Write electromagnetic field tensor in terms of four-vector potentialProof that 4-potential exists from...
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Write electromagnetic field tensor in terms of four-vector potential
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$begingroup$
How can we know that the electromagnetic tensor $F^{munu}$ can be written in terms of a four-vector potential $A^mu$ as $F^{mu nu} = partial^mu A^nu - partial^nu A^mu$? In the vector calculus approach, this is not really hard to see (under reasonable 'smoothness' conditions on the fields), but I would like to know how one would see this in the four-vector approach.
More specifically, how can we prove (mathematically) that given the electromagnetic tensor, there exists a four-vector such that $F^{munu} = partial^mu A^nu - partial^nu A^mu$.
electromagnetism tensor-calculus
asked 5 hours ago
Lucas L.Lucas L.
335
New contributor
Lucas L. 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 |
$begingroup$
How can we know that the electromagnetic tensor $F^{munu}$ can be written in terms of a four-vector potential $A^mu$ as $F^{mu nu} = partial^mu A^nu - partial^nu A^mu$? In the vector calculus approach, this is not really hard to see (under reasonable 'smoothness' conditions on the fields), but I would like to know how one would see this in the four-vector approach.
More specifically, how can we prove (mathematically) that given the electromagnetic tensor, there exists a four-vector such that $F^{munu} = partial^mu A^nu - partial^nu A^mu$.
electromagnetism tensor-calculus
asked 5 hours ago
Lucas L.Lucas L.
335
New contributor
Lucas L. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
The EM field is actually a two-form $F$ satisfying Maxwell's equations, one of which is in form notation $dF = 0$. By definition this says that $F$ is a closed form. A form which is $F = dA$ for some $A$ is said exact. Now all exact forms are closed because $d^2 = 0$. On the other hand, Poincare's lemma says that all closed forms are exact if the domain is contractible. Assuming a contractible spacetime implies the existence of the potential from Poincare's lemma
$endgroup$
– user1620696
4 hours ago
$begingroup$
This was what I was looking for. Thank you, I will look up Poincare's lemma.
$endgroup$
– Lucas L.
4 hours ago
add a comment |
$begingroup$
How can we know that the electromagnetic tensor $F^{munu}$ can be written in terms of a four-vector potential $A^mu$ as $F^{mu nu} = partial^mu A^nu - partial^nu A^mu$? In the vector calculus approach, this is not really hard to see (under reasonable 'smoothness' conditions on the fields), but I would like to know how one would see this in the four-vector approach.
More specifically, how can we prove (mathematically) that given the electromagnetic tensor, there exists a four-vector such that $F^{munu} = partial^mu A^nu - partial^nu A^mu$.
electromagnetism tensor-calculus
asked 5 hours ago
Lucas L.Lucas L.
335
New contributor
Lucas L. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
How can we know that the electromagnetic tensor $F^{munu}$ can be written in terms of a four-vector potential $A^mu$ as $F^{mu nu} = partial^mu A^nu - partial^nu A^mu$? In the vector calculus approach, this is not really hard to see (under reasonable 'smoothness' conditions on the fields), but I would like to know how one would see this in the four-vector approach.
More specifically, how can we prove (mathematically) that given the electromagnetic tensor, there exists a four-vector such that $F^{munu} = partial^mu A^nu - partial^nu A^mu$.
electromagnetism tensor-calculus
electromagnetism tensor-calculus
asked 5 hours ago
Lucas L.Lucas L.
335
New contributor
Lucas L. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Lucas L.Lucas L.
335
New contributor
Lucas L. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
Lucas L.
asked 5 hours ago
Lucas L.Lucas L.
335
New contributor
Lucas L. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Lucas L.Lucas L.
335
asked 5 hours ago
Lucas L.Lucas L.
