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Applying the Euler-Lagrange equations to Maxwell's Theory
Derivation of Maxwell's equations from field tensor lagrangianHow the boundary term in the variation of the action vanishesCoefficient matrix of quadratic LagrangianWhich transformations *aren't* symmetries of a Lagrangian?Problems while doing $dfracpartialpartial(partial_mu phi)$ and $dfracpartialpartial(partial_mu A_mu)$Problem to find field equations with Euler-Lagrange in field theoryProblem with Lagrangian densityEuler-Lagrange for simple scalar field (Peskin & Shroeder)Derivatives in Euler-Lagrange for fieldsDeriving Euler-Lagrange for Electrodynamics LagrangianBox form of the kinetic term and Euler-Lagrange equation
$begingroup$
In Prof. David Tong's notes, specifically on page 10, he gives the Lagrangian of Maxwell's theory to be
$$
mathcalL = -frac12(partial_mu A_nu)(partial^mu A^nu) + frac12(partial_mu A^mu)^2
$$
and then he computes the following
$$
fracpartialmathcalLpartial(partial_mu A_nu) = -partial_mu A_nu + (partial_rho A^rho)eta^munu.
$$
I can see how the first term in the derivative is computed but am having problems with the second term. Any help is appreciated!
homework-and-exercises electromagnetism lagrangian-formalism classical-electrodynamics variational-calculus
edited Mar 18 at 7:26
Qmechanic♦
106k121961224
asked Mar 18 at 5:05
LimzyLimzy
304
$endgroup$
add a comment |
$begingroup$
In Prof. David Tong's notes, specifically on page 10, he gives the Lagrangian of Maxwell's theory to be
$$
mathcalL = -frac12(partial_mu A_nu)(partial^mu A^nu) + frac12(partial_mu A^mu)^2
$$
and then he computes the following
$$
fracpartialmathcalLpartial(partial_mu A_nu) = -partial_mu A_nu + (partial_rho A^rho)eta^munu.
$$
I can see how the first term in the derivative is computed but am having problems with the second term. Any help is appreciated!
homework-and-exercises electromagnetism lagrangian-formalism classical-electrodynamics variational-calculus
edited Mar 18 at 7:26
Qmechanic♦
106k121961224
asked Mar 18 at 5:05
LimzyLimzy
304
$endgroup$
$begingroup$
Possible duplicates: physics.stackexchange.com/q/3005/2451 and links therein.
$endgroup$
– Qmechanic♦
Mar 18 at 11:09
add a comment |
$begingroup$
In Prof. David Tong's notes, specifically on page 10, he gives the Lagrangian of Maxwell's theory to be
$$
mathcalL = -frac12(partial_mu A_nu)(partial^mu A^nu) + frac12(partial_mu A^mu)^2
$$
and then he computes the following
$$
fracpartialmathcalLpartial(partial_mu A_nu) = -partial_mu A_nu + (partial_rho A^rho)eta^munu.
$$
I can see how the first term in the derivative is computed but am having problems with the second term. Any help is appreciated!
homework-and-exercises electromagnetism lagrangian-formalism classical-electrodynamics variational-calculus
edited Mar 18 at 7:26
Qmechanic♦
106k121961224
asked Mar 18 at 5:05
LimzyLimzy
304
$endgroup$
In Prof. David Tong's notes, specifically on page 10, he gives the Lagrangian of Maxwell's theory to be
$$
mathcalL = -frac12(partial_mu A_nu)(partial^mu A^nu) + frac12(partial_mu A^mu)^2
$$
and then he computes the following
$$
fracpartialmathcalLpartial(partial_mu A_nu) = -partial_mu A_nu + (partial_rho A^rho)eta^munu.
$$
I can see how the first term in the derivative is computed but am having problems with the second term. Any help is appreciated!
homework-and-exercises electromagnetism lagrangian-formalism classical-electrodynamics variational-calculus
homework-and-exercises electromagnetism lagrangian-formalism classical-electrodynamics variational-calculus
edited Mar 18 at 7:26
Qmechanic♦
106k121961224
asked Mar 18 at 5:05
LimzyLimzy
304
edited Mar 18 at 7:26
Qmechanic♦
106k121961224
asked Mar 18 at 5:05
LimzyLimzy
304
edited Mar 18 at 7:26
Qmechanic♦
106k121961224
edited Mar 18 at 7:26
Qmechanic♦
106k121961224
edited Mar 18 at 7:26
Qmechanic♦
106k121961224
106k121961224
asked Mar 18 at 5:05
LimzyLimzy
304
asked Mar 18 at 5:05
LimzyLimzy
304
asked Mar 18 at 5:05
LimzyLimzy
304
304
$begingroup$
Possible duplicates: physics.stackexchange.com/q/3005/2451 and links therein.
$endgroup$
– Qmechanic♦
Mar 18 at 11:09
add a comment |
$begingroup$
Possible duplicates: physics.stackexchange.com/q/3005/2451 and links therein.
$endgroup$
– Qmechanic♦
Mar 18 at 11:09
$begingroup$
Possible duplicates: physics.stackexchange.com/q/3005/2451 and links therein.
