# D3x d3p invariant?

1. Nov 13, 2014

### Vrbic

1. The problem statement, all variables and given/known data
Hello, I have probably quit easy task, but I dont know how show that d3x d3p is a Lorentz invariant.

2. Relevant equations

3. The attempt at a solution
I mean I have to show that d3x d3p = d3x' d3p', where ' marks other system. I can prove ds2=ds'2, but I am not sure what with p?
Assume boost in x-way:
x'=g(x-vt) => dx'=g(dx-vdt) , where g=1/(1-(v/c)2)...is it right?

2. Nov 18, 2014

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Nov 19, 2014

### BiGyElLoWhAt

yes that's right. What does p=? You've got dx right, can you write dp in terms of dx? (or rather right p in terms of dx)? If you can, then you should be able to handle the situation.

Remember, you're not trying to show that $d^3(xp)$ is invariant, you're trying to show that $d^3x*d^3p$ is invariant, which is sufficiently easier than the former.

4. Nov 19, 2014

### Vrbic

p is momentum, well p=mv. $v=\frac{dx}{dt}$. $t'=g(t-\frac{V}{c^2}x)$->$dt'=g(dt-\frac{V}{c^2}dx)$, where V is movement between o and o' and is constant. So can I say $v'=\frac{dx'}{dt'}=\frac{g(dx-Vdt)}{g(dt-\frac{V}{c^2}dx)}$?

5. Nov 19, 2014

### BiGyElLoWhAt

No reason to carry the t through. You're trying to show that $\frac{d^3x}{dt^3}\frac{d^3p}{dt^3} = \frac{d^3x'}{dt'^3}\frac{d^3p'}{dt'^3} \rightarrow d^3xd^3p=d^3x'd^3p'$
But yes, that's the right idea.

6. Nov 19, 2014

### Vrbic

Can I prove it just for one dimension and suppose the others will be same? It means prove dx'dp'=dxdp

7. Nov 20, 2014

### BiGyElLoWhAt

You mean for momentum?
I would assume that the momentum was p in the x direction and 0 in y and z. Otherwise you need to do the lorentz transformations for a 3d velocity. The standard transforms you see online (i think) tend to be for a velocity in the x direction.