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Applying the Euler-Lagrange equations to Maxwell's Theory

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4

1

$begingroup$

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

share|cite|improve this question

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 | 

4

1

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

edited Mar 18 at 7:26

Qmechanic♦

106k121961224

asked Mar 18 at 5:05

LimzyLimzy

304

$endgroup$

share|cite|improve this question

edited Mar 18 at 7:26

Qmechanic♦

106k121961224

asked Mar 18 at 5:05

LimzyLimzy

304

share|cite|improve this question

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

asked Mar 18 at 5:05

LimzyLimzy

304

asked Mar 18 at 5:05

LimzyLimzy

304

1 Answer
1

active

oldest

votes

4

$begingroup$

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.

share|cite|improve this answer

answered Mar 18 at 5:13

DwaggDwagg

619113

$endgroup$

add a comment | 

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4

$begingroup$

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.

share|cite|improve this answer

answered Mar 18 at 5:13

DwaggDwagg

619113

$endgroup$

add a comment | 

4

$begingroup$

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.

share|cite|improve this answer

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.

share|cite|improve this answer

answered Mar 18 at 5:13

DwaggDwagg

619113

$endgroup$

share|cite|improve this answer

answered Mar 18 at 5:13

DwaggDwagg

619113

answered Mar 18 at 5:13

DwaggDwagg

619113

answered Mar 18 at 5:13

DwaggDwagg

619113