335
335
New contributor
Lucas L. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Lucas L. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
The EM field is actually a two-form $F$ satisfying Maxwell's equations, one of which is in form notation $dF = 0$. By definition this says that $F$ is a closed form. A form which is $F = dA$ for some $A$ is said exact. Now all exact forms are closed because $d^2 = 0$. On the other hand, Poincare's lemma says that all closed forms are exact if the domain is contractible. Assuming a contractible spacetime implies the existence of the potential from Poincare's lemma
$endgroup$
– user1620696
4 hours ago
$begingroup$
This was what I was looking for. Thank you, I will look up Poincare's lemma.
$endgroup$
– Lucas L.
4 hours ago
add a comment |
$begingroup$
The EM field is actually a two-form $F$ satisfying Maxwell's equations, one of which is in form notation $dF = 0$. By definition this says that $F$ is a closed form. A form which is $F = dA$ for some $A$ is said exact. Now all exact forms are closed because $d^2 = 0$. On the other hand, Poincare's lemma says that all closed forms are exact if the domain is contractible. Assuming a contractible spacetime implies the existence of the potential from Poincare's lemma
$endgroup$
– user1620696
4 hours ago
$begingroup$
This was what I was looking for. Thank you, I will look up Poincare's lemma.
$endgroup$
– Lucas L.
4 hours ago
$begingroup$
The EM field is actually a two-form $F$ satisfying Maxwell's equations, one of which is in form notation $dF = 0$. By definition this says that $F$ is a closed form. A form which is $F = dA$ for some $A$ is said exact. Now all exact forms are closed because $d^2 = 0$. On the other hand, Poincare's lemma says that all closed forms are exact if the domain is contractible. Assuming a contractible spacetime implies the existence of the potential from Poincare's lemma
$endgroup$
– user1620696
4 hours ago
$begingroup$
The EM field is actually a two-form $F$ satisfying Maxwell's equations, one of which is in form notation $dF = 0$. By definition this says that $F$ is a closed form. A form which is $F = dA$ for some $A$ is said exact. Now all exact forms are closed because $d^2 = 0$. On the other hand, Poincare's lemma says that all closed forms are exact if the domain is contractible. Assuming a contractible spacetime implies the existence of the potential from Poincare's lemma
$endgroup$
– user1620696
4 hours ago
$begingroup$
This was what I was looking for. Thank you, I will look up Poincare's lemma.
$endgroup$
– Lucas L.
4 hours ago
$begingroup$
This was what I was looking for. Thank you, I will look up Poincare's lemma.
$endgroup$
– Lucas L.
4 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The Bianchi identity $mathrm{d}F~=~0$ together with Poincare lemma guarantee the local existence of $A$ in contractible regions of spacetime. See also this related Phys.SE post.
answered 4 hours ago
Qmechanic♦Qmechanic
109k122051270
$endgroup$
add a comment |
$begingroup$
One way to write the homogenous Maxwell's equations is
with the Levi-Civita symbol $epsilon$:
$$epsilon^{alphabetamunu} partial_beta F_{munu} = 0$$
Solution to this is obviously (with arbitrary potential $A$):
$$F_{munu} = partial_mu A_nu - partial_nu A_mu$$
It is easy to verify using the antisymmetry of $epsilon^{alphabetamunu}$
upon swapping any 2 indexes, together with $partial_mupartial_nu = partial_nupartial_mu$.
answered 5 hours ago
Thomas FritschThomas Fritsch
1,8301016
$endgroup$
add a comment |
$begingroup$
You are asking "how we know...". This may not be a fair question. We created this formalism. You could also ask how do we know that we can write Maxwell's equations using vectors. Long ago they were not, they were written as a large set of coupled scalar (scalar type) equations. The formalism of vector notation was still evolving and being accepted and one has to PROVE that a set of quantities actually behaves like a vector under coordinate transformations.
This is a key to understanding the 4-vector approach. There is a scalar E field potential, Phi, and a vector potential A, in classical electrodynamics.
Once Einstein presented Lorentz invariance in physics (I'm not going to write extensively about that history here) we started on the path of putting all equations in a covariant format. It is the nature of light that governs this invariance, and the postulate that the speed of light is the same for all relatively interital observers. Even Newton's laws of mechanics were elevated to a covariant 4-vector form, as was momentum (E, p) where Energy, E, was thought to be a scalar.
