# The phase space integral for Compton scattering

Hello fellow physicists

I'm trying the calculate the phase space integral for Compton scattering in the Lab frame where the electron is initially at rest. I follow the derivation I found here (equation 4.1) but the derivation can also be found in Peskin and Schroeder page 163.

In the calculation of the phase space integral they end up with \begin{align} \int d\Pi_2 &= \int\frac{d\omega^\prime d\Omega}{(2\pi)^2}\frac{\omega^\prime}{4E^\prime}\delta(\omega^\prime+E^\prime-\omega-m) \\ &= \int\frac{d\omega^\prime d\Omega}{(2\pi)^2}\frac{\omega^\prime}{4E^\prime}\delta(\omega^\prime+\sqrt{m^2+\omega^2+(\omega^\prime)^2-2\omega\omega^\prime\cos\theta}-\omega-m)\\ \end{align} Where they have used the following equality to get to the last row $E^\prime=\sqrt{|\mathbf{p}^\prime|^2-m^2}=\sqrt{|\mathbf{k}-\mathbf{k}^\prime|^2-m^2}=\sqrt{m^2+\omega^2+(\omega^\prime)^2-2\omega\omega^\prime\cos\theta}$

We are now supposed to use the identity $$\int f(x)\delta(g(x))dx=\sum_i\frac{f(x_i)}{|g'(x_i)|} \tag{1}$$ Where $x_i$ are the roots to $g(x)$ to perform the integral over $(\omega^\prime)$. When they do this they end up with $$\int d\Pi_2=\int\frac{d\Omega}{(2\pi)^2}\frac{\omega^\prime}{4E^\prime}\frac{1}{\Big|1+\dfrac{\omega^\prime-\omega\cos\theta}{E^\prime}\Big|}$$ But If I do it and in my case call $$g(\omega^\prime)=\omega^\prime+\sqrt{m^2+\omega^2+(\omega^\prime)^2-2\omega\omega^\prime\cos\theta}-\omega-m$$ I find that the root to $g(\omega^\prime)$ is $$\omega^\prime\rightarrow \frac{\omega}{1+\dfrac{\omega}{m}(1-\cos\theta)}$$ and the derivative of $g(\omega^\prime )$ is $$g'(\omega^\prime)=1+\frac{\omega^\prime-\omega\cos\theta}{E^\prime}$$ I get using equation (1) that the phase space integral is $$\int d\Pi_2=\int\frac{d\Omega}{(2\pi)^2}\frac{\dfrac{\omega}{1+\dfrac{\omega}{m}(1-\cos\theta)}}{4E^\prime}\frac{1}{\Big|1+\dfrac{\dfrac{\omega}{1+\dfrac{\omega}{m}(1-\cos\theta)}-\omega\cos\theta}{E^\prime}\Big|}$$ Which is an ugly ass expression to say the least and doesn't contain any $(\omega^\prime)$. It seems to me like they have instead written the identity (1) as $$\int f(x)\delta(g(x))dx=\frac{f(x)}{|g'(x)|}$$ Which I think is odd because you're supposed to have a sum over the roots of $g(x)$ in the expression. And also by definition: when we integrate over $(\omega^\prime)$ the expression should not contain $(\omega^\prime)$ anymore.

Can somebody please explain to me what is wrong?

Edit: Or is it just possible to make the substitution $$\omega^\prime=\frac{\omega}{1+\dfrac{\omega}{m}(1-\cos\theta)}$$ to get to the correct answer?

Start with $$\int d \Pi_2 = \int \frac{\omega'^2 d \omega' d \Omega'}{(2 \pi)^3} \frac{1}{4 \omega E'} (2 \pi) \delta( \omega' + E'(\omega') - \omega -m)$$ and use $\int d \Omega ' = 2 \pi \int d \cos \theta$ to write $$\int d \Pi_2 = \frac{1}{8 \pi} \int d \omega' d \cos \theta \frac{\omega'}{E'} \delta( \omega' + E'(\omega') - \omega -m) \label{eq:dp2}$$ We can now use the following $\delta$ function identity: $$\int \delta f(x)\lbrack g(x) \rbrack dx = \sum_i \frac{f(x_i)}{|g'(x_i)|}$$ where the sum is over all zeroes. Here $g(x) = g(\omega') = \omega' + E'(\omega') - \omega -m$ and hence $g'(\omega') = d g(\omega')/d \omega' = 1 + dE'(\omega')/d \omega'$. So we need to find an expression for $dE' (\omega')/d\omega'$. We can find this by differentiating the square of $$E'(\omega') = \sqrt{ m^2 + \omega'^2 + \omega^2 - 2 \omega\omega' \cos \theta}$$ w.r.t. $\omega'$: $$\frac{d}{d\omega'} \lbrack E'^2 \rbrack = \frac{d}{d\omega'} \lbrack m^2 + \omega'^2 + \omega^2 - 2 \omega\omega' \cos \theta\rbrack$$ to find $$E' \frac{dE'}{d \omega'} = \omega' - \omega \cos \theta$$ or $$\frac{dE'}{d \omega'} = \frac{\omega' - \omega \cos \theta}{E'}$$ Using this we find $$\int d\Pi_2 = \frac{1}{8\pi} \int d \cos \theta \frac{1}{|1 + \frac{\omega' - \omega \cos \theta}{E'}|}$$ Taking the zeroes of $g$ here is ensuring momentum conservation $m+\omega = E'+\omega'$. It also ensures we can drop the absolute signs.

Yes, you just make this substitution to get the correct answer. The relation between omegas is also exactly the expression for which the delta function is zero - such that 4-momentum is conserved.