Are wave equations equivalent to Maxwell's equations in free space?Maxwell's equation in free space from wave...

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Are wave equations equivalent to Maxwell's equations in free space?


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In free space, do Maxwell's equations contain the same amount of information regarding electric and magnetic fields as is contained in the wave equations derived from them? If so, how?










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Jeevesh Juneja is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 1




    $begingroup$
    One way to prove the non-equivalence is by constructing a set of equations which is different from Maxwell's, but for which the ${bf E}$ and ${bf B}$ fields satisfy a wave-equation. For example we can consider the toy-theory of electromagnetism that is, in the absence of charges, described by the two equations $dot{bf E} = k{bf B},~dot{bf B} = frac{c^2}{k}nabla^2{bf E},$ where $k$ is a free constant. Not hard to show that this implies wave equations and the theory is obviously different from Maxwells.
    $endgroup$
    – Winther
    20 hours ago
















8












$begingroup$


In free space, do Maxwell's equations contain the same amount of information regarding electric and magnetic fields as is contained in the wave equations derived from them? If so, how?










share|cite|improve this question









New contributor




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







$endgroup$








  • 1




    $begingroup$
    One way to prove the non-equivalence is by constructing a set of equations which is different from Maxwell's, but for which the ${bf E}$ and ${bf B}$ fields satisfy a wave-equation. For example we can consider the toy-theory of electromagnetism that is, in the absence of charges, described by the two equations $dot{bf E} = k{bf B},~dot{bf B} = frac{c^2}{k}nabla^2{bf E},$ where $k$ is a free constant. Not hard to show that this implies wave equations and the theory is obviously different from Maxwells.
    $endgroup$
    – Winther
    20 hours ago














8












8








8


2



$begingroup$


In free space, do Maxwell's equations contain the same amount of information regarding electric and magnetic fields as is contained in the wave equations derived from them? If so, how?










share|cite|improve this question









New contributor




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







$endgroup$




In free space, do Maxwell's equations contain the same amount of information regarding electric and magnetic fields as is contained in the wave equations derived from them? If so, how?







electromagnetism waves maxwell-equations vacuum






share|cite|improve this question









New contributor




Jeevesh Juneja 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




Jeevesh Juneja 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 6 hours ago









Nat

3,48241831




3,48241831






New contributor




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









asked yesterday









Jeevesh JunejaJeevesh Juneja

442




442




New contributor




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





New contributor





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






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








  • 1




    $begingroup$
    One way to prove the non-equivalence is by constructing a set of equations which is different from Maxwell's, but for which the ${bf E}$ and ${bf B}$ fields satisfy a wave-equation. For example we can consider the toy-theory of electromagnetism that is, in the absence of charges, described by the two equations $dot{bf E} = k{bf B},~dot{bf B} = frac{c^2}{k}nabla^2{bf E},$ where $k$ is a free constant. Not hard to show that this implies wave equations and the theory is obviously different from Maxwells.
    $endgroup$
    – Winther
    20 hours ago














  • 1




    $begingroup$
    One way to prove the non-equivalence is by constructing a set of equations which is different from Maxwell's, but for which the ${bf E}$ and ${bf B}$ fields satisfy a wave-equation. For example we can consider the toy-theory of electromagnetism that is, in the absence of charges, described by the two equations $dot{bf E} = k{bf B},~dot{bf B} = frac{c^2}{k}nabla^2{bf E},$ where $k$ is a free constant. Not hard to show that this implies wave equations and the theory is obviously different from Maxwells.
    $endgroup$
    – Winther
    20 hours ago








1




1




$begingroup$
One way to prove the non-equivalence is by constructing a set of equations which is different from Maxwell's, but for which the ${bf E}$ and ${bf B}$ fields satisfy a wave-equation. For example we can consider the toy-theory of electromagnetism that is, in the absence of charges, described by the two equations $dot{bf E} = k{bf B},~dot{bf B} = frac{c^2}{k}nabla^2{bf E},$ where $k$ is a free constant. Not hard to show that this implies wave equations and the theory is obviously different from Maxwells.
$endgroup$
– Winther
20 hours ago




