I have this general wave equation:
\begin{equation}
\dfrac{\partial^2 \psi}{\partial x^2}+\dfrac{\partial^2 \psi}{\partial y^2}-\dfrac{1}{c^2}\dfrac{\partial^2 \psi}{\partial t^2}=0
\end{equation}
And the following transformation : $t'=t$ ; $x'=x-Vt$ and $y'=y$
The solution to this has to be :
$$\dfrac{\partial^2 \psi}{\partial x'^2}\left( 1-\frac{V^2}{c^2}\right)+\dfrac{\partial^2 \psi}{\partial y'^2}+\dfrac{2V}{c^2}\dfrac{\partial^2 \psi}{\partial x' \partial t'^2}-\dfrac{1}{c^2}\dfrac{\partial^2 \psi}{\partial t^{'2}}=0$$
This proves that the velocity of the wave depends on the direction you are looking at. I don't know how to get to this? If you just substitute it in the equation you get $x'+Vt$ in the partial derivative. Is there another way to do this, or which rule do I have to use to solve it? I was thinking about the chain rule or something, but how do I apply it on partial derivatives?
Best Answer
You must first rewrite the old partial derivatives in terms of the new ones. A priori, they're some linear combinations with coefficients that could depend on the spacetime coordinates in general but here they don't depend because the transformation is linear. The rules $$ t'=t, \quad x'=x-Vt,\quad y'=y $$ get translated to $$ \frac{\partial}{\partial t} = \frac{\partial}{\partial t'} - V \frac{\partial}{\partial x'}$$ $$ \frac{\partial}{\partial x} = \frac{\partial}{\partial x'}$$ $$ \frac{\partial}{\partial y} = \frac{\partial}{\partial y'}$$ If you write the coefficients in front of the right-hand-side primed derivatives as a matrix, it's the same matrix as the original matrix of derivatives $\partial x'_i/\partial x_j$. If you don't want to work with matrices, just verify that all the expressions of the type $\partial x/\partial t$ are what they should be if you rewrite these derivatives using the three displayed equations and if you use the obvious partial derivatives $\partial y'/\partial t'$ etc.
If you simply rewrite the (second) derivatives with respect to the unprimed coordinates in terms of the (second) derivatives with respect to the primed coordinates, you will get your second, Galilean-transformed form of the equation. I've verified it works – up to the possible error in the sign of $V$ which only affects the sign of the term with the mixed $xt$ second derivative.
I guess that if this explanation won't be enough, you should re-ask this question on the math forum.