How to Prove Lorentz Invariance of Volume Element in Momentum Space?

[Leeds student]

Simple question.. How do you prove the volume element of momentum space (d3k/Ek) is Lorentz Invariant?

I tried making it proportional to "velocity volume element" derived from the Lorentz transformations but didn't seem to get very far.

 

[pmb_phy]

Simple question.. How do you prove the volume element of momentum space (d3k/Ek) is Lorentz Invariant?

I tried making it proportional to "velocity volume element" derived from the Lorentz transformations but didn't seem to get very far.

Are you sure its supposed to be invariant?

Pete

 

[meopemuk]

Simple question.. How do you prove the volume element of momentum space (d3k/Ek) is Lorentz Invariant?

Basically, what you need to prove is

\frac{d^3k'}{E_{k'}} = \frac{d^3k}{E_{k}}

where 4-vectors (\mathbf{k'}, E_{k'}) and (\mathbf{k}, E_{k}) are related to each other by a Lorentz transformation. To do that you just need to calculate the Jacobian of the transformation \mathbf{k'} \to \mathbf{k}.

Eugene.

 

[Leeds student]

Basically, what you need to prove is

\frac{d^3k'}{E_{k'}} = \frac{d^3k}{E_{k}}

where 4-vectors (\mathbf{k'}, E_{k'}) and (\mathbf{k}, E_{k}) are related to each other by a Lorentz transformation. To do that you just need to calculate the Jacobian of the transformation \mathbf{k'} \to \mathbf{k}.

Eugene.

Yes, that's exactly what I was trying to show.
I looked up Jacobian's and for that transformation I calculated the determinant to be: \gamma (Lorentz factor). So I tried another approach and got a 3x4 matrix which I can't find how to solve.

I know it should be quite trivial and only a few lines but I keep getting stuck. Please point me in the right direction to complete this. Do I even need the velocity transforms? Or does k_{x'} transform like x' etc?

 

[meopemuk]

Yes, that's exactly what I was trying to show.
I looked up Jacobian's and for that transformation I calculated the determinant to be: \gamma (Lorentz factor).

I think you are on the right track here. The momentum-energy Lorentz transformations are (if the velocity v of the moving reference frame is along the x-axis)

k'_x = k_x \cosh \theta - E_k/c \sinh \theta
k'_y = k_y
k'_z = k_z
E'_k = E_k \cosh \theta - k_x c \sinh \theta

where v = c \tanh \theta; \gamma \equiv \cosh \theta

Eugene.

 

[pmb_phy]

Upon thinking this over more it occurred to me that what you called "volume element of momentum space" and the quantity (d^3k/Ek) are not the same. If it was then it'd be

momentum density = d^3p/dV

That quantity is the compent of a 4-tensor and is not invariant. I don't see the connection between what you say and the expressions you write. And you never answered my question: Why do you think it should be invariant? Was this a homework assignment?

Pete

 

[Leeds student]

Upon thinking this over more it occurred to me that what you called "volume element of momentum space" and the quantity (d^3k/Ek) are not the same. If it was then it'd be

momentum density = d^3p/dV

That quantity is the compent of a 4-tensor and is not invariant. I don't see the connection between what you say and the expressions you write. And you never answered my question: Why do you think it should be invariant? Was this a homework assignment?

Pete

This was not homework but a practice question. To show (what they called a volume element) \frac{d^{3}k}{E_{k}} is Lorentz invariant. The post above answers my question, thanks to both of you.

 

[pmb_phy]

This was not homework but a practice question. To show (what they called a volume element) \frac{d^{3}k}{E_{k}} is Lorentz invariant. The post above answers my question, thanks to both of you.

You would be doing me a great service if you post the question exactly as it was stated in your source. I find this to be an interesting topic and I'd like to familiarize myself with it. Thanks.

Best wishes

Pete

 

[lebashad]

Simple question.. How do you prove the volume element of momentum space (d3k/Ek) is Lorentz Invariant?

I tried making it proportional to "velocity volume element" derived from the Lorentz transformations but didn't seem to get very far.

dans les transformation de lorentz V'=V/(1-v2/c2)^3/2

 

[lebashad]

dans les transformation de lorentz V'=V/(1-v2/c2)^3/2

Sorry I am French!
In the transformation of the volume is lorentz V '= V / (1-v2/c2) ^ 3 / 2

 

[pmb_phy]

My question still remains to be answered, i.e. to Leeds student

You would be doing me a great service if you post the question exactly as it was stated in your source. I find this to be an interesting topic and I'd like to familiarize myself with it. Thanks.

Best wishes

Pete

 

[reilly]

The easist way to show the invariance of d3k/E(k) is via a delta function,
delta(E^^2 - P^^2 - M^^2) ==D, which is clearly invariant under LTs. Evaluate

Integral( d4k D) and you will get the answer you want. (Note, this is a fairly standard approach, and can be found in many texts on E&M and QFT.)
Regards,
Reilly Atkinson

 

[Leeds student]

My question still remains to be answered, i.e. to Leeds student


Best wishes

Pete

I did, it was:
Show that the volume element \frac{d^{3}k}{E_{k}} is Lorentz invariant

 

[reilly]

The easist way to show the invariance of d3k/E(k) is via a delta function,
delta(E^^2 - P^^2 - M^^2) ==D, which is clearly invariant under LTs. Evaluate

Integral( d4k D) and you will get the answer you want. (Note, this is a fairly standard approach, and can be found in many texts on E&M and QFT.)
Regards,
Reilly Atkinson

In fact, for those who are interested, the difference between old-fashioned QFT perturbation theory -- Weisskopf, Pauli and Heisenberg, Dirac, etc -- and modern covariant perturbation theory -- Feynman, Schwinger, Tomonaga -- is exactly the two versions of d3k/E(k) and d4k delta(E^^2- P^^2 - M^^2); The non-covariant approach uses the d3k approach, while the covariant approach uses the d4k approach. I'm quite sure that this difference is discussed by Zee in his field theory book.

Regards,
Reilly Atkinson

 

[pmb_phy]

I did, it was:
Show that the volume element \frac{d^{3}k}{E_{k}} is Lorentz invariant

Hmmm. That simple huh? Okay. Thank you.

By the way, Welcome to the Forum!

Best wishes

Pete