We know that electromagnetism needs to be Lorentz invariant and this motivates elevating Phi to the time component of a 4-vector, just like E is the time component of 4-momentum. This is also indicated by seeing how the equations transform under Lorentz. The 4-potential (Phi, A) must be as is to obey this symmetry. Then the rest follows.
answered 5 hours ago
ggcgggcg
1,626114
$endgroup$
$begingroup$
I think you misinterpreted my question. I wanted to ask: how can we prove (mathematically) that given the electromagnetic tensor, there exists a four-vector such that $F^{munu} = partial^mu A^nu - partial^nu A^mu$. I will edit my question accordingly.
$endgroup$
– Lucas L.
5 hours ago
$begingroup$
I think I did elude to it. Maxwell's equations will actually convert to the field tensor by collecting terms, using (Phi, A).
$endgroup$
– ggcg
4 hours ago
$begingroup$
Okay, I am 180 degrees from your intent. Sorry. But it's the same logic as div(B) = 0 and curl(E) = 0. Namely that del(F) = 0 --> F has a potential. Of course you need to get the correct form of the equation, the source free version.
$endgroup$
– ggcg
4 hours ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The Bianchi identity $mathrm{d}F~=~0$ together with Poincare lemma guarantee the local existence of $A$ in contractible regions of spacetime. See also this related Phys.SE post.
answered 4 hours ago
Qmechanic♦Qmechanic
109k122051270
$endgroup$
add a comment |
$begingroup$
The Bianchi identity $mathrm{d}F~=~0$ together with Poincare lemma guarantee the local existence of $A$ in contractible regions of spacetime. See also this related Phys.SE post.
answered 4 hours ago
Qmechanic♦Qmechanic
109k122051270
$endgroup$
add a comment |
$begingroup$
The Bianchi identity $mathrm{d}F~=~0$ together with Poincare lemma guarantee the local existence of $A$ in contractible regions of spacetime. See also this related Phys.SE post.
answered 4 hours ago
Qmechanic♦Qmechanic
109k122051270
$endgroup$
The Bianchi identity $mathrm{d}F~=~0$ together with Poincare lemma guarantee the local existence of $A$ in contractible regions of spacetime. See also this related Phys.SE post.
answered 4 hours ago
Qmechanic♦Qmechanic
109k122051270
edited 4 hours ago
answered 4 hours ago
Qmechanic♦Qmechanic
109k122051270
answered 4 hours ago
Qmechanic♦Qmechanic
109k122051270
answered 4 hours ago
Qmechanic♦Qmechanic
109k122051270
109k122051270
add a comment |
add a comment |
$begingroup$
One way to write the homogenous Maxwell's equations is
with the Levi-Civita symbol $epsilon$:
$$epsilon^{alphabetamunu} partial_beta F_{munu} = 0$$
Solution to this is obviously (with arbitrary potential $A$):
$$F_{munu} = partial_mu A_nu - partial_nu A_mu$$
It is easy to verify using the antisymmetry of $epsilon^{alphabetamunu}$
upon swapping any 2 indexes, together with $partial_mupartial_nu = partial_nupartial_mu$.
answered 5 hours ago
Thomas FritschThomas Fritsch
1,8301016
$endgroup$
add a comment |
$begingroup$
One way to write the homogenous Maxwell's equations is
with the Levi-Civita symbol $epsilon$:
$$epsilon^{alphabetamunu} partial_beta F_{munu} = 0$$
Solution to this is obviously (with arbitrary potential $A$):
$$F_{munu} = partial_mu A_nu - partial_nu A_mu$$
It is easy to verify using the antisymmetry of $epsilon^{alphabetamunu}$
upon swapping any 2 indexes, together with $partial_mupartial_nu = partial_nupartial_mu$.
