# Wave equation and Galilean Transformation

1. Apr 29, 2010

Hi! I was reading some notes on relativity (Special relativity) (http://teoria-de-la-relatividad.blogspot.com/2009/03/3-la-fisica-es-parada-de-cabeza.html) and it says that the classical wave equation is not Galilean Invariant. I tried to show it by myself, but I think there is some point that I'm missing.

Given two coordinate frames, if $$x=\bar{x}+vt$$ the classical wave equation transforms to:

$$\frac{\partial^2 \phi}{\partial \bar{x}^2} + \frac{\partial^2 \phi}{\partial \bar{y}^2} + \frac{\partial^2 \phi}{\partial \bar{z}^2} - \frac{1}{c^2} \frac{\partial^2 \phi}{\partial \bar{t}^2} + \frac{1}{c^2} \left( 2v \frac{\partial^2 \phi}{\partial \bar{x} \partial \bar{t}}- v^2 \frac{\partial^2 \phi}{\partial \bar{x}^2} \right) = 0$$

But i can really get that answer. Supose I want to compute

$$\frac{\partial \phi}{\partial \bar{x}} = \frac{\partial \phi}{\partial x} \frac{\partial x }{\partial \bar{x}} + \frac{\partial \phi}{\partial y } \frac{\partial y }{\partial \bar{x}} + \frac{\partial \phi }{\partial z } \frac{\partial z }{\partial \bar{x}} + \frac{\partial \phi }{\partial t } \frac{\partial t }{\partial \bar{x}}$$

And

$$t = \bar{t} = \frac{x-\bar{x}}{v}$$

should be my hints to get to that result?

Thanks!

Last edited: Apr 29, 2010
2. Apr 29, 2010

### starthaus

No, this is incorrect. You need to use

$$t = \bar{t}$$

and

$$x=\bar{x}+v \bar{t}$$

3. Apr 29, 2010

So the right thing is $$\frac{\partial t}{\partial \bar{x}} = 0$$ ? and every partial derivative involving time except $$\frac{\partial t}{\partial \bar{t}} = \frac{\partial \bar{t}{\partial t}$$ wich are one

4. Apr 29, 2010

### starthaus

Yes, of course

5. Apr 29, 2010

Ok, thank you! Let's see if I can do it now

6. Apr 29, 2010

Ok, I'm almost done, because now I have problems with the indices.

I started by $$x=\bar{x} + vt$$ where $$v$$ is a constant. Now the wave equation $$\nabla^2 \phi = \frac{1}{c^2} \frac{\partial^2 \phi}{\partial t}$$.

So i start writing the wave equation in the "\bar" coordinates. The Laplacian is the same $$\nabla_{bar}^2 \phi = \nabla^2 \phi$$ because

$$\frac{\partial \phi}{\partial \bar{x}} = \frac{\partial \phi}{\partial x} \frac{\partial x}{\partial \bar{x}} + \frac{\partial \phi}{\partial t} \frac{\partial t}{\partial \bar{x}} = \frac{\partial \phi}{\partial x}$$

For time:

$$\frac{\partial \phi}{\partial \bar{t}} = \frac{\partial \phi}{\partial t} \frac{\partial t}{\partial \bar{t}} + \frac{\partial \phi}{\partial x} \frac{\partial x}{\partial \bar{t}} = \frac{\partial \phi}{\partial t} + v \frac{\partial \phi}{\partial x}$$

$$\frac{\partial^2 \phi}{\partial \bar{t}^2} = \frac{\partial }{\partial \bar{t}} \frac{\partial \phi}{\partial \bar{t}} = \frac{\partial \phi^2}{\partial t ^2} + 2v \frac{\partial ^2 \phi}{\partial t \partial x} + v^2 \frac{\partial ^2 \phi}{\partial x^2}$$

And the wave equation reads what I said in the first post but without bars. It should be a very silly thing, but I can't see it... Thanks for your time.

7. Apr 29, 2010

### starthaus

What happens if you start by calculating the partial derivatives wrt (x,t) instead of (x_bar,t)?

8. Apr 29, 2010

I was thinking that should be that... in fact, a very very silly thing.

PS: Btw the equation you have calculating with "bar" instead with normal is not the one in my first post. Actually is calculating with normal instead of bars.

Absolutely clear now, thanks again!!

9. Apr 29, 2010

### starthaus

You are welcome, glad that I could help.
Now, if you do the same exercise by replacing the Galilean transforms with the Lorentz ones, you should get the famous invariance of the wave equation.

10. Apr 29, 2010