Maxwell Tensor Identity [duplicate]Expanding electromagnetic field Lagrangian in terms of gauge fieldNoether current for the Yang-Mills-Higgs LagrangianRiemann tensor in 2d and 3dDerivation of the quadratic form of the Dirac equationEnergy-momentum tensor for dustDielectric tensor vs. conductivity tensor in (cold) plasmasExpanding electromagnetic field Lagrangian in terms of gauge fieldHow can I see where this formula for a general vertex factor comes from?Equation of Motion for non-linear sigma model (WZW)What is meant by the coupling term $g_munuT^munu$ in Supergravity?Electromagnetic energy stress tensor with non zero current
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Maxwell Tensor Identity [duplicate]
Expanding electromagnetic field Lagrangian in terms of gauge fieldNoether current for the Yang-Mills-Higgs LagrangianRiemann tensor in 2d and 3dDerivation of the quadratic form of the Dirac equationEnergy-momentum tensor for dustDielectric tensor vs. conductivity tensor in (cold) plasmasExpanding electromagnetic field Lagrangian in terms of gauge fieldHow can I see where this formula for a general vertex factor comes from?Equation of Motion for non-linear sigma model (WZW)What is meant by the coupling term $g_munuT^munu$ in Supergravity?Electromagnetic energy stress tensor with non zero current
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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
edited 11 hours ago
Qmechanic♦
106k121961222
asked 12 hours ago
EthanTEthanT
382110
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marked as duplicate by knzhou, John Rennie electromagnetism
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7 hours ago
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.
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
edited 11 hours ago
Qmechanic♦
106k121961222
asked 12 hours ago
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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7 hours ago
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
12 hours ago
$begingroup$
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
12 hours ago
2
$begingroup$
Possible duplicate of Expanding electromagnetic field Lagrangian in terms of gauge field
$endgroup$
– knzhou
12 hours ago
$begingroup$
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
11 hours ago
$begingroup$
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
11 hours ago
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
edited 11 hours ago
Qmechanic♦
106k121961222
asked 12 hours ago
EthanTEthanT
382110
$endgroup$
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?
This question already has an answer here:
Expanding electromagnetic field Lagrangian in terms of gauge field
1 answer
homework-and-exercises electromagnetism lagrangian-formalism
homework-and-exercises electromagnetism lagrangian-formalism
edited 11 hours ago
Qmechanic♦
106k121961222
asked 12 hours ago
EthanTEthanT
382110
edited 11 hours ago
Qmechanic♦
106k121961222
asked 12 hours ago
EthanTEthanT
382110
edited 11 hours ago
Qmechanic♦
106k121961222
edited 11 hours ago
Qmechanic♦
106k121961222
edited 11 hours ago
Qmechanic♦
106k121961222
106k121961222
asked 12 hours ago
EthanTEthanT
382110
asked 12 hours ago
EthanTEthanT
382110
asked 12 hours ago
EthanTEthanT
382110
382110
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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7 hours ago
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.
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7 hours ago
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
12 hours ago
$begingroup$
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
12 hours ago
2
$begingroup$
Possible duplicate of Expanding electromagnetic field Lagrangian in terms of gauge field
$endgroup$
– knzhou
12 hours ago
$begingroup$
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
11 hours ago
$begingroup$
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
11 hours ago
add a comment |
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
12 hours ago
$begingroup$
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
12 hours ago
2
$begingroup$
Possible duplicate of Expanding electromagnetic field Lagrangian in terms of gauge field
$endgroup$
– knzhou
12 hours ago
$begingroup$
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
11 hours ago
$begingroup$
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
11 hours ago
4
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
12 hours ago
$begingroup$
Does Schwartz include an integral? If so, he might be integrating out certain terms to the boundary and setting them to zero.
$endgroup$
– Aditya
12 hours ago
$begingroup$
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
12 hours ago
$begingroup$
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
12 hours ago
2
2
$begingroup$
Possible duplicate of Expanding electromagnetic field Lagrangian in terms of gauge field
$endgroup$
– knzhou
12 hours ago
$begingroup$
Possible duplicate of Expanding electromagnetic field Lagrangian in terms of gauge field
$endgroup$
– knzhou
12 hours ago
$begingroup$
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
11 hours ago
$begingroup$
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
11 hours ago
$begingroup$
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
11 hours ago
$begingroup$
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
11 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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.
answered 10 hours ago
PaulPaul
513
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.
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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.
answered 12 hours ago
AdityaAditya
354113
$endgroup$
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.
answered 10 hours ago
my2ctsmy2cts
5,5692718
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 10 hours ago
PaulPaul
513
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.
answered 10 hours ago
PaulPaul
513
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.
answered 10 hours ago
PaulPaul
513
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.
$endgroup$
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.
answered 10 hours ago
PaulPaul
513
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 10 hours ago
PaulPaul
513
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 10 hours ago
PaulPaul
513
answered 10 hours ago
PaulPaul
513
513
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.
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.
Paul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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.
answered 12 hours ago
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.
answered 12 hours ago
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.
answered 12 hours ago
AdityaAditya
354113
$endgroup$
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.
answered 12 hours ago
AdityaAditya
354113
answered 12 hours ago
AdityaAditya
354113
answered 12 hours ago
AdityaAditya
354113
answered 12 hours ago
AdityaAditya
354113
354113
add a comment |
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.
answered 10 hours ago
my2ctsmy2cts
5,5692718
$endgroup$
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.
answered 10 hours ago
my2ctsmy2cts
5,5692718
$endgroup$
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.
answered 10 hours ago
my2ctsmy2cts
5,5692718
$endgroup$
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.
answered 10 hours ago
my2ctsmy2cts
5,5692718
answered 10 hours ago
my2ctsmy2cts
5,5692718
answered 10 hours ago
my2ctsmy2cts
5,5692718
answered 10 hours ago
my2ctsmy2cts
5,5692718
5,5692718
add a comment |
add a comment |
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
12 hours ago
$begingroup$
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
12 hours ago
2
$begingroup$
Possible duplicate of Expanding electromagnetic field Lagrangian in terms of gauge field
$endgroup$
– knzhou
12 hours ago
$begingroup$
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
11 hours ago
$begingroup$
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
11 hours ago