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Based on my understanding, the differential cross section $\frac{d\sigma}{d\Omega}$ gives us the differential area of incident particles that projects onto a differential piece of solid angle on the detector. When one calculates total cross section, one integrates over the entire solid angle; i.e. $\sigma = \int d\sigma = \int \frac{d\sigma}{d\Omega}d\Omega$. I do not comprehend this. The total cross section is indicative of the likelihood of an interaction between the incident beam and the target, and let us assume that the incident beam is a beam ranging across all impact parameters. If at impact parameter large enough the incident particles do not deflect and thus has a scattering angle of close to zero, wouldn't there be an unbounded range of impact parameter scattering onto near zero angles, and thus any integration that passes through zero will result in a blow up? (not just for rutherford scattering)

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  • $\begingroup$ There is a difference between 'close to zero' and 'zero'. To contrast with Rutherford, calculate for hard sphere scattering where scattering probability truly is zero above some impact parameter. $\endgroup$ – Jon Custer Jul 6 '16 at 11:10
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It should be noted that the differential cross section at $\theta = 0$ is not really defined - what is the difference between a particle being scattered by angle zero and not being scattered at all? However, it's completely legitimate to ask about the behavior of $\frac{d\sigma}{d\Omega}$ as $\theta$ approaches close to zero.

For Rutherford scattering, the total cross-section blows up as we integrate $\frac{d\sigma}{d\Omega}$ closer and closer to $\theta=0$. This is because the scattering probability decreases too slowly as the impact parameter grows, so the contribution from larger and larger impact parameters cannot be neglected (the integral diverges as $b \to \infty $). Fundamentally, this is because the photon is massless, so the EM interaction strength falls as $1/r^2$ (known as a long-range interaction).

For interactions mediated by massive particles, the interaction strength falls as $\frac{e^{-mr}}{r^2}$ (known as a short-range interaction). There is a non-zero scattering probability for every impact parameter, but this probability decays exponentially, so the integral converges at $b \to \infty$.

For classical hard-sphere scattering, the situation is even simpler - the scattering probability actually hits zero for impact parameters above a threshold. Therefore the integral also converges as $b \to \infty$.

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  • $\begingroup$ I am with you for hard-sphere scattering. However, for any force (short or long range) acting over an infinite distance, wouldn't any particle of any impact parameter be scattered? As such wouldn't any integration over the whole circle (even if we exclude 0) result in infinite cross section? $\endgroup$ – Zhanfeng Lim Jul 8 '16 at 15:11
  • $\begingroup$ For a force acting over an infinite distance, particles of every impact parameter have some probability of being scattered. However, that probability may decay to zero as the impact parameter grows. If it decays to zero quickly enough, then the integral will converge. $\endgroup$ – Sergei Patiakin Jul 8 '16 at 15:20
  • $\begingroup$ I do not see how that is possible, by 'conservation of number of particles'. Are you referring to the non-classical case? $\endgroup$ – Zhanfeng Lim Jul 11 '16 at 14:14
  • $\begingroup$ Yes, I was referring to quantum-mechanical scattering, which is probabilistic. For classical scattering of a particle in a potential well, I think you're right: any force acting over an infinite distance will have an infinite total cross section, no matter how quickly it decays with distance. $\endgroup$ – Sergei Patiakin Jul 12 '16 at 8:21
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I think any cross section is just a number, which hints at how more/less number of particles (electrons/protons etc) can be ejected/absorbed(interact) with the application of an external field. So are the terminologies like absorption cross section, photoionization cross section etc.

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