Obviously, mass comes in elementary chunks, like all elementary charges (electric, color, and assumed weak). All elementary particles have an inherent mass. Unlike charges though, these masses are not rational multiples of one another (which is why I think they are not truly elementary, but that's another question). In calculating the ratio between the electric and gravitational force one usually calculates this by reference to two electrons. See for example this question. The ratio, in this case, is about $10^{42}$. Dirac used for this ratio a proton and an electron, which gives about $10^{39}$. See this video, in which he explains his large number hypothesis: in Natural units, the age of the universe is about $10^{39}$ too, from which he (in my eyes wrongly) concludes that at the beginning of the universe gravity was much stronger (or the electric force weaker?).
But why don't use the mass of one of the quarks? Or that of a neutrino? Or even a photon? What is the real ratio? It's clear that is a large number.
Can we even compare the strengths of two fundamentally different kinds of interaction (one mediated by spacetime curvature and one by part particle exchange)?

  • 1
    $\begingroup$ A note, masses of the neutrinos are not known. And mass of the photon is zero. $\endgroup$
    – Triatticus
    Commented May 23, 2021 at 16:17
  • $\begingroup$ @Triatticus But the mass of the neutrino is close to zero for sure. Indeed the mass of the photon is zero. Which would mean that gravity due to mass is zero... $\endgroup$ Commented May 23, 2021 at 16:23
  • $\begingroup$ What is the real ratio? There isn’t one. $\endgroup$
    – G. Smith
    Commented May 23, 2021 at 17:27

1 Answer 1


When you are dealing with ratios on the order of $10^{42}$, does it really matter if you use $m_e$ or $M_p$? (on the order of $10^3$)?

The reason to not use proton/anti-proton is two-fold: we don't make bound states of them, and, they interact strongly.

Quarks are even worse, as they don't exist in a free-state. Nevertheless, you could consider the energy required to hold $udd$ together with-in a range of 1.7 fm (the nucleon diameter), and then compare that with the mass of neutron minus the small quark masses.

But this is about gravity, not the strong force. One uses electrons because they are color neutral, and only interact via EM, gravity, and the weak interaction (which is negligible at low energy). You can look at positronium's energy levels and go from there. (You can also compare E and M separately via ortho and para positronium).

It may be more natural to look at the work-horse of bound states: the hydrogen atom, via:

$$ G\frac{m_eM_p}{a_0^2} \ne \frac 1 {4\pi\epsilon_0}\frac{e^2}{a_0^2}$$


$$a_0 = \frac{\hbar}{m_ec\alpha} $$

is the Bohr radius.

Photons don't have mass, so there's no Newtonian gravity, but that is only an approximation. Spacetime curvature ($G_{\mu\nu}$) is due to stress-energy ($T_{\mu\nu}$) and dark energy $(-\Lambda g_{\mu\nu})$. Whether two parallel photons travel in parallel lines would be a good PSE question. Enough photons in a small enough volume yields the kugelblitz black hole, after all.

Regarding the weak force, it could compare $ee$ scattering with $\nu\nu$ scattering. Bound states are discussed here: Can the weak force create a bound state?.

It is indeed difficult to create a formal definition of the relative strength of gravity and EM, as there is no fundamental unit of mass, while the color force is complicated by a lack of free-states, and the strong force is residual. Really only the weak and electromagnetic forces have a formal (energy dependent) relation provided by electroweak unification. The other comparisons are back-of-the-napkin stuff.

  • $\begingroup$ Exactly the question about two photons (actually about the Nature of the gravitational field of a beam of light, the Bonner beam), I was about to ask! $\endgroup$ Commented May 23, 2021 at 19:39
  • $\begingroup$ What if the fundamental unit of mass were zero (say that three massless particles make up the elementary particles)? $\endgroup$ Commented May 23, 2021 at 19:41

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