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Solutions to Maxwell's equations, I don't understand the notes provided

  1. Mar 12, 2010 #1

    fluidistic

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    I just finished the first week of the term. It was very loaded I must say. I also ended my introductory courses and I'm now taking upper level undergraduate physics courses.
    My professor wrote notes for us because he doesn't really like Jackson's book, but I'm having a very hard time to understand the most basics of his notes. There are a lot of typos and redundancies, but this is not the reason I don't understand. Here there are: http://www.famaf.unc.edu.ar/~reula/Docencia/Electromagnetismo/part1.pdf.
    What I understand from the notes (first part of the first chapter): 2 of Maxwell's vacuum equations are equations of evolution. Then it introduces us a theorem that basically states that if I have 2 fields satisfying certain conditions, then there exists a unique solution of Maxwell's evolution equations. I realize that in reality these 2 fields are the electric and magnetic fields.
    Now what's the deal with the wave equation? It's only used to prove the theorem? For one moment I've been thinking the wave equation was a solution of Maxwell's equations, but of course it doesn't make any sense, it's an equation... So I thought that the solution to the wave equation was also the solution to Maxwell's evolution equations. Now I realize it's false.
    Also, I don't understand the detour into the wave equation taken. What is the purpose of it? I got lost. If someone could explain a bit these things in a few lines (compared to the notes), I'd be grateful. Until now, I didn't even developed an intuition on the material seen yet.
     
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  3. Mar 12, 2010 #2

    ideasrule

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    The wave equation is a solution of Maxwell's equations for a region of space free of charge. The fact that Maxwell's equations can be turned into the wave equation means that electromagnetic waves can exist. Even more significantly, the term in the wave equation that indicates the wave's speed is equal to the speed of light, suggesting to 19th century scientists that light is the same thing as electromagnetism.
     
  4. Mar 12, 2010 #3

    fluidistic

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    I'm not understanding well. The wave equation is a solution to Maxwell's equation in vacuum, ok. Now that the Maxwell's equations can be turned into the wave equation... I find hard to believe an equation can be turned into its solution. This is probably due to my almost total lack of knowledge.
    So it seems I wasn't wrong thinking that the wave equation was a solution to Maxwell's equations. I will try to read through the notes once again, maybe after 10⁴ times of reading I'll understand much better. I'm leaving, I'm having a terrible wisdom teeth pain (they're growing! ouch!). Thanks for your reply, greatly appreciated.
     
  5. Mar 13, 2010 #4
    The inhomogenous wave equation works for non-vaccuum as well. You can think of the wave equations as equal to the maxwell equations. However, because they we written in terms of potentials, and not fields, they are not unique. So they are not quite equivalent.
     
  6. Mar 13, 2010 #5

    Pengwuino

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    I think he wasn't clear, the wave equation is the form of the differential equation you can find using Maxwell's Equations. The SOLUTIONS, on the other hand, have to be solved just ike you solve any other differential equation.
     
  7. Mar 13, 2010 #6

    fluidistic

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    I appreciate the answers.
    To Pengwuino, could you explain a bit more what do you mean by
    What differential equation?
     
  8. Mar 13, 2010 #7

    Pengwuino

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    For example, Newton's second law is actually a differential equation. What you probably know best as [tex]F = ma[/tex] can actually be thought of as a differential equation. For example, given a particle in a constant gravitational field, you see the differential equation as [tex]\frac{d^2x}{dt^2}m = -g[/tex] and you can solve for the position of a particle.

    Maxwell's equations work just like that, except instead of equations of motions for a particle, you're looking at the fields. So for example, take two of Maxwell's equations:

    [tex]\[
    \begin{array}{l}
    \nabla \times \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
    \over E} = \frac{{\partial \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
    \over B} }}{{\partial t}} \\
    \nabla \times \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
    \over B} = \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
    \over J} + \frac{{\partial \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
    \over E} }}{{\partial t}} \\
    \end{array}
    \]
    [/tex]

    These are coupled partial differential equations. Now, using some vector identities (which I'm not awake enough to remember/figure out) you can get something that looks like:

    [tex]\[
    \frac{{\partial ^2 E_x }}{{\partial x^2 }} + \frac{{\partial ^2 E_x }}{{\partial y^2 }} + \frac{{\partial ^2 E_x }}{{\partial z^2 }} - \frac{1}{{c^2 }}\frac{{\partial ^2 E_x }}{{\partial t^2 }}=0
    \]
    [/tex]

    which is a partial differential equation. There's 5 more to take into account as well as constraint equations, but the general idea is that that is (part of) the wave equation and that is what you're looking for solutions for.
     
    Last edited: Mar 13, 2010
  9. Mar 13, 2010 #8

    fluidistic

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    Thanks once again pengwuino. I think there is a typo error because you talk about a partial differential equation while there's just a sum of partial derivatives, in other words there is no equation. But I think I get the idea. And yeah, I do know about DE's. Don't know much about PDE's though, but I have to self study those for now.
     
  10. Mar 13, 2010 #9

    Pengwuino

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    Sorry, I fixed it, it was meant to equal 0 (with the assumption that we're working with no free charges).

    To hopefully be clearer, the field E is a function of the coordinates and time, that is [tex]E=E(x,y,z,t)[/tex]. What I had as an example earlier, the particles motion in a constant gravitational field was a ordinary differential equation because [tex]X=X[t][/tex], that is, the position X is only a function of one variable. With electrodynamics, your fields are in terms of multiple coordinatesand in this case, even a time, so you need partial differential equations.
     
  11. Mar 13, 2010 #10

    fluidistic

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    I see, thanks a lot!
     
  12. Mar 14, 2010 #11

    ideasrule

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    Last edited by a moderator: Apr 24, 2017
  13. Mar 14, 2010 #12

    fluidistic

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    Last edited by a moderator: Apr 24, 2017
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