Tensors and 4-Vectors: The Pathway of SRQM, or [SR→QM]

1) In General Relativity (GR), the metric tensor g^μν encodes the geometry of spacetime.
The general spacetime metric g^μν reduces to the "flat" spacetime Minkowski metric η^μν in the limit-case of zero curvature.
Show the SR Minkowski Metric in the time-positive signature.
Show Answer

η^μν = Diag[+1,-1,-1,-1] =
[ +1  0  0  0 ]
[ 0  -1  0  0 ]
[ 0  0  -1  0 ]
[ 0  0  0  -1 ]

Shown here:
Temporal components are in blue
Spatial components are in red
Mixed SpaceTime components are in purple

1a) All tensor types with indices can be modified using tensor raising/lowering rules: ex. η^μ_ν = η^μα η_αν or V^μ = η^μα V_α
Show the Raised, Lowered, and Mixed SR Minkowski Metric in the time-positive signature.
Show Answer

Raised
η^μν = Diag[+1,-1,-1,-1] =
[ +1  0  0  0 ]
[ 0  -1  0  0 ]
[ 0  0  -1  0 ]
[ 0  0  0  -1 ]

Lowered
η_μν = Diag[+1,-1,-1,-1] =
[ +1  0  0  0 ]
[ 0  -1  0  0 ]
[ 0  0  -1  0 ]
[ 0  0  0  -1 ]

Mixed
η^μ_ν = δ^μ_ν = Diag[+1,+1,+1,+1] =
[ +1  0  0  0 ]
[ 0  +1  0  0 ]
[ 0  0  +1  0 ]
[ 0  0  0  +1 ]

Note that the mixed Minkowski Metric is identical to the Kronecker Delta δ^μ_ν.

2) In Special Relativity (SR), 4-Vectors are the geometric objects of spacetime. They unite temporal and spatial quantities into a single object.
Show all components of the 4-Position R = R^μ.
Show Answer

4-Position R = R^μ = (r⁰,r¹,r²,r³) = (ct,r) = (ct,x,y,z)

4-Vectors can be shown in vector R or tensor R^μ formats.
Temporal components in blue
Spatial components in red
I always name the 4-Vector after the spatial 3-vector component.

2a) There is also a tensor-lowered form of 4-Position, called a 4D Position One-Form R_μ
Show the 4D Position One-Form R_μ and that the standard 4-Vector is just the tensor-raised form.
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4D Position One-Form R_μ = η_μνR^ν = (r₀,r₁,r₂,r₃) = (r⁰,-r¹,-r²,-r³) = (ct,-r) = (ct,-x,-y,-z)

4-Position R = R^μ = η^μνR_ν = (r⁰,r¹,r²,r³) = (r₀,-r₁,-r₂,-r₃) = (ct,r) = (ct,x,y,z)

Since the upper and lower tensor versions are related by the Minkowski Metric, the spatial parts change sign on raising/lowering.

2b) The upper and lower index notation leads to Einstein Summation, a convenient abbreviation in tensor formulas.
Show the Einstein Summation rule.
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A·B = A^μη_μνB^ν = A^μB_μ = ∑_μ=0..3[A^μB_μ] = (a⁰)(b₀)+(a¹)(b₁)+(a²)(b₂)+(a³)(b₃) = (a⁰)(+b⁰)+(a¹)(-b¹)+(a²)(-b²)+(a³)(-b³) = (a⁰b⁰) - a·b

The rule is that you sum over the index values of paired upper/lower indices. The summation is implied, so you can omit the explicit Summation sign ∑_μ=0..3.
In any equation, there should be only paired "dummy" indices or separate (non-repeating) "true" indices. (i.e. never triple or more identical)

3) SR 4-Vectors have a tensor invariant known as the Lorentz Scalar Product.
Show the Lorentz Scalar Product of the 4-Position.
Show Answer

R·R = R^μη_μνR^ν = R^μR_μ = (ct,r)·(ct,r) = (ct)² - r·r = (ct_o)² = (cτ)²

with

Proper Time τ
Rest Time t_o

The is the well-known Invariant Interval.

