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Comparing the strength of gravity and electricity

Physics Asked by ByoTic on December 31, 2020

I often see and hear people claiming that "the gravitational force is much weaker than the electromagnetic force".
Usually, they justify it by comparing the universal gravity constant to Coulomb’s constant. But obviously, such comparison is meaningless, as they differ in dimensions.
I’ll make myself clear: of course you can say it is true for electron-electron interaction, but I’m talking about whether they can be compared fundamentally somehow in any area of physics.

3 Answers

Yes, they can. Both interactions can be modeled using perturbative quantum field theory, where their strength is parametrized by a dimensionless coupling constant.

Electromagnetic repulsion between two electrons can be written as a power series in $alpha$, the fine structure constant, which is dimensionless and has a value of roughly 1/137.

Meanwhile, the gravitational attraction between two electrons can be expanded in a similar way in a power series in $alpha_G$, which is a dimensionless constant with a value of roughly $10^{-45}$.

The precise value of $alpha_G$ depends somewhat on which particle you're comparing, since ultimately it's the square of the ratio of the particle's mass to the Planck mass. However, for fundamental particles, this ratio does not vary by more than ten orders of magnitude, which still places $alpha_G$ far smaller than $alpha$ no matter which fundamental particle you choose to compare.

Correct answer by probably_someone on December 31, 2020

You may not be able to compare the constants directly, but you can compare the resulting forces. For instance, when you put two electrons $1mathrm m$ apart, you can calculate their gravitational attraction and their electrostatic repulsion. Guess which of the two forces dwarves the other...

Answered by cmaster - reinstate monica on December 31, 2020

Welcome to the Physics SE site.

The forces can be compared on a fundamental level with reference to the universal gravitational constant and the Coulomb constant.
These constants have different units indeed but this is only logical as these constants deal with mass ($kg$) and charge combined with mass ($C$ and $kg$, as will become clear). Only the values of these constants are compared.

The universal gravitational constant:

$$frac{m^3}{s^2kg},$$

And the Coulomb constant:

$$frac{Nm^2}{C^2}=frac{m^3kg}{s^2C^2}.$$

Both constants give $F=ma$ if we fill them in in the corresponding formulae of the corresponding forces (the Coulomb force and Newton's law of gravity, taking $C$ for the charges $q_1$ and $q_2$ in the Coulomb force and one kilogram for the masses in Newton's law).

Obviously, these units have a factor of $frac{m^3}{s^2}$ in common. So what we actually compare are $frac{1}{kg}$ and $frac{kg}{C^2}$, which means only the cause of the electric field ($C$) and the cause of the gravitational field ($kg$) are involved in the comparison. So this leads to (setting $frac{1}{kg}$ equal to $frac{kg}{C^2}$) $kg^2=C^2.$

Comparing the two (though units differ, we compare the values), and filling in for the Coulomb $6,2times 10^{18}$ (see here), and for one kilogram, well, 1:

$$1=38,4times 10^{36}$$

We have to divide the right side still by $4pi$ which means that the electric force is about $10^{36}$ times as strong as the gravitational force, as it should be.

Answered by Deschele Schilder on December 31, 2020

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