One of the first things one learns when comparing the fundamental interactions is that gravity is much, much weaker than electromagnetism. Generally, the story goes like "you can easily lift this apple, although the whole Earth is actually pulling down on it". The number $10^{36}$ gets thrown around a lot (e.g. in this answer) and usually this comes from comparing the strengths of the Coulomb with the Newtonian gravitational force using the electron charge and mass, respectively.

Is using the charge and mass of the electron really the correct comparison, though? I would assume that an alternative comparison would be more suited, one that involves comparing the quantum of charge, being the electron charge${}^\textrm{[1]}$, with some sort of "quantum of mass", which certainly isn't the electron mass.

Instead of the electron you could also take the proton, which changes that ratio slightly, since it is heavier than the electron, but also carries one elementary charge.

What is the correct way to compare the strengths of gravity and electromagnetism? Is there even one?

${}^\textrm{[1]}$ One could of course argue here that the quantum of charge is actually $\frac13 e$, since that's the charge the quarks carry.

  • $\begingroup$ Can you think of anything of nonzero mass that's both lighter than the electron and present in nonrelativistic situations? $\endgroup$ – probably_someone Jul 5 '17 at 19:08
  • $\begingroup$ @probably_someone Hm, good insight - is the neutrino considered a cheap cop-out in this case? $\endgroup$ – Wojciech Morawiec Jul 5 '17 at 19:26
  • 1
    $\begingroup$ Neutrinos are weird because they don't have a definite mass due to neutrino flavor mixing; rather, they exist as a superposition of mass states. $\endgroup$ – probably_someone Jul 5 '17 at 20:00
  • 1
    $\begingroup$ The coupling strengths of gravity are dimensionless quantities, just as in electromagnetism or the other interactions. These dimensionless couplings are masses measured in Planck mass units. It is unclear what puzzles you about that answer or the standard theory. $\endgroup$ – Cosmas Zachos Jul 5 '17 at 21:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.