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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2022.08.26 Friday (c) All rights reserved by ACHK