Scattering Formula is Δλ = λC(1-cos[θ]) or Δλ = λC(1-cos[θ]) and is typically used to calculate the wavelength shift from the scattering of a photon off of an electron.
The effect was important for proving that light could not have a "wave-only" description, but that light as individual photons with momentum was necessary.
Compton Scattering Derivation
P·P = (moc)2 generally == >0 for photons
Pphot1·Pphot2 = ћ2K1·K2 = (ћ2ω1ω2/c2)(1-1·2) = (ћ2ω1ω2/c2)(1-cos[ø])
Pphot·Pmass = ћK·P = (ћω/c)(1,)·(E/c,p) = (ћω/c)(E/c-n̂·p) = (ћωEo/c2) = (ћωmo)
Pphot + Pmass = P'phot + P'mass :Conservation of 4-Momentum in Photon·Mass Interaction
===
Pphot + Pmass - P'phot = P'mass  :rearrange
(Pphot + Pmass - P'phot)2 = (P'mass)2 :square to get scalars
(Pphot·Pphot + 2 Pphot·Pmass - 2 Pphot·P'phot + Pmass·Pmass - 2 Pmass·P'phot + P'phot·P'phot) = (P'mass)2
(0 + 2 Pphot·Pmass - 2 Pphot·P'phot + (moc)2 - 2 Pmass·P'phot + 0) = (moc)2
Pphot·Pmass - Pmass·P'phot = Pphot·P'phot
(ћωmo)-(ћω'mo) = (ћ2ωω'/c2)(1-cos[ø])
(ω-ω')/(ωω') = (ћ/moc2)(1-cos[ø])
(1/ω'-1/ω) = (ћ/moc2)(1-cos[ø])
(λ' - λ) = (ћ/moc)(1-cos[ø])
Δλ = λC(1-cos[ø]): The Compton Effect
=====================


= -iK {for  f = ae^-i(K·X)}
The SR Plane-Wave Equation... among other things...
Up to this point, we have noticed that the various SR 4-Vectors have been related to one another by simple Lorentz Scalar constants.
Now, let's do a mathematical experiment, let's see if the 4-Gradient can be related to the other SR 4-Vectors with a simple Lorentz Scalar.
Which 4-Vector might it be related to?  Well, both and K have dimensional units of [length-1].
Also, we showed earlier that { K = -[Φ] }, i.e. the 4-WaveVector is defined by the negative 4-Gradient of the SR Phase.
Let's take this a bit further.
Let f = f(Φ) be some function of the SR Phase.
Then [f] = f ' [Φ] = f ' (-K) = (-K) f '
Is there a general function ( f ) that solves this?
Try { f(Φ) = ae^i(Φ) }, a plane wave equation which gives { f ' = i f }
[f] = (i)ae^i(Φ)[Φ] = (i)ae^i(Φ)(-K) = (-iK)ae^i(Φ) = (-iK)f
[f] = (-iK)f
= -iK   {for  f = ae^i(Φ), and this is manifestly Lorentz Invariant}

Now, showing the same argument slightly differently...
K·X = (ω/c,k)·(ct,x) = (ωt - k·x) = -Φ
Try {f = ae^b(K·X) = ae^b(-Φ)}, which is just a simple exponential function of 4-Vectors, specifically the SR Phase Φ in this case...
Then [f] = [ae^b(K·X)] = (bK)ae^b(K·X) = (bK)f
And ∂·∂[f] = b2(K·K)f = (bωo/c)2f
Note that { b = -i } is an interesting choice --> it leads to SR Plane Waves, which we observe empirically, e.g. EM Plane Waves = Photons...
Also, since K is the 4-WaveVector, this is a totally reasonable choice.
This gives:
[f] = (-iK)ae^-i(K·X)
[f] = (-iK)f
= -iK   {for  f = ae^-i(K·X), and this is manifestly Lorentz Invariant}
This is nice because the (-i) ends up being a Lorentz Scalar invariant between the two 4-Vectors, and K.
This is the same thing that was used in the relativistic description of the 4-WaveVector
The relativistic condition {K = -∂[Φ]} is that which chooses {f = ae^i(Φ)}, and thus { = -iK}
=====================



Ok, so for all the relations above, everything was found via pure mathematics or via empirical experimentation.
They are all manifestly covariant, not to mention extremely simple - Many of the 4-Vectors are simple Lorentz scalar proportional to one another.
In no case was knowledge of QM or QM axioms required.
R = (ct,r)
U = (d/dτ) R
P = (mo) U
K = (1/ћ) P
= (-i) K
What are the consequences of = -iK?


Answer: We just derived Quantum Mechanics (QM) from Special Relativity (SR) + a few empirical observations.
This is the SRQM Interpretation.  We did NOT assume any Quantum Axioms.

Quantum Mechanics from SR 4-Vector Relations

What are usually known as QM Axioms are instead SR derived Principles of QM...


The Relativistic Klein-Gordon Quantum Wave Equation:
Form a chain of Lorentz Scalar Equations...
U·U = (c)2
P·P = (moc)2
K·K = (moc/ћ)2
∂·∂ = (-imoc/ћ)2 = -(moc/ћ)2 = -(ωo/c)2
The last is the Klein-Gordon Equation, the Relativistic Quantum Wave Equation for Spin-0 Particles.
This is RQM = Relativistic Quantum Mechanics.
It is interesting to note that one can do the KG Equation with just a rest frequency (ωo) and the speed of light (c).
Planck's constant (h) is not actually required.



Operator Formalism: [] = -iK
Unitary Evolution: = [-i]K
Wave Structure: = -i[K]
These three are straight from the basic equation for plane waves.
1st, the 4-Gradient is already an operator in SR.  We don't need an extra QM axiom for it.
2nd, the wave here evolves unitarily - it's a plane wave. We observe plane waves in photon phenomena.
3rd, it's a particle, but it's also a wave. The 4-WaveVector is a standard SR 4-Vector, as the SR Doppler Effect indicates.



The Schrödinger QM Relations:
P = ћK
K = i
Thus P = (E/c,p) = i ћ = i ћ (t/c,-).
The temporal eqn. is (E = i ћ t) and the spatial eqn. is (p = -i ћ ), the Schrödinger QM Energy and 3-Momentum Relations.
These were generated by the SR Lorentz scalar product relations between the physical SR 4-Vectors { P, K, }
P = ћ K is required to exist since both P and K are Lorentz scalar proportional to U.
K = i is just the noticing photon plane waves.




Non-zero Commutation Relation between position and momentum:
4-Position R = (ct,r)
4-Gradient = (t/c,-)
4-Momentum P = (E/c,p)

Let f be an arbitrary SR function.
X[f] = Xf
[f] = [f]
X[[f]] = X∂[f]
[X[f]] = [Xf] = [X]f + X∂[f]
[Xf] - X∂[f] = [X]f
[X[f]] - X[[f]] = [X]f

Now with commutator notation...
[,X]f = [X]f

And since f was an arbitrary SR function, we can remove it, which leaves...
[,X] = [X] = (∂t/c,-)[(ct,r)] = (∂t/c,-∂x,-∂y,-∂z)[(ct,x,y,z)] = Diag[+1,-1,-1,-1] = ημν = Minkowski Metric

hence
[∂,X] = ημν = Minkowski Metric
=======================
At this point, we have established purely mathematically, that there is a non-zero commutation relation between the SR 4-Gradient and SR 4-Position

Now, using { = -iK and P = ћK }, which we derived from above...
[∂,X] = ημν
[K,X] = i ημν
[P,X] = i ћ ημν

[Xμ,Pν] = - i ћ ημν and, looking at just the spatial part...