4) Calculus can be applied in the standard way to 4-Vectors.
Show the infintesimal and discrete differentials of the 4-Position, dR and ΔR.
Show Answer

dR = dR^μ = (cdt,dr) = (cdt,dx,dy,dz)

ΔR = R₂ - R₁ = ΔR^μ = (cΔt,Δr) = (cΔt,Δx,Δy,Δz)

5) When taking the derivative of a 4-Vector, one should use an invariant 4-Scalar to keep the tensor equation invariant.
The Proper Time τ (which is the same as rest time t_o) is one such invariant. Show the generic derivative with respect to Proper Time τ.
Show Answer

Proper Time Derivative = d/dτ = γd/dt

with Lorentz Gamma Factor γ = 1/√[1-u·u/c²]
The Lorentz Gamma Factor γ appears in time dilation t = γt_o and length contraction L = L_o/γ, as well as relativistic quantites such as total energy E = γE_o.

6) Show the derivative of 4-Position R with respect to Proper Time τ.
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dR/dτ = d/dτ[R] = γd/dt[R] = γd/dt[(ct,r)] = γ(cdt/dt,dr/dt) = γ(c,u) = 4-Velocity

7) Show the Lorentz Scalar Product of the 4-Velocity U.
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U·U = U^μη_μνU^ν = γ(c,u)·γ(c,u) = γ²[c² - u·u] = c²

(c) is a universal invariant. It is the speed of causality. It is used in the temporal component of almost all SR 4-vectors to maintain dimensional-analysis units.

8) Show the 4-Momentum P and how it is related to the 4-Velocity U via the rest mass and rest energy.
Show Answer

4-Momentum P = (E/c,p) = m_oU = γm_o(c,u) = m(c,u) = (mc,mu)

with

Rest mass m_o = E_o/c²

4-Momentum P = (E/c,p) = (E_o/c²)U = γ(E_o/c²)(c,u) = (E/c²)(c,u) = (E/c,Eu/c²) = (E/c,p)

with E = γE_o = γm_oc² = mc²
Internal components: {E,m,p,u} are relativistic and depend on frame.
Tensor 4-Scalars: {E_o,m_o,c} are 4D invariants and frame-independent. The nought (_o) refers to a rest-frame.

9) Show the Lorentz Scalar Product of the 4-Momentum P.
Show Answer

P·P = P^μη_μνP^ν = (E/c,p)·(E/c,p) = (E/c)² - p·p = (E_o/c)²

This is often rearranged to get the relativistic energy relation: E = √[E_o² + p·pc²] showing that the energy E of a particle can be thought of as a rest energy E_o and a momentum energy |p|c.

9a) Show the rest frame and massless limit-cases of relativistic energy relation: E = √[E_o² + p·pc²].
Show Answer

Generally: E = √[E_o² + p·pc²]
Rest frame: E_rest → √[E_o² + 0·0c²] = E_o
Massless frame: E_massless → √[0 + p·pc²] = |p|c. This is the energy of a photon.

9b) Show the relation between the 4-Momentum P and the 4-Velocity U via the Lorentz Scalar Product.
Show Answer

P·U = P^μη_μνU^ν = (E/c,p)·γ(c,u) = γ[E - p·u] = γ[E - Eu·u/c²] = Eγ[1 - u·u/c²] = E/γ = E_o = m_oc²

giving E = γE_o

9c) Show that a massless particle must move at speed c.
Show Answer

P·U = P^μη_μνU^ν = (E/c,p)·γ(c,u) = γ[E - p·u] = γ[E - Eu·u/c²] = Eγ[1 - u·u/c²] = E/γ = E_o = m_oc²

Since m_o = E_o/c², choose E_o = 0.
Then Eγ[1 - u·u/c²] = E_o = 0. Since (γ >= 1), then either (E = 0), in which case there is no particle, or [1 - u·u/c²] = 0, which gives |u| = c.
The photon (an EM particle of light) is such a particle.

