I'm trying to follow the derivation of the Boltzmann equation in my Theory of Heat script, but have a little trouble understanding the following:

The cross-section $d\sigma$ is defined as: The amount of particles which get scattered into the solid angle $d\Omega$ per time over the incoming flux intensity In my script they write: The amount of particles getting scattered into the region $[\Omega,\Omega + d\Omega]$ is equal to $|db|d\phi \cdot I$ where $I$ is the flux, and $db$ is the infinitesimal impact parameter. I already have trouble understanding that. How can I intuitively see that this is the correct formulation? I don't even know that $\phi$ is, because they use $\theta$ as the angle between the path of the incoming particle, and its outgoing path. The calculation for the crosssection goes on as follows: $d\sigma = |db|d\phi = b |\frac{db}{d\theta}| d\phi d\theta = b |\frac{db}{d\theta}|\frac{1 }{sin\theta}d\Omega$ which gives us the differential crossection: $\frac{d\sigma}{d\Omega} = \frac{b}{sin \theta}|\frac{db}{d\theta}|$

I can follow the calculation, I just can't understand what I mentioned above.

Can someone clear this up for me or point me to a good derivation of this?

Cheers and thanks in advance!


Here's a self-contained (hopefully clear) derivation.

Step 1. Setup and definition of differential scattering cross-section

Let $\mathcal L$ denote the incident luminosity (number of incident particles per unit area, per unit time) of a beam to be scattered. We assume that we have a spherical detector at infinity with which to measure scattered particles, and we assume that the detector's center is at the positon of the target. We further assume that the direction of travel of the beam is along the positive $z$-axis so that we can measure position on the spherical dector by the standard spherical coordinates $(\theta, \phi)$, where $\theta$ is the polar angle and $\phi$ is the azimuthal angle. Finally, we assume that the system is cylindrically symmetric about the $z$-axis, and we let $\eta(\theta)$ denote the number of particles per unit solid angle, per unit time hitting the detector at the angular position $\theta$. Then as you noted, the differential scattering cross-section, which I will denote by $D(\theta)$ is defined as $$ D(\theta) = \frac{\eta(\theta)}{\mathcal L} $$

Step 2. Understanding the function $b(\theta)$

Now, we assume that there is a differentiable function $b(\theta)$ defined on $[0,\pi]$ such that if a particle scatters through an angle $\theta$ then $b(\theta)$ gives the impact parameter at which it was incident on the target. Since no two particles incident at the same impact parameter can have different scattering angles, the function $b$ has the property that for all $\theta_1, \theta_2\in[0,\pi]$, $$ \text{If}\quad b(\theta_1) = b(\theta_2)\quad \text{then} \quad\theta_1=\theta_2 $$ In other words, $b$ is a one-to-one function of $\theta$. This means, in particular, that given any angles $\theta_1<\theta_2$, if a particle is scattered at some angle $\theta$ between $\theta_1$ and $\theta_2$, then it must have had impact parameter $b(\theta)$ between $b(\theta_1) $ and $b(\theta_2)$.

Step 3. Derivation of the standard expression for the cross-section

With this in mind, we consider some small range of scattering angles between $\theta$ and $\theta+\Delta\theta$. From our observation above, we note that particles scattered in this range must have had impact parameters between $b(\theta)$ and $b(\theta+\Delta\theta)$. On the other hand, the number of particles per unit time scattering through this range of impact parameters is $$ \Delta \nu = 2\pi b(\theta)|b(\theta+\Delta\theta) - b(\theta)|\cdot \mathcal L +\mathcal O(\Delta\theta^2) $$ This comes from just imagining the cross section of the beam and noting that the area of the ring of impact parameters between $b(\theta)$ and $b(\theta + \Delta\theta)$ is (to first order in $\Delta\theta$) simply the expression above that is multiplying the luminosity. To see this, recall that the area of a ring between two circles have radii $r$ and $r+\Delta r$ is \begin{align} |\pi(r+\Delta r)^2 - \pi r^2| &= 2\pi r|\Delta r| + \mathcal O(\Delta r^2) \end{align} The absolute value is necessary because area must be positive while it's possible that $\Delta r$ is negative. Now just set $r = b(\theta)$ and $\Delta r = b(\theta + \Delta\theta) - b(\theta) =b'(\theta)\Delta\theta + \mathcal O(\Delta\theta^2)$.

On the other hand, we know that all of these particles will be scattered into the solid angle subtended by the polar angles $\theta$ and $\theta + \Delta\theta$ which is given by $$ \Delta\Omega = \int_0^{2\pi}d\phi \int_\theta^{\theta+\Delta\theta}d\theta\sin\theta = 2\pi\sin\theta\Delta\theta + \mathcal O(\Delta\theta^2) $$

Combining these results, we see that the number of particles per unit solid angle, per unit time hitting the detector in the range of angles between $\theta$ and $\theta +\Delta\theta$ is $$ \frac{\Delta\nu}{\Delta\Omega} = \frac{b(\theta)}{\sin\theta}\frac{|b(\theta+\Delta\theta) - b(\theta)|}{\Delta\theta}\cdot \mathcal L + \mathcal O(\Delta\theta) $$ The $\Delta\theta\to 0$ limit of this gives the number of particles per unit time, per unit solid angle scattered at the angle $\theta$, which is precisely what we called $\eta(\theta)$. In this limit, the $\mathcal O(\Delta\theta)$ term vanishes, and the second fraction becomes the absolute value of the derivative of $b$, so we get $$ \eta(\theta) = \frac{b(\theta)}{\sin\theta}\left|b'(\theta)\right|\cdot \mathcal L $$ where here I use a prime for differentiation. It follows then, by dividing by $\mathcal L$, that the differential scattering cross-section is precisely the standard expression; $$ \boxed{D(\theta) = \frac{b(\theta)}{\sin\theta}\left|b'(\theta)\right|} $$

  • $\begingroup$ Wow thanks a lot, I'm gonna read through this tomorrow carefully! $\endgroup$
    – user17574
    Jun 26 '13 at 21:14
  • $\begingroup$ @user17574 My pleasure. Let me know if you have any questions or comments so that I can increase the clarity of the response. $\endgroup$ Jun 26 '13 at 21:31
  • $\begingroup$ OK I finally managed to read this carefully now. I have several questions: How do you derive the expression for $\Delta \nu$? I was trying to follow that by calculating the volume of a cylinder shell of thickness $|b(\theta + \Delta \theta) - b(\theta)|$, but then I don't know where the $b(\theta)$ in front of the absolute value comes from... care to elaborate on this? $\endgroup$
    – user17574
    Jul 10 '13 at 8:12
  • $\begingroup$ Oh god, I'm not capable to properly calculate the area of a circle shell... give me a few more minutes rofl. $\endgroup$
    – user17574
    Jul 10 '13 at 8:31
  • $\begingroup$ Nah I dont understand how to compute that. What I'm trying is, I imagine a cylindrical shell around de z-axis w. thickness $b(\Delta \theta)$. Since we are looking for the amount of particles per time in the area between $b(\theta)$ and $b(\theta + \Delta \theta)$ I thought simply multiplying the area of the circle shell ($\pi \cdot (b(\theta + \Delta \theta)^2 - b(\theta)^2))$ and multiplying by $L$ would give me the desired result. But it doesn't... can you help? $\endgroup$
    – user17574
    Jul 10 '13 at 8:58

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