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Worked current boost law and coordinate versus rest density

Example

Worked computation. In an inertial frame with coordinates $(x^{0}, x^{1}, x^{2}, x^{3})$ consider the constant vector field $U$ with components $(U^{0}, U^{1}, U^{2}, U^{3}) = (5 / 4, 3 / 4, 0, 0)$ on an open set $Ω$ (C^k tensor fields on affine Minkowski space and their inertial components), so that $η_{0} (U, U) = - 25 / 16 + 9 / 16 = - 1$ with $U^{0} > ∣ U_{v e c} ∣ = 3 / 4$ ; $U$ is a future timelike unit vector field. Let $ρ_{0}$ be a constant scalar field and put $J : = ρ_{0} U$ . Choose the primed inertial chart $x' = B_{x} x$ , where $B_{x}$ is the matrix of $B (\frac{3}{5} e_{1})$ ; it exists by Inertial charts are related by proper orthochronous affine Lorentz maps, and every such map is a chart transition. Thus the primed frame basis has $e'_{a} = B (- \frac{3}{5} e_{1}) e_{a}$ and the primed component columns are $[U]_{x'} = B_{x} [U]_{x}$ , $[J]_{x'} = B_{x} [J]_{x}$ ; these are changes of components of the same fields. Then:

(i) $J$ has components $(J^{0}, J^{1}) = (5 ρ_{0} / 4, 3 ρ_{0} / 4)$ , and $\partial_{a} J^{a} = 0$ because the components are constant; the continuity equation of Transformation of tensor, covector, derivative and current components under affine Lorentz maps reads $\partial_{0} J^{0} + \partial_{1} J^{1} = 0$ identically.

(ii) Under the passive boost $B = B (\frac{3}{5} e_{1})$ one computes $B U = e_{0}$ : the first component of $B U$ is $(5 / 4) (5 / 4 - (3 / 5) (3 / 4)) = (5 / 4) (16 / 20) = 1$ , the second is $(5 / 4) (3 / 4 - (3 / 5) (5 / 4)) = (5 / 4) \cdot 0 = 0$ , and the transverse components vanish. Correspondingly $[J]_{x'} = (ρ_{0}, 0, 0, 0)$ : in the primed chart the field $J$ has those components. Hence the coordinate density $J^{0}$ is $5 ρ_{0} / 4$ in the original chart and $ρ_{0}$ in the chart in which $U = e'_{0}$ , while the scalar multiplier $ρ_{0}$ is the same in both; this is the worked instance of clause (vii) of Transformation of tensor, covector, derivative and current components under affine Lorentz maps.

(iii) The general boost law of clause (vi) of Transformation of tensor, covector, derivative and current components under affine Lorentz maps reproduces (ii): $J^{' 0} = γ (J^{0} - β J^{1}) = (5 / 4) (5 ρ_{0} / 4 - (3 / 5) (3 ρ_{0} / 4)) = (5 / 4) (16 ρ_{0} / 20) = ρ_{0}$ and $J^{' 1} = γ (J^{1} - β J^{0}) = (5 / 4) (3 ρ_{0} / 4 - (3 / 5) (5 ρ_{0} / 4)) = 0$ .

The example carries no physical charge assignment: it illustrates the transformation of a specified vector field, and the identification of $J$ with a physical current is a model premise of the electromagnetic framework.

Facts & Assumptions

Given: An inertial frame with coordinates $(x^{0}, x^{1}, x^{2}, x^{3})$ and $η_{0} = diag (- 1, 1, 1, 1)$ , an open set $Ω$ , the constant future unit timelike vector field $U$ with components $(5 / 4, 3 / 4, 0, 0)$ , a constant scalar field $ρ_{0}$ , and $J : = ρ_{0} U$ .

[L1]

A $C^{k}$ vector field on $Ω$ is a map assigning to each point a vector, with components in the inertial chart; for a constant field all partial derivatives of the components vanish, and for a $C^{1}$ vector field $J$ in an inertial chart the continuity equation $\partial_{a} J^{a} = 0$ is equivalent to $\partial_{0} J^{0} + \sum_{i} \partial_{i} J^{i} = 0$ ; under the passive boost with parameter $β$ , $Λ = B (β)$ , the components transform as $J^{' 0} = γ (J^{0} - β \cdot J_{c o v})$ and $J_{c o v}^{'} = J_{c o v} + \frac{γ - 1}{∣ β ∣^{2}} (β \cdot J_{c o v}) β - γ β J^{0}$ (Transformation of tensor, covector, derivative and current components under affine Lorentz maps).

[L2]

For a $C^{1}$ future timelike vector field $U$ and a $C^{1}$ scalar field $ρ_{0}$ , $J = ρ_{0} U$ is a $C^{1}$ vector field with first component $J^{0} = ρ_{0} U^{0}$ ; the normalized chart direction $u : = U / U^{0}$ satisfies $u^{0} = 1$ and $ρ : = J^{0} = ρ_{0} U^{0}$ is the coordinate density, while $ρ_{0}$ is the scalar multiplier; $ρ = ρ_{0}$ holds exactly when $ρ_{0} (U^{0} - 1) = 0$ , hence at points with $ρ_{0} \neq 0$ exactly when $U^{0} = 1$ (Transformation of tensor, covector, derivative and current components under affine Lorentz maps).

