Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I often hear that gravitational force is much "weaker" than electroweak and strong forces. But how can you compare the strength of interactions without the parameters like mass, charge on which it depends? Surely the gravitational force can be made as (practically) strong as one wants by increasing the mass?

share|cite|improve this question
In a sense you are right: were you able to excite and scatter particles with mass around M_pl, then the gravitational interaction would be as important as the other forces. The point is that he mass spectrum involve things like the electron though, that is many order of magnitude below M_pl, whereas the electric charges are quantized and below 1/3 there aren't any. – TwoBs Jun 26 '14 at 19:18
Related: , and links therein. – Qmechanic Jun 26 '14 at 21:02
up vote 9 down vote accepted

For forces that can be expressed in terms of a quantum field theory, you can compare the size of the coupling constants (which are dimensionless). In short this means that perturbative expansions of weaker forces are well represented by a small number of leading terms because the series converges quickly, while those of stronger forces require more terms (or in the case of the strong interaction don't converge at all in the naive application).

This spares us the philosophical quandary of comparing dimensional coefficients in force laws.

So far, however, there is no quantum field theory of gravity. But it is possible to work out the scale of the gravitational coupling constant by observing scattering problems for which there is a close analogy in E&M. I suppose that this in effect this means that we are comparing the force-law coefficients written in natural units ($c = \hbar = G = 1$).

share|cite|improve this answer
There is a field theory of gravity. It's general relativity. – Matt Reece Jun 26 '14 at 22:01
Quantized is implied I believe. – BMS Jun 26 '14 at 22:15
@Matt As BMS says, I'm using the sloppy language of particle physics where "field theory" is too often used when "quantum field theory" is meant. I'll fix it. – dmckee Jun 27 '14 at 6:46
General relativity works perfectly well as a quantum field theory of gravity as well. For instance, there are completely finite calculable quantum corrections to Newton's law that are insensitive to the UV completion. – Matt Reece Jun 27 '14 at 10:20
It seems that the dimensionless coupling constant is more of a measure of "scale" than "strength" of the interaction... – ashpool Jun 27 '14 at 11:40

You compare the size of the constants in the force law. For example $$F_g = \frac{GMm}{r^2},$$ and $$F_E = \frac{kQq}{r^2}.$$ In SI units $k= 9 x 10^9$, while $G = 6.67 x 10^{-11}$. That means to have an equivalent force, you have to assemble a mass twenty orders of magnitude (in SI units) larger than a collection of charges.

Basically, this, along with the fact the electric force has both negative and positive charges is responsible for most of the interactions we see in everyday life, and in astronomy. You wonder why gravity is the dominant force on large scales? It's because, even though it's smaller than the E/M force, you cannot create a large collection of like charges because they all repel each other while you can assemble large massive bodies. However, on small scales, like the atomic scale, a few charges are enough to create molecules without being pulled together by their mutual gravity.

share|cite|improve this answer
Are the units of k and G the same? – Andrew Jun 26 '14 at 18:27
@jhobbie then how do you compare ? Basically, then the ratio can vary numerically.(?) if units are changed. – Isomorphic Jun 26 '14 at 18:31
Scales. All scales where the E/M force has meaning it is larger. That's the point of my second paragraph is to iterate that you can have more force from gravity. At different scales, there's a particular force that's dominant. The only way you can really achieve a stronger gravitational force is by adding up mass so that it is neutrally or near neutrally charged. It is extremely difficult to find something that is big enough to have a gravitational effect on us, but also charged enough to have an E/M force. – jhobbie Jun 26 '14 at 18:35
I still don't understand. The reason the EM force between protons is greater than the gravitation force between protons is due to the accident that the masses and charges of the particles have the values they do. Why is it said that gravity is inherently weaker? How is the strong force "stronger" than the EM force? There must be some way to compare the forces on some equal footing. – garyp Jun 26 '14 at 19:10
It seems to be what you're asking is "Does mass have any effect on charge" to which the answer is no. You have to look at things contextually. The strong force has the most noticeable effect at certain distance scales, the E/M force at other scales, and gravity at the largest scales. Why? Well, I'm not sure I understand the strong force, but for the E/M force to be stronger than the gravitational force, you require either a small build-up of mass or a large build-up of charge. The second is impossible, but the first happens all the time at atomic scales. Comparing forces is all about context. – jhobbie Jun 26 '14 at 19:17

The proposition ''gravitational force is much weaker than electroweak and strong forces'' is wrong , at least if you mean is ALWAYS weaker. For example, for a charged objects like an electron you can simply make the ratio between the force of gravity $ F_g=\frac{GmM}{r^2} $ and the electric force $ F_e=\frac{kqQ}{r^2} $. You find a difference of $ \simeq 40 $ orders of magnitude. But the result is inverted for a very massive object or for a very energetic one.

The point is that you don't have some adimensional coupling constant to compare, like with the other forces. So the comparing is energy dependent (or equivalently mass dependent).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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