Are wave equations equivalent to Maxwell's equations in free space?Maxwell's equation in free space from wave equations of electric and magnetic fieldSolution to Maxwell's equations in free space that are not plane wavesAre Lorentz force and maxwell's equations independent?When studying electrodynamics do we assume Maxwell's Equations or derive them?Integration constants in Maxwell's equations (ambiguousness?)What causes electromagnetic waves to propagate in free space?Are there super-maxwell's equations?Laser beam in terms of maxwell's equationsMaxwell's Equations in differential form… what is the advantage?How to read Maxwell's Equations?Wave Equations from Decoupling Maxwell's Equations in Bianisotropic MediaMaxwell's equation in free space from wave equations of electric and magnetic field
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Are wave equations equivalent to Maxwell's equations in free space?
Maxwell's equation in free space from wave equations of electric and magnetic fieldSolution to Maxwell's equations in free space that are not plane wavesAre Lorentz force and maxwell's equations independent?When studying electrodynamics do we assume Maxwell's Equations or derive them?Integration constants in Maxwell's equations (ambiguousness?)What causes electromagnetic waves to propagate in free space?Are there super-maxwell's equations?Laser beam in terms of maxwell's equationsMaxwell's Equations in differential form… what is the advantage?How to read Maxwell's Equations?Wave Equations from Decoupling Maxwell's Equations in Bianisotropic MediaMaxwell's equation in free space from wave equations of electric and magnetic field
$begingroup$
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?
electromagnetism waves maxwell-equations vacuum
edited Mar 8 at 9:54
Nat
3,48841831
asked Mar 7 at 9:11
Jeevesh JunejaJeevesh Juneja
716
$endgroup$
add a comment |
$begingroup$
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?
electromagnetism waves maxwell-equations vacuum
edited Mar 8 at 9:54
Nat
3,48841831
asked Mar 7 at 9:11
Jeevesh JunejaJeevesh Juneja
716
$endgroup$
2
$begingroup$
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 $dotbf E = kbf B,~dotbf B = fracc^2knabla^2bf E,$ 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
Mar 7 at 20:19
add a comment |
$begingroup$
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?
electromagnetism waves maxwell-equations vacuum
edited Mar 8 at 9:54
Nat
3,48841831
asked Mar 7 at 9:11
Jeevesh JunejaJeevesh Juneja
716
$endgroup$
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
electromagnetism waves maxwell-equations vacuum
edited Mar 8 at 9:54
Nat
3,48841831
asked Mar 7 at 9:11
Jeevesh JunejaJeevesh Juneja
716
edited Mar 8 at 9:54
Nat
3,48841831
asked Mar 7 at 9:11
Jeevesh JunejaJeevesh Juneja
716
edited Mar 8 at 9:54
Nat
3,48841831
edited Mar 8 at 9:54
Nat
3,48841831
edited Mar 8 at 9:54
Nat
3,48841831
3,48841831
asked Mar 7 at 9:11
Jeevesh JunejaJeevesh Juneja
716
asked Mar 7 at 9:11
Jeevesh JunejaJeevesh Juneja
716
asked Mar 7 at 9:11
Jeevesh JunejaJeevesh Juneja
716
716
2
$begingroup$
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 $dotbf E = kbf B,~dotbf B = fracc^2knabla^2bf E,$ 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
Mar 7 at 20:19
add a comment |
2
$begingroup$
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 $dotbf E = kbf B,~dotbf B = fracc^2knabla^2bf E,$ 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
Mar 7 at 20:19
2
2
$begingroup$
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 $dotbf E = kbf B,~dotbf B = fracc^2knabla^2bf E,$ 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
Mar 7 at 20:19
$begingroup$
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 $dotbf E = kbf B,~dotbf B = fracc^2knabla^2bf E,$ 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
Mar 7 at 20:19
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
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.)