[xj,pk] = i ћ δjk Hence, we have derived the standard QM commutator rather than assume it as an axiom...
The non-zero commutation is not about the size of (ћ) or the imaginary (i), the source is the Minkowski Metric, which gives non-zero commutation relations between the SR 4-Gradient and SR 4-Position.
=======================

Going back to the original argument, this time using a function of 4-Position
Let f be an arbitrary SR function, and let g = g(X).
g(X)[f] = g(X)f
[f] = [f]
g(X)[[f]] = g(X)[f]
[g(X)[f]] = [g(X)f] = [g(X)]f + g(X)[f]
[g(X)f] - g(X)[f] = [g(X)]f

Now with commutator notation...
[,g(X)]f = [g(X)]f

And since f was an arbitrary SR function...
[,g(X)] = [g(X)] = g'(X)[X] = g'(Xuv
The usual commutation results using a function of position...
=======================


We can actually continue this using the Wave-space or Momentum-space representation

4-WaveVector K = (ω/c,k)
4-Gradient = X = (t/c,-x)
4-WaveGradient K = (c∂ω,-k)
4-MomentumGradient P = (c∂E,-p)

Let f be an arbitrary SR function.
K[f] = Kf
K[f] = K[f]
K[K[f]] = K∂K[f]
K[K[f]] = K[Kf] = K[K]f + K∂K[f]
K[Kf] - K∂K[f] = K[K]f

Now with commutator notation...
[K,K]f = K[K]f

And since f was an arbitrary SR function...
[K,K] = K[K] = (c∂ω,-k)[(ω/c,k)] = (c∂ω,-∂kx,-∂ky,-∂kz)[(ω/c,kx,ky,kz)] = Diag[1,-1,-1,-1] = ημν = Minkowski Metric

hence
[K,K] = ημν = Minkowski Metric
Once again, the non-zero commutation is due to the Minkowski Metric, not Planck's constant (ћ), not the imaginary unit (i)
=========================
Now then, we arbitrary chose (-i) from the SR phase definition of plane waves: = -iK
or, to be careful here, X = -iK

[X,X] = ημν = [K,K]
[X,X] = [K,K]
[X,X] = -[K,∂K]
[-iK,X] = -[K,∂K]
-i[K,X] = -[K,∂K]
[K,X] = -i[K,∂K]
[K,X] = [K,-iK]

X = -iK
K = iX

We see that the Fourier Transform correctly pops out automatically
X = (-i)K
K = (+i)X

And continuing to momentum-space ( P = ћK , K = ћP )
X = -iK = -iћP

P = (E/c,p) = i ћ X = i ћ (t/c,-x ) = i ћ ( t/c,-x ) giving ( E = iћt and p = -iћx ) in the Position-Space Representation
X = (ct,x) = -i ћ P = -i ћ ( c∂E,-p ) = i ћ (-c∂E,k ) giving ( t = -iћ∂E and x =   iћp ) in the Momentum-Space Representation

Note: Given that we started with the SR spacetime formulation, there is no valid reason to discount the ( t = -iћ∂E ) formula.
Space and time are on an equal footing in SR. It is just as valid as the other 3 formulas.

Those derivations that do discount (t = -iћ∂E ) as a valid statement started with classical assumptions, not relativistic, which is incorrect thinking:
(i.e. Relativistic is *NOT* the generalization of Classical/Newtonian,
       Classical/Newtonian is the limiting case approximation of Relativistic for |v| <<c )

So, the Lorentz Scalar (i) is the physical "thing" connecting particles to waves, since it is this relation that gives the Fourier Transforms and wave relations.
(ћ) is more a connection between the light-cone boundary (massless photonic behavior) and the light-cone interior (massive particle behavior).



Heisenberg Uncertainty:
The core of the uncertainty equation comes from the non-zero commutator, which was proved above.
The Generalized Uncertainty Relation: σf2σg2 = (ΔF) * (ΔG) > = (1/2)|< i[F,G] >|
The uncertainty relation is a general mathematical property, which applies to both classical or quantum systems.

The Cauchy–Schwarz inequality asserts that (for all vectors f and g of an inner product space, with either real or complex numbers):
σf2σg2 = [< f | f > · < g | g >]   > =   |< f | g >| 2

But first, let's back up a bit; Using standard complex number math, we have:
z = a+ib
z* = a-ib
Re(z) = a = (z + z*)/(2)
Im(z) = b = (z - z*)/(2i)
z*z = |z|2 = a2 + b2 = [Re(z)]2 + [Im(z)]2 = [(z + z*)/(2)]2 + [(z - z*)/(2i)]2
or
|z|2 = [(z + z*)/(2)]2 + [(z - z*)/(2i)]2

Generically, based on the rules of a complex inner product space we can arbitrarily assign:
z = < f | g >, z* = < g | f >

Which allows us to write:
|< f | g >| 2  =   [(< f | g> + < g | f >)/(2)]2 + [(< f | g> - < g | f >)/(2i)]2
*Note* This is not a QM axiom - This is just pure math.  At this stage we already see the hints of commutation and anti-commutation.

We can also note that:
| f  > = F| ψ > and | g > = G| ψ >

Thus,
|< f | g >| 2 = [(< ψ |F* G| ψ> + < ψ |G* F| ψ >)/(2)]2 + [(< ψ |F* G| ψ> - < ψ |G* F| ψ >)/(2i)]2

For Hermetian Operators...
F* = +F, G* = +G

For Anti-Hermetian (Skew-Hermetian) Operators...
F* = -F, G* = -G

Assuming that F and G are either both Hermetian, or both anti-Hermetian...

|< f | g >| 2 = [(< ψ | (±)FG| ψ> + < ψ |(±)GF| ψ >)/(2)]2 + [(< ψ |(±)FG| ψ> - < ψ |(±)GF| ψ >)/(2i)]2

|< f | g >| 2 = [(±)(< ψ |FG| ψ> + < ψ |GF| ψ >)/(2)]2 + [(±)(< ψ |FG| ψ> - < ψ |GF| ψ >)/(2i)]2

We can write this in commutator and anti-commutator notation...

|< f | g >| 2 = [(±)(< ψ |{F,G}| ψ >)/(2)]2 + [(±)(< ψ |[F,G]| ψ >)/(2i)]2

Due to the squares, the (±)'s go away, and we can also multiply the commutator by an ( i2 )

|< f | g >| 2 = [(< ψ |{F,G}| ψ >)/2]2 + [(< ψ |i[F,G]| ψ >)/2]2

|< f | g >| 2 = [(< {F,G} >)/2]2 + [(< i[F,G] >)/2]2

The Cauchy–Schwarz inequality again...
σf2σg2 = [< f | f > · < g | g >]   > =   |< f | g >| 2 = [(< {F,G} >)/2]2 + [(< i[F,G] >)/2]2

Taking the root:
σfσg > =   (1/2)|< i[F,G] >|
Which is what we had for the generalized Uncertainty Relation.


Now for a few comments:
The Uncertainty Principle *IS NOT* about properties, it *IS* about measurements.
The commutator, and hence the Heisenberg Uncertainty Equation, is specifically about "sequential" measurements,
  not "simultaneous" measurements and not "simultaneous" properties.
This commutator has to do with the order of the operations acting on an event's worldline, which indicates timelike intervals, not spacelike intervals.
Thus, it makes no statement about whether a given event's "properties" can be simultaneous or not.
It also makes no statement about "simultaneous measurements".
It does make a statement about the effects of "sequential measurements" along the timelike intervals of individual worldlines.
The SRQM interpretation is that non-commuting measurements cannot be simultaneous, which makes way more sense than non-commuting properties not being able to be simultaneous.

Examples:
Consider a coin, a pencil with eraser, a light.  The properties of the coin are as follows: heads up/tails up, visible side marked/unmarked, lighted/un-lighted.
Possible operations are: Flip Coin, Mark/Unmark Coin Face, Light On/Off

Example of commuting operations: [Flip Coin , Light Switch] = 0
Start Heads Up & No Light:  Then Flip to Tail followed by Light On = Light On followed by Flip to Tail.

Example of non-commuting operations: [Flip Coin, Mark Coin] = / = 0
Start Heads Up & No Mark:  Then Flip to Tail followed by Mark Coin Face = / = Mark Coin Face followed by Flip to Tail.

Note that some of these properties are with respect to the observer, following the whole idea of the Relativistic Derivation.
The coin always has each of these binary properties (1 of  2 sides up, face marked or unmarked, face lighted/un-lighted), it is the measurement process that is possibly non-commuting.

Which operation can be simultaneous? Try Light and Flip...  These are independent, they commute.
Which operations cannot be simultaneous? Try Mark and Flip...  You can't do both at the same time, they are non-commuting.

So, Uncertainty is not a fundamentally quantum thing...
Also, again note that the non-commutivity comes from the operations, the measurements, not the properties.

