Solving the D3x D3p Invariant Puzzle

[Vrbic]

Homework Statement

Hello, I have probably quit easy task, but I don't know how show that d³x d³p is a Lorentz invariant.

Homework Equations

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?
what about p?

 

[BiGyElLoWhAt]

x'=g(x-vt) => dx'=g(dx-vdt) , where g=1/(1-(v/c)²)...is it right?
what about p?

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.

 

[Vrbic]

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.

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

 

[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.

 

[Vrbic]

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.

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

 

[BiGyElLoWhAt]

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

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.