10) One may do more advanced calculus with 4-Vectors.
Show the 4-Gradient ∂, which is the partial derivative with respect to 4-Position R
Show Answer

4-Gradient ∂ = ∂^μ = ∂/∂R_μ = (∂/∂r₀,∂/∂r₁,∂/∂r₂,∂/∂r₃) = (+∂/∂r⁰,-∂/∂r¹,-∂/∂r²,-∂/∂r³) = (+∂/∂ct,-∇) = (+∂/∂ct,-∂/∂x,-∂/∂y,-∂/∂z)

The 4-Gradient is a natural SR 4-Vector operator.
Its related One-Form is:

4D Gradient One-Form ∂_μ = ∂/∂R^μ = (∂/∂r⁰,∂/∂r¹,∂/∂r²,∂/∂r³) = (+∂/∂r⁰,+∂/∂r¹,+∂/∂r²,+∂/∂r³) = (+∂/∂ct,+∇) = (+∂/∂ct,+∂/∂x,+∂/∂y,+∂/∂z)

10a) Show that the 4-Velocity U combined with the 4-Gradient ∂, gives another way to define the derivative with respect to proper time τ.
Show Answer

Note the total derivative of a function f[t,x,y,z] = df = (∂f/∂t)dt + (∂f/∂x)dx + (∂f/∂y)dy + (∂f/∂z)dz

U·∂ = γ(c,u)·(+∂/∂ct,-∇) = γ[∂/∂t + u·∇] = γ[(dt/dt)(∂/∂t) + (dx/dt)(∂/∂x) + (dy/dt)(∂/∂y) + (dz/dt)(∂/∂z)] = γd/dt = d/dτ

An interesting side derivation is as follows:
U·∂ = d/dτ
(dR/dτ)·∂ = d/dτ
(dR)·∂ = d
dR·∂ = d
The differential operator d[ ] is itself a Lorentz 4-Scalar, with the 4-Gradient ∂ = ∂_R here implied to be with respect to 4-Position R.

dR·∂_R = d

Likewise, one could use the 4-Velocity U instead to get a 4-VelocityGradient ∂_U

dU·∂_U = d

10b) Using the tensor properties of the calculus differential d[ ], derive the relativistic Euler-Lagrange Equation.
Show Answer

d=d
(dR·∂_R) = (dU·∂_U)
(dR·∂_R) = [d(dR/dτ)·∂_U]
(dR·∂_R) = [(d)(dR)/dτ)·∂_U]
(dR·∂_R) = [(dR)(d/dτ)·∂_U]
(dR·∂_R) = [dR·(d/dτ)∂_U]

∂_R = (d/dτ)∂_U

Just apply a Lagrangian L, which is basically a boundary condition.

∂_R[L] = (d/dτ)∂_U[L]

This is the Relativistic Euler-Lagrange Equation
Note: This is an SR operator equation, which implies the existence of a Lagrangian solution.

11) We can find several interesting properties of the 4-Gradient ∂ by doing various combinations with the 4-Position R
Show how these two 4-Vectors can derive SpaceTime Dimension=4, the Minkowski Metric η^μν, and the Lorentz Transforms Λ^μ_ν.
Show Answer

∂·R = ∂^μη_μνR^ν = ∂^μR_μ = (+∂/∂ct,-∇)·(ct,r) = (∂/∂ct[ct]) - -∇·r = 1+3 = 4 = SR SpaceTime Dimension

