Question about the solution to Maxwell's equations in vacuum

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fluidistic
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The maxwell's equations in vacuum are satisfied by a non trivial solution involving [tex]\vec E (t,\vec x)[/tex] and [tex]\vec B (t, \vec x)[/tex]. Correct me if I'm wrong.
I don't really understand the physical interpretation of the solution. I know that if I'm given an initial condition then I can know the solution for all t and [tex]\vec x[/tex]. Assuming I'm given an initial condition... then I'd have that a varying electromagnetic field satisfies the Maxwell's equations for all the space and at any time. Does this mean that vacuum doesn't contain any charge (pretty likely) and further is like a soup of electromagnetic waves (which is what strikes me)?
I mean, to my understanding, there's a non vanishing electromagnetic field in all the space at any time, all this in vacuum. Am I understanding this well?
 

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Dale
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I mean, to my understanding, there's a non vanishing electromagnetic field in all the space at any time, all this in vacuum. Am I understanding this well?
If you have a "chunk" of vacuum that you are analyzing then you would take the boundary conditions of the fields and then from that you could determine what the fields are on the interior. If the boundary conditions are 0 then the fields on the interior would also be 0 (0 is a valid solution to Maxwell's equations). So there is not necessarily a non-vanishing EM field in all space at any time.
 
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fluidistic
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If you have a "chunk" of vacuum that you are analyzing then you would take the boundary conditions of the fields and then from that you could determine what the fields are on the interior. If the boundary conditions are 0 then the fields on the interior would also be 0 (0 is a valid solution to Maxwell's equations). So there is not necessarily a non-vanishing EM field in all space at any time.
Ok thank you very much. I had this question in mind (namely that if the boundaries are 0, would the field be 0 everywhere inside the boundaries? I was convinced that no. To be fully convinced, I think I'll have to do the calculation, but I fully trust you.)
Question answered.
 
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fluidistic
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I've a question arising. I've seen a theorem saying that knowing the initial conditions at one moment is enough to determine the whole solution to these equations at any time; that is, if we already have the general form of the solution of the equations.

But what if I considered say a 10 cm x 10 cm space of vacuum? Say there's a charge situated at 300,000 km from these imaginary box. At time t=0 I'd have that the boundary conditions are [tex]\vec E(t=0,\vec x_0)=0[/tex] and [tex]\vec B(t=0, \vec x_0)=0[/tex] (I'd have E and B equal to zero for any further time, the solution of Maxwell's equations will be the trivial solution). In other words, I've no information of the far away charge.
Of course E and B won't be 0 at any time. After a second for example, the system should be aware of the far away charge... But I can I know this from the initial time t=0?
 
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Dale
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Maxwell's equations are partial differential equations, so you generally need what are called "boundary conditions", not just initial conditions. The boundary conditions are an essential part of the problem, and include things like the introduction of energy at some later point in time.
 
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fluidistic
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Maxwell's equations are partial differential equations, so you generally need what are called "boundary conditions", not just initial conditions. The boundary conditions are an essential part of the problem, and include things like the introduction of energy at some later point in time.
I understand what you mean.
However I'm confused on this: page 13 of http://www.famaf.unc.edu.ar/~reula/Docencia/Electromagnetismo/part1.pdf.
It doesn't seem to say that boundary conditions are necessary, only initial ones.
But I understand you're right. The notes would have been correct if they had involved boundary conditions I believe, otherwise the case of my example would have no explanation.
Thanks once again. I think I won't pop up with other questions for now.
 
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Dale
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On that page the x is infinite, i.e. it runs over all space. If you have such an initial condition then, yes, it is deterministic. Essentially that is the same as setting a "non-reflecting" boundary condition.
 
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Born2bwire
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If you want to know more about the underlying mathematics of this, you should read up on the Uniqueness Theorem. The assignment of the appropriate boundary conditions is a significant point in seemingly mundane problems.
 

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