Proving the Lorentz invariance of an integration measure? QFT related?

[jeebs]

So, first off, I'm thinking Lorentz invariant quantities are the same in any inertial frames S and S' regardless of their relative velocity.

I'm thinking I need to show that
$\frac{d^3k}{(2\pi)^32E(\vec{k})} = \frac{d^3k'}{(2\pi)^32E'(\vec{k'})}$ where the primed & unprimed quantities denote different frames.

We also have $E(\vec{k}) = \sqrt{\vec{k}^2 + m^2}$ in the denominator. This isn't Lorentz invariant - the mass term is, but the 3 - momentum $\vec{p} = \hbar \vec{k}$ is not, therefore E is not.

$E \neq E'$ where $E'(\vec{k'}) = \sqrt{\vec{k}'^2 + m^2}$.

This is making me think that neither E nor d³k are Lorentz invariant, so that when they are used in this fraction, the Lorentz invariance from each one somehow cancels out overall.

Then we come to the hint. I 'm really not sure what to make of this, I mean, I can write $\delta(x^2 - x_0^2) = \frac{1}{2|x|}(\delta(x-x_0) + \delta(x+x_0))$ with k's and m's: $\delta(k^2 - m^2) = \frac{1}{2|k|}(\delta(k-m) + \delta(k+m))$
to get an apparently Lorentz invariant expression: $\frac{d^4k}{(2\pi)^3}\delta(k^2 - m^2)\theta(k_0) = \frac{d^4k}{(2\pi)^3}\frac{1}{2|k|}(\delta(k-m) + \delta(k+m))\theta(k_0)$
but I'm not sure where this gets me, other than guessing that
$\delta(k^2 - m^2)\theta(k_0)$ is also Lorentz invariant seeing as I am told d⁴k is.

I have no idea how to proceed with this. I don't know what to make of the delta functions either.

 

[vela]

You should recognize that $d^4k\ \delta(k^2-m^2)\theta(k_0)$ is manifestly Lorentz invariant because $k^2 = k^\mu k_\mu$ is invariant.

The idea here is to note that $k^2-m^2 = k_0^2-\mathbf{k}^2-m^2$ and to do the integral over k₀.

Intuitively, you can think of the volume $d^3k$ being reduced by a factor $\gamma$ due to length contraction while E(k) is reduced by the same factor due to time dilation, so their ratio is invariant.

 

[jeebs]

You should recognize that $d^4k\ \delta(k^2-m^2)\theta(k_0)$ is manifestly Lorentz invariant because $k^2 = k^\mu k_\mu$ is invariant.

right, so, if I'm not mistaken, $k^\mu = (k_0, k_1, k_2, k_3)$ and $k_\mu = (k_0, -k_1, -k_2, -k_3)$. So, $k^2 = k^\mu k_\mu = (k_0, k_1, k_2, k_3).(k_0, -k_1, -k_2, -k_3) = k_0^2 - k_1^2 -k_2^2 -k_3^2 = k_0^2 - \vec{k}^2$.
Is there some reason why I should instantly recognise this as an invariant quantity?
Do you always get an invariant quantity if you take any 4-vector and do this with it?
Also, what am I supposed to make of the $\theta(k_0)$ or the d⁴k? What about them says Lorentz invariant?

The idea here is to note that $k^2-m^2 = k_0^2-\mathbf{k}^2-m^2$ and to do the integral over k₀.

I'm meant to be computing $\frac{1}{(2\pi)^3}∫ \delta(k_0^2 - \vec{k}^2 - m^2)\theta(k_0)d^4k$ right?
I'm struggling to be change the $d^4k$ into something involving a dk₀.
Can I start off by saying $dk^\mu = (dk_0, dk_1, dk_2, dk_3)$?

Actually, can I say that $d^4k = d^3k dk_0$?

 

[vela]

right, so, if I'm not mistaken, $k^\mu = (k_0, k_1, k_2, k_3)$ and $k_\mu = (k_0, -k_1, -k_2, -k_3)$. So, $k^2 = k^\mu k_\mu = (k_0, k_1, k_2, k_3).(k_0, -k_1, -k_2, -k_3) = k_0^2 - k_1^2 -k_2^2 -k_3^2 = k_0^2 - \vec{k}^2$.
Is there some reason why I should instantly recognise this as an invariant quantity?
Do you always get an invariant quantity if you take any 4-vector and do this with it?

Yes, the product of any two four-vectors, including a four-vector with itself, is an invariant. It's analogous to the regular dot product being invariant under rotations.

Also, what am I supposed to make of the $\theta(k_0)$ or the d⁴k? What about them says Lorentz invariant?

Think about the coordinate transformation $k'^\mu = \Lambda^\mu{}_\nu k^\nu$ where $\Lambda^\mu{}_\nu$ is a Lorentz transformation. How are the volume elements related? What about the signs of k₀ of k'₀?

I'm meant to be computing $\frac{1}{(2\pi)^3}∫ \delta(k_0^2 - \vec{k}^2 - m^2)\theta(k_0)d^4k$ right?
I'm struggling to be change the $d^4k$ into something involving a dk₀.
Can I start off by saying $dk^\mu = (dk_0, dk_1, dk_2, dk_3)$?

Actually, can I say that $d^4k = d^3k dk_0$?

Yes, $d^4k = d^3k\:dk_0$

 

[paranormal]

Are they supposed to be Lorentz invariant?

I would approach this problem by first understanding the concept of Lorentz invariance and its relevance in quantum field theory (QFT). Lorentz invariance refers to the principle that the laws of physics should be the same in all inertial frames of reference. In other words, the physical quantities and measurements should not depend on the observer's velocity or frame of reference.

In QFT, it is important to ensure that the integration measures used to calculate physical quantities are also Lorentz invariant. This is because the integration measure is used to integrate over all possible momentum states of particles, and if it is not Lorentz invariant, it can lead to incorrect results.

In this case, we are trying to prove the Lorentz invariance of the integration measure \frac{d^3k}{(2\pi)^32E(\vec{k})}. As you correctly pointed out, the energy term E(\vec{k}) is not Lorentz invariant due to the momentum term \vec{p} = \hbar \vec{k}. However, we can use the hint provided to us and write the measure in terms of delta functions.

The delta function \delta(k^2 - m^2) can be interpreted as a measure of the energy of a particle with momentum \vec{k} and mass m. This delta function is Lorentz invariant, as it only depends on the magnitude of the momentum \vec{k} and not its direction. Therefore, when we multiply this delta function with the Lorentz non-invariant measure \frac{d^3k}{(2\pi)^32E(\vec{k})}, the non-invariance of the energy term cancels out, and we are left with a Lorentz invariant measure.

Furthermore, the theta function \theta(k_0) ensures that we only integrate over positive energy states, which is also Lorentz invariant. Therefore, we can conclude that the expression \frac{d^4k}{(2\pi)^3}\delta(k^2 - m^2)\theta(k_0) is Lorentz invariant.

In summary, the use of delta functions in the integration measure ensures that it is Lorentz invariant, despite the non-invariance of the energy term. This is a crucial step in proving the Lorentz invariance of the integration measure in QFT