Let's examine the simultaneous vs. sequential aspect of it:

Consider the following:
Operator A acts on System | Ψ > at SR Event A:  A| Ψ > →| Ψ' >
Operator B acts on System | Ψ' > at SR Event B:  B| Ψ' > →| Ψ'' >
or BA| Ψ > = B| Ψ' > = | Ψ'' >

If measurement Events A & B are space-like separated, then there are observers who can see {A before B, A
simultaneous with B, A after B}, which of course does not match the quantum description of how Operators act on
Kets

If Events A & B are time-like separated, then all observers will always see A before B.  This does match how the
operators act on Kets, and also matches how | Ψ > would be evolving along its worldline, starting out as | Ψ >,
getting hit with operator A at Event A to become | Ψ' >, then getting hit with operator B at Event B to become | Ψ'' >.

The Uncertainty Relation here does NOT refer to simultaneous (space-like separated) measurements nor simultaneous properties,
it refers to sequential (time-like separated) measurements along individual worldlines.
This removes the need for ideas about the particles not having simultaneous properties.
There are simply no “simultaneous measurements” on an individual system, on a single worldline.
They are sequential, and the first measurement places the system in such a state that the outcome of
the second measurement will be altered wrt. if the order of the operations had been reversed.



Resolution of the EPR Paradox:
The EPR paradox does not refute Heisenberg Uncertainty, nor locality, nor realism.
The EPR paradox *IS NOT* about measurements, it *IS* about properties.
In the SRQM Interpretation, it shows that particles *CAN* have simultaneous properties, which are local and real.
As stated before, the Heisenberg Uncertainty Principle is about the non-commutative properties of the Minkowski Metric with regard to sequential measurement acts.
This commutator has to do with the order of the operations/measurements on a single particle/event, which indicates timelike intervals, not spacelike intervals.

Thus, it makes no statement about whether a given event's "properties" can be simultaneous or not.
It also makes no statement about "simultaneous measurements" on an individual worldline.
It does make a statement about the effects of "sequential measurements" along the timelike intervals of individual worldlines.

In the EPR Paradox, the correlated particles are space-like separated, and along their own individual worldlines.
The observer correlates particles A and B at some local spacetime event, and then allows them to space-like separate onto their own individual worldlines.
At the point of correlation, the particles have identically opposite ket vectors due to conservation laws. (P) + (-P) = 0. Conservation of property P.
As the particles separate, the ket vectors go along with their own particles.
Each particle has it's own ket vector.  It is a simply an informational state.
A measurement act on one particle does not affect the other space-like separated particle.
Even though they share the same informationally correlated state, the measurement on A has no instantaneous effect on B, or vice-versa.
The only thing that happens is that the correlated state is (usually) removed from the particle under observation due to the measurement act.
In fact, there are some possible observers who see event A before B, some who see A simultaneous with B, and some who see A after B.
The measurement of one cannot have any operational effect on the other, as this would violate the rules of SR, and we have shown above that QM is directly derivable from SR.

Thus, the EPR Paradox does indeed prove Einstein correct, that a particle does have simultaneous properties.
The interesting fact of nature is that apparently only certain commuting properties can be simulatenously measured on the same particle.
The Heisenberg Uncertainty is about sequential measurements along individual worldlines.



Quantum Superpostion:
Superposition is the simple mathematical consequence of the Klein-Gordon Equation being a linear PDE.
Any linear PDE (partial differential equation) will obey the principle of superposition.
This is a mathematical artifact of the KG Equation itself.  It is not necessary as an additional axiom.
Klein-Gordon Equation: ∂·∂ = (-imoc/ћ)2 = -(moc/ћ)2 = -(ωo/c)2
Straight from standard calculus of PDĖs... find the particular solution and the homogeneous solutions...
K·K     = (ω/c)2 - k·k = (ωo/c)2: The particular solution (with rest mass or rest frequency)
Kn·Kn = (ωn/c)2 - kn·kn = 0 : The homogenous solution for a (virtual photon?) microstate
Note that Kn·Kn = 0 is a null 4-vector (photonic).
Let Ψp = ape^-i(K·X), then ∂·∂p] = (-i)2(K·Kp = -(ωo/c)2Ψp
which is the Klein-Gordon Equation particular solution...
Let Ψn = ane^-i(Kn·X), then ∂·∂n] = (-i)2(Kn·Knn = (0)Ψn
which is the Klein-Gordon Equation homogeneous solution for a microstate (n).
We may now take Ψ = Ψp + Σn Ψn
Note that the amplitude ap and an's can be complex valued constants.
∂·∂[Ψ] = (-i)2[(K·K)+Σn(Kn·Kn)]Ψ = (-i)2[(K·K)+(0)]Ψ = -(ωo/c)2Ψ
Thus there are any number of possible microstates (n) which solve the KG Equation.
Also note that since the homogeneous solutions are null, this could explain the apparent "extent" of a wavefunction beyond that of the particle.

The possibility of having an expression of the form KTotal = Σn(Kn) arises.
We can multiply each term by ћ.
PTotal = Σn(Pn) = (ћ)KTotal = (ћ)Σn(Kn) = ΣnKn)
Showing that a 4-TotalMomentum 4-vector can be constructed from a sum of individual 4-Momentuṁs.



Quantum Hilbert Space:
Hilbert Space is the simple mathematical consequence of the Klein-Gordon Equation being a linear PDE.
As with superposition, Hilbert spaces occur as a consequence of the equation.  They occur in contexts besides those of QM as well.
They are not necessary as an additional axiom.



The Schrödinger Quantum Wave Equation:
The Schrödinger QM Equation is simply the low-velocity limit of the KG Equation.
It is formally similar to a diffusion equation.

4-Momentum P = (E/c,p)
4-Gradient = (t/c,-)
P = ħK = iħ: The Schrödinger Relation which we derived from SR above
(E/c,p) = iħ(∂t/c,-), or in component form, [E = iħ ∂t] and [p = -iħ ]

∂·∂
= (t/c)2 - 2 = (-imoc/ћ)2 : The Klein-Gordon Equation
P·P = (E/c)2 - p2    = (moc)2: Einstein Mass-Energy-Momentum Equivalence

E2 = (moc2)2 + c2p2: Einstein Relativistic Energy Relation
E = Sqrt[(moc2)2+ c2p2]
E = (moc2)Sqrt[1 + (p2/(moc)2)]
E ~ (moc2)  [1 + (p2/2(moc)2)] : Low velocity (|v| <<c) Newtonian Limiting Case using {  Sqrt[1+x] ~ (1 + x/2 + ...O[x2] )  }
E ~ [ (moc2) + p2/2mo] : Newtonian approximation gives Total Energy = Rest Energy + Kinetic Energy

(iħ∂t) ~ [ (moc2) + (-iħ)2/2mo] : The Free Particle Schrödinger Equation
(iħ∂t) ~ [ (moc2) - (ħ)2/2mo] : The Free Particle Schrödinger Equation



The Schrödinger Quantum Wave Equation including the effects of a potential V:
The Schrödinger QM Equation is simply the low-velocity limit of the KG Equation.
It is formally similar to a diffusion equation.
Here, I have used the Minimal Coupling relations, themselves based on 4-vectors

4-Momentum P = (E/c,p)
4-Gradient = (t/c,-)
4-VectorPotential A = (φ/c,a)
4-TotalMomentum PT = (H/c,pT) = (ET/c + qφ/c , p + qa)

*Note*
The SR Phase Φ = - K·X = -(ω/c,k)·(ct,x) = -(ωt - k·x) = (-ωt + k·x)
The Scalar Potential φ is the temporal component of the 4-VectorPotential A = (φ/c,a)
These are not the same thing.

