# Tree-level elastic scattering in phi-4

(this is not homework)

I'm trying to solve some simple exercises in order to get back into gauge theory. I have some problems and solutions from 'Gauge Theory' by Cheng and Li. Right now I am stuck trying to calculate the differential cross section for tree-level elastic scattering in $\phi^4$. I am trying to evaluate the following: $$\rho = \int (2\pi)^4\delta^{(4)}(\bar{p}-p_3-p_4)\frac{d^3\mathbf{p}_3}{(2\pi)^32E_3}\frac{d^3\mathbf{p}_4}{(2\pi)^32E_4},$$ where $\bar{p} = p_1 + p_2$ are the incoming momenta and $p_3, p_4$ are the outgoing. The matrix element is $-i\lambda$. After carrying out the $d^3\mathbf{p}_4$ integral and defining (we are in the cms frame) $p_1 = (E,\mathbf{p}), p_2 = (E,-\mathbf{p}), p_3 = (E^\prime,\mathbf{p}^\prime), p_4 = (E^\prime,-\mathbf{p}^\prime)$ we are left with $$\rho = \int (2\pi)^{-2}\delta(2E-2E^\prime)\frac{d^3\mathbf{p}^\prime}{4E^{\prime 2}}$$

Here is where I am stuck. The solution suggests that we can transform the integrand to this: $$\rho = \int (2\pi)^{-2}\delta(2E-2E^\prime)\underbrace{\frac{p^\prime E^\prime dE^\prime}{4E^{\prime 2}} d\Omega}_{\text{How do we get to this?}}$$ I assume that the new $p^\prime$ is a 4-vector and that $E^\prime = p^{\prime 0}$, but I really don't see how to arrive to this. Any help would be appreciated.

• Please note that homework-like questions and check-my-work questions are generally considered off-topic here. Note that these policies concern the type of question and not whether or not this is actual homework. – ACuriousMind Jul 17 '17 at 9:26
• Additionally, the formula you ask about doesn't make sense - if $p'$ is a 4-vector, then the whole r.h.s. is a 4-vector, but $\rho$ is meant to be a number, not a vector. So unless you define what $p'$ is, it's unclear what you're asking about, since $\mathrm{d}^3 \mathbf{p'} = E' \mathrm{d} E' \mathrm{d}\Omega$ (without the $p'$!) would just be using polar coordinates and $E'$ as the radius. – ACuriousMind Jul 17 '17 at 9:29
• Thanks for clarifying, I'll keep that in mind. Regarding $p^\prime$, I don't actually know what it is, since it isn't defined. However, I see now that the rest is indeed polar coordinates. Thanks. – koldrakan Jul 17 '17 at 9:37

I think the trick you are looking for after is to first write first in $\mathbf p$ in polar coordinates $d^3\mathbf p=p^2\,dp\,d\Omega$ and then to use $E^2 = p^2 +m^2 \Rightarrow E\,dE=p\,dp$ to change the integral into an integral over energy.