# Why do inverse square forces not rise asymptotically for $r\to 0$?

An example of an inverse square force is the gravitational force:

$$F = \dfrac{G m_1 m_2}{r^2}$$

For large bodies I can understand why the force does not rise asymptotically as you get closer. For example if I drill into a planet of uniform density, the gravitational force will actually decrease according to a linear relationship (as shown by application of Newton's Shell Theorem). But at the subatomic level, will the gravity due to a single elementary particle (an electron perhaps)rise asymptotically as you approach it? I have read that subatomic particles are in fact not spheres as we generally perceive them in diagrams, but if they do not have definite shape, (i.e. if they are points) then the field must increase asymptotically as you approach? So does this actually happen, or is there something wrong?

-the same logic can be applied to other forces, such as the electrostatic force, which also follows an inverse square law.

## 1 Answer

This logic when applied to the electrostatic force of 1/r^2 led to the discovery of quantum mechanics. When nuclei and electrons around them were visualized, because that is what the data said, the big question was "why the electrons do not drop on the nucleus and the whole thing becomes neutral"?

An electron dropping into the nucleus would classically emit continuous radiation, but the radiation spectra observed from atoms were quantized. Thus the Bohr atom was proposed , where the electrons had only stable orbits . Further observations and studies led to the development of Quantum Mechanics.

In the hierarchy of fundamental forces the gravitational force is very much weaker than the electromagnetic one , about 10^-37 or so. Thus the gravitational field an electron sees in the lowest orbit around hydrogen is very very weak, and it can never reach the proton to test gravity because it is constrained in its stable orbit quantum mechanically.

In scattering experiments, e- on proton when a lot of energy is supplied and the quantization of the atom is orders of magnitude away from the energy supplied, the electron gets close enough but still strong and electromagnetic interactions dominate by orders of magnitude and the gravitational effect is not measurable. Instead lots of new particles will be produced by the excess energies provided in the experiment.

Neutrinos are chargeless and have a small mass. Could two neutrinos approach each other at distances that will allow the gravitational interaction to dominate with its 1/r? This is a moot question explored here .

In conclusion quantum mechanics saves atoms from gravitational collapse too. It is also quite probable that once a unified theory with quantized gravity is established the gravitational singularity at r=0 will also theoretically be defused.

• Umm... electrostatic force of 1/r??? What on earth...?? Am I missing something? – N Unnikrishnan Oct 17 '15 at 19:32
• @NUnnikrishnan you caught an error, should have said potential. thanks – anna v Oct 18 '15 at 4:44
• I didn't have the guts to pull off a correction with you. :) But now, the question remains for the electrostatic case - does the electrostatic force grow asymptotically for two free particles? Of course, we can dodge the question by appealing to the quantum picture, but is there an explanation possible in the classical setting? – N Unnikrishnan Oct 18 '15 at 17:37
• Yes it grows without limit, it would be a singularity for point particles . That is why quantum mechanics comes to the rescue. – anna v Oct 18 '15 at 18:20