P = ħK = iħ: The Schrödinger Relation which we derived from SR above
PT =   P + qA
: Minimal Coupling Relation, essentially the Total 4-Momentum = 4-Momentum of the Particle + 4-Momentum of the Potential

P = (E/c,p)     = PT - qA      = (ET/c  -  qφ/c       , pT - qa)
= (∂t/c,-) = T+ (iq/ħ)A = (∂tT/c + (iq/ħ)φ/c , -T+ (iq/ħ)a)

∂·∂ = (t/c)2 - 2 = (-imoc/ћ)2
P·P
= (E/c)2 - p2    = (moc)2: Einstein Mass-Energy-Momentum Equivalence

E2 = (moc2)2 + c2p2: Einstein Relativistic Energy Relation
E = Sqrt[(moc2)2+ c2p2]
E = (moc2)Sqrt[1 + (p2/(moc)2)]
E ~ (moc2)  [1 + (p2/2(moc)2)] : Low velocity (|v| <<c) Newtonian Limiting Case using {  Sqrt[1+x] ~ (1 + x/2 + ...O[x2] )  }
E ~ [ (moc2) + p2/2mo]

E2 = (moc2)2 + c2p2: Einstein Relativistic Energy Relation
E ~ [ (moc2) + p2/2mo]: Newtonian Energy Relation (the low velocity limiting case)

(ET-qφ)2 = (moc2)2+ c2(pT-qa)2: Relativistic w Minimal Coupling
(ET-qφ) ~ [ (moc2) + (pT-qa)2/2mo] : Low velocity w Minimal Coupling

(iħ∂tT-qφ)2 = (moc2)2+ c2(-iħT-qa)2: Relativistic w Minimal Coupling
(iħ∂tT-qφ) ~ [ (moc2) + (-iħT-qa)2/2mo] : Low velocity w Minimal Coupling

(iħ∂tT) ~ [ qφ +(moc2) + (-iħT-qa)2/2mo] : Low velocity w Minimal Coupling
(iħ∂tT) ~ [ V + (-iħT-qa)2/2mo] : with [ V = qφ + (moc2) ]
(iħ∂tT) ~ [ V - (ħT)2/2mo] : Typically the vector potential a ~ 0 in many physical situations, esp. low velocity/low energy situations
(iħ∂tT)| Ψ> ~ [ V - (ħT)2/2mo]| Ψ> : The Schrödinger Equation with potential ( = standard non-relativistic QM)

Note that most books don't mention that the vector potential ( a ) is actually required by Lorentz Invariance,
since it's effects are typically small in the low-energy limit.



Conservation of Probability, Conservation of Probability Current, Conservation of Charge Current:
Consider the following purely mathematical argument (based on Green's Vector Identity):
·( f [g] - [f] g ) = f ∂·∂[g] - ∂·∂[f] g, with f and g as SR Lorentz Scalar functions

·( f [g] - [f] g )
= ·( f [g] ) - ·( [f] g )
= (f ∂·∂[g] + [f]·[g]) - ([f]·[g] + ∂·∂[f] g)
= f ∂·∂[g] - ∂·∂[f] g

So, f ∂·∂[g] - ∂·∂[f] g = ·( f [g] - [f] g )
I can also multiply this by a Lorentz Invariant Scalar Constant s
s (f ∂·∂[g] - ∂·∂[f] g) = s ·( f [g] - [f] g ) =   ·s( f [g] - [f] g )
Ok, so we have the math that we need...


Now, on to the physics...
∂·∂ = (-imoc/ћ)2 = -(moc/ћ)2: The Klein-Gordon Eqn
∂·∂ + (moc/ћ)2 = 0

Let it act on Lorentz Invariant function g
∂·∂ [g]+ (moc/ћ)2 [g] = 0 [g]
Then pre-multiply by f
[f] ∂·∂ [g]+ [f] (moc/ћ)2 [g] = [f] 0 [g]
Resulting in
[f] ∂·∂ [g]+ (moc/ћ)2[f][g] = 0

Now, do similarly with Lorentz Invariant function f
∂·∂ [f]+ (moc/ћ)2 [f] = 0 [f]
Then post-multiply by g
∂·∂ [f][g]+ (moc/ћ)2 [f][g] = 0 [f][g]
Resulting in
∂·∂ [f][g]+ (moc/ћ)2 [f][g] = 0

Now, subtract the two equations
{[f] ∂·∂ [g]+ (moc/ћ)2[f][g] = 0} - {∂·∂ [f][g]+ (moc/ћ)2 [f][g] = 0}
Resulting in
[f] ∂·∂ [g]+ (moc/ћ)2[f][g] - ∂·∂ [f][g]- (moc/ћ)2 [f][g] = 0
[f] ∂·∂ [g] - ∂·∂ [f][g] = 0

And as we noted from the mathematical identity at the start...
[f] ∂·∂ [g] - ∂·∂ [f][g] = ·( f [g] - [f] g ) = 0
s ([f] ∂·∂ [g] - ∂·∂ [f][g]) = s ·( f [g] - [f] g ) = s 0 = 0

Therefore,
s ·( f [g] - [f] g ) = 0
·s( f [g] - [f] g ) = 0

There is a conserved current 4-vector, Jprob = s ( f [g] - [f] g ), for which ∂·Jprob = 0, and which also solves the Klein-Gordon equation.

Let's choose as before ( = -i K) with a plane wave function f = ae^-i(K·X), and choose g = f* = ae^i(K·X) as it's complex conjugate.
At this point, I am going to choose s = (iћ/2mo), which is Lorentz Scalar Invariant, in order to make the probability have dimensionless units and be normalized to unity in the rest case.

Then Jprob = s ( f [g] - [f] g ) = s{f (-i) K g} - s{i K f g} = s (-2i) K fg
Jprob = -2is K fg
Jprob = (ρprobc,jprob) = -2is K fg = -2is (ω/c,k) fg

Now, take the temporal component
ρprobc = -2is (ω/c) fg
ρprob = -2is (ω/c2) fg
ρprob = -2i (iћ/2mo)(γωo/c2) fg
ρprob = (-i2 )(2/2)(ћωo/moc2)(γ) fg
ρprob = (γ) fg

Now, to put it in a more obvious form...  f = ψ*, g = ψ
I have put off using psi until now to point out that (f and g) could be any kind of SR Lorentz Invariant functions for this derivation.
ρprob = (γ)(ψ*ψ) = (γ)(ρo)

o_prob) = (ψ*ψ) = fg
The probability density ρprob = γρo_prob is only equal to (ψ*ψ) in the Newtonian low-energy limiting case with γ --> 1
Thus, the Born Probability Interpretation of (ψ*ψ) only applies in the Newtonian low-energy limiting case

Jprob = s ( f [g] - [f] g ) = (iћ/2mo)( ψ* [ψ] - [ψ*] ψ )
Jprob = (iћ/2mo)( ψ* [ψ] - [ψ*] ψ )

Also, we know that J = ρoU and K = P/ћ = (mo/ћ)U
Jprob = (ρo_prob)(ћ/mo)K
Jprob = (fg)(-2is) K
So, everything matches correctly

Jprob = ρo_probU = (ψ*ψ)U
Jcharge = qJprob

Jcharge = qρo_probU = q(ψ*ψ)U
If we do the calculation including an EM potential, then the 4-ProbabilityCurrentDensity gains an extra component

From before: J = -2is K fg

Minimal Coupling is just observation that the 4-TotalMomentum is the sum of the particle 4-Momentum and the field 4-PotentialMomentum
We have conservation of the Total 4-Momentum PT
PT = P + qA
P = PT - qA
K = KT - (q/ћ)A

Jprob = -2is( K )fg
Jprob = -2is( KT - (q/ћ)A )fg
Jprob = -2is( KT )fg + 2is(q/ћ)A )fg

Jprob = (iћ/2mo)( ψ* [ψ] - [ψ*] ψ ) + 2is(q/ћ)A fg
Jprob = (iћ/2mo)( ψ* [ψ] - [ψ*] ψ ) + 2i(iћ/2mo)(q/ћ)A fg
Jprob = (iћ/2mo)( ψ* [ψ] - [ψ*] ψ ) - (q/mo)A fg

Jprob = (iћ/2mo)( ψ* [ψ] - [ψ*] ψ ) - (q/mo)A (ψ*ψ)

Jcharge = qJprob
Jcharge = q[(iћ/2mo)( ψ* [ψ] - [ψ*] ψ ) - (q/mo)A (ψ*ψ)]
Jcharge = (iћq/2mo)( ψ* [ψ] - [ψ*] ψ ) - (q2/mo)A (ψ*ψ)
4-ProbabilityCurrentDensity Jprob = (iћ/2mo)( ψ* ∂[ψ] - ∂[ψ*] ψ ) - (q/mo)A (ψ*ψ)
4-ChargeCurrentDensity Jcharge = qJprob