$begingroup$
One way to prove the non-equivalence is by constructing a set of equations which is different from Maxwell's, but for which the ${bf E}$ and ${bf B}$ fields satisfy a wave-equation. For example we can consider the toy-theory of electromagnetism that is, in the absence of charges, described by the two equations $dot{bf E} = k{bf B},~dot{bf B} = frac{c^2}{k}nabla^2{bf E},$ where $k$ is a free constant. Not hard to show that this implies wave equations and the theory is obviously different from Maxwells.
$endgroup$
– Winther
20 hours ago










4 Answers
4






active

oldest

votes


















14












$begingroup$

No, they're not. The wave equations for the force fields contain a strict subset of the information contained in the full set of Maxwell's equations. In particular, it's important to note that you need the Gauss-type equations,
$$
nablacdot mathbf E = 0 = nablacdotmathbf B,
$$

to ensure the transversality of the waves. If all you had to go was the wave equations in the form
$$
left[partial_t^2 - c^2 nabla^2 right]mathbf E = 0
$$

then you'd have no way of knowing that longitudinal EM waves are forbidden. (Though, to be clear, the transversality conditions are not sufficient, either.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    physics.stackexchange.com/q/465124
    $endgroup$
    – Kavita Juneja
    21 hours ago





















12












$begingroup$

The wave equations for electromagnetic waves in free space can be derived from Maxwell's equations. However, Maxwell's equations can be used to describe much more. For example, you can derive from them, how an electromagnetic wave is launched from an antenna. Or you can treat electrostatic and magnetostatic phenomena. You can learn from them how electric motors work, and how we can convert mechanical into electric energy in generators. There is a huge wealth of physics in these four equations, which has enormous importance for most of the phenomena we observe around us, and for much of today's modern technology.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I am talking of maxwell's equations in free space. No antennas or motors etc. I am asking when we write maxwell's equations in free space and then from those equations derive the wave equation , is any information lost in the derivation??
    $endgroup$
    – Kavita Juneja
    yesterday












  • $begingroup$
    You should really explain why the two are non-equivalent, though (the reason being, not all solutions to wave equations are physical solutions for electric and magnetic field); this said, if you accidentally picked those wave solutions that do represent the physical fields, then all you have described is automatically incorporated.
    $endgroup$
    – gented
    23 hours ago



















1












$begingroup$

Remaining to the case of Maxwell's equations (ME) in vacuum, there is no equivalence between the wave equations for the fields and the original set. As already pointed out, solutions of ME are a subset of the solutions of the two three-dimensional wave equations.



The case made by Emilio Pisanty (one loses information about the transversality) has to be taken as just one example of the non-equivalence. Another information which gets lost is the phase relation between magnetic and electric field.



From a mathematical point of view it is not difficult to understand the reason of the non-equivalence: in order to derive the wave equations one has to




  1. take the curl of one of the equations containing the time derivative of a field;

  2. use the other equation to rewrite the time derivative of a curl as a second time derivative.


It is clear that the additional derivative implied by step 1 may eliminate some information. It is quite well known that if one takes additional derivatives of a differential equations, the resulting equation usually has more solutions than the original one and one has to choose among them those satisfying the original equation.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It's not just the phase relation between the two fields - it's the loss of any connection at all between them. You can zero out one of the fields, or replace it with some completely different solution, and the wave equations won't blink an eye.
    $endgroup$
    – Emilio Pisanty
    6 hours ago



















0












$begingroup$

In the so-called Lorenz gauge Maxwell's equations take the form of a set of inhomogeneous wave equations in terms of the potential. All of the physics is described at least as well by these. Maxwell's equations can be written in covariant notation as $partial_mu F^{munu} = partial_mu partial^mu A^nu - partial_mu partial^nu A^mu = - j^nu /epsilon_0 $. Choosing the Lorenz gauge, $partial_mu A^mu = 0$ reduces this to the inhomogeneous wave equations, $ partial_mu partial^mu A^nu = - j^nu /epsilon_0 $.






share|cite|improve this answer











$endgroup$













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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    14












    $begingroup$

    No, they're not. The wave equations for the force fields contain a strict subset of the information contained in the full set of Maxwell's equations. In particular, it's important to note that you need the Gauss-type equations,
    $$
    nablacdot mathbf E = 0 = nablacdotmathbf B,
    $$

    to ensure the transversality of the waves. If all you had to go was the wave equations in the form
    $$
    left[partial_t^2 - c^2 nabla^2 right]mathbf E = 0
    $$

    then you'd have no way of knowing that longitudinal EM waves are forbidden. (Though, to be clear, the transversality conditions are not sufficient, either.)






