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:

$x' = \frac{x - vt}{\sqrt{1 - \frac{v^2}{c^2}}}$

$t' = \frac{t - \frac{v}{c^2}x}{\sqrt{1 - \frac{v^2}{c^2}}}$

$y' = y$

$z' = z$

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

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

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?