Maxwell Tensor Identity [duplicate]
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This question already has an answer here:
Expanding electromagnetic field Lagrangian in terms of gauge field
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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 Mar 17 at 16:30
Qmechanic♦
108k122001246
asked Mar 17 at 15:05
EthanTEthanT
382110
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marked as duplicate by knzhou, John Rennie electromagnetism
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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.
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 Mar 17 at 16:30
Qmechanic♦
108k122001246
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
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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
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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
edited Mar 17 at 16:30
Qmechanic♦
108k122001246
asked Mar 17 at 15:05
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 Mar 17 at 16:30
Qmechanic♦
108k122001246
asked Mar 17 at 15:05
EthanTEthanT
382110
edited Mar 17 at 16:30
Qmechanic♦
108k122001246
asked Mar 17 at 15:05
EthanTEthanT
382110
edited Mar 17 at 16:30
Qmechanic♦
108k122001246
edited Mar 17 at 16:30
Qmechanic♦
108k122001246
edited Mar 17 at 16:30
Qmechanic♦
108k122001246
108k122001246
asked Mar 17 at 15:05
EthanTEthanT
382110
asked Mar 17 at 15:05
EthanTEthanT
382110
asked Mar 17 at 15:05
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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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.
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 |
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
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
Mar 17 at 15:15
$begingroup$
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
$begingroup$
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
2
$begingroup$
Possible duplicate of Expanding electromagnetic field Lagrangian in terms of gauge field
$endgroup$
– knzhou
Mar 17 at 15:50
$begingroup$
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
$begingroup$
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
$begingroup$
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 |
3 Answers
3
active
oldest
votes
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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 Mar 17 at 17:59
PaulPaul
1818
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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 Mar 17 at 15:21
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 Mar 17 at 17:04
my2ctsmy2cts
5,7962719
$endgroup$
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 Mar 17 at 17:59
PaulPaul
1818
$endgroup$
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 Mar 17 at 17:59
PaulPaul
1818
$endgroup$
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 Mar 17 at 17:59
PaulPaul
1818
$endgroup$
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 Mar 17 at 17:59
PaulPaul
1818
answered Mar 17 at 17:59
PaulPaul
1818
answered Mar 17 at 17:59
PaulPaul
1818
answered Mar 17 at 17:59
PaulPaul
1818
1818
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 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.
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.
answered Mar 17 at 15:21
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 Mar 17 at 15:21
AdityaAditya
354113
answered Mar 17 at 15:21
AdityaAditya
354113
answered Mar 17 at 15:21
AdityaAditya
354113
answered Mar 17 at 15:21
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 Mar 17 at 17:04
my2ctsmy2cts
5,7962719
$endgroup$
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 Mar 17 at 17:04
my2ctsmy2cts
5,7962719
$endgroup$
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 Mar 17 at 17:04
my2ctsmy2cts
5,7962719
$endgroup$
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 Mar 17 at 17:04
my2ctsmy2cts
5,7962719
answered Mar 17 at 17:04
my2ctsmy2cts
5,7962719
answered Mar 17 at 17:04
my2ctsmy2cts
5,7962719
answered Mar 17 at 17:04
my2ctsmy2cts
5,7962719
5,7962719
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
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