answered Mar 7 at 13:58
Emilio PisantyEmilio Pisanty
85.5k23211429
$endgroup$
add a comment |
$begingroup$
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.
answered Mar 7 at 9:43
flaudemusflaudemus
1,811213
$endgroup$
$begingroup$
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
Mar 7 at 14:45
$begingroup$
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
Mar 7 at 16:52
add a comment |
$begingroup$
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
- take the curl of one of the equations containing the time derivative of a field;
- 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.
edited Mar 8 at 10:14
Emilio Pisanty
85.5k23211429
answered Mar 7 at 16:27
GiorgioPGiorgioP
3,9301527
$endgroup$
$begingroup$
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
Mar 8 at 10:16
$begingroup$
Sure. I wrote "another" information. Your point goes directly at the core of the problem: one could have a wave "only magnetic" or "only electric". Although I know some crackpot who thinks this would be possible...
$endgroup$
– GiorgioP
Mar 8 at 17:14
add a comment |
$begingroup$
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 $.
answered Mar 7 at 10:41
my2ctsmy2cts
5,5692718
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.)
answered Mar 7 at 13:58
Emilio PisantyEmilio Pisanty
85.5k23211429
$endgroup$
add a comment |
$begingroup$
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.)
answered Mar 7 at 13:58
Emilio PisantyEmilio Pisanty
85.5k23211429
$endgroup$
add a comment |
$begingroup$
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.)
answered Mar 7 at 13:58
Emilio PisantyEmilio Pisanty
85.5k23211429
$endgroup$
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.)
answered Mar 7 at 13:58
Emilio PisantyEmilio Pisanty
85.5k23211429
edited Mar 7 at 18:09
answered Mar 7 at 13:58
Emilio PisantyEmilio Pisanty
85.5k23211429
answered Mar 7 at 13:58
Emilio PisantyEmilio Pisanty
85.5k23211429
answered Mar 7 at 13:58
Emilio PisantyEmilio Pisanty
85.5k23211429
85.5k23211429
add a comment |
add a comment |
$begingroup$
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.
answered Mar 7 at 9:43
flaudemusflaudemus
1,811213
$endgroup$
$begingroup$
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
Mar 7 at 14:45
$begingroup$
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
Mar 7 at 16:52
add a comment |
$begingroup$
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.
answered Mar 7 at 9:43
flaudemusflaudemus
1,811213
$endgroup$
$begingroup$
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
Mar 7 at 14:45
$begingroup$
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
Mar 7 at 16:52
add a comment |
$begingroup$
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.
answered Mar 7 at 9:43
flaudemusflaudemus
1,811213
$endgroup$
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.
answered Mar 7 at 9:43
flaudemusflaudemus
1,811213
answered Mar 7 at 9:43
flaudemusflaudemus
1,811213
answered Mar 7 at 9:43
flaudemusflaudemus
1,811213
answered Mar 7 at 9:43
flaudemusflaudemus
1,811213
1,811213
$begingroup$
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
Mar 7 at 14:45
$begingroup$
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
Mar 7 at 16:52
add a comment |
$begingroup$
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
Mar 7 at 14:45
$begingroup$
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
Mar 7 at 16:52
$begingroup$
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
Mar 7 at 14:45
$begingroup$
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
Mar 7 at 14:45
$begingroup$
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
Mar 7 at 16:52
$begingroup$
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
Mar 7 at 16:52
add a comment |
$begingroup$
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
- take the curl of one of the equations containing the time derivative of a field;
- 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.
edited Mar 8 at 10:14
Emilio Pisanty
85.5k23211429
answered Mar 7 at 16:27
GiorgioPGiorgioP
3,9301527
$endgroup$
$begingroup$
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
Mar 8 at 10:16
$begingroup$
Sure. I wrote "another" information. Your point goes directly at the core of the problem: one could have a wave "only magnetic" or "only electric". Although I know some crackpot who thinks this would be possible...