answered 5 hours ago
Thomas FritschThomas Fritsch
1,8301016
$endgroup$
add a comment |
$begingroup$
One way to write the homogenous Maxwell's equations is
with the Levi-Civita symbol $epsilon$:
$$epsilon^{alphabetamunu} partial_beta F_{munu} = 0$$
Solution to this is obviously (with arbitrary potential $A$):
$$F_{munu} = partial_mu A_nu - partial_nu A_mu$$
It is easy to verify using the antisymmetry of $epsilon^{alphabetamunu}$
upon swapping any 2 indexes, together with $partial_mupartial_nu = partial_nupartial_mu$.
answered 5 hours ago
Thomas FritschThomas Fritsch
1,8301016
$endgroup$
One way to write the homogenous Maxwell's equations is
with the Levi-Civita symbol $epsilon$:
$$epsilon^{alphabetamunu} partial_beta F_{munu} = 0$$
Solution to this is obviously (with arbitrary potential $A$):
$$F_{munu} = partial_mu A_nu - partial_nu A_mu$$
It is easy to verify using the antisymmetry of $epsilon^{alphabetamunu}$
upon swapping any 2 indexes, together with $partial_mupartial_nu = partial_nupartial_mu$.
answered 5 hours ago
Thomas FritschThomas Fritsch
1,8301016
edited 3 hours ago
answered 5 hours ago
Thomas FritschThomas Fritsch
1,8301016
answered 5 hours ago
Thomas FritschThomas Fritsch
1,8301016
answered 5 hours ago
Thomas FritschThomas Fritsch
1,8301016
1,8301016
add a comment |
add a comment |
$begingroup$
You are asking "how we know...". This may not be a fair question. We created this formalism. You could also ask how do we know that we can write Maxwell's equations using vectors. Long ago they were not, they were written as a large set of coupled scalar (scalar type) equations. The formalism of vector notation was still evolving and being accepted and one has to PROVE that a set of quantities actually behaves like a vector under coordinate transformations.
This is a key to understanding the 4-vector approach. There is a scalar E field potential, Phi, and a vector potential A, in classical electrodynamics.
Once Einstein presented Lorentz invariance in physics (I'm not going to write extensively about that history here) we started on the path of putting all equations in a covariant format. It is the nature of light that governs this invariance, and the postulate that the speed of light is the same for all relatively interital observers. Even Newton's laws of mechanics were elevated to a covariant 4-vector form, as was momentum (E, p) where Energy, E, was thought to be a scalar.
We know that electromagnetism needs to be Lorentz invariant and this motivates elevating Phi to the time component of a 4-vector, just like E is the time component of 4-momentum. This is also indicated by seeing how the equations transform under Lorentz. The 4-potential (Phi, A) must be as is to obey this symmetry. Then the rest follows.
answered 5 hours ago
ggcgggcg
1,626114
$endgroup$
$begingroup$
I think you misinterpreted my question. I wanted to ask: how can we prove (mathematically) that given the electromagnetic tensor, there exists a four-vector such that $F^{munu} = partial^mu A^nu - partial^nu A^mu$. I will edit my question accordingly.
$endgroup$
– Lucas L.
5 hours ago
$begingroup$
I think I did elude to it. Maxwell's equations will actually convert to the field tensor by collecting terms, using (Phi, A).
$endgroup$
– ggcg
4 hours ago
$begingroup$
Okay, I am 180 degrees from your intent. Sorry. But it's the same logic as div(B) = 0 and curl(E) = 0. Namely that del(F) = 0 --> F has a potential. Of course you need to get the correct form of the equation, the source free version.
$endgroup$
– ggcg
4 hours ago
add a comment |
$begingroup$
You are asking "how we know...". This may not be a fair question. We created this formalism. You could also ask how do we know that we can write Maxwell's equations using vectors. Long ago they were not, they were written as a large set of coupled scalar (scalar type) equations. The formalism of vector notation was still evolving and being accepted and one has to PROVE that a set of quantities actually behaves like a vector under coordinate transformations.
This is a key to understanding the 4-vector approach. There is a scalar E field potential, Phi, and a vector potential A, in classical electrodynamics.
