# Galilean transformation of the wave equation

I have this general wave equation:

$$\dfrac{\partial^2 \psi}{\partial x^2}+\dfrac{\partial^2 \psi}{\partial y^2}-\dfrac{1}{c^2}\dfrac{\partial^2 \psi}{\partial t^2}=0$$

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?

• See this. It does not provide an answer directly, but provides the needed input. Other related issues are also discussed. May 24 '12 at 18:19
• Is the sign in the middle term, $-\dfrac{2V}{c^2}\dfrac{\partial^2 \psi}{\partial x'\partial t'}$ correct? Or should it be positive? May 24 '12 at 20:39
• It should be positive. Sep 8 '15 at 13:50

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.

• Thaks alot! I've checked, and it works. I had some troubles with the transformation of differential operators. I apologize for posting this mathematical question in the physics category, although the meaning of the solution is appropriate. May 24 '12 at 19:26
• Hi ... shouldn't $\frac{\partial }{\partial x'} = \frac{\partial }{\partial x} - \frac{1}{V}\frac{\partial }{\partial t}$?? could you elaborate why just $\frac{\partial}{\partial x} = \frac{\partial}{\partial x'}$ ?? Nov 30 '13 at 8:51
• @SantoshLinkha because $\partial_x(\psi(x'))=\partial_x(\psi(x-vt))=\partial_{x'}\psi * \partial_x(x-Vt)=\partial_{x'}\psi$ Jul 19 '16 at 23:15
• In case anyone else accidentally falls into the same trap @SantoshLinkha (easily) did, a slightly more obvious way to see the mistake is that using the chain (transformation) rule for partial derivatives we we get a term that is $\frac{\partial t'}{\partial x}$, which is actually $0$, since $x$ does not depend directly on $t'$, rather only on $t$ (it would be $1/V$ if we were considering the total derivative, of course). Feb 13 at 1:36

transformation rule for partial derivatives:

$$\frac{\partial}{\partial x_{\mu}} = \sum_{\nu} \frac{\partial x'_{\nu}}{\partial x_\mu} \frac{\partial}{\partial x'_{\nu}}$$