Invariance of the Lorentz transform

[tina21]

Homework Statement

Prove the invariance of the electromagnetic wave equation by showing that the corresponding differential operator is an invariant.

Relevant Equations

d^2/dx^2+d^2/dy^2+d^2/dz^2-1/c^2d^2/dt^2 = d^2/dx^2+d^2/dy^2+d^2/dz^2-1/c^2d^2/dt^2

of course y and z terms are invariant but for the x and t terms I am getting an additional factor of 1/1-v^2/c^2

 

[PeroK]

Homework Statement: Prove the invariance of the electromagnetic wave equation by showing that the corresponding differential operator is an invariant.
Homework Equations: d^2/dx^2+d^2/dy^2+d^2/dz^2-1/c^2d^2/dt^2 = d^2/dx^2+d^2/dy^2+d^2/dz^2-1/c^2d^2/dt^2

of course y and z terms are invariant but for the x and t terms I am getting an additional factor of 1/1-v^2/c^2

Can you use Latex to show what you got?

https://www.physicsforums.com/help/latexhelp/

 

[tina21]

I haven't used latex before. I hope these images are okay

 

[PeroK]

That doesn't look right. I prefer to differentiate a trial function. Imagine we have a function $f(t', x')$, where $t' = \gamma(t - vx/c^2), \ x' = \gamma(x - vt)$.

Now, we differentiate $f$ with respect to $x$ using the chain rule:

$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial t'}\frac{\partial t'}{\partial x} + \frac{\partial f}{\partial x'}\frac{\partial x'}{\partial x} = \frac{\partial f}{\partial t'}(-\gamma v/c^2) + \frac{\partial f}{\partial x'}(\gamma)$

Now you need to take the second derivative of $f$ by repeating this process - and remember that you have to differentiate both terms using the chain rule, so you will get cross terms in $\frac{\partial^2 f}{\partial t' \partial x'}$.

At the end, you can remove the trial function to leave, for example:

$\frac{\partial}{\partial x} = (-\gamma v/c^2)\frac{\partial }{\partial t'} + (\gamma)\frac{\partial}{\partial x'}$

That's what you get if you only wanted the first derivative.

Note that you can do the chain rule on differentials, but I prefer to have a function to differentiate, and then take the function away when I'm finished.

 

[tina21]

yes, I tried doing it too. Made it a whole lot easier and I got the answer. Thanks so much !