∂^μ[R^ν] = (+∂/∂ct,-∇)[(ct,r)] = Diag[+1,-1,-1,-1] = η^μν = SR Minkowski Metric

∂_ν[R^μ] = ∂/∂R^ν[R^μ] = ∂R^μ/∂R^ν = Λ^μ_ν = SR Lorentz Transform

Can also do by pure tensor math:
∂·R = ∂^μη_μνR^ν = ∂^μR_μ = ∂/∂R_μ[R_μ] = ∂R_μ/∂R_μ = 1+1+1+1 = 4 (An Einstein Summation to a scalar due to paired indices of same R components)
∂^μ[R^ν] = ∂/∂R_μ[R^ν] = ∂R^ν/∂R_μ = Diag[+1,-1,-1,-1] = η^μν (Goes to 4-Tensor due to independent indices, and to Minkowski Metric due to lowered and raised of same R components)
∂_ν[R'^μ] = ∂/∂R^ν[R'^μ] = ∂R'^μ/∂R^ν = Λ^μ_ν (Goes to 4-Tensor due to independent indices, and to Lorentz Transform due to different R and R' components)

∂·R = ∂^μη_μνR^ν = ∂^μ[η_μνR^ν] = ∂^μ[η_μν]R^ν+η_μν∂^μ[R^ν] = 0+η_μν∂^μ[R^ν] = η_μνη^μν = +1+1+1+1 = 4
So η_μνη^μν = 4. This can be thought of as the Tensor Inner Product[η^μν] or the Trace[η^μν].
The Tensor Inner Product IP[T^μν] = T_μνT^μν = (Lowered)*(Raised)
The Trace Tr[T^μν] = η_μνT^μν.

12) Show the Lorentz Scalar Product of the 4-Gradient ∂, which gives the d'Alembertian ∂·∂.
Show Answer

∂·∂ = ∂^μη_μν∂^ν = (+∂/∂ct,-∇)·(+∂/∂ct,-∇) = (∂/∂ct)² - ∇·∇

This is the SR relativistic 4D wave equation, also known as the d'Alembertian.
It implies the existence of a wave ψ, and a 4-WaveVector K, which is a part of the solution of the wave equation ∂·∂.
The wave has a general form of ψ = A * e^±i(K·R), in which ψ and A are tensors of the same kind.
A → A can be a 4-Scalar, as in scalar waves
A → A^μ can be a 4-Vector, as in EM waves
A → A^μν can be a 4-Tensor, as in gravitational waves
∂·∂ψ = ∂·∂A * e^±i(K·R) = A * (±i)^2 * K·K e^±i(K·R) = -(K·K) A * e^±i(K·R) = -(K·K)ψ
∂·∂ψ = -(K·K)ψ
This also gives a direct relation of:

∂ = ±iK

13) Show the correct unit-dimensional components of the 4-WaveVector K, whose existence is implied by the d'Alembertian.
Show Answer

4-WaveVector K = (ω/c,k) = (ω/c,ω/v_phase) = (ω/c,ωu/c²)

with units of [angle/length] → SI units of [rad/meter]
u * v_phase = c²
If (|u| = c), then K_null = (ω/c,ω/c) is a null vector, as expected.

13a) Show that the Lorentz Scalar Product of the 4-WaveVector K with the 4-Position R gives an invariant phase.
Show Answer

K·R = (ω/c,k)·(ct,r) = ωt - k·r = -φ

Wavefunction phases φ are Lorentz invariants.

14) Show the Lorentz Scalar Product of the 4-WaveVector K.
Show Answer

K·K = K^μη_μνK^ν = (ω/c,k)·(ω/c,k) = (ω/c)² - k·k = (ω_o/c)²

14a) Show the relation between the 4-WaveVector K and the 4-Velocity U.
Show Answer

K·U = K^μη_μνU^ν = (ω/c,k)·γ(c,u) = γ[ω - k·u] = γ[ω - ωu·u/c²] = ωγ[1 - u·u/c²] = ω/γ = ω_o

giving ω = γω_o
Note the similarity to E = γE_o

This also gives:

4-WaveVector K = (ω/c,k) = (ω/c,ω/v_phase) = (ω_o/c²)U = γ(ω_o/c²)(c,u) = (ω/c²)(c,u) = (ω/c,ωu/c²)