Harmonic Oscillation of Position and Momentum Operators:
I'm not sure of the validity of this, but it's interesting...
U·∂
= γ(∂/∂t + u·∇) = γ d/dt = d/dτ
d/dτ = U·∂
d/dτ = (-i K)
d/dτ = (-i/ћ P)
d/dτ = (-imo/ћ U)
d/dτ = (-imo/ћ )U·U
d/dτ = (-imoc2/ћ)
d/dτ = (-iωomoc2/ћωo)
d/dτ = (-iωo)
d2/dτ2 = -(ωo)2
Now, apply this to the 4-Position...
d2X/dτ2 = -(ωo)2X
This is the differential equation of a relativistic harmonic oscillator.
Quantum events oscillate at their rest-frequency ωo.
Likewise for the 4-Momentum:
d2P/dτ2 = -(ωo)2P

One can then apply the {Creation/Annihilation} Operator Formalism:
This is purely mathematical definition, it is not a QM axiom.
One simply defines an SR function of X and
At = √[1/2](X/lo+ lo )
A  = √[1/2]( X/lo - lo )
with ( ћ/moωo = ћωoc2/moωo2c2 = c2o2 = lo2 = {length2})
Note that this makes the A and At 4-Vectors dimensionless
Also, I need to double-check minus sign conventions here..., I might have them backwards

solving for X and gives:
X = lo/(√2) (At+A) = √[ћ/(2moωo)] (At+A)
= 1/(lo√2) (At-A) = √[moωo/(2ћ)] (At-A)
P = iћ = i√[moωoћ/2] (At-A)

from before:
[,X] = [X] = (∂t/c,-)[(ct,r)] = (∂t/c,-∂x,-∂y,-∂z)[(ct,x,y,z)] = Diag[1,-1,-1,-1] = ηuv = Minkowski Metric
one can show that:
AtA = (1/2)(X2/lo2 - ηuv - lo22) =
AAt = (1/2)(X2/lo2 + ηuv - lo22)

[A,At] = (1/2)(2ηuv ) = ηuv
[,At] = [AtA,At] = (AtA)(At)-(At)(AtA) = (At)(AAt)-(At)(AtA) = (At)(AAt-AtA) = (At)[A,At] = (Atuv
[,At] = (At) ηuv
[,A] = -(A) ηuv

more to come...

Now for fermions, we use the anti-commutation relations from the Dirac eqn. instead:
0p0 - γ·p)Ψ = (Γ·P)Ψ = (moc)Ψ
(Γ·P) = (ΓμPμ) = iћ(Γμμ) = (moc)

μν} = 2ημν
See derivation of Dirac Equation from 4-Vector formalism further below...


Next to show Gordon Decomposition of Spin stuff...


Quantum Mysteries:
There are a number of mysterious quantum effects which might now have reasonable and logical explanations:

Hypothesis: All mysterious quantum phenomena which up till now have been attributed simply to the axiomatic weirdness of quantum mechanics are in fact manifestations of relativistic effects.  A couple that come to mind are radioactive decay with it's associated half-life rule, and quantum tunneling.

Let's examine quantum radioactive decay:
A macroscopic sample of a pure substance, meaning all the atoms should be identical, has a "half-life", whereby approximately half of the sample decays into new products within a specified timeframe.  Another half of the remaining will again have decayed in the next specified time-frame, and so on and so on.  The question is why does it occur this way.  If all of the atoms are identical, why don't they all decay at the same time?

Special Relativity to the rescue...

Consider the twin paradox of relativity.  Two individuals start at event A with identical ages.  One individual takes a trip while one remains at rest.  At some future point the individual that had left returns to the the one at rest at event B.  There is a difference in the ages of the two individuals.  While paradoxical, the mathematics behind the difference in aging is fully explained by relativistic laws.  It is only a paradox if you consider to all particles age at the same rate.  Relativity tells us that particles on separate worldlines age according to their own clocks, and that the actual path taken through spacetime determines the age difference upon the reuniting of the particles.

Now, back to the radioactive sample.  Each atom, indeed each subatomic particle is cruising along on it's own worldline.  As the particles "jiggle around" one another, there will be differences in the aging rates due to the twin paradox.  Two particles initially at the same age at event A with be at different ages at event B, thus leading to states in which the system is no longer "in-sync", so to speak.  This could be the physical reason why the atom then undergoes spontaneous decay.

Now, consider a bunch of particles in the sample.  There will be some distribution of ages that build up as the entire sample "ages", with each individual part of it aging at slightly different rates.  Those that get "out-of-sync" undergo spontaneous decay.  I propose that the distribution is that of the half-life rule.  I will attempt to prove this mathematically at some point, but for now it's just a hypothesis.

Consider also that the difference in half-life rates may be a function of "how relativistic" the individual particles get.  Those with large relativistic velocities may get "out-of-sync" more quickly, and thus have shorter half-lives.  This will have to be examined.



One other major quantum mystery is quantum tunneling.  I am currently thinking this might be the "pole in barn" relativistic paradox.
More to come...

Let's summarize a bit:
We used the following relations: (particle/location-->movement/velocity-->mass/momentum-->wave duality-->spacetime structure)
With the exception of 4-Velocity being the derivative of 4-Position, all of these relations are just constants times other 4-Vectors.
R = (ct,r particle/location
U = dR/dτ movement/velocity
P = moU mass/momentum
K = (1/ћ) P wave/particle duality
= -iK spacetime/wave structure


(c) connects the time dimension to the space dimensions
(ћ) connects the light-cone boundary (massless photonic behavior) to the light-cone interior (massive particle behavior)
(i) connects wave to particle


By applying the Scalar Product law to these relations, we get:
U·U = (c)2
P·P = (moc)2
K·K = (moc/ћ)2
∂·∂ = (-imoc/ћ)2 = -(moc/ћ)2
t2/c2 = ∇·∇-(moc/ћ)2
This is the RQM Klein-Gordon Relativistic Wave Eqn (derived from SR with no QM axioms assumed)

More SR Physical 4-Vectors


Going further,

Consider the following more comprehensive set of SR 4-Vectors:

4-Position R = (ct,r) = > (ct,r,θ,z); X = (ct,x) = > (ct,x,y,z)
4-Velocity U = γ(c,u)
4-Acceleration A = γ(cγ̇,γ̇u) = γ(cγ̇,γ̇u + γa) = (cγ4uu̇,γ4uu̇u + γ2a)
4-Differential dR = (cdt,dr); dX = (cdt,dx)
4-Displacement ΔR = (cΔt,Δr); ΔX = (cΔt,Δx)
4-Momentum P = (E/c,p) = (mc,p)
4-MomentumDensity G = (Eden/c,pden) = (ue/c,g)
4-Force F = γ(Ė/c,) = γ(Ė/c,f) = γ(ṁc,f)
4-ForcePure Fp = γ(f·u/c,f)
4-ForceEM FEM = γq( (u·E)/c , (E) + (u x B) )
4-ForceHeat Fh = γṁ(c,u) = γ2o(c,u)
4-ForceScalar Fs = k(∂t[Φ],-[Φ])
4-ForceDensity Fden = γ(Ėden/c,fden)
4-NumberFlux N = (cn,nu) = Σa [ ∫
δ4(X-Xa(τ))dxa/dτ ]
4-CurrentDensity J = (cρ,ρu) = (cρ,j)
4-VectorPotential A = (φ/c,a) = A[X] = A[(ct,x)]  = (φ[(ct,x)]/c, a[(ct,x)]), often used as AEM
4-PotentialMomentum Q = (U/c,q) = q(φ/c,a)
4-TotalMomentum PT = (ET,pT) = (H,pT) = (E/c+U/c,p+q) = (E/c+qφ/c,p+qa)
4-WaveVector K = (ω/c,k) = (ω/c,ω/vphase) = (ω/c,ωu/c2) = (ω/c)(1,β) = (1/cT,/λ)
4-Gradient = X = (t/c,-) = (t/c,-del) = > (t/c,-∂x,-∂y,-z) = (∂/c∂t,-∂/∂x,-∂/∂y,-∂/∂z)
4-Polarization Ε = (ε0,ε) = > (ε·β,ε), with (ε·β = 0) *Note* this can have complex coefficients, and comes from standard EM theory (non-QM)

***
4-WaveGradient K = (c∂ω,-k)
4-MomentumGradient P = (c∂E,-p)
4-Momentum P = (E/c,p) = (H/c - U/c , pT - q) = (H/c - U/c , pT - q) = (H/c - qφ/c , pT - qa)
4-MomentumIncSpin Ps = (ps0,ps) = (σ0 E/c , σ·p) = ( σ0(ET/c - qφ/c) , σ·(pT-qa) )  {can have complex components due to Pauli Spin Matrices}
4-ProbabilityCurrentDensity Jprob = (iћ/2mo)( ψ* ∂[ψ] - ∂[ψ*] ψ ) - (q/mo)A (ψ*ψ)  {can have complex components}
4-ChargeCurrentDensity Jcharge = qJprob
4-LatticePosition RK = (ctK,rK)
4-Polarization Ε = (ε0,ε) = > (ε·β,ε)  {can have complex components due to circular polarization - see Jones Vector}
====


The combination of a Lorentz Invariant Charge*4-VectorPotential leads to a 4-PotentialMomentum.

