(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.

  • 1
    $\begingroup$ 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. $\endgroup$ – ACuriousMind Jul 17 '17 at 9:26
  • $\begingroup$ 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. $\endgroup$ – ACuriousMind Jul 17 '17 at 9:29
  • $\begingroup$ 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. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.