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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?

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Start with \begin{equation} \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) \end{equation} and use $\int d \Omega ' = 2 \pi \int d \cos \theta$ to write \begin{equation} \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} \end{equation} We can now use the following $\delta$ function identity: \begin{equation} \int \delta f(x)\lbrack g(x) \rbrack dx = \sum_i \frac{f(x_i)}{|g'(x_i)|} \end{equation} 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 \begin{equation} E'(\omega') = \sqrt{ m^2 + \omega'^2 + \omega^2 - 2 \omega\omega' \cos \theta} \end{equation} w.r.t. $\omega'$: \begin{equation} \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 \end{equation} to find \begin{equation} E' \frac{dE'}{d \omega'} = \omega' - \omega \cos \theta \end{equation} or \begin{equation} \frac{dE'}{d \omega'} = \frac{\omega' - \omega \cos \theta}{E'} \end{equation} Using this we find \begin{equation} \int d\Pi_2 = \frac{1}{8\pi} \int d \cos \theta \frac{1}{|1 + \frac{\omega' - \omega \cos \theta}{E'}|} \end{equation} Taking the zeroes of $g$ here is ensuring momentum conservation $m+\omega = E'+\omega'$. It also ensures we can drop the absolute signs.

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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.

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