The 4-TotalMomentum = 4-Momentum + 4-PotentialMometum

Potentials/Fields:

Let's back up to the 4-Momentum equation.  Momentum is not just a property of individual particles, but also of fields.
These fields can be described by 4-vectors as well.
One such relativistically invariant field is the 4-VectorPotential A, which is itself a function of 4-Position X.
Typically, we deal with the  Electromagnetic (EM) 4-VectorPotential, but it could be any kind of relativistic charge potential...
4-VectorPotential A [X] = A [(ct,x)] = (φ/c, a) = (φ[(ct,x)]/c, a[(ct,x)]), with the [(ct,x)] meaning it is a function of time t and position x.
While a particle exists as a worldline over spacetime, the 4-VectorPotential exists over all spacetime.
The 4-VectorPotential can carry energy and momentum, and interact with particles via their charge q.

PotentialMomentum:

One may obtain the PotentialMomentum 4-vector by multiplying by a charge q, Q = qA
4-PotentialMomentum Q = qA = q(φ/c, a) = (U/c, q)
The 4-TotalMomentum is then given by PT = P + Q
This includes the momentum of particle and field, and it is the locally conserved quantity.
4-TotalMomentum PT = (H/c,pT), for which these are the TotalEnergy = Hamiltonian and 3-TotalMomentum.
P = PT - Q = moU
Now working back, we can make our dynamic 4-Momentum more general, including the effects of potentials.
4-Momentum P = (E/c,p) = (H/c - U/c,pT - pEM) = (H/c - qφ/c,pT - qa)
The dynamic 4-momentum of a particle thus now has a component due to the 4-VectorPotential,
and reverts back to the usual definition of 4-momentum in the case of zero 4-VectorPotential.
Likewise, following the same path as before...
K = P/ ћ
4-WaveVector K = (ωT/c - (q/ћ)Φ/c, kT - (q/ћ)a)
= -iK
4-Gradient = (T/ct - (iq/ћ)Φ/c,-T - (iq/ћ)a) = (t /c,-)
Define 4-TotalGradient D = + iq/ћ A
This is the concept of "Minimal Coupling"
Minimal Coupling can be extended all the way to non-Abelian gauge theories and can be used to write down all the interactions of the Standard Model of elementary particles physics between spin-1/2 "matter particles" and spin-1 "force particles"
Minimal Coupling applied to the Dirac Eqn. leads to the Spin Magnetic Moment-External Magnetic Field coupling W = -γeS·B, with γe = qe/me, the gyromagnetic ratio.
The corrections to the anomalous magnetic moment come from minimal coupling applied to QED


In addition, we can go back to the velocity formula:

u = c2 (p)/(E) = c2 (pT - qa)/(H - qΦ)

Spacetime Lattice, Direct Quantization, Angular Momentum


Let's examine an interesting lattice concept... see Wikipedia "Reciprocal Lattice", also Bravais Lattice, Pontryagin Duality

4-LatticePosition ΔRK = (cΔtK,ΔrK)
4-WaveVector K = (ω/c,k) = (ω/c,ω/vphase) = (ω/c)(1,β) = (1/cT,/λ) = (2π/cT,2π/λ)

Let [ (ΔRK·K) = -n ], with (n) as an integer. This gives a Lorentz Invariant with magnitude in a spatial direction.
Then e^[i2π(ΔRK·K)] = e^[-i2πn] = 1, a unit phase factor for integer(n).
Oddly, this is the crystallographer's definition rather than the physics definition, which would normally be e^[i(ΔRK·K)] = 1.
I suspect that this has to do with the form of periodicity: I will be applying circular periodicity rather than linear periodicity...
However, let's see what happens based on this definition...

One can think of this as a particle with a particular wavevector (K).
There exists a series (n) of certain intervals (ΔRK) based on this wavevector (K) for which the phase factor is unity,
and thus transformations between these leaves the phase state unchanged.

**Note that scalar invariant (n), and the unit normal wave direction (), are different things in the following derivation**

(ΔRK·K) = (cΔtKrK)·(ω/c,k) = (cΔtKrK)·(1/cT,/λ) = (cΔtK/cT - ΔrK·n̂/λ) = (ΔtK/T - ΔrK·n̂/λ) = -n
If we examine the time ΔtK = 0 {simultaneity} case, we get:
rK·n̂/λ = -n
ΔrK/λ = n
ΔrK = nλ
2πΔrK = nλ : Note that this is the condition that an integer number (n) of wavelengths (λ) fit the circumference of a circle with radius ΔrK
(i.e. it is de BrogliĖs Standing Wave Hypothesis, with the additional implied relativistic criterion that it is the case for simultaneity)

Now, combine this with P = ћ K = (E/c,p) = ћ(1/cT,/λ)
p = ћ/λ
p = ћ/λ
λp = ћ
nλp = nћ: We can multiply both sides by the integer (n)
ΔrKp = nћ
L = nћ: the condition for quantization of angular momentum
(i.e. the Bohr Condition for Quantization)

Similarly, we can examine the ΔrK = 0 {stationarity, maybe locality - along a worldline} case:
ΔtKE = -nћ,
which is totally analagous to ΔrKp = nћ

Thus, [ (ΔRK·K) = -n ], is an SR Lorentz Invariant method of Quantization.
Not really sure that this is convincing, but it is at least interesting...

The Dirac Equation Derived


It can be shown that spin comes from the Poincare Symmetry of SR, not from QM axiom.
Writing a 4-SpinMomentum then leads naturally to the Dirac Equation.

All of the relativistic wave equations can be derived from a common source, the relativistic mass-energy relation, inc. spin, in an EM field.
Note that this formalism fits well with the Stern-Gerlach experiment.


4-Momentum inc. Spin Ps
Ps = Σ·P = Σμν Pν = ηαβ Σμα Pβ = Psμ
Σμν is a Pauli Spin Matrix Tensor = Diag[σ0,σ]
Σμν is a Pauli Spin Matrix Tensor = Diag[σ0,-σ]

Ps = Diag[σ0,-σ]·P = Diag[σ0,-σ]·(E/c,p) = (σ0E/c,σ·p)

Ps = (ps0,ps) = (σ0E/c,σ·p)
with σ0 as an identity matrix of appropriate spin dimension and σ is the Pauli Spin Matrix Vector


4-Momentum inc. Spin in External Field
with:
H = ET = Hamiltonian = TotalEnergyOfSystem
pT = Total3-MomentumOfSystem

4-TotalMomentum PT = P + qA
4-Momentum P = PT - qA

4-MomentumIncSpin Ps = (ps0,ps) = (σ0 E/c , σ·p) = ( σ0(ET/c - qφ/c) , σ·(pT-qa) )
Ps·Ps = (ps0)2 - (ps)2 = [σ0(E/c)]2 - [σ·(p)]2 = [σ0(Et/c-qφ/c)]2 - [σ·(pt-qa)]2  = (moc)2 = (Eo/c)2

The 4-TotalMomentum (inc. External Field Minimal-Coupling and Spin)
Ps = Σ·P = Σ·(PT-qA) =  [σ0(ET/c-qφ/c),σ·(pT-qa)]
with Σ = Σμν as the Pauli Spin Matrices, and taking the Einstein summation gives the σ0 and σ

Ps·Ps = ( Σ·P )2 = [ Σ·(PT-qA)]2 =  [σ0(ET/c-qφ/c)]2 - [σ·(pT-qa)]2 = (moc)2
( Σ·P )2 = (moc)2
( Σ·∂ )2 = -(moc/ћ)2
( Σ·∂ )2 + (moc/ћ)2 = 0
( Σ·(D-(i/h)qA) )2 + (moc/ћ)2 = 0


