# Phase space measure in positron-electron annihilation calculation

I'm still trying to calculate the cross-section of the $e^- e^+ \rightarrow \mu^- \mu^+$ interaction in first order. This time I'm struggling with the phase space measure. Note that I have two questions, which are not exactly related but since they deal with the same equation I'll post them both at once.

1) So generally one has to integrate over the unknown momenta, i.e. the momenta of the outgoing muons, while keeping in mind that energy and momentum conservation needs to hold:

$$d\Phi_{12 \rightarrow 34} = [dp_3][dp_4] (2 \pi)^4 \delta(p_1 + p_2 - p_3 - p_4) = \frac{d^4p_3}{(2 \pi )^3}\delta(p_3^2 - m_{\mu}^2)\frac{d^4p_4}{(2\pi)^3}\delta(p_4^2 - m_{\mu}^2)(2\pi)^4\delta(p_1 + p_2 - p_3 - p_4)$$

In my script they jump from this point to this expression:

$$\frac{d^4p_3}{(2\pi)^2}\delta(p_3^2 - m_{\mu}^2)\delta((p_1 + p_2 - p_3)^2 - m_{\mu}^2)$$

Can somebody explain to me what happened here? Where did the integration over $p_4$ vanish to?

2) In the expression below:

$$\frac{|\vec{p_3}|^2 d|\vec{p_3}| d(\cos\theta) d\phi}{(2\pi)^2} \frac{1}{4E} \delta(E_3^2 - |\vec{p_3}|^2 - m_{\mu}^2) = \frac{|\vec{p_3}|d(\cos\theta)d\phi}{(2\pi)^2}\frac{1}{8E}$$

How was the Delta function manipulated here to give make the two terms equal? I have trouble seeing this.

• Review your terms again because I see some mistakes in the equations. Once you do that I can guide you in the right direction. – PhotonBoom Jan 7 '15 at 14:32
• I forgot some brackets and a sqaure above one muon mass. I conciously disregarded the Heaviside function that ensures that the arguments of the Delta function are greater than zero. Is that a grave mistake? Otherwise I simply copied the script... – user17574 Jan 7 '15 at 14:55
• For the last equation, should the momentum inside the delta function be squared? – PhotonBoom Jan 7 '15 at 15:13
• Damnit, yes it should be! Thought I caught everything :) – user17574 Jan 7 '15 at 15:21
• related question by the OP: physics.stackexchange.com/q/156969/58382 – glS Jan 7 '15 at 15:40

Generally, a Dirac delta function is there to ensure conservation of momentum. In the first case where you have:

$$\delta(p_1 + p_2 - p_3 - p_4)$$

This is telling you that the momentum of the incoming particles must necessarily be equal to the momentum of the outgoing particles. Hence from the delta function above, we know that

$$p_1 + p_2 = p_3 + p_4$$

From this it follows that $p_4 = p_1 + p_2 - p_3$. All you do next is perform some cancellations of factors of $2\pi$ and replace everywhere you see $p_4$ with the expression above.

Its the same with the second expression but its a bit trickier because of the $d|\vec{p_3}|$. To remove the delta function, you need

$$|\vec{p_3}| = \sqrt{E_3^2 - m_{\mu}^2}$$

Performing the differential:

$$d|\vec{p_3}| = d(\sqrt{E_3^2 - m_{\mu}^2}) = \frac{1}{2}(E_3^2 - m_{\mu}^2)^{-1/2} = \frac{1}{2}(|\vec{p_3}|^2)^{-1/2} = \frac{1}{2}(|\vec{p_3}|)^{-1}$$

Perform the cancellations and you are done.

• Thanks a lot for the answer! That is a much easier answer than I expected, which is slightly embarrassing :) Thanks again! – user17574 Jan 7 '15 at 15:58
• @user17574 My pleasure! – PhotonBoom Jan 7 '15 at 15:59
• After going over this a bit more carefully I'm actually a bit confused: By replacing $p_4$ with $p_1 + p_2 - p_3$ I get the right Delta function, but why does $d^4(p_1+p_2-p_3)$ vanish? Also how exactly do you perform the differential? I'd say $\frac{d|\vec{p_3}|}{d(\sqrt{E_3^2 - m_{\mu}^2})} = 1$, but you seem to be deriving $|\vec{p_3}|$ by the argument of the square root to get the expressions on the right. – user17574 Jan 14 '15 at 15:41
• Oh, does the integration over $dp_4$ simply vanish because we basically explicitly calculate the integral over it? I.e. by enforcing the momentum conservation we have already calculated that integral, right? – user17574 Jan 14 '15 at 15:47
• Yes! As for your second question, we replace $|\vec{p}_3|$ with the new differential imposed by our delta function. So we are stuck with it! We need to calculate the new differential, which is evaluated as performed in my answer. – PhotonBoom Jan 14 '15 at 15:49