Once Einstein presented Lorentz invariance in physics (I'm not going to write extensively about that history here) we started on the path of putting all equations in a covariant format. It is the nature of light that governs this invariance, and the postulate that the speed of light is the same for all relatively interital observers. Even Newton's laws of mechanics were elevated to a covariant 4-vector form, as was momentum (E, p) where Energy, E, was thought to be a scalar.
We know that electromagnetism needs to be Lorentz invariant and this motivates elevating Phi to the time component of a 4-vector, just like E is the time component of 4-momentum. This is also indicated by seeing how the equations transform under Lorentz. The 4-potential (Phi, A) must be as is to obey this symmetry. Then the rest follows.
answered 5 hours ago
ggcgggcg
1,626114
$endgroup$
$begingroup$
I think you misinterpreted my question. I wanted to ask: how can we prove (mathematically) that given the electromagnetic tensor, there exists a four-vector such that $F^{munu} = partial^mu A^nu - partial^nu A^mu$. I will edit my question accordingly.
$endgroup$
– Lucas L.
5 hours ago
$begingroup$
I think I did elude to it. Maxwell's equations will actually convert to the field tensor by collecting terms, using (Phi, A).
$endgroup$
– ggcg
4 hours ago
$begingroup$
Okay, I am 180 degrees from your intent. Sorry. But it's the same logic as div(B) = 0 and curl(E) = 0. Namely that del(F) = 0 --> F has a potential. Of course you need to get the correct form of the equation, the source free version.
$endgroup$
– ggcg
4 hours ago
add a comment |
$begingroup$
You are asking "how we know...". This may not be a fair question. We created this formalism. You could also ask how do we know that we can write Maxwell's equations using vectors. Long ago they were not, they were written as a large set of coupled scalar (scalar type) equations. The formalism of vector notation was still evolving and being accepted and one has to PROVE that a set of quantities actually behaves like a vector under coordinate transformations.
This is a key to understanding the 4-vector approach. There is a scalar E field potential, Phi, and a vector potential A, in classical electrodynamics.
Once Einstein presented Lorentz invariance in physics (I'm not going to write extensively about that history here) we started on the path of putting all equations in a covariant format. It is the nature of light that governs this invariance, and the postulate that the speed of light is the same for all relatively interital observers. Even Newton's laws of mechanics were elevated to a covariant 4-vector form, as was momentum (E, p) where Energy, E, was thought to be a scalar.
We know that electromagnetism needs to be Lorentz invariant and this motivates elevating Phi to the time component of a 4-vector, just like E is the time component of 4-momentum. This is also indicated by seeing how the equations transform under Lorentz. The 4-potential (Phi, A) must be as is to obey this symmetry. Then the rest follows.
answered 5 hours ago
ggcgggcg
1,626114
$endgroup$
You are asking "how we know...". This may not be a fair question. We created this formalism. You could also ask how do we know that we can write Maxwell's equations using vectors. Long ago they were not, they were written as a large set of coupled scalar (scalar type) equations. The formalism of vector notation was still evolving and being accepted and one has to PROVE that a set of quantities actually behaves like a vector under coordinate transformations.
This is a key to understanding the 4-vector approach. There is a scalar E field potential, Phi, and a vector potential A, in classical electrodynamics.
Once Einstein presented Lorentz invariance in physics (I'm not going to write extensively about that history here) we started on the path of putting all equations in a covariant format. It is the nature of light that governs this invariance, and the postulate that the speed of light is the same for all relatively interital observers. Even Newton's laws of mechanics were elevated to a covariant 4-vector form, as was momentum (E, p) where Energy, E, was thought to be a scalar.
We know that electromagnetism needs to be Lorentz invariant and this motivates elevating Phi to the time component of a 4-vector, just like E is the time component of 4-momentum. This is also indicated by seeing how the equations transform under Lorentz. The 4-potential (Phi, A) must be as is to obey this symmetry. Then the rest follows.
answered 5 hours ago
ggcgggcg
1,626114
answered 5 hours ago
ggcgggcg
1,626114
answered 5 hours ago
ggcgggcg
1,626114
answered 5 hours ago
ggcgggcg
1,626114
1,626114
$begingroup$
I think you misinterpreted my question. I wanted to ask: how can we prove (mathematically) that given the electromagnetic tensor, there exists a four-vector such that $F^{munu} = partial^mu A^nu - partial^nu A^mu$. I will edit my question accordingly.
