Show EM Wave equation invariant under a Lorentz Transformation

[Antic_Hay]

Homework Statement

Show that the electromagnetic wave equation

\frac{\partial^{2}\phi}{\partial x^{2}} +<br /> \frac{\partial^{2}\phi}{\partial y^{2}} +<br /> \frac{\partial^{2}\phi}{\partial z^{2}} -<br /> \frac{1}{c^2}\frac{\partial^{2} \phi}{\partial t^2}

is invariant under a Lorentz transformation.

Homework Equations

Lorentz Transformations:

x&#039; = \frac{x - vt}{\sqrt{1 - \frac{v^2}{c^2}}}

t&#039; = \frac{t - \frac{v}{c^2}x}{\sqrt{1 - \frac{v^2}{c^2}}}

y&#039; = y

z&#039; = z

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 \frac{\partial x&#039;}{\partial x}, then use the chain rule ( \frac{\partial\phi}{\partial x} = \frac{\partial\phi}{\partial x&#039;}\frac{\partial x&#039;}{\partial x} ), 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 \frac{\partial x&#039;}{\partial x} is...in fact, I do know, since I have been told, that \frac{\partial x&#039;}{\partial x} = \frac{1}{\sqrt{1 - v^2/c^2}}, 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 \partial x&#039; = \frac{\partial x - v\partial t&#039;}{1 - v^2/c^2}, but then dividing that by \partial x&#039; I'd get \frac{1 - \frac{\partial t&#039;}{\partial t}}{\sqrt{1-v^2/c^2}}, 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...

 

[JaWiB]

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

 

[Antic_Hay]

Ah, that made sense, I knew it would be something trivial, I was just approaching it weirdly.

Great, have the answer now, thanks :)

 

[Antic_Hay]

OK, I thought I had the answer but this but it turns out I don't...

I want to calculate \frac{\partial t&#039;}{\partial t}

So I have t&#039; = \frac{t - \frac{vx}{c^2}}{\sqrt{1 - v^2/c^2}}

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

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

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

\frac{\partial}{\partial t} x = \frac{\partial x}{\partial t} = v so \frac{\partial}{\partial t} (t - vx/c^2) = 1 - v^2/c^2 for the derivative of the top line?