Looking at the spatial part:
ω/v_phase = ωu/c²
v_phase*u = c²

15) Show various Lorentz Scalar Products of the 4-Momentum P and 4-WaveVector K, and how they imply an invariant tensor relation.
Show Answer

P·P = P^μη_μνP^ν = (E/c,p)·(E/c,p) = (E/c)² - p·p = (E_o/c)²

P·K = P^μη_μνK^ν = (E/c,p)·(ω/c,k) = (Eω)/c² - p·k = (E_o)(ω_o)/c²

K·K = K^μη_μνK^ν = (ω/c,k)·(ω/c,k) = (ω/c)² - k·k = (ω_o/c)²

These Lorentz Invariants show that there is an invariant tensor relation between the 4-Momentum and the 4-WaveVector.

P = (E_o/ω_o)K = (γE_o/γω_o)K = (E/ω)K = (E/c,p) = (E/ω)(ω/c,k)

This is ** particle:wave duality **. No quantum axiom required.
The temporal part is the Einstein/Planck photonic relation E = ћω.
The spatial part is the de Broglie matter-wave relation p = ћk.

16) REVIEW: Show all the 4-Vectors that have appeared so far, for compare and contrast.
Show Answer

4-Position R = R^μ = (r⁰,r¹,r²,r³) = (ct,r)

4-Velocity U = U^μ = (u⁰,u¹,u²,u³) = γ(c,u)

4-Momentum P = P^μ = (p⁰,p¹,p²,p³) = (E/c,p) = (mc,mu)

4-WaveVector K = K^μ = (k⁰,k¹,k²,k³) = (ω/c,k)

4-Gradient ∂ = ∂^μ = (∂⁰,∂¹,∂²,∂³) = (∂/∂ct,-∇) = (∂/∂ct,-∂/∂x,-∂/∂y,-∂/∂z)

Note that each 4-Vector unites a scalar temporal variable with a 3-vector spatial variable, with (c) providing correct dimensional units across the whole 4-Vector.

17) REVIEW: Show all the Lorentz Scalar Products of the 4-Vectors that have appeared so far, for compare and contrast.
Show Answer

R·R = R^μη_μνR^ν = R^μR_μ = (ct,r)·(ct,r) = (ct)² - r·r = (ct_o)² = (cτ)²

U·U = U^μη_μνU^ν = γ(c,u)·γ(c,u) = γ²[c² - u·u] = c²

P·P = P^μη_μνP^ν = (E/c,p)·(E/c,p) = (E/c)² - p·p = (E_o/c)² = (m_oc)²

K·K = K^μη_μνK^ν = (ω/c,k)·(ω/c,k) = (ω/c)² - k·k = (ω_o/c)²

∂·∂ = ∂^μη_μν∂^ν = (+∂/∂ct,-∇)·(+∂/∂ct,-∇) = (∂/∂ct)² - ∇·∇

Note that each gives a new invariant scalar: the proper time τ, the speed of light c, the rest energy E_o, the rest angular frequency ω_o, the d'Alembertian (∂·∂)

18) REVIEW: Show the there is an invariant tensor relation between the various 4-Vectors.
Show Answer

4-Position R = (ct,r)

4-Velocity U = γ(c,u) = (d/dτ)R

4-Momentum P = (E/c,p) = (m_o)U = (E_o/c²)U

4-WaveVector K = (ω/c,k) = (ω_o/E_o)P

4-Gradient ∂ = (∂/∂ct,-∇) = (±i)K

Note that each 4-Vector unites a scalar temporal variable with a 3-vector spatial variable, with (c) providing correct dimensional units across the whole 4-Vector.