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      physics.stackexchange.com/q/465124
      $endgroup$
      – Kavita Juneja
      21 hours ago


















    14












    $begingroup$

    No, they're not. The wave equations for the force fields contain a strict subset of the information contained in the full set of Maxwell's equations. In particular, it's important to note that you need the Gauss-type equations,
    $$
    nablacdot mathbf E = 0 = nablacdotmathbf B,
    $$

    to ensure the transversality of the waves. If all you had to go was the wave equations in the form
    $$
    left[partial_t^2 - c^2 nabla^2 right]mathbf E = 0
    $$

    then you'd have no way of knowing that longitudinal EM waves are forbidden. (Though, to be clear, the transversality conditions are not sufficient, either.)






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      physics.stackexchange.com/q/465124
      $endgroup$
      – Kavita Juneja
      21 hours ago
















    14












    14








    14





    $begingroup$

    No, they're not. The wave equations for the force fields contain a strict subset of the information contained in the full set of Maxwell's equations. In particular, it's important to note that you need the Gauss-type equations,
    $$
    nablacdot mathbf E = 0 = nablacdotmathbf B,
    $$

    to ensure the transversality of the waves. If all you had to go was the wave equations in the form
    $$
    left[partial_t^2 - c^2 nabla^2 right]mathbf E = 0
    $$

    then you'd have no way of knowing that longitudinal EM waves are forbidden. (Though, to be clear, the transversality conditions are not sufficient, either.)






    share|cite|improve this answer











    $endgroup$



    No, they're not. The wave equations for the force fields contain a strict subset of the information contained in the full set of Maxwell's equations. In particular, it's important to note that you need the Gauss-type equations,
    $$
    nablacdot mathbf E = 0 = nablacdotmathbf B,
    $$

    to ensure the transversality of the waves. If all you had to go was the wave equations in the form
    $$
    left[partial_t^2 - c^2 nabla^2 right]mathbf E = 0
    $$

    then you'd have no way of knowing that longitudinal EM waves are forbidden. (Though, to be clear, the transversality conditions are not sufficient, either.)







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 22 hours ago

























    answered yesterday









    Emilio PisantyEmilio Pisanty

    84.8k22209427




    84.8k22209427












    • $begingroup$
      physics.stackexchange.com/q/465124
      $endgroup$
      – Kavita Juneja
      21 hours ago




















    • $begingroup$
      physics.stackexchange.com/q/465124
      $endgroup$
      – Kavita Juneja
      21 hours ago


















    $begingroup$
    physics.stackexchange.com/q/465124
    $endgroup$
    – Kavita Juneja
    21 hours ago






    $begingroup$
    physics.stackexchange.com/q/465124
    $endgroup$
    – Kavita Juneja
    21 hours ago













    12












    $begingroup$

    The wave equations for electromagnetic waves in free space can be derived from Maxwell's equations. However, Maxwell's equations can be used to describe much more. For example, you can derive from them, how an electromagnetic wave is launched from an antenna. Or you can treat electrostatic and magnetostatic phenomena. You can learn from them how electric motors work, and how we can convert mechanical into electric energy in generators. There is a huge wealth of physics in these four equations, which has enormous importance for most of the phenomena we observe around us, and for much of today's modern technology.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I am talking of maxwell's equations in free space. No antennas or motors etc. I am asking when we write maxwell's equations in free space and then from those equations derive the wave equation , is any information lost in the derivation??
      $endgroup$
      – Kavita Juneja
      yesterday












