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Maxwell Tensor Identity [duplicate]

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1

$begingroup$

This question already has an answer here:

• 
  Expanding electromagnetic field Lagrangian in terms of gauge field
  
  1 answer
  
  

In Schawrtz, Page 116, formula 8.23, he seems to suggest that the square of the Maxwell tensor can be expanded out as follows:

$$-frac14F_mu nu^2=frac12A_musquare A_mu-frac12A_mupartial_mupartial_nuA_nu$$

where:

$$F_munu=partial_mu A_nu - partial_nuA_mu$$

For the life of me, I can't seem to derive this. I get close, but always with an extra unwanted term, or two.

Anyone have a hint on the best way to proceed?

homework-and-exercises electromagnetism lagrangian-formalism

share|cite|improve this question

edited Mar 17 at 16:30

Qmechanic♦

106k121961224

asked Mar 17 at 15:05

EthanTEthanT

382110

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marked as duplicate by knzhou, John Rennie electromagnetism
Users with the  electromagnetism badge can single-handedly close electromagnetism questions as duplicates and reopen them as needed.

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Mar 17 at 20:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Does Schwartz include an integral? If so, he might be integrating out certain terms to the boundary and setting them to zero.
  $endgroup$
  – Aditya
  Mar 17 at 15:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  He does not, and I thought I had derived this in the past sans integral. I'll try that out, though. At the very least, I learn a new way of deriving this
  $endgroup$
  – EthanT
  Mar 17 at 15:37
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Expanding electromagnetic field Lagrangian in terms of gauge field
  $endgroup$
  – knzhou
  Mar 17 at 15:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, this was really easy keeping it under the integral of S. However, I thought there was a way to achieve the same thing, w/ just tensor manipulation. Maybe I am not remembering correctly, though
  $endgroup$
  – EthanT
  Mar 17 at 16:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Products of zeroth and second derivatives are not generally equal to products of first derivatives in any identity... What you need is a context where a derivative of a product is zero, as $d(x~dx)=dx~dx + x~d^2x.$ Getting the left hand side to vanish in this context might be possible with antisymmetry but looks non-trivial—maybe it amounts to a boundary term in some integral though?
  $endgroup$
  – CR Drost
  Mar 17 at 16:22
  

  
  
  
  

add a comment | 

1

$begingroup$

This question already has an answer here:

• 
  Expanding electromagnetic field Lagrangian in terms of gauge field
  
  1 answer
  
  

In Schawrtz, Page 116, formula 8.23, he seems to suggest that the square of the Maxwell tensor can be expanded out as follows:

$$-frac14F_mu nu^2=frac12A_musquare A_mu-frac12A_mupartial_mupartial_nuA_nu$$

where:

$$F_munu=partial_mu A_nu - partial_nuA_mu$$

For the life of me, I can't seem to derive this. I get close, but always with an extra unwanted term, or two.

Anyone have a hint on the best way to proceed?

homework-and-exercises electromagnetism lagrangian-formalism

share|cite|improve this question

edited Mar 17 at 16:30

Qmechanic♦

106k121961224

asked Mar 17 at 15:05

EthanTEthanT

382110

$endgroup$

marked as duplicate by knzhou, John Rennie electromagnetism
Users with the  electromagnetism badge can single-handedly close electromagnetism questions as duplicates and reopen them as needed.

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Mar 17 at 20:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Does Schwartz include an integral? If so, he might be integrating out certain terms to the boundary and setting them to zero.
  $endgroup$
  – Aditya
  Mar 17 at 15:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  He does not, and I thought I had derived this in the past sans integral. I'll try that out, though. At the very least, I learn a new way of deriving this
  $endgroup$
  – EthanT
  Mar 17 at 15:37
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Expanding electromagnetic field Lagrangian in terms of gauge field
  $endgroup$
  – knzhou
  Mar 17 at 15:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, this was really easy keeping it under the integral of S. However, I thought there was a way to achieve the same thing, w/ just tensor manipulation. Maybe I am not remembering correctly, though
  $endgroup$
  – EthanT
  Mar 17 at 16:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Products of zeroth and second derivatives are not generally equal to products of first derivatives in any identity... What you need is a context where a derivative of a product is zero, as $d(x~dx)=dx~dx + x~d^2x.$ Getting the left hand side to vanish in this context might be possible with antisymmetry but looks non-trivial—maybe it amounts to a boundary term in some integral though?
  $endgroup$
  – CR Drost
  Mar 17 at 16:22
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Expanding electromagnetic field Lagrangian in terms of gauge field
  
  1 answer
  
  

In Schawrtz, Page 116, formula 8.23, he seems to suggest that the square of the Maxwell tensor can be expanded out as follows:

$$-frac14F_mu nu^2=frac12A_musquare A_mu-frac12A_mupartial_mupartial_nuA_nu$$

where:

$$F_munu=partial_mu A_nu - partial_nuA_mu$$

For the life of me, I can't seem to derive this. I get close, but always with an extra unwanted term, or two.

