# How can one fundamental force be stronger than another?

This doesn't seem to make any sense. Gravity depends on mass or spacetime curvature but the electromagnetic force depends on electric charge. For any particular particle we could say that the force of gravity is weaker than the electric force. But for a particle of arbitrary mass and charge, neither force is stronger. What does it mean when a physicist says the X force is b times stronger than the Y force. It's probably already clear what I'm saying, but here's my analogy. Saying gravity is weaker than the EM force is like saying that perimeter is greater than width. They are both the same kind of measure but they depend on different aspects of the same object.

• It makes sense once you appreciate the technical term "relative strength'. Can you express your question in language consistent with WP? Jan 28, 2020 at 20:11
• Jan 28, 2020 at 20:19
• This question basically reduces to calculating Coupling Constants for the Fundamental Forces Jan 28, 2020 at 20:30
• Of course, you could argue that for any 2-d object in euclidean space, its perimeter will always be greater than its width Jan 29, 2020 at 22:28

For any particular particle we could say that the force of gravity is weaker than the electric force. But for a particle of arbitrary mass and charge, neither force is stronger.

Actually, the electrostatic force between two chargeless neutrons would be 0, so, weaker than gravitation—but they still have nonvanishing magnetic moments, whose long distance interaction lags by two powers of r the Coulomb power feature assumed in the comparison. (Astrophysicists probably use the gravitational features of neutral neutrinos, as well, but I doubt that their masses enter directly and significant, there.)

You might appreciate that the concept of relative strength comes for particle physics, and represents a comparison of dimensionless couplings, times an integer charge eigenvalue (or fraction thereof), except for gravity, where the charge×coupling is an arbitrary particle mass over the Planck mass.

The reason is the charges of SU(3), SU(2)xU(1) and U(1) for the standard particle interactions are related to the integer/fractional indices of the extant low lying representations; the electron charge is an evidently fine zero-point. Whereas there is no rhyme or reason for the hierarchy of masses we are familiar with. (But people are trying to figure it out. Don't hold your breath for a mainstream theory, though.)

In that sense, your question is valid and there is no answer to it. An unregenerately moot issue, where "...the mere asking of a question causes it to disappear and merge into something else." (EMF)

Normally, however, people focus an a particular particle, and use its mass and charge for the charges multiplying the couplings. The electric charge is one-ish. They may use the electron mass, or the proton mass, and, perhaps perversely, the highest mass around, the top mass.

They are in ratios of 0.51 : 938 : 173 000 . This gives a factor of 340 000 as slop in the masses, so a slop factor of the square of that, eleven orders of magnitude, in the ratio of forces, validating your point, a little.

But we are talking a force ratio of 40 orders of magnitude here... (Recall the proton mass, our zero point here, is close to a.u., the mass yardstick for constituents of matter, so, then, in the Coulomb regime, $$F_g/F_e=(m_p/m_{Pl})^2/\alpha = 137\cdot (0.938/1.2)^2 \cdot 10^{38}\approx 0.84 \cdot 10^{40}$$.)

The other two interactions lack the infinite range feature allowing EM-Gravity comparisons, so you have to fix a short standardized distance scale to make such comparisons meaningful (cf. WP) and not let the apple to papayas comparison overwhelm you.

Like many of such educational tools, this type of comparison is a lark to drive home the point. By the time you've parsed out what it means, it has lost its meaning, while proving its utility.