Now, to prove that this "Relativistic Pauli" Energy-Momentum equation can lead to the Dirac equation
Ps·Ps = [σ0(ET/c-qφ/c)]2 - [σ·(pT-qa)]2 = (ps0)2 - (ps)2 = (moc)2 = (Eo/c)2
Ps·Ps = I(ET/c-qφ/c)]2 - [σ·(pT-qa)]2 = (ps0)2 - (ps)2 = (moc)2 = (Eo/c)2
Ps·Ps = (ps0)2 - (ps)2 = (moc)2
(ps0 + ps) (ps0 - ps) = (moc)2

Multiply both sides by any arbitrary function, ψχ
(ps0 + ps) (ps0 - ps)ψχ = (moc)2ψχ

We can also split the arbitrary function into two parts, and it still solves the original equation
let   (ps0 + ps)ψ = (moc)χ
and (ps0ps)χ = (moc)ψ

now add and subtract these...
(ps0 + ps)ψ + (ps0 - ps)χ = (moc)χ + (moc)ψ
(ps0 + ps)ψ - (ps0 - ps)χ = (moc)χ  - (moc)ψ

gather terms...
ps0 (ψ+χ) + ps (ψ-χ) = (moc)(χ+ψ)
ps0 (ψ-χ) + ps (ψ+χ) = (moc)(χ-ψ)

rearrange a bit...
ps0 (ψ+χ) + ps (ψ-χ) = (moc)(ψ+χ)
ps (ψ+χ) + ps0 (ψ-χ) = -(moc)(ψ-χ)

we are free to relabel our variables
let (ψ+χ) = a and (ψ-χ) = -b
ps0 a - ps b = (moc)a
ps a - ps0 b = (moc)b

put in matrix form...
[ps0 - ps ][a] =   (moc)[a]
[ps  - ps0][b] =            [b]

put in suggestive matrix form...
([1  0 ]ps0 + [0 -1]ps)[a] =   (moc)[a]
([0 -1]         [1  0 ]    )[b] =            [b]

put in even more suggestive matrix form...
([1  0 ]σ0p0 + [0 -1]σ·p)[a] =   (moc)[a]
([0 -1]            [1  0 ]     )[b] =            [b]

put in highly suggestive matrix form...
([σ0  0 ]p0 - [0    σ]·p)[a] =   (moc)[a]
([0 -σ0]       [-σ  0 ]    )[b] =            [b]

let Spinor Ψ = [a]
                       [b]

and note that σ0 = I

this is equivalent to Dirac Gamma Matrices...
([I  0 ]p0 - [0    σ]·p)Ψ =   (moc)Ψ
([0 -I]       [-σ  0 ]   )

0p0 - γ·p)Ψ = (moc)Ψ

(Γ·P)Ψ = (moc)Ψ

(Γ·P) = (moc)

μPμ)Ψ = (moc)Ψ

iћ(Γμμ)Ψ = (moc)Ψ
The Dirac Relativistic Quantum Equation for spin 1/2 particles

Photon Polarization

Derivation of the SR reason that the QM photon polarization only has 2 independent states, instead of 3 or 4.

4-Polarization Ε = (ε0,ε) = > (ε·β,ε)
4-WaveVector K = (ω/c,k) = (ω/c,ω/vphase) = (ω/c)(1,β) = (1/cT,/λ)

4-Polarization E = (ε0,ε): Normally as a typical 4-vector it would have 4 independent components
However, we know empirically that the 4-Polarization E is orthogonal to the direction of propagation 4-WaveVector K
Ε·K = 0 =  (ε0,ε)·(ω/c)(1,β) = (ω/c)(ε0 - ε·β) = 0: This is a Lorentz Invariant restriction
So,  ε0 = ε·β
We can go ahead and simplify the 4-Polarization a bit with this new info
4-Polarization E = (ε·β,ε): Now it only has 3 independent components (ε)

However, we also know that the  4-Polarization is normalized to unity along a spatial direction.
This one is a bit trickier because we know that the 4-Polarization can have complex components (those representing circular polarization).
See the Jones Vector, Stokes Vector, Mueller Calculus, etc.
But, this can be accommodated using using standard complex notation...
Ε·E* = -1, Normalization to Spatial Unity, which is also a Lorentz Invariant Restriction

Ε·E* = -1 imposes
(ε·β)(ε*·β) -ε·ε* = -1
(ε·β)2 -1 = -1
(ε·β)2 = 0
ε·β = 0, so that the spatial components must be orthogonal, which is an additional constraining formula.

For a massive particle, there is always a rest frame with β = 0, so ε can have 3 independent components.
For a photonic particle, there is no rest frame.  | β| = 1 always. ( ε·β = ε·n̂ = 0 ) is therefore an additional constraint,
limiting ε to 2 independent components, with polarization ε orthogonal to direction of photon motion .

So, ( Ε·K = 0 ) and ( Ε·E* = -1 ) are both Lorentz Invariant Relations, which apply to all observers.

This appears to be related to the Ward-Takahashi Identity and Gauge-Fixing.

∂·U = 0 :Advection Equation (with γ as the advected quantity)

∂·J = 0 :Continuity Equation

∂·∂
= (-imoc/ћ)2 = -(moc/ћ)2  :Wave Equation
Low Energy Limiting case Schrödinger: Diffusion Equation

CPT & SR Phase Connection


The Phase is a Lorentz Scalar Invariant - all observers must agree on its value.

K·X = (ω/c,k)·(ct,x) = (ωt -k·x) = -Φ: Phase of SR Wave


We take the point of view of an observer operating on a particle at 4-Position X which has an initial 4-WaveVector K.
The 4-Position X of the particle, the operation's event, will not change: we are applying the various operations only to the particlĖs 4-Momentum K.


Note that for matter particles:
K = (ωo/c)T, with T as the Unit-Temporal 4-Vector T = γ(1,β), which defines the particlĖs worldline at each point. The gamma factor ( γ ) will be unaffected in the following operations since it uses the square of β:
γ = 1/Sqrt(1-β·β).

For photonic particles, K = (ω/c)
, with N as the "Unit"-Null 4-Vector = (1,) and as a unit-spatial 3-vector. All operations listed below
work similarly on the 4-Null 4-Vector.

=====

Do a Time Reversal Operation: T
The particlĖs temporal direction is reversed & complex-conjugated:
TT = -T* = γ(-1,β)*

Do a Parity Operation (Space Reflection): P
Only the spatial directions are reversed:
TP = γ(1,-β)

Do a Charge Conjugation Operation: C
Feynman showed this is the equivalent of world-line reversal & complex-conjugation
This is the Feynman–Stueckelberg Interpretation, it changes all internal quantum #'s - charge, lepton #, etc.:
TC = γ(-1,-β)*

=====
Pairwise combinations:
TTP = TPT = TC = γ(-1,-β)*
TTC = TCT = TP = γ(1,-β)
TPC = TCP = TT = γ(-1,β)*,

which shows that a CP event is mathematically the same as a T event

TCPT = T = γ(1,β)
TCC = T = γ(1,β)
TPP = T = γ(1,β)
TTT = T = γ(1,β)

It is only the combination of all three ops, in any order: (C,P,T) , or repeats of the same op,
that leave the Unit-Temporal 4-Vector, and thus the Phase, Invariant

Lagrangian/Hamiltonian Formalisms:

Relativistic Beta Factor β = u/c
Relativistic Gamma Factor γ = 1/√[1-β2]

The whole Lagrangian/Hamiltonian connection is given by the relativistic identity:
γ = 1/√[1 - β2]
γ2 = 1/(1 - β2)
(1 - β22 = 1
2 - γ2β2) = 1
2 - 1) = γ2β2
( γ - 1/γ ) = ( γβ2 )
Now multiply by your favorite Lorentz Scalars... In this case for a free relativistic particle...
( γ - 1/γ )(P·U) = ( γβ2 )(P·U)
( γ - 1/γ )(moc2) = ( γβ2 )(moc2)
( γmoc2 - moc2/γ ) = γmoc2β2
( γmoc2 - moc2/γ ) = γmou2
( γmoc2 ) + (- moc2/γ ) = γmou·u
( γmoc2 ) + (- moc2/γ ) = ( p·u )
    ( H )   +       ( L )     = ( p·u )
The Hamiltonian/Lagrangian connection falls right out.
Note that neither (H) nor (L) are scalar invariants, due to the extra (γ) factors.