    • $begingroup$
      You should really explain why the two are non-equivalent, though (the reason being, not all solutions to wave equations are physical solutions for electric and magnetic field); this said, if you accidentally picked those wave solutions that do represent the physical fields, then all you have described is automatically incorporated.
      $endgroup$
      – gented
      23 hours ago
















    12












    $begingroup$

    The wave equations for electromagnetic waves in free space can be derived from Maxwell's equations. However, Maxwell's equations can be used to describe much more. For example, you can derive from them, how an electromagnetic wave is launched from an antenna. Or you can treat electrostatic and magnetostatic phenomena. You can learn from them how electric motors work, and how we can convert mechanical into electric energy in generators. There is a huge wealth of physics in these four equations, which has enormous importance for most of the phenomena we observe around us, and for much of today's modern technology.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I am talking of maxwell's equations in free space. No antennas or motors etc. I am asking when we write maxwell's equations in free space and then from those equations derive the wave equation , is any information lost in the derivation??
      $endgroup$
      – Kavita Juneja
      yesterday












    • $begingroup$
      You should really explain why the two are non-equivalent, though (the reason being, not all solutions to wave equations are physical solutions for electric and magnetic field); this said, if you accidentally picked those wave solutions that do represent the physical fields, then all you have described is automatically incorporated.
      $endgroup$
      – gented
      23 hours ago














    12












    12








    12





    $begingroup$

    The wave equations for electromagnetic waves in free space can be derived from Maxwell's equations. However, Maxwell's equations can be used to describe much more. For example, you can derive from them, how an electromagnetic wave is launched from an antenna. Or you can treat electrostatic and magnetostatic phenomena. You can learn from them how electric motors work, and how we can convert mechanical into electric energy in generators. There is a huge wealth of physics in these four equations, which has enormous importance for most of the phenomena we observe around us, and for much of today's modern technology.






    share|cite|improve this answer









    $endgroup$



    The wave equations for electromagnetic waves in free space can be derived from Maxwell's equations. However, Maxwell's equations can be used to describe much more. For example, you can derive from them, how an electromagnetic wave is launched from an antenna. Or you can treat electrostatic and magnetostatic phenomena. You can learn from them how electric motors work, and how we can convert mechanical into electric energy in generators. There is a huge wealth of physics in these four equations, which has enormous importance for most of the phenomena we observe around us, and for much of today's modern technology.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered yesterday









    flaudemusflaudemus

    1,582213




    1,582213












    • $begingroup$
      I am talking of maxwell's equations in free space. No antennas or motors etc. I am asking when we write maxwell's equations in free space and then from those equations derive the wave equation , is any information lost in the derivation??
      $endgroup$
      – Kavita Juneja
      yesterday












    • $begingroup$
      You should really explain why the two are non-equivalent, though (the reason being, not all solutions to wave equations are physical solutions for electric and magnetic field); this said, if you accidentally picked those wave solutions that do represent the physical fields, then all you have described is automatically incorporated.
      $endgroup$
      – gented
      23 hours ago


















    • $begingroup$
      I am talking of maxwell's equations in free space. No antennas or motors etc. I am asking when we write maxwell's equations in free space and then from those equations derive the wave equation , is any information lost in the derivation??
      $endgroup$
      – Kavita Juneja
      yesterday












    • $begingroup$
      You should really explain why the two are non-equivalent, though (the reason being, not all solutions to wave equations are physical solutions for electric and magnetic field); this said, if you accidentally picked those wave solutions that do represent the physical fields, then all you have described is automatically incorporated.
      $endgroup$
      – gented
      23 hours ago
















    $begingroup$
    I am talking of maxwell's equations in free space. No antennas or motors etc. I am asking when we write maxwell's equations in free space and then from those equations derive the wave equation , is any information lost in the derivation??
    $endgroup$
    – Kavita Juneja
    yesterday






    $begingroup$
    I am talking of maxwell's equations in free space. No antennas or motors etc. I am asking when we write maxwell's equations in free space and then from those equations derive the wave equation , is any information lost in the derivation??
    $endgroup$
    – Kavita Juneja
    yesterday














