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In the Schwartz's quantum field theory, Lorentz-invariant phase space (LIPS) is defined by $$d\Pi_{LIPS}=\prod_{\text{final states } j}\frac{d^3p_j}{(2\pi)^3}\frac{1}{2E_j}(2\pi)^4\delta^4(\Sigma p)\tag{5.22}$$

For 2->2 scattering in center of mass frame $$d\Pi_{LIPS}=\frac{d^3p_3}{(2\pi)^3}\frac{1}{2E_3}\frac{d^3p_4}{(2\pi)^3}\frac{1}{2E_4}(2\pi)^4\delta^4(\Sigma p)\tag{5.26}$$ $$d\Pi_{LIPS}=\frac{1}{16\pi^2}d\Omega\int dp_f\frac{p^2_f}{E_3E_4}\delta(E_3+E_4-E_{CM})\tag{5.27}$$ I don't understand how come there is an integral?

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  • $\begingroup$ Hi KJH, welcome to Physic SE! Your question is a little unclear, perhaps try to edit it. What exactly don't you understand? $\endgroup$ – Andrea Jan 19 '16 at 16:01
  • $\begingroup$ Thank you for your comment. My question is How can get third eq from second eq I know $dp=p^2dp$ but i can't understand how can get integral $\endgroup$ – KJH Jan 19 '16 at 16:10
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EDIT: With the risk of giving away to much, here is another attempt.

For a scattering event where $a+b\rightarrow c+d$, the Lorenz invariant phase space volume is given by

$$d\mathrm{LIPS}=\frac{d^{3}\mathbf{p}_{c}\, d^{3}\mathbf{p}_{d}}{\left(2\pi\right)^{3}\left(2\pi\right)^{3}}\frac{1}{4E_{c}E_{d}} $$ the cross section part from the quantum amplitude is

$$ \rho_{a}\rho_{b}\left|v_{ab}\right|\sigma=\int\frac{d^{3}\mathbf{p}_{c}\, d^{3}\mathbf{p}_{d}}{\left(2\pi\right)^{6}}\frac{1}{4E_{c}E_{d}}\left(2\pi\right)^{4}\delta^{4}\left(p_{f}-p_{i}\right)\left|T\right|^{2}. $$

Here $T$ is the quantum amplitude from your QFT. The $\delta$-functions technically also comes from here.

We see that we can use the momentum delta function $\delta^{\left(3\right)}\left(\mathbf{p}_{i}-\mathbf{p}_{f}\right)$ to remove the $d^{3}\mathbf{p}_{d}$ integral and lock $\mathbf{p}_{c}$and $\mathbf{p}_{d}$ together. After this integral the cross section is \begin{eqnarray*} \rho_{a}\rho_{b}\left|v_{ab}\right|\sigma & = & \int\frac{d^{3}\mathbf{p}_{c}}{\left(2\pi\right)^{2}}\frac{1}{4E_{c}E_{d}}\delta\left(E_{f}-E_{i}\right)\left|T\right|^{2}\\ & = & \int\frac{\left|\mathbf{p}_{c}\right|^{2}d\left|\mathbf{p}_{c}\right|\, d\Omega}{\left(2\pi\right)^{2}4E_{c}E_{d}}\delta\left(E_{f}-E_{i}\right)\left|T\right|^{2} \end{eqnarray*} where we changed variables to polar coordinates and $d\Omega=d\phi\,d\cos\theta$ is the solid angle integrated over. Here $\theta$ is the angle between $\mathbf{p}_{a}$ and $\mathbf{p}_{c}$ such that $\mathbf{p}_{a}\cdot\mathbf{p}_{c}=\left|\mathbf{p}_{a}\right|\left|\mathbf{p}_{c}\right|\cos\theta$.

Note: Since we have not performed the integral over the solid angle yet, we can interpret that is under the integral sign as precisely the differential cross section

$$ \rho_{a}\rho_{b}\left|v_{ab}\right|\frac{d\sigma}{d\Omega} = \frac{\left|\mathbf{p}_{c}\right|^{2}d\left|\mathbf{p}_{c}\right|}{16\pi^2E_{c}E_{d}}\delta\left(E_{f}-E_{i}\right)\left|T\right|^{2}. $$

For completeness, we continue a bit further. Since the last $\delta$-function is in energy we notice that we can change variables to $E_{c}$ as $E_{c}^{2}=m_{c}^{2}+\left|\mathbf{p}_{c}\right|^{2}$. Differencing both terms directly gives $E_{c}dE_{c}=\left|\mathbf{p}_{c}\right|d\left|\mathbf{p}_{c}\right|$. We should note here that since $\mathbf{p}_{c}$ and $\mathbf{p}_{d}$ are locked that automatically implies that there is a relation between $E_{c}$ and $E_{d}$. The cross section becomes

$$ \rho_{a}\rho_{b}\left|v_{ab}\right|\sigma=\int\frac{\left|\mathbf{p}_{c}\right|E_{c}dE_{c}\, d\Omega}{\left(2\pi\right)^{2}4E_{c}E_{d}}\delta\left(E_{c}+E_{d}\left(E_{c}\right)-E_{i}\right)\left|T\right|^{2} $$ Note that since $E_{c}$ and $E_{d}$ are dependent the $\delta$ function behavior complicated. However, remember that $$ \int dx\, g\left(x\right)\delta\left(f\left(x\right)\right)=g\left(x_{0}\right)\frac{1}{f^{\prime}\left(x_{0}\right)} $$ where $x_{0}$ solves $f\left(x_{0}\right)=0$. This means that our cross section becomes

$$ \rho_{a}\rho_{b}\left|v_{ab}\right|\sigma=\int\frac{\left|\mathbf{p}_{c}\right|d\Omega}{\left(2\pi\right)^{2}4E_{d}}\left[1+E_{d}^{\prime}\left(E_{c}\right)\right]^{-1}\left|T\right|^{2} $$

This is as far as you can reasonably go without specifying a frame or knowing about $T$.

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    $\begingroup$ You should still use the $d$ actually. You haven't really integrated all the variables. The integral over $\Omega$ is still left over. Remember that this is simply the measure you are talking about. The full answer of course requires you to integrate the amplitude w.r.t. this measure. For 2->2 scattering, the amplitude only depends on the angle $\Omega$. Thus, we can determine a reduced measure wherein all but the angles have been integrated over. Your final result is therefore a reduced measure, and should really contain the $d$ (differential) $\endgroup$ – Prahar Jan 19 '16 at 17:34
  • $\begingroup$ I understand answer's point but i have same think comment... I want get differential cross section $\frac{d\sigma}{d\Omega}$. But there is not {\Omega}. How can I get integral and $\Omega$? $\endgroup$ – KJH Jan 19 '16 at 18:22
  • $\begingroup$ I have appended on the answer to show how the differential cross section comes about. $\endgroup$ – Mikael Fremling Jan 20 '16 at 9:48

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