# 3.1 Lorentz covariance for motion in electromagnetic fields, 2

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Is $\displaystyle{\frac{d p_\mu}{ds} = \frac{q}{c} F_{\mu \nu} \frac{d x^\nu}{ds}}$ gauge invariant?

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… the defining property of a Lorentz transformation, $\Lambda^\mu_{\;\;\nu}$:

$\eta_{\mu\nu} \Lambda^{\mu}_{\;\;\alpha} \Lambda^\nu_{\;\;\beta} = \eta_{\alpha\beta}$

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… 4-vectors and (Lorentz)-tensors are transformed like this:

$U'^\mu = \Lambda^\mu_{\;\;\nu}U^\nu$

and

$F'_{\mu\nu} = \Lambda_\mu^{\;\;\alpha} \Lambda_\nu^{\;\;\beta}F_{\alpha\beta}= \Lambda_\mu^{\;\;\alpha} F_{\alpha\beta} (\Lambda^{-1})^{\beta}_{\;\;\nu}$

where we have used the conventional notation

$\displaystyle{\Lambda_\nu^{\;\;\mu} = (\Lambda^{-1})^\mu_{\;\;\nu}}$

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Let us then take your equation and apply $\displaystyle{\Lambda_\sigma^{\;\;\mu}}$ on both sides (recall this Lorentz transformation does not depend on $\displaystyle{\tau}$), and try rewriting everything in terms of prime quantities:

\displaystyle{ \begin{aligned} m \Lambda_\sigma^{\;\;\mu}\frac{{\rm d}U_\mu}{{\rm d}\tau} &= e \Lambda_\sigma^{\;\;\mu} F_{\mu\nu} U^\nu \\ &= e \Lambda_\sigma^{\;\;\mu} F_{\mu\nu}\eta^{\nu\alpha} U_\alpha\\ m\frac{{\rm d}U'_\sigma}{{\rm d}\tau} &= e \Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\alpha} ((\Lambda^{-1})^{\beta}_{\;\;\nu} \Lambda^\alpha_{\;\;\beta}) U_\alpha\\ &= e \Big(\Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\alpha}(\Lambda^{-1})^{\beta}_{\;\;\nu}\Big) \Big(\Lambda^\alpha_{\;\;\beta} U_\alpha\Big)\\ &= e \Big(\Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu\Big) U'_\beta\\ &= e \Big(\Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu\Big) U'_\beta\\ &= e F'_{\sigma\nu}\eta^{\nu\beta} U'_\beta \\ \\ m\frac{{\rm d}U'_\sigma}{{\rm d}\tau} &= e F'_{\sigma\nu} U'^\nu \\ \end{aligned}}

This “game” can always be done with contracted indices, …

— answered Jul 7, 2020 at 15:06

— ohneVal

— Lorentz invariance of the Lorentz force law

— Physics StackExchange

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How come \displaystyle{ \begin{aligned} \eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu &= \eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu \\ \end{aligned}}?

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Note that

\displaystyle{ \begin{aligned} \sum ... \sum \Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu &= \sum ... \sum \Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu \\ \end{aligned}}

does not mean that

\displaystyle{ \begin{aligned} \eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu &= \eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu \\ \end{aligned}}

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Also

\displaystyle{ \begin{aligned} \sum \eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu &= \sum \eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu \\ \end{aligned}}

does not mean that

\displaystyle{ \begin{aligned} \eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu &= \eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu \\ \end{aligned}}

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\displaystyle{ \begin{aligned} A = \sum_{\alpha, \beta, \nu} F_{\mu\alpha}\eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu U_\beta' &= \sum_{\alpha, \beta} F_{\mu\alpha}\eta^{\alpha\alpha}\Lambda^{\;\;\beta}_{\nu=\alpha} U_\beta' \\ &= \sum_{\beta} \left( - F_{\mu 0} \Lambda^{\;\;\beta}_{0} + F_{\mu 1} \Lambda^{\;\;\beta}_{1} + F_{\mu 2} \Lambda^{\;\;\beta}_{2} + F_{\mu 3} \Lambda^{\;\;\beta}_{3} \right) U_\beta' \\ \\ \end{aligned}}