[L3]

For $β = β e_{1}$ with $∣ β ∣ < 1$ the boost acts by $x^{' 0} = γ (x^{0} - β x^{1})$ , $x^{' 1} = γ (x^{1} - β x^{0})$ on the $(e_{0}, e_{1})$ plane with $γ = (1 - β^{2})^{- 1 / 2}$ , and $B (β)$ is a proper orthochronous Lorentz map (The passive Lorentz boost with parameter beta, The passive boost is a proper orthochronous Lorentz map with inverse B(-beta)).

[L4]

In the frame, $e_{0} \in C^{+}$ and the coordinates identify the timelike sets: $T_{+} = {v : v^{0} > ∣ v_{v e c} ∣}$ , $T_{-} = {v : v^{0} < - ∣ v_{v e c} ∣}$ , with $C^{+} = T_{+}$ ; a nonzero vector with $η (v, v) < 0$ is timelike, and $η (v, v) = - (v^{0})^{2} + (v^{1})^{2} + (v^{2})^{2} + (v^{3})^{2}$ (Inertial frames and inertial charts of the affine Minkowski model, The timelike set has exactly two convex components, described in orthonormal coordinates, Time orientation, future and past timelike/null/causal vectors, and the closed cones, Timelike, null, spacelike and causal vectors of a Lorentz vector space).

Proof

technique · direct

1.1L1L2L4algebra

By [L4], $η_{0} (U, U) = - (5 / 4)^{2} + (3 / 4)^{2} = (- 25 + 9) / 16 = - 1$ , so $U$ is timelike, and $U^{0} = 5 / 4 > 3 / 4 = ∣ U_{v e c} ∣$ places $U$ in $T_{+} = C^{+}$ : $U$ is future timelike unit. Since $ρ_{0}$ is constant, [L2] gives $J = ρ_{0} U$ with components $(J^{0}, J^{1}, J^{2}, J^{3}) = ρ_{0} (5 / 4, 3 / 4, 0, 0) = (5 ρ_{0} / 4, 3 ρ_{0} / 4, 0, 0)$ , and all partial derivatives of the components vanish; the sum $\partial_{0} J^{0} + \sum_{i} \partial_{i} J^{i}$ is then a sum of vanishing terms, so the continuity equation $\partial_{a} J^{a} = 0$ holds by [L1].

1.2L2L3algebra

By [L3] with $β = (3 / 5, 0, 0)$ , $γ = 5 / 4$ : the first component of $B U$ is $(5 / 4) (5 / 4 - (3 / 5) (3 / 4)) = (5 / 4) (5 / 4 - 9 / 20) = (5 / 4) (16 / 20) = 1$ , the second is $(5 / 4) ((3 / 4) - (3 / 5) (5 / 4)) = (5 / 4) (3 / 4 - 3 / 4) = 0$ , and the transverse components of $U$ vanish, so $B U = e_{0}$ . By linearity $B J = ρ_{0} e_{0}$ as an active image in $V$ , and by the primed-chart component law [L1], $[J]_{x'} = B_{x} [J]_{x} = (ρ_{0}, 0, 0, 0)$ . Likewise $[U]_{x'} = (1, 0, 0, 0)$ , so $U = e'_{0}$ and $J = ρ_{0} e'_{0}$ intrinsically. This is clause (ii).

2.1L1step 1.1step 1.2algebra

For $β = \frac{3}{5} e_{1}$ the general law of [L1] reduces, since $J^{2} = J^{3} = 0$ and $β \cdot J_{c o v} = \frac{3}{5} J^{1}$ , to $J^{' 0} = γ (J^{0} - \frac{3}{5} J^{1})$ and $J^{' 1} = γ (J^{1} - \frac{3}{5} J^{0})$ (the correction term $(γ - 1) J^{1} e_{1}$ combining with $J^{1}$ to give $γ J^{1}$ ). Substituting $γ = 5 / 4$ , $J^{0} = 5 ρ_{0} / 4$ and $J^{1} = 3 ρ_{0} / 4$ : $J^{' 0} = (5 / 4) (5 ρ_{0} / 4 - (3 / 5) (3 ρ_{0} / 4)) = (5 / 4) (25 ρ_{0} / 20 - 9 ρ_{0} / 20) = (5 / 4) (16 ρ_{0} / 20) = ρ_{0}$ and $J^{' 1} = (5 / 4) (3 ρ_{0} / 4 - (3 / 5) (5 ρ_{0} / 4)) = (5 / 4) (3 ρ_{0} / 4 - 3 ρ_{0} / 4) = 0$ , reproducing the components of step 1.2; this is clause (iii).

3.1step 1.1step 1.2step 2.1∎

Clause (i) is step 1.1, clause (ii) is step 1.2, and clause (iii) is step 2.1. Every entry is an exact rational evaluation of the boost matrix on the displayed components, and no physical identification of $J$ with a charge-current is made.

Worked current boost law and coordinate versus rest density

Example

Facts & Assumptions

Proof

Depends on

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