Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
    
Maths?......... –  Ashu May 24 '12 at 17:58
    
See this. It does not provide an answer directly, but provides the needed input. Other related issues are also discussed. –  Vijay Murthy 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? –  stupidity May 24 '12 at 20:39
add comment

2 Answers

up vote 5 down vote accepted

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.

share|improve this answer
1  
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. –  ComputerSaysNo 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 $\frac{\partial}{\partial x} = \frac{\partial}{\partial x'}$ ?? –  Santosh Linkha Nov 30 '13 at 8:51
add comment

transformation rule for partial derivatives:

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

share|improve this answer
    
Thanks, actually I should have known this.. –  ComputerSaysNo May 24 '12 at 20:05
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.