13

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.

Deschele Schilder
  • 1
ByoTic
  • 241
  • 3
    For known particles the gravity force is much weaker then electromagnetic force. They would be roughly equal for hypothetical particles with 1 electron charge and about $3.5\times10^{15}$ protons mass – Peter Sep 29 '20 at 23:19
  • 6
    Here's a simple demonstration. Get yourself a decent fridge magnet. Now use it to pick up a paper clip. Congratulations, you've just overcome the gravitational force of all 6e24 kg of the Earth with the electromagnetic force of 100 grams of cheap magnet. – Schwern Sep 30 '20 at 08:14
  • 1
    +1 I might understand the typical arguments, such as those in the answers here, if mass were quantized, and there was a universal "mass unit" in the same way charge is quantized and we have a universal "charge unit". But I've never heard of mass being quantized. – garyp Sep 30 '20 at 12:45
  • @Schwern: I get your point, but isn't it a little misleading because, on average, the 6e24 kg is at a distance of 4000 miles from the paperclip? I wonder how much mass would be required to overcome the magnet if the mass was concentrated a few millimeters from the paperclip? – James Sep 30 '20 at 12:47
  • 2
    @Schwern: that's not an entirely fair comparison, because most of the earth's mass is far away. For example, if you jump up, at noon, then you'll be jumping towards the sun... and then you'll fall back down to earth. So then you might say that the puny little rock that we call home somehow overcame the gravitational force of the entire sun! – acdr Sep 30 '20 at 13:48
  • @James Why did not you calculate this? Assuming 1 cm for the magnet paperclip distance r$^-2^ is 36$\cdot 10^{16}$ times larger, so a sphere of 17 million kg with radius smaller than 1 cm at a distance of 1 cm would do the job. – my2cts Sep 30 '20 at 14:24
  • @my2cts: Thanks for the calculation. I didn't do the calculation because I wasn't sure how. By the way, your equation is garbled (at least on my screen). I don't think this site is interpreting it correctly. – James Sep 30 '20 at 14:28
  • "Usually, they justify it by comparing the universal gravity constant to Coulomb's constant." I've never seen anyone doing that. (If they did, masses being in different units from charges would also compound it.) I've only seen them compare the two forces on a given pair of bodies, as you describe, or derive the fine-structure constant & its gravitational counterpart, as @probably_someone describes. – J.G. Sep 30 '20 at 14:36
  • @James Be bold. Edit: Assuming 1 cm for the magnet paperclip distance, r$^{−2}$ is 36$\cdot 10^{16}$ times larger than for an Earth radius, so a sphere of 17 million kg with radius smaller than 1 cm at a distance of 1 cm would do the job. – my2cts Sep 30 '20 at 15:46
  • @garyp mass may not be quantised, but rest mass is! The smallest amount of rest mass is the mass of one of the neutrinos. – Andrea Oct 27 '20 at 10:40
  • @Andrea Are you saying that because the neutrinos are the lightest known particles? Does that imply the the mass of the electron is $m_e = nm_\nu$ where $n$ is an integer? – garyp Oct 27 '20 at 12:17
  • @garyp yes, as far as we know you can’t have arbitrarily low rest mass. No, it does not imply that all masses are multiples of the neutrino mass, like it is not the case that all angular momenta are multiples of hbar. – Andrea Oct 30 '20 at 17:03
  • Related: https://physics.stackexchange.com/q/755038/226902 , https://physics.stackexchange.com/q/145518/226902 – Quillo Mar 13 '23 at 09:16

3 Answers3

25

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.

probably_someone
  • 35,459
  • 1
    This link http://hydrogen.physik.uni-wuppertal.de/hyperphysics/hyperphysics/hbase/forces/couple.html gives a smaller value for gravity's coupling constant – anna v Sep 30 '20 at 03:57
  • Is it then fair to say that we can equally view the situation as: why are the masses of fundamental particles so much lower than their charges? I.e. we can either view this from the forces themselves, or the properties of the particles – gardenhead Sep 30 '20 at 15:17
  • @gardenhead Not quite. The reason is the equivalence principle: the mass of a fundamental particle is both the "gravitational charge" and the rest energy. It's totally possible for there to be a mechanism entirely separate from gravitation to fix the rest energies of the particles, which therefore also fixes the "gravitational charges". This would leave the question unanswered as to why the relation between "gravitational charge" and gravitational force is so weak. – probably_someone Sep 30 '20 at 15:27
  • @annav As I said, it depends on which reference you're comparing. Theirs is roughly 6 orders of magnitude above mine, because they use the force between two protons, each of which is 3 orders of magnitude more massive than an electron, with the same electric charge. – probably_someone Sep 30 '20 at 18:59
  • 1
    that clears it . thanks. Teaches me to look further then statements in links I trust – anna v Oct 01 '20 at 03:13
1

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

cmaster - reinstate monica
  • 4,003
  • 1
    This was addressed in the question, and could just as well be seen as a comparison of the charge of an electron to its mass. It does not, fundamentally, compare the forces. The question is whether a "true comparison of the forces" is something that even makes sense. – Arthur Sep 30 '20 at 13:41
  • @Arthur The answer does compare the forces . It does not compare the fields, which anyway cannot be done. – my2cts Sep 30 '20 at 14:28
  • @my2cts Which fields are you talking about? – Deschele Schilder Sep 30 '20 at 16:46
  • @my2cts About which fields are you talking (typing)? – Deschele Schilder Sep 30 '20 at 22:14
-4

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 $6,2\times 10^{18}$ (see here), and for one kilogram, well, 1:

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

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

Deschele Schilder
  • 1