Wave equation and Galilean Transformation

[Advent]

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!

[starthaus]

And

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

No, this is incorrect. You need to use

t = \bar{t}

and

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

[Advent]

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

[starthaus]

So the right thing is \frac{\partial t}{\partial \bar{x}} = 0

Yes, of course

[Advent]

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

[Advent]

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.

[starthaus]

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.

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

[Advent]

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!!

[starthaus]

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!!

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.

[Advent]

I know, in fact i should do it for my speclal relativity exam, but there is some little work and theory before Lorentz Transformations.