$endgroup$
– Lucas L.
5 hours ago
$begingroup$
I think I did elude to it. Maxwell's equations will actually convert to the field tensor by collecting terms, using (Phi, A).
$endgroup$
– ggcg
4 hours ago
$begingroup$
Okay, I am 180 degrees from your intent. Sorry. But it's the same logic as div(B) = 0 and curl(E) = 0. Namely that del(F) = 0 --> F has a potential. Of course you need to get the correct form of the equation, the source free version.
$endgroup$
– ggcg
4 hours ago
add a comment |
$begingroup$
I think you misinterpreted my question. I wanted to ask: how can we prove (mathematically) that given the electromagnetic tensor, there exists a four-vector such that $F^{munu} = partial^mu A^nu - partial^nu A^mu$. I will edit my question accordingly.
$endgroup$
– Lucas L.
5 hours ago
$begingroup$
I think I did elude to it. Maxwell's equations will actually convert to the field tensor by collecting terms, using (Phi, A).
$endgroup$
– ggcg
4 hours ago
$begingroup$
Okay, I am 180 degrees from your intent. Sorry. But it's the same logic as div(B) = 0 and curl(E) = 0. Namely that del(F) = 0 --> F has a potential. Of course you need to get the correct form of the equation, the source free version.
$endgroup$
– ggcg
4 hours ago
$begingroup$
I think you misinterpreted my question. I wanted to ask: how can we prove (mathematically) that given the electromagnetic tensor, there exists a four-vector such that $F^{munu} = partial^mu A^nu - partial^nu A^mu$. I will edit my question accordingly.
$endgroup$
– Lucas L.
5 hours ago
$begingroup$
I think you misinterpreted my question. I wanted to ask: how can we prove (mathematically) that given the electromagnetic tensor, there exists a four-vector such that $F^{munu} = partial^mu A^nu - partial^nu A^mu$. I will edit my question accordingly.
$endgroup$
– Lucas L.
5 hours ago
$begingroup$
I think I did elude to it. Maxwell's equations will actually convert to the field tensor by collecting terms, using (Phi, A).
$endgroup$
– ggcg
4 hours ago
$begingroup$
I think I did elude to it. Maxwell's equations will actually convert to the field tensor by collecting terms, using (Phi, A).
$endgroup$
– ggcg
4 hours ago
$begingroup$
Okay, I am 180 degrees from your intent. Sorry. But it's the same logic as div(B) = 0 and curl(E) = 0. Namely that del(F) = 0 --> F has a potential. Of course you need to get the correct form of the equation, the source free version.
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– ggcg
4 hours ago
$begingroup$
Okay, I am 180 degrees from your intent. Sorry. But it's the same logic as div(B) = 0 and curl(E) = 0. Namely that del(F) = 0 --> F has a potential. Of course you need to get the correct form of the equation, the source free version.
$endgroup$
– ggcg
4 hours ago
add a comment |
Lucas L. is a new contributor. Be nice, and check out our Code of Conduct.
Lucas L. is a new contributor. Be nice, and check out our Code of Conduct.
Lucas L. is a new contributor. Be nice, and check out our Code of Conduct.
Lucas L. is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
The EM field is actually a two-form $F$ satisfying Maxwell's equations, one of which is in form notation $dF = 0$. By definition this says that $F$ is a closed form. A form which is $F = dA$ for some $A$ is said exact. Now all exact forms are closed because $d^2 = 0$. On the other hand, Poincare's lemma says that all closed forms are exact if the domain is contractible. Assuming a contractible spacetime implies the existence of the potential from Poincare's lemma
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– user1620696
4 hours ago
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This was what I was looking for. Thank you, I will look up Poincare's lemma.
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– Lucas L.
4 hours ago