An alternate, diagrammatical way to show this is:

4-Position
R = (ct,r)

-->

(U·∂)
= (d/dτ)
= γ(d/dt)

-->

4-Velocity
U = γ(c,u)

-->

(m_o)
=(E_o/c²)

-->

4-Momentum
P = (E/c,p)

-->

(ω_o/E_o)
=1/ћ

-->

4-WaveVector
K = (ω/c,k)

-->

(±i)

-->

4-Gradient
∂ = (∂/∂ct,-∇)

19) Look at the d'Alembertian again... Based on what the relations between 4-Vectors are what does it imply?
Show Answer

∂·∂ = ∂^μη_μν∂^ν = (+∂/∂ct,-∇)·(+∂/∂ct,-∇) = (∂/∂ct)² - ∇·∇

∂·∂ = (±i)²K·K = -K·K = -(ω_o/E_o)²P·P

∂·∂ = -(ω_o/E_o)²(E_o/c)² = -(ω_o/c)²

∂·∂ = -(ω_o/E_o)²(m_oc)² = -(m_oc)²/(E_o/ω_o)²

Just admit it. In every known empirical experiment, E_o/ω_o = ћ, a physical constant. It doesn't require an axiom for its existence.
Arguments saying SR doesn't give a value are irrelevant. ћ as a QM axiom doesn't give the value either. It must be experimentally measured.

∂·∂ = -(m_oc/ћ)²

This is the relativistic Klein-Gordon quantum equation. All from tensor analysis of physical SR 4-Vectors, no quantum assumptions.

20) Show the relation between the 4-Momentum P and the 4-Gradient ∂. Based on what the relations between these 4-Vectors are what does it imply?
Show Answer

4-Momentum P = (E_o/ω_o)K

4-WaveVector K = (±i)∂

Therefore:

P = (±iE_o/ω_o)∂

Again, just admit it:

P = (±iћ)∂ = (E/c,p) = (±iћ)(∂/∂ct,-∇)

These are the Schrödinger Quantum Relations. All from tensor analysis of physical SR 4-Vectors, no quantum assumptions.
E = ±iћ∂/∂t
p = ∓iћ∇

21) Show the commutation relations between the 4-Position R and the 4-Gradient ∂.
Show Answer

Let f be an SR function of spacetime variables (ct,x,y,z)
Standard calculus of a partial differential on multiple variables.

∂^μ[R^νf] = ∂^μ[R^ν]f + R^ν∂^μ[f]

Rearrange:

∂^μ[R^νf] - R^ν∂^μ[f] = ∂^μ[R^ν]f

Recognize:

∂^μ[R^ν[f]] - R^ν[∂^μ[f]] = ∂^μ[R^ν]f

as a commutation relation:

[∂^μ,R^ν]f = ∂^μ[R^ν]f = η^μνf

Now, show in abstract form:

[∂^μ,R^ν] = η^μν

This is a non-zero commutation relation, purely from within SR tensor relations. No quantum assuptions.

22) Use Green's Vector Identity to show there is a Conserved 4-CurrentDensity J
Show Answer

Green's Vector Identity is a purely mathematical identity:

∂·(f∂[g] - ∂[f]g) = f∂·∂[g] - ∂·∂[f]g

with (f) and (g) as SR functions
One can also multiply this by a Lorentz 4-Scalar constant s

s*(f∂·∂[g] - ∂·∂[f]g) = ∂·s*(f∂[g] - ∂[f]g) = ∂·J

with

J = s*(f∂[g] - ∂[f]g)

∂·∂ + (m_oc/ћ)² = 0

Let it act on SR function g, then pre-multiply by f
Let it act on SR function f, then post-multiply by g
Take (1st group) - (2nd group)

f∂·∂[g] - ∂·∂[f]g = 0

Which gives a conserved

4-CurrentDensity J = (ρc,j)

with

∂·J = 0

At this point, we can choose s = (iћ/2m_o) = (ic²/2ω_o), which is Lorentz Scalar Invariant, in order to make the ChargeDensity have [dimensionless units = #] and be normalized to unity in the rest case.
This is then identical to a dimensionless 4-NumberFlux or dimensionless 4-ProbabilityCurrentDensity.