    $begingroup$
    You should really explain why the two are non-equivalent, though (the reason being, not all solutions to wave equations are physical solutions for electric and magnetic field); this said, if you accidentally picked those wave solutions that do represent the physical fields, then all you have described is automatically incorporated.
    $endgroup$
    – gented
    23 hours ago




    $begingroup$
    You should really explain why the two are non-equivalent, though (the reason being, not all solutions to wave equations are physical solutions for electric and magnetic field); this said, if you accidentally picked those wave solutions that do represent the physical fields, then all you have described is automatically incorporated.
    $endgroup$
    – gented
    23 hours ago











    1












    $begingroup$

    Remaining to the case of Maxwell's equations (ME) in vacuum, there is no equivalence between the wave equations for the fields and the original set. As already pointed out, solutions of ME are a subset of the solutions of the two three-dimensional wave equations.



    The case made by Emilio Pisanty (one loses information about the transversality) has to be taken as just one example of the non-equivalence. Another information which gets lost is the phase relation between magnetic and electric field.



    From a mathematical point of view it is not difficult to understand the reason of the non-equivalence: in order to derive the wave equations one has to




    1. take the curl of one of the equations containing the time derivative of a field;

    2. use the other equation to rewrite the time derivative of a curl as a second time derivative.


    It is clear that the additional derivative implied by step 1 may eliminate some information. It is quite well known that if one takes additional derivatives of a differential equations, the resulting equation usually has more solutions than the original one and one has to choose among them those satisfying the original equation.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      It's not just the phase relation between the two fields - it's the loss of any connection at all between them. You can zero out one of the fields, or replace it with some completely different solution, and the wave equations won't blink an eye.
      $endgroup$
      – Emilio Pisanty
      6 hours ago
















    1












    $begingroup$

    Remaining to the case of Maxwell's equations (ME) in vacuum, there is no equivalence between the wave equations for the fields and the original set. As already pointed out, solutions of ME are a subset of the solutions of the two three-dimensional wave equations.



    The case made by Emilio Pisanty (one loses information about the transversality) has to be taken as just one example of the non-equivalence. Another information which gets lost is the phase relation between magnetic and electric field.



    From a mathematical point of view it is not difficult to understand the reason of the non-equivalence: in order to derive the wave equations one has to




    1. take the curl of one of the equations containing the time derivative of a field;

    2. use the other equation to rewrite the time derivative of a curl as a second time derivative.


    It is clear that the additional derivative implied by step 1 may eliminate some information. It is quite well known that if one takes additional derivatives of a differential equations, the resulting equation usually has more solutions than the original one and one has to choose among them those satisfying the original equation.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      It's not just the phase relation between the two fields - it's the loss of any connection at all between them. You can zero out one of the fields, or replace it with some completely different solution, and the wave equations won't blink an eye.
      $endgroup$
      – Emilio Pisanty
      6 hours ago














    1












    1








    1





    $begingroup$

    Remaining to the case of Maxwell's equations (ME) in vacuum, there is no equivalence between the wave equations for the fields and the original set. As already pointed out, solutions of ME are a subset of the solutions of the two three-dimensional wave equations.



    The case made by Emilio Pisanty (one loses information about the transversality) has to be taken as just one example of the non-equivalence. Another information which gets lost is the phase relation between magnetic and electric field.



    From a mathematical point of view it is not difficult to understand the reason of the non-equivalence: in order to derive the wave equations one has to




    1. take the curl of one of the equations containing the time derivative of a field;

    2. use the other equation to rewrite the time derivative of a curl as a second time derivative.


    It is clear that the additional derivative implied by step 1 may eliminate some information. It is quite well known that if one takes additional derivatives of a differential equations, the resulting equation usually has more solutions than the original one and one has to choose among them those satisfying the original equation.






    share|cite|improve this answer











    $endgroup$



    Remaining to the case of Maxwell's equations (ME) in vacuum, there is no equivalence between the wave equations for the fields and the original set. As already pointed out, solutions of ME are a subset of the solutions of the two three-dimensional wave equations.



    The case made by Emilio Pisanty (one loses information about the transversality) has to be taken as just one example of the non-equivalence. Another information which gets lost is the phase relation between magnetic and electric field.