$endgroup$
– GiorgioP
Mar 8 at 17:14
add a comment |
$begingroup$
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
- take the curl of one of the equations containing the time derivative of a field;
- 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.
edited Mar 8 at 10:14
Emilio Pisanty
85.5k23211429
answered Mar 7 at 16:27
GiorgioPGiorgioP
3,9301527
$endgroup$
$begingroup$
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
Mar 8 at 10:16
$begingroup$
Sure. I wrote "another" information. Your point goes directly at the core of the problem: one could have a wave "only magnetic" or "only electric". Although I know some crackpot who thinks this would be possible...
$endgroup$
– GiorgioP
Mar 8 at 17:14
add a comment |
$begingroup$
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
- take the curl of one of the equations containing the time derivative of a field;
- 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.
edited Mar 8 at 10:14
Emilio Pisanty
85.5k23211429
answered Mar 7 at 16:27
GiorgioPGiorgioP
3,9301527
$endgroup$
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
- take the curl of one of the equations containing the time derivative of a field;
- 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.
edited Mar 8 at 10:14
Emilio Pisanty
85.5k23211429
answered Mar 7 at 16:27
GiorgioPGiorgioP
3,9301527
edited Mar 8 at 10:14
Emilio Pisanty
85.5k23211429
edited Mar 8 at 10:14
Emilio Pisanty
85.5k23211429
edited Mar 8 at 10:14
Emilio Pisanty
85.5k23211429
85.5k23211429
answered Mar 7 at 16:27
GiorgioPGiorgioP
3,9301527
answered Mar 7 at 16:27
GiorgioPGiorgioP
3,9301527
answered Mar 7 at 16:27
GiorgioPGiorgioP
3,9301527
3,9301527
$begingroup$
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
Mar 8 at 10:16
$begingroup$
Sure. I wrote "another" information. Your point goes directly at the core of the problem: one could have a wave "only magnetic" or "only electric". Although I know some crackpot who thinks this would be possible...
$endgroup$
– GiorgioP
Mar 8 at 17:14
add a comment |
$begingroup$
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
Mar 8 at 10:16
$begingroup$
Sure. I wrote "another" information. Your point goes directly at the core of the problem: one could have a wave "only magnetic" or "only electric". Although I know some crackpot who thinks this would be possible...
$endgroup$
– GiorgioP
Mar 8 at 17:14
$begingroup$
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
Mar 8 at 10:16
$begingroup$
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
Mar 8 at 10:16
$begingroup$
Sure. I wrote "another" information. Your point goes directly at the core of the problem: one could have a wave "only magnetic" or "only electric". Although I know some crackpot who thinks this would be possible...
$endgroup$
– GiorgioP
Mar 8 at 17:14
$begingroup$
Sure. I wrote "another" information. Your point goes directly at the core of the problem: one could have a wave "only magnetic" or "only electric". Although I know some crackpot who thinks this would be possible...
$endgroup$
– GiorgioP
Mar 8 at 17:14
add a comment |
$begingroup$
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 $.
answered Mar 7 at 10:41
my2ctsmy2cts
5,5692718
$endgroup$
add a comment |
$begingroup$
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 $.
answered Mar 7 at 10:41
my2ctsmy2cts
5,5692718
$endgroup$
add a comment |
$begingroup$
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 $.
answered Mar 7 at 10:41
my2ctsmy2cts
5,5692718
$endgroup$
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 $.
answered Mar 7 at 10:41
my2ctsmy2cts
5,5692718
edited Mar 7 at 21:35
answered Mar 7 at 10:41
my2ctsmy2cts
5,5692718
answered Mar 7 at 10:41
my2ctsmy2cts
5,5692718
answered Mar 7 at 10:41
my2ctsmy2cts
5,5692718
5,5692718
add a comment |
add a comment |
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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 $dotbf E = kbf B,~dotbf B = fracc^2knabla^2bf E,$ 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
Mar 7 at 20:19