$endgroup$
– Qmechanic♦
Mar 18 at 11:09
$begingroup$
Possible duplicates: physics.stackexchange.com/q/3005/2451 and links therein.
$endgroup$
– Qmechanic♦
Mar 18 at 11:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We have $frac12 (partial_muA^mu)^2 = frac12 (partial_alpha A^alpha)(partial_betaA^beta)= frac12 (partial_alpha A_sigma) eta^sigmaalpha(partial_betaA_rho) eta^rhobeta$ so the derivative w.r.t. $partial_mu A_nu$ is
$$frac12delta_alpha^mu delta_sigma^nu eta^sigmaalpha(partial_betaA_rho) eta^rhobeta+frac12(partial_alpha A_sigma) eta^sigmaalphadelta_beta^mu delta_rho^nu eta^rhobeta= frac12 eta^munu (partial_beta A^beta)+frac12 (partial_alphaA^alpha) eta^munu = (partial_rhoA^rho)eta^munu $$
where I've freely labeled and relabeled dummy indices.
answered Mar 18 at 5:13
DwaggDwagg
619113
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
We have $frac12 (partial_muA^mu)^2 = frac12 (partial_alpha A^alpha)(partial_betaA^beta)= frac12 (partial_alpha A_sigma) eta^sigmaalpha(partial_betaA_rho) eta^rhobeta$ so the derivative w.r.t. $partial_mu A_nu$ is
$$frac12delta_alpha^mu delta_sigma^nu eta^sigmaalpha(partial_betaA_rho) eta^rhobeta+frac12(partial_alpha A_sigma) eta^sigmaalphadelta_beta^mu delta_rho^nu eta^rhobeta= frac12 eta^munu (partial_beta A^beta)+frac12 (partial_alphaA^alpha) eta^munu = (partial_rhoA^rho)eta^munu $$
where I've freely labeled and relabeled dummy indices.
answered Mar 18 at 5:13
DwaggDwagg
619113
$endgroup$
add a comment |
$begingroup$
We have $frac12 (partial_muA^mu)^2 = frac12 (partial_alpha A^alpha)(partial_betaA^beta)= frac12 (partial_alpha A_sigma) eta^sigmaalpha(partial_betaA_rho) eta^rhobeta$ so the derivative w.r.t. $partial_mu A_nu$ is
$$frac12delta_alpha^mu delta_sigma^nu eta^sigmaalpha(partial_betaA_rho) eta^rhobeta+frac12(partial_alpha A_sigma) eta^sigmaalphadelta_beta^mu delta_rho^nu eta^rhobeta= frac12 eta^munu (partial_beta A^beta)+frac12 (partial_alphaA^alpha) eta^munu = (partial_rhoA^rho)eta^munu $$
where I've freely labeled and relabeled dummy indices.
answered Mar 18 at 5:13
DwaggDwagg
619113
$endgroup$
add a comment |
$begingroup$
We have $frac12 (partial_muA^mu)^2 = frac12 (partial_alpha A^alpha)(partial_betaA^beta)= frac12 (partial_alpha A_sigma) eta^sigmaalpha(partial_betaA_rho) eta^rhobeta$ so the derivative w.r.t. $partial_mu A_nu$ is
$$frac12delta_alpha^mu delta_sigma^nu eta^sigmaalpha(partial_betaA_rho) eta^rhobeta+frac12(partial_alpha A_sigma) eta^sigmaalphadelta_beta^mu delta_rho^nu eta^rhobeta= frac12 eta^munu (partial_beta A^beta)+frac12 (partial_alphaA^alpha) eta^munu = (partial_rhoA^rho)eta^munu $$
where I've freely labeled and relabeled dummy indices.
answered Mar 18 at 5:13
DwaggDwagg
619113
$endgroup$
We have $frac12 (partial_muA^mu)^2 = frac12 (partial_alpha A^alpha)(partial_betaA^beta)= frac12 (partial_alpha A_sigma) eta^sigmaalpha(partial_betaA_rho) eta^rhobeta$ so the derivative w.r.t. $partial_mu A_nu$ is
$$frac12delta_alpha^mu delta_sigma^nu eta^sigmaalpha(partial_betaA_rho) eta^rhobeta+frac12(partial_alpha A_sigma) eta^sigmaalphadelta_beta^mu delta_rho^nu eta^rhobeta= frac12 eta^munu (partial_beta A^beta)+frac12 (partial_alphaA^alpha) eta^munu = (partial_rhoA^rho)eta^munu $$
where I've freely labeled and relabeled dummy indices.
answered Mar 18 at 5:13
DwaggDwagg
619113
answered Mar 18 at 5:13
DwaggDwagg
619113
answered Mar 18 at 5:13
DwaggDwagg
619113
answered Mar 18 at 5:13
DwaggDwagg
619113
619113
add a comment |
add a comment |
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$begingroup$
Possible duplicates: physics.stackexchange.com/q/3005/2451 and links therein.
$endgroup$
– Qmechanic♦
Mar 18 at 11:09