1
$\begingroup$

G. Feinberg and J. Sucher showed that neutrino-pair exchange can induce long-range repulsive forces between electrons that is proportional to $r^{-6}$. In the paper, they also stated that changing a particle to its antiparticle changes the sign of the force. However, in case of neutrino-pair exchange, one cannot induce attractive forces. So i was thinking about exchange of hydrogen atom (as it can be considered as a boson, particularly "composed boson" ), could it be induce attractive forces between $e^-e^-$? And if it couldn't induce attractive forces, which bosons do?

$\endgroup$

1 Answer 1

1
$\begingroup$

First, a hydrogen atom is not an exchange particle. It is a real particle. An exchange particle is termed a virtual particle. A virtual particle mediates an interaction between two other particles. The reason "virtual" is used is because it is usually "off mass shell" and not observable. The lifetime for such particles is ridiculously small and this amount of time $t$ and their energy $E$ are determined by the Heisenberg Uncertainty relation, namely

$\Delta E \Delta t \geq \frac{\hbar}{2} $

But if we worked with it anyway, and we applied this to the hydrogen atom (using the mass/energy of a proton) this places its range at roughly $0.11$ fermi or about 1% of its own 'radius'.

$\endgroup$
2
  • $\begingroup$ Thank you Dr jh for you useful corrections. So which properties a boson must have to induce stronger-than-electromagnetic forces between $e^-e^-$? $\endgroup$ Oct 19, 2020 at 13:34
  • $\begingroup$ Hi again Elsayed. The photon, a boson, is the exchange particle responsible for the electromagnetic interaction. This includes electron-electron scattering. I’m not sure I understand what you mean by stronger-than-electromagnetic. cheers. $\endgroup$
    – joseph h
    Oct 19, 2020 at 22:55

Your Answer

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

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