Applying the Euler-Lagrange equations to Maxwell's Theory
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In Prof. David Tong's notes, specifically on page 10, he gives the Lagrangian of Maxwell's theory to be
$$
mathcal{L} = -frac{1}{2}(partial_mu A_nu)(partial^mu A^nu) + frac{1}{2}(partial_mu A^mu)^2
$$
and then he computes the following
$$
frac{partialmathcal{L}}{partial(partial_mu A_nu)} = -partial_mu A_nu + (partial_rho A^rho)eta^{munu}.
$$
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♦
107k121981229
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
$$
mathcal{L} = -frac{1}{2}(partial_mu A_nu)(partial^mu A^nu) + frac{1}{2}(partial_mu A^mu)^2
$$
and then he computes the following
$$
frac{partialmathcal{L}}{partial(partial_mu A_nu)} = -partial_mu A_nu + (partial_rho A^rho)eta^{munu}.
$$
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♦
107k121981229
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
$$
mathcal{L} = -frac{1}{2}(partial_mu A_nu)(partial^mu A^nu) + frac{1}{2}(partial_mu A^mu)^2
$$
and then he computes the following
$$
frac{partialmathcal{L}}{partial(partial_mu A_nu)} = -partial_mu A_nu + (partial_rho A^rho)eta^{munu}.
$$
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♦
107k121981229
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
$$
mathcal{L} = -frac{1}{2}(partial_mu A_nu)(partial^mu A^nu) + frac{1}{2}(partial_mu A^mu)^2
$$
and then he computes the following
$$
frac{partialmathcal{L}}{partial(partial_mu A_nu)} = -partial_mu A_nu + (partial_rho A^rho)eta^{munu}.
$$
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♦
107k121981229
asked Mar 18 at 5:05
LimzyLimzy
304
edited Mar 18 at 7:26
Qmechanic♦
107k121981229
asked Mar 18 at 5:05
LimzyLimzy
304
edited Mar 18 at 7:26
Qmechanic♦
107k121981229
edited Mar 18 at 7:26
Qmechanic♦
107k121981229
edited Mar 18 at 7:26
Qmechanic♦
107k121981229
107k121981229
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
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We have $frac12 (partial_{mu}A^{mu})^2 = frac12 (partial_{alpha} A^{alpha})(partial_{beta}A^{beta})= frac12 (partial_{alpha} A_{sigma}) eta^{sigmaalpha}(partial_{beta}A_{rho}) eta^{rhobeta}$ so the derivative w.r.t. $partial_{mu} A_{nu}$ is
$$frac12delta_{alpha}^{mu} delta_{sigma}^{nu} eta^{sigmaalpha}(partial_{beta}A_{rho}) eta^{rhobeta}+frac12(partial_{alpha} A_{sigma}) eta^{sigmaalpha}delta_{beta}^{mu} delta_{rho}^{nu} eta^{rhobeta}= frac12 eta^{munu} (partial_{beta} A^{beta})+frac12 (partial_{alpha}A^{alpha}) eta^{munu} = (partial_{rho}A^{rho})eta^{munu} $$
where I've freely labeled and relabeled dummy indices.
answered Mar 18 at 5:13
DwaggDwagg
619113
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1 Answer
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active
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1 Answer
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$begingroup$
We have $frac12 (partial_{mu}A^{mu})^2 = frac12 (partial_{alpha} A^{alpha})(partial_{beta}A^{beta})= frac12 (partial_{alpha} A_{sigma}) eta^{sigmaalpha}(partial_{beta}A_{rho}) eta^{rhobeta}$ so the derivative w.r.t. $partial_{mu} A_{nu}$ is
$$frac12delta_{alpha}^{mu} delta_{sigma}^{nu} eta^{sigmaalpha}(partial_{beta}A_{rho}) eta^{rhobeta}+frac12(partial_{alpha} A_{sigma}) eta^{sigmaalpha}delta_{beta}^{mu} delta_{rho}^{nu} eta^{rhobeta}= frac12 eta^{munu} (partial_{beta} A^{beta})+frac12 (partial_{alpha}A^{alpha}) eta^{munu} = (partial_{rho}A^{rho})eta^{munu} $$
where I've freely labeled and relabeled dummy indices.
answered Mar 18 at 5:13
DwaggDwagg
619113
$endgroup$
add a comment |
$begingroup$
We have $frac12 (partial_{mu}A^{mu})^2 = frac12 (partial_{alpha} A^{alpha})(partial_{beta}A^{beta})= frac12 (partial_{alpha} A_{sigma}) eta^{sigmaalpha}(partial_{beta}A_{rho}) eta^{rhobeta}$ so the derivative w.r.t. $partial_{mu} A_{nu}$ is
$$frac12delta_{alpha}^{mu} delta_{sigma}^{nu} eta^{sigmaalpha}(partial_{beta}A_{rho}) eta^{rhobeta}+frac12(partial_{alpha} A_{sigma}) eta^{sigmaalpha}delta_{beta}^{mu} delta_{rho}^{nu} eta^{rhobeta}= frac12 eta^{munu} (partial_{beta} A^{beta})+frac12 (partial_{alpha}A^{alpha}) eta^{munu} = (partial_{rho}A^{rho})eta^{munu} $$
where I've freely labeled and relabeled dummy indices.
answered Mar 18 at 5:13
DwaggDwagg
619113
$endgroup$
add a comment |
$begingroup$
We have $frac12 (partial_{mu}A^{mu})^2 = frac12 (partial_{alpha} A^{alpha})(partial_{beta}A^{beta})= frac12 (partial_{alpha} A_{sigma}) eta^{sigmaalpha}(partial_{beta}A_{rho}) eta^{rhobeta}$ so the derivative w.r.t. $partial_{mu} A_{nu}$ is
$$frac12delta_{alpha}^{mu} delta_{sigma}^{nu} eta^{sigmaalpha}(partial_{beta}A_{rho}) eta^{rhobeta}+frac12(partial_{alpha} A_{sigma}) eta^{sigmaalpha}delta_{beta}^{mu} delta_{rho}^{nu} eta^{rhobeta}= frac12 eta^{munu} (partial_{beta} A^{beta})+frac12 (partial_{alpha}A^{alpha}) eta^{munu} = (partial_{rho}A^{rho})eta^{munu} $$
where I've freely labeled and relabeled dummy indices.
answered Mar 18 at 5:13
DwaggDwagg
619113
$endgroup$
We have $frac12 (partial_{mu}A^{mu})^2 = frac12 (partial_{alpha} A^{alpha})(partial_{beta}A^{beta})= frac12 (partial_{alpha} A_{sigma}) eta^{sigmaalpha}(partial_{beta}A_{rho}) eta^{rhobeta}$ so the derivative w.r.t. $partial_{mu} A_{nu}$ is
$$frac12delta_{alpha}^{mu} delta_{sigma}^{nu} eta^{sigmaalpha}(partial_{beta}A_{rho}) eta^{rhobeta}+frac12(partial_{alpha} A_{sigma}) eta^{sigmaalpha}delta_{beta}^{mu} delta_{rho}^{nu} eta^{rhobeta}= frac12 eta^{munu} (partial_{beta} A^{beta})+frac12 (partial_{alpha}A^{alpha}) eta^{munu} = (partial_{rho}A^{rho})eta^{munu} $$
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