As we know the total cross section can always be obtained from the differential cross section:

$$\sigma = \int_0 ^{2 \pi } \int_{0}^{\pi } \frac{d \sigma}{ d \Omega} d \Omega $$

I understand how the integration is done.

For example, sometimes I see the differential cross section formula is written as: $$ \frac{d \sigma}{ d^4 q \ d \Omega} = K (1 + \cos ^2 \theta ) $$

The $d^4q = \frac{1}{2}dq_0 \ dq_t^2 \ dq_z$

How would I write the total cross section? $$\sigma = \int dq_0 \ dq_t^2 \ dq_z \int_0 ^{2 \pi } \int_{0}^{\pi } \frac{d \sigma}{ d \Omega} d \Omega \ \sin \phi \ K \ (1 + \cos ^2 \theta ) \ \ \ \ \ ?$$


1 Answer 1


Your first formula makes no sense. You're integrating over the solid angle, which is ok but then you're integrating another time over no variable in the $[\pi,\pi]$ interval, which again makes no sense.

In general the differential cross section can be given in a number of different ways depending on how one treats the phase space integral. In fact, the differential cross section for a two particle scattering process is given by $$d\sigma = \frac{1}{4\sqrt{(p_1p_2)^2-m_1^2m_2^2}}(2\pi)^4\delta^4\left(\sum_f p_f-\sum_i p_i\right)|\mathcal{M}_{fi}|^2\prod_{i=1}^n\frac{d^4p_i}{(2\pi)^3 2E_i}$$

and as you can see there are $n$ integrations to be done, one for every phase space of every particle. One could even not bother to integrate over any phase space and instead study the differential cross section $$\frac{d\sigma}{dp_1\,dp_2\,\dots dp_n} = \frac{1}{4\sqrt{(p_1p_2)^2-m_1^2m_2^2}}(2\pi)^4\delta^4\left(\sum_f p_f-\sum_i p_i\right)|\mathcal{M}_{fi}|^2\prod_{i=1}^n\frac{1}{(2\pi)^3 2E_i}$$ or instead, which is often done, integrate over some specific phase space and leave behind others. One example of that is exactly the differential cross section over the solid angle.

As you gave, one often uses the differential cross section $$\frac{d\sigma}{d\Omega}$$ which is what one gets in a two particle scattering process $a+b\to c+d$ integrating over the phase space of one variable and then integrating the other variable only over the energy. To get back the full cross section from this, one needs to integrate over the solid angle which in three dimensions gives $d\Omega= \sin\theta d\theta d\phi$ and so $$\sigma = \int_0^{4\pi} \frac{d\sigma}{d\Omega} = \int_0^\pi\int_0^{2\pi}\frac{d\sigma}{d\Omega}\sin\theta\, d\phi\, d\theta$$

The same goes for the other integral. Moreover $d^4q$ is not what you gave, but rather $d^4 = dq^0 d^3q$ where $d^3q$ is the normal euclidean measure and $dq^0 = dE$.

Just a little post scriptum: I'm not sure the second differential cross section you gave us makes any sense. But i could be wrong on this one.

  • $\begingroup$ Sorry, for the typo in the integral limit. I fixed that. Second, the $d^3 q = \frac{1}{2} dq_t^2 dq_z $, which I have treated the same way you wrote. The $q_t$ is the transverse momentum here which is written in the polar coordinate. And third, you are right, the example cross section is not from any decay process, but an example. That is that I used a proportional term K. $\endgroup$
    – user12906
    Jul 6, 2020 at 8:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.