Now, including the effects of a potential:
4-VectorPotential A = (φ/c, a) { = (φEM/c, aEM) for EM potential }
4-PotentialMomentum Q = qA due to 4-VectorPotential acting on charge q
4-TotalMomentum of System PT = P + Q = P + qA = moU + qA = (H/c,pT) = (γmoc+qφ/c,γmou+qa)
A·U = γ(φ - a·u ) = φo
P·U = γ(E - p·u ) = Eo
PT·U = Eo+ qφo = moc2 + qφo
I assume the following as a valid SR relation:
A = (φo/c2)U = (φ/c,a) = (φo/c2)γ(c,u) = (γφo/c, γφo/c2u) giving (φ = γφo and a = γφo/c2u)
This is analogous to P = (Eo/c2)U

( γ - 1/γ )(PT·U) = ( γβ2 )(PT·U)
γ(PT·U) + -(PT·U)/γ  = ( γβ2 )(PT·U)
γ(PT·U) + -(PT·U)/γ  = (pT·u)
    ( H )   +      ( L )     = (pT·u)
L = -(PT·U)/γ H = γ(PT·U) H + L = pT·u = γ(PT·U) - (PT·U)/γ
L = -(PT·U)/γ
L = -((P + QU)/γ
L = -(P·U + Q·U)/γ
L = -P·U/γ - Q·U
L = -moU·U/γ - qA·U
L = -moc2/γ - qA·U
L = -moc2/γ - q(φ/c, a)·γ(c, u)/γ
L = -moc2/γ - q(φ/c, a)·(c,u)
L = -moc2/γ - q(φ - a·u)
L = -moc2/γ - qφ + qa·u
L = -moc2/γ - qφo
L = -(moc2 + qφo)/γ
H = γ(PT·U)
H = γ((P + QU)
H = γ(P·U + Q·U)
H = γP·U + γQ·U
H = γmoU·U + γqA·U
H = γmoc2 + qγφo
H = γmoc2 + qφ
H = ( γβ2 + 1/γ )moc2 + qφ
H = ( γmoβ2c2 + moc2/γ) + qφ
H = ( γmou2 + moc2/γ) + qφ
H = p·u + moc2/γ  + qφ
H = E + qφ
H = ± c√[mo2c2+p2] + qφ
H = ± c√[mo2c2+(pT-qa)2] + qφ
H + L = γ(PT·U) - (PT·U)/γ
H + L = (γ - 1/γ)(PT·U)
H + L = ( γβ2 )(PT·U)
H + L = ( γβ2 )((P + Q)·U)
H + L = ( γβ2 )(P·U + Q·U)
H + L = ( γβ2 )(moc2 + qφo)
H + L = (γmoβ2c2 + qγφoβ2)
H + L = (γmou·uc2/c2 + qφoγu·u/c2)
H + L = (γmou·u + qa·u)
H + L = (p·u + qa·u)
H + L = pT·u

The non-relativistic Hamiltonian H is an approximation of the relativistic H:
H = γ(moc2 + qΦo)
H = (1/√[1-(v/c)2])(moc2 + qΦo) ~ [1+(v/c)2/2])(moc2 + qΦoEM) = (moc2 + qΦo)+(1/2)(moc2v2/c2 + qΦov2/c2)  ~ (moc2 + qΦo) -+(1/2)(mov2 + 0)
H ~ (1/2)(mov2) + (moc2 + qΦo)
H ~ (Kinetic) + (Rest+Potential) = T + V {for |v| << c}

The non-relativistic Lagrangian L is an approximation of the relativistic L:
L = -(moc2 + qΦo)/γ
-L = (moc2 + qΦo)/γ = √[1-(v/c)2](moc2 + qΦo) ~ (moc2 + qΦo) - (1/2)(moc2v2/c2 + qΦov2/c2)  ~ (moc2 + qΦo) - (1/2)(mov2 + ~0 )
L ~ (1/2)(mov2) - (moc2 + qΦo)
L ~ (Kinetic) - (Rest+Potential) = T - V {for |v| << c}

Thus, (H ~ T + V) and (L ~ T - V) only in the non-relativistic limit (|v| << c)
H + L ~ (T + V) + (T - V) = 2T = 2 (1/2 mou·u) = p·u
Thus, ( H ) + ( L ) = ( p·u ) is always true, in both the relativistic and non-relativistic case.

More on the Group Properties of Minkowski Space --> Poincaré Group
from "Representations of the Symmetry Group of Spacetime" by Kyle Drake, Michael Feinberg, David Guild, Emma Turetsky

Minkowski spacetime is the mathematical model of flat (gravity-less) space and time.
The transformations on this space are the Lorentz transformations, known as O(1,3).
The identity component of O(1,3) is SO+(1,3).
The component SO+(1,3) taken with the translations R1,3 is the Poincaré group, which is the symmetry group of Minkowski spacetime.
However, physical experiments show that a connected double cover of the Poincaré is more appropriate in creating the symmetry group actual spacetime.
Why the double cover of the Poincaré group?
What is so special about this group in particular that it describes the particles?
The answer lies in the way the group is defined.
Simply, it is the set of all transformations that preserve the Minkowski metric of Minkowski spacetime.
In other words, it is the set of all transformations of spacetime that preserve the speed of light.
SL(2,C) is the connected double cover of the identity component of SO+(1,3).
This is the symmetry group of spacetime: the double cover of the Poincaré group, SL(2,C) x R1,3.

In 1939, Eugene Wigner classified the fundamental particles using the irreducible representations of the double cover of the Poincaré group.
Wigner was motivated by the idea that symmetry underlies all physical laws.
In particular, a physical experiment should come up with the same results regardless of where, when, or what orientation the experiment is done in.
An experiment's results should also be invariant whether the experiment is at rest or moving at a constant velocity.
It turns out the symmetries of physics go further than this.
It is common to combine several systems together such as when protons and electrons combine to form atoms.
When this is done, the overall symmetry of the system should be related to the individual symmetries of its components.
These components are the irreducible representations [3].
The double cover of the Poincaré group acts to classify the fundamental particles in physics and explain patterns in their behaviors.
In particular, the particles are most easily classified by the irreducible representations of the double cover of the Poincaré group.
The representations are determined by the different orbits in the group.
These orbits serve to classify types of particles.
The first two types of orbits correspond to the value m2 > 0.
The value m2 is equivalent to the idea of mass in physics.
Thus elements in the first two orbits correspond to massive particles which travel slower than the speed of light.
The particles that correspond to the light cones are those with m2 = 0, which are particles that travel at the speed of light.
Examples of these particles are photons, from the which the light cone gains its name, and gravitons, which currently only exist in theory.
The particles on the single-sheet hyperboloid with values m2 < 0 are called tachyons.
They travel faster than the speed of light and have imaginary mass.
As tachyons have never been observed, we will not find the representations corresponding this orbit.
Finally, the orbit in which m2 = 0 and the particles are not moving through time corresponds to vacuum.
Vacuum is devoid of all particles, and thus is uninteresting to the purposes of this paper.
It is somewhat surprising that the correspondences between the orbits of the double cover of the Poincaré group and types of particles line up as well as they do.
The spin values of particles predicted by physicists are exactly the values that come as a result of finding the irreducible representations of the double cover of the Poincaré group.
This fact is further explained by the fact that the Poincaré group's covering space is connected and has nice properties that emulate those observed in quantum spin.

Since QM is derivable from SR, this explains why you can't just "quantize gravity".
SR is a limiting case of GR.  RQM is directly derivable from SR.  QM is a limiting case of RQM.
Quantum Mechanics is not a "separate system" from Relativity.  It is a special case of SR phenomena.  As such, one cannot assume that SR or QM or SRQM "rules of quantization" apply to the more the general GR, except perhaps in the local tangent spaces which are locally Minkowskian.
This makes sense actually - QM is the science of really small things, i.e. you have to zoom in to the region for which the spacetime is locally flat... --> Minkowskian...
The hint for QM has been in the local SR tangent space all along.   :)

This remains a work in progress.

Reference papers/books can be found in the see 4-Vectors & Lorentz Scalars Reference.

Email me, especially if you notice errors or have interesting comments.
Please, send comments to John

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