Write electromagnetic field tensor in terms of four-vector potentialProof that 4-potential exists from Gauss-Faraday field equationHistory of Electromagnetic Field TensorElectromagnetic field tensor via tensor products?Is electromagnetic vector field a sum of E and B?Can a static electric field have a vector potential field?contravariant components of electromagnetic field tensor under lorentz transformationWhy is the electromagnetic field strength $F_munu=partial_nu A_mu-partial_mu A_nu$ a tensor?Electromagnetic field tensorRapid question about Electromagnetic TensorQuestion about derivation of four-velocity vector4-Gradient vector and the Field strength tensor

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Write electromagnetic field tensor in terms of four-vector potential


Proof that 4-potential exists from Gauss-Faraday field equationHistory of Electromagnetic Field TensorElectromagnetic field tensor via tensor products?Is electromagnetic vector field a sum of E and B?Can a static electric field have a vector potential field?contravariant components of electromagnetic field tensor under lorentz transformationWhy is the electromagnetic field strength $F_munu=partial_nu A_mu-partial_mu A_nu$ a tensor?Electromagnetic field tensorRapid question about Electromagnetic TensorQuestion about derivation of four-velocity vector4-Gradient vector and the Field strength tensor













2












$begingroup$


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 $F^munu = partial^mu A^nu - partial^nu A^mu$.










share|cite|improve this question









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
    3 hours ago










  • $begingroup$
    This was what I was looking for. Thank you, I will look up Poincare's lemma.
    $endgroup$
    – Lucas L.
    3 hours ago















2












$begingroup$


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 $F^munu = partial^mu A^nu - partial^nu A^mu$.










share|cite|improve this question









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
    3 hours ago










  • $begingroup$
    This was what I was looking for. Thank you, I will look up Poincare's lemma.
    $endgroup$
    – Lucas L.
    3 hours ago













2












2








2





$begingroup$


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 $F^munu = partial^mu A^nu - partial^nu A^mu$.










share|cite|improve this question









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






share|cite|improve this question









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.










share|cite|improve this question









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.








share|cite|improve this question




share|cite|improve this question








edited 3 hours ago







Lucas L.













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 4 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
    3 hours ago










  • $begingroup$
    This was what I was looking for. Thank you, I will look up Poincare's lemma.
    $endgroup$
    – Lucas L.
    3 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
    3 hours ago










  • $begingroup$
    This was what I was looking for. Thank you, I will look up Poincare's lemma.
    $endgroup$
    – Lucas L.
    3 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
3 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
3 hours ago












$begingroup$
This was what I was looking for. Thank you, I will look up Poincare's lemma.
$endgroup$
– Lucas L.
3 hours ago




$begingroup$
This was what I was looking for. Thank you, I will look up Poincare's lemma.
$endgroup$
– Lucas L.
3 hours ago










3 Answers
3






active

oldest

votes


















2












$begingroup$

The Bianchi identity $mathrmdF~=~0$ together with Poincare lemma guarantee the local existence of $A$ in contractible regions of spacetime. See also this related Phys.SE post.






share|cite|improve this answer











$endgroup$




















    2












    $begingroup$

    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 $partial_mupartial_nu = partial_nupartial_mu$.






    share|cite|improve this answer











    $endgroup$




















      0












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$












      • $begingroup$
        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.
        3 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
        3 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
        3 hours ago











      Your Answer








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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      The Bianchi identity $mathrmdF~=~0$ together with Poincare lemma guarantee the local existence of $A$ in contractible regions of spacetime. See also this related Phys.SE post.






      share|cite|improve this answer











      $endgroup$

















        2












        $begingroup$

        The Bianchi identity $mathrmdF~=~0$ together with Poincare lemma guarantee the local existence of $A$ in contractible regions of spacetime. See also this related Phys.SE post.






        share|cite|improve this answer











        $endgroup$















          2












          2








          2





          $begingroup$

          The Bianchi identity $mathrmdF~=~0$ together with Poincare lemma guarantee the local existence of $A$ in contractible regions of spacetime. See also this related Phys.SE post.






          share|cite|improve this answer











          $endgroup$



          The Bianchi identity $mathrmdF~=~0$ together with Poincare lemma guarantee the local existence of $A$ in contractible regions of spacetime. See also this related Phys.SE post.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 3 hours ago

























          answered 3 hours ago









          QmechanicQmechanic

          109k122051270




          109k122051270





















              2












              $begingroup$

              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 $partial_mupartial_nu = partial_nupartial_mu$.






              share|cite|improve this answer











              $endgroup$

















                2












                $begingroup$

                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 $partial_mupartial_nu = partial_nupartial_mu$.






                share|cite|improve this answer











                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  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 $partial_mupartial_nu = partial_nupartial_mu$.






                  share|cite|improve this answer











                  $endgroup$



                  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 $partial_mupartial_nu = partial_nupartial_mu$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 2 hours ago

























                  answered 3 hours ago









                  Thomas FritschThomas Fritsch

                  1,8201016




                  1,8201016





















                      0












                      $begingroup$

                      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.






                      share|cite|improve this answer









                      $endgroup$












                      • $begingroup$
                        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.
                        3 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
                        3 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
                        3 hours ago















                      0












                      $begingroup$

                      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.






                      share|cite|improve this answer









                      $endgroup$












                      • $begingroup$
                        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.
                        3 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
                        3 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
                        3 hours ago













                      0












                      0








                      0





                      $begingroup$

                      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.






                      share|cite|improve this answer









                      $endgroup$



                      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.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 3 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.
                        3 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
                        3 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
                        3 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.
                        3 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
                        3 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
                        3 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.
                      3 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.
                      3 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
                      3 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
                      3 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
                      3 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
                      3 hours ago










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