\displaystyle{ \begin{aligned} B = \sum_{\alpha, \beta, \nu} F_{\mu\alpha}\eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu U_\beta' &= \sum_{\alpha, \beta} F_{\mu\alpha} \eta^{\beta \beta}\Lambda^{\;\;\alpha}_{\nu=\beta} U_\beta \\ &= \sum_{\alpha} F_{\mu\alpha} \left( - \Lambda^{\;\;\alpha}_{0} U_0' + \Lambda^{\;\;\alpha}_{1} U_1' + \Lambda^{\;\;\alpha}_{2} U_2' + \Lambda^{\;\;\alpha}_{3} U_3' \right) \\ \end{aligned}}

At the first glance, it seems to be unlikely that

\displaystyle{ \begin{aligned} \sum_{\alpha, \beta, \nu} F_{\mu\alpha}\eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu U_\beta' &= \sum_{\alpha, \beta, \nu} F_{\mu\alpha}\eta^{\nu\beta}\Lambda^{\;\;\alpha}_\nu U_\beta' \\ \end{aligned}},

because while in $A$, for any $\beta$, $U_\beta'$‘s have visible negative terms; in $B$, only $U_0'$‘s do.

Without additional mathematical properties among those physical quantities, the identity is impossible to prove.

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Just for future reference:

From the invariance of the spacetime interval it follows

$\displaystyle{\eta =\Lambda ^{\mathrm {T} }\eta \Lambda}$

— Wikipedia on Lorentz transformation

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\displaystyle{ \begin{aligned} \eta_{\mu \nu} = \eta'_{\mu \nu} &= \Lambda^{\alpha}{}_{\mu} \Lambda^{\beta}{}_{\nu} \eta_{\alpha \beta} \\ &= (\Lambda^T)_{\mu}{}^{\alpha} \eta_{\alpha \beta} \Lambda^{\beta}{}_{\nu} \\ \\ \eta_{\alpha\beta} &= \Lambda^{\mu}_{\;\;\alpha} \Lambda^{\nu}_{\;\;\beta}\eta_{\mu\nu} \\ \eta^{\alpha\beta} &= \Lambda^{\alpha}_{\;\;\mu} \Lambda^{\beta}_{\;\;\nu}\eta^{\mu\nu} \\ \end{aligned}}

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\displaystyle{ \begin{aligned} m \Lambda_\sigma^{\;\;\mu}\frac{{\rm d}U_\mu}{{\rm d}\tau} &= e \Lambda_\sigma^{\;\;\mu} F_{\mu\nu} U^\nu \\ & = e \Lambda_\nu^{\;\;\mu} F_{\mu\nu}\eta^{\nu\alpha} U_\alpha\\ \end{aligned}}

Inserting the identity $\sum_\beta (\Lambda^{-1})^{\beta}_{\;\;\nu} \Lambda^\alpha_{\;\;\beta}$ into the expression:

\displaystyle{ \begin{aligned} m\frac{{\rm d}U'_\sigma}{{\rm d}\tau} &= e \Big(\Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\alpha}(\Lambda^{-1})^{\beta}_{\;\;\nu}\Big) \Big(\Lambda^\alpha_{\;\;\beta} U_\alpha\Big) \\ &= e \Big(\Lambda_\sigma^{\;\;\mu} F_{\mu\alpha}\eta^{\nu\alpha}\Lambda^{\;\;\beta}_\nu\Big) U'_\beta\\ \end{aligned}}

This path does not work. Also, the formula $\displaystyle{ \Lambda^\alpha_{\;\;\beta} U_\alpha = U'_\beta }$ is plain wrong!

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Inserting the identity $\sum_\beta (\Lambda^{-1})^{\beta}_{\;\;\nu} \Lambda^\alpha_{\;\;\beta}$ into the expression before (actually without) lowering the index:

\displaystyle{ \begin{aligned} m \Lambda_\sigma^{\;\;\mu}\frac{{\rm d}U_\mu}{{\rm d}\tau} &= e \Lambda_\sigma^{\;\;\mu} F_{\mu\nu} U^\nu \\ &= e \Lambda_\sigma^{\;\;\mu} F_{\mu\alpha} ((\Lambda^{-1})^{\alpha}_{\;\;\beta} \Lambda^\beta_{\;\;\nu}) U^\nu \\ m \frac{{\rm d}U_\sigma'}{{\rm d}\tau} &= e (\Lambda_\sigma^{\;\;\mu} F_{\mu\alpha} (\Lambda^{-1})^{\alpha}_{\;\;\beta}) (\Lambda^\beta_{\;\;\nu} U^\nu) \\ &= e {F'}_{\sigma \beta} {U'}^\beta \\ \end{aligned}}

— Me@2022-08-23 12:03:54 PM

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