4
$\begingroup$

The Bhabha scattering differential cross section is given by $$\frac{d\sigma}{d\Omega}=\frac{\alpha}{2s}\left(\frac{3+\cos^{2}\theta}{1-\cos\theta}\right)^{2}$$ where $\theta$ denotes the angle of the outgoing states. However, the total cross section, obtained by integration over all angles diverges. Why is this so?

$\endgroup$
1
  • 1
    $\begingroup$ All scattering cross sections between pointlike charged particles diverge at $\theta=0$, because the long distance interaction scatters every particle in the beam, no matter how far away it is. $\endgroup$
    – Buzz
    Commented Aug 23, 2023 at 2:53

1 Answer 1

6
$\begingroup$

However, the total cross section, obtained by integration over all angels diverges. Why is this so?

The total cross section is infinite because the interaction has infinite range and does not fall off "fast enough." The interaction falls off quite slowly (as $1/r$) which is not fast enough to make the total cross section finite.

Remember that the total cross section is basically a measure of the size of the scatterer, which has no sharp boundary in this case since the "size" is determined by the range of the coulomb potential.


N.b., this is not some esoteric effect due to Quantum Field Theory. It is rather banal. The same thing happens in single-particle non-relativistic scattering off a fixed coulomb potential.

$\endgroup$
9
  • 1
    $\begingroup$ It's not even a quantum mechanical effect. The same kind of divergence shows up in the Rutherford cross section, whether the Coulomb interaction attractive or repulsive. $\endgroup$
    – Buzz
    Commented Aug 23, 2023 at 2:54
  • $\begingroup$ @hft: in the lab, would you really measure such an undefined total cross section? $\endgroup$
    – Yair
    Commented Aug 23, 2023 at 6:23
  • 1
    $\begingroup$ @Yair: in the lab, would you really perform totally unscreened Bhabha scattering? $\endgroup$
    – hft
    Commented Aug 23, 2023 at 15:30
  • 1
    $\begingroup$ ....in the lab, would your detector really be capable of integrating over the entire solid angle? $\endgroup$
    – hft
    Commented Aug 23, 2023 at 15:32
  • $\begingroup$ Anyways. The answer is no, you would never measure infinity "in the lab." But nevertheless the total cross section is infinite. $\endgroup$
    – hft
    Commented Aug 23, 2023 at 15:33

Not the answer you're looking for? Browse other questions tagged or ask your own question.