    From a mathematical point of view it is not difficult to understand the reason of the non-equivalence: in order to derive the wave equations one has to




    1. take the curl of one of the equations containing the time derivative of a field;

    2. use the other equation to rewrite the time derivative of a curl as a second time derivative.


    It is clear that the additional derivative implied by step 1 may eliminate some information. It is quite well known that if one takes additional derivatives of a differential equations, the resulting equation usually has more solutions than the original one and one has to choose among them those satisfying the original equation.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 6 hours ago









    Emilio Pisanty

    84.8k22209427




    84.8k22209427










    answered 23 hours ago









    GiorgioPGiorgioP

    3,6151526




    3,6151526












    • $begingroup$
      It's not just the phase relation between the two fields - it's the loss of any connection at all between them. You can zero out one of the fields, or replace it with some completely different solution, and the wave equations won't blink an eye.
      $endgroup$
      – Emilio Pisanty
      6 hours ago


















    • $begingroup$
      It's not just the phase relation between the two fields - it's the loss of any connection at all between them. You can zero out one of the fields, or replace it with some completely different solution, and the wave equations won't blink an eye.
      $endgroup$
      – Emilio Pisanty
      6 hours ago
















    $begingroup$
    It's not just the phase relation between the two fields - it's the loss of any connection at all between them. You can zero out one of the fields, or replace it with some completely different solution, and the wave equations won't blink an eye.
    $endgroup$
    – Emilio Pisanty
    6 hours ago




    $begingroup$
    It's not just the phase relation between the two fields - it's the loss of any connection at all between them. You can zero out one of the fields, or replace it with some completely different solution, and the wave equations won't blink an eye.
    $endgroup$
    – Emilio Pisanty
    6 hours ago











    0












    $begingroup$

    In the so-called Lorenz gauge Maxwell's equations take the form of a set of inhomogeneous wave equations in terms of the potential. All of the physics is described at least as well by these. Maxwell's equations can be written in covariant notation as $partial_mu F^{munu} = partial_mu partial^mu A^nu - partial_mu partial^nu A^mu = - j^nu /epsilon_0 $. Choosing the Lorenz gauge, $partial_mu A^mu = 0$ reduces this to the inhomogeneous wave equations, $ partial_mu partial^mu A^nu = - j^nu /epsilon_0 $.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      In the so-called Lorenz gauge Maxwell's equations take the form of a set of inhomogeneous wave equations in terms of the potential. All of the physics is described at least as well by these. Maxwell's equations can be written in covariant notation as $partial_mu F^{munu} = partial_mu partial^mu A^nu - partial_mu partial^nu A^mu = - j^nu /epsilon_0 $. Choosing the Lorenz gauge, $partial_mu A^mu = 0$ reduces this to the inhomogeneous wave equations, $ partial_mu partial^mu A^nu = - j^nu /epsilon_0 $.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        In the so-called Lorenz gauge Maxwell's equations take the form of a set of inhomogeneous wave equations in terms of the potential. All of the physics is described at least as well by these. Maxwell's equations can be written in covariant notation as $partial_mu F^{munu} = partial_mu partial^mu A^nu - partial_mu partial^nu A^mu = - j^nu /epsilon_0 $. Choosing the Lorenz gauge, $partial_mu A^mu = 0$ reduces this to the inhomogeneous wave equations, $ partial_mu partial^mu A^nu = - j^nu /epsilon_0 $.






        share|cite|improve this answer











        $endgroup$



        In the so-called Lorenz gauge Maxwell's equations take the form of a set of inhomogeneous wave equations in terms of the potential. All of the physics is described at least as well by these. Maxwell's equations can be written in covariant notation as $partial_mu F^{munu} = partial_mu partial^mu A^nu - partial_mu partial^nu A^mu = - j^nu /epsilon_0 $. Choosing the Lorenz gauge, $partial_mu A^mu = 0$ reduces this to the inhomogeneous wave equations, $ partial_mu partial^mu A^nu = - j^nu /epsilon_0 $.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 18 hours ago

























        answered yesterday









        my2ctsmy2cts

        5,4722718




        5,4722718






















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