Anyone have a hint on the best way to proceed?

homework-and-exercises electromagnetism lagrangian-formalism

share|cite|improve this question

edited Mar 17 at 16:30

Qmechanic♦

106k121961224

asked Mar 17 at 15:05

EthanTEthanT

382110

$endgroup$

share|cite|improve this question

edited Mar 17 at 16:30

Qmechanic♦

106k121961224

asked Mar 17 at 15:05

EthanTEthanT

382110

share|cite|improve this question

edited Mar 17 at 16:30

Qmechanic♦

106k121961224

asked Mar 17 at 15:05

EthanTEthanT

382110

edited Mar 17 at 16:30

Qmechanic♦

106k121961224

edited Mar 17 at 16:30

Qmechanic♦

106k121961224

asked Mar 17 at 15:05

EthanTEthanT

382110

asked Mar 17 at 15:05

EthanTEthanT

382110

3 Answers
3

active

oldest

votes

4

$begingroup$

Your expression is part of a Lagrangian. As the physics remains the same as long as the action remains the same, one can always do partial integration in the action integral over the Lagrangian to derive alternative Lagrangians describing the same physics.

share|cite|improve this answer

answered Mar 17 at 17:59

PaulPaul

1498

New contributor

Paul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

3

$begingroup$

Hint: Try introducing an integral to the expression so it becomes $$-frac14int F_munuF^munutextd^d x$$
and take the total derivative terms to vanish at infinity. A much more careful argument can be made here in the presence of boundaries, but this should get you started.

share|cite|improve this answer

answered Mar 17 at 15:21

AdityaAditya

354113

$endgroup$

add a comment | 

0

$begingroup$

The relation as you state it does not hold. Only the space time integral of both hands of the equation is equal under suitable boundary conditions. So this would be an error.

share|cite|improve this answer

answered Mar 17 at 17:04

my2ctsmy2cts

5,6572718

$endgroup$

add a comment | 

4

$begingroup$

Your expression is part of a Lagrangian. As the physics remains the same as long as the action remains the same, one can always do partial integration in the action integral over the Lagrangian to derive alternative Lagrangians describing the same physics.

share|cite|improve this answer

answered Mar 17 at 17:59

PaulPaul

1498

New contributor

Paul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

4

$begingroup$

Your expression is part of a Lagrangian. As the physics remains the same as long as the action remains the same, one can always do partial integration in the action integral over the Lagrangian to derive alternative Lagrangians describing the same physics.

share|cite|improve this answer

answered Mar 17 at 17:59

PaulPaul

1498

New contributor

Paul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

Your expression is part of a Lagrangian. As the physics remains the same as long as the action remains the same, one can always do partial integration in the action integral over the Lagrangian to derive alternative Lagrangians describing the same physics.

share|cite|improve this answer

answered Mar 17 at 17:59

PaulPaul

1498

New contributor

Paul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

answered Mar 17 at 17:59

PaulPaul

1498

New contributor

Paul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 17 at 17:59

PaulPaul

1498

New contributor

Paul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 17 at 17:59

PaulPaul

1498

3

$begingroup$

Hint: Try introducing an integral to the expression so it becomes $$-frac14int F_munuF^munutextd^d x$$
and take the total derivative terms to vanish at infinity. A much more careful argument can be made here in the presence of boundaries, but this should get you started.

share|cite|improve this answer

answered Mar 17 at 15:21

AdityaAditya

354113

$endgroup$

add a comment | 

3

$begingroup$

Hint: Try introducing an integral to the expression so it becomes $$-frac14int F_munuF^munutextd^d x$$
and take the total derivative terms to vanish at infinity. A much more careful argument can be made here in the presence of boundaries, but this should get you started.

share|cite|improve this answer

answered Mar 17 at 15:21

AdityaAditya

354113

$endgroup$

add a comment | 

$begingroup$

Hint: Try introducing an integral to the expression so it becomes $$-frac14int F_munuF^munutextd^d x$$
and take the total derivative terms to vanish at infinity. A much more careful argument can be made here in the presence of boundaries, but this should get you started.

share|cite|improve this answer

answered Mar 17 at 15:21

AdityaAditya

354113

$endgroup$

share|cite|improve this answer

answered Mar 17 at 15:21

AdityaAditya

354113

answered Mar 17 at 15:21

AdityaAditya

354113

answered Mar 17 at 15:21

AdityaAditya

354113

0

$begingroup$

The relation as you state it does not hold. Only the space time integral of both hands of the equation is equal under suitable boundary conditions. So this would be an error.

share|cite|improve this answer

answered Mar 17 at 17:04

my2ctsmy2cts

5,6572718

$endgroup$

add a comment | 

0

$begingroup$

The relation as you state it does not hold. Only the space time integral of both hands of the equation is equal under suitable boundary conditions. So this would be an error.

share|cite|improve this answer

answered Mar 17 at 17:04

my2ctsmy2cts

5,6572718

$endgroup$

add a comment | 

$begingroup$

The relation as you state it does not hold. Only the space time integral of both hands of the equation is equal under suitable boundary conditions. So this would be an error.

share|cite|improve this answer

answered Mar 17 at 17:04

my2ctsmy2cts

5,6572718

$endgroup$

share|cite|improve this answer

answered Mar 17 at 17:04

my2ctsmy2cts

5,6572718

answered Mar 17 at 17:04

my2ctsmy2cts

5,6572718

answered Mar 17 at 17:04

my2ctsmy2cts

5,6572718