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Show EM Wave equation invariant under a Lorentz Transformation

  1. Jul 3, 2011 #1
    1. The problem statement, all variables and given/known data

    Show that the electromagnetic wave equation

    [itex]\frac{\partial^{2}\phi}{\partial x^{2}} +
    \frac{\partial^{2}\phi}{\partial y^{2}} +
    \frac{\partial^{2}\phi}{\partial z^{2}} -
    \frac{1}{c^2}\frac{\partial^{2} \phi}{\partial t^2} [/itex]

    is invariant under a Lorentz transformation.

    2. Relevant equations

    Lorentz Transformations:

    [itex]x' = \frac{x - vt}{\sqrt{1 - \frac{v^2}{c^2}}}[/itex]

    [itex]t' = \frac{t - \frac{v}{c^2}x}{\sqrt{1 - \frac{v^2}{c^2}}}[/itex]

    [itex]y' = y[/itex]

    [itex]z' = z[/itex]


    3. The attempt at a solution

    Well I know exactly what I'm supposed to do here, transform the equations from x to x', y to y' etc. Then rearrange terms and show that the wave equation with x,y,z, and t is equal to the wave equation with x', y', z', and t'.

    I understand the method is to get [itex]\frac{\partial x'}{\partial x}[/itex], then use the chain rule ( [itex]\frac{\partial\phi}{\partial x} = \frac{\partial\phi}{\partial x'}\frac{\partial x'}{\partial x}[/itex] ), and similarly for t -> t', then a bit of simple algebra and the answer should pop out the other end.

    My only problem is that I have no idea what [itex]\frac{\partial x'}{\partial x}[/itex] is...in fact, I do know, since I have been told, that [itex]\frac{\partial x'}{\partial x} = \frac{1}{\sqrt{1 - v^2/c^2}}[/itex], but I have no idea how that follows from the Lorentz transforms, could someone give me a hint in deriving it?

    (My attempt was simply saying [itex]\partial x' = \frac{\partial x - v\partial t'}{1 - v^2/c^2}[/itex], but then dividing that by [itex]\partial x'[/itex] I'd get [itex]\frac{1 - \frac{\partial t'}{\partial t}}{\sqrt{1-v^2/c^2}}[/itex], which isn't right...

    As you can tell, I'm not really very comfortable with partial derivatives or differential calculus, so I'm probably doing something really silly in that derivation above...
     
    Last edited: Jul 3, 2011
  2. jcsd
  3. Jul 3, 2011 #2
    Simply take the partial derivative of x' with respect to x. Same as taking a normal derivative of x' but treat all the variables except x as constants
     
  4. Jul 3, 2011 #3
    Ah, that made sense, I knew it would be something trivial, I was just approaching it weirdly.

    Great, have the answer now, thanks :)
     
  5. Jul 3, 2011 #4
    OK, I thought I had the answer but this but it turns out I don't...

    I want to calculate [itex]\frac{\partial t'}{\partial t}[/itex]

    So I have [itex]t' = \frac{t - \frac{vx}{c^2}}{\sqrt{1 - v^2/c^2}}[/itex]

    The answer is supposedly [itex]\frac{1}{\sqrt{1 - v^2/c^2}}[/itex]

    I don't see how this would be...the bottom line is a constant term so looking at the top line alone,

    [itex]\frac{\partial}{\partial t} t[/itex] is obviously 1, fine, but I don't see how [itex]\frac{\partial}{\partial t} (-\frac{vx}{c^2}) = 0[/itex], surely x is a function of t, so we get

    [itex]\frac{\partial}{\partial t} x = \frac{\partial x}{\partial t} = v[/itex] so [itex]\frac{\partial}{\partial t} (t - vx/c^2) = 1 - v^2/c^2[/itex] for the derivative of the top line?
     
    Last edited: Jul 3, 2011
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