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In Section 5.1.2 of Schwartz's when he changes variables from Equation 5.27 to Equation 5.29 he left a piece of old variables and treat it as constant. Shouldn't $p_f$ also be replaced with some expression $p_f(x)$?

The relevant equations are reproduced below; $E_3=\sqrt{m_3^2+p_f^2}$ and $E_4=\sqrt{m_4^2+p_f^2}$ \begin{align} \tag{5.27}\mathrm d\Pi_{LIPS}&=\frac{1}{16\pi^2}\mathrm d\Omega \int\mathrm dp_f \frac{p_f^2}{E_3E_4}\delta(E_3+E_4-E_\text{CM})\\ \text{Introducing}\qquad x(p_f)&=E_3(p_f)+E_4(p_f)-E_\text{CM}\\ \tag{5.28}\frac{\mathrm dx}{\mathrm dp_f}=\frac{\mathrm d\ \ }{\mathrm dp_f}(E_3+E_4-E_\text{CM})&=\frac{p_f}{E_3}+\frac{p_f}{E_4}=\frac{E_3+E_4}{E_3E_4}p_f\qquad\text{(Jacobian)}\\ \mathrm d\Pi_{LIPS}&=\frac{1}{16\pi^2}\mathrm d\Omega\int\mathrm dx \frac{p_f}{E_{CM}}\delta(x)\\ \tag{5.29}&=\frac1{16\pi^2}\mathrm d\Omega\frac{p_f}{E_\text{CM}}\theta(E_\text{CM}-m_3-m_4) \end{align}

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    $\begingroup$ Welcome. Please don't post images containing text or equations that are essential to understand the question, please transcribe everything that's necessary (You can use MathJax for equations). $\endgroup$
    – Amit
    Commented Jun 3, 2023 at 14:23
  • $\begingroup$ I can see why the OP did it in this case. $\endgroup$
    – mmesser314
    Commented Jun 3, 2023 at 14:47

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It's elementary calculus, or, rather, notation. $p_f$ is not a real variable, as it is constraint to the constant enforced by the δ-function constraint: the integration is fake/illusory! The author is aiming to collapse the δ-function, so he switches, locally invertibly, to $x(p_f)$ from $p_f$. The expression he left, as you noticed, is actually its inverse function, $p_f(x)$, a function of x in reality, with the original name retained, but, as originally, it is fixed to a constant by the vanishing of the δ-function argument.

He can now collapse the δ-function, given the transformed lower limit of the integral, into a step function (some xs are disallowed as being out of that range!). If you wish, you may now "remember" the function $p_f(0)$ as the original constant $p_f$, as the variable x has disappeared with the integration, and you need not recall it anymore. You may now luxuriate in the simplicity of equation (5.32).

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