An electron has a mass of $9.10938291(40) \times 10^{−31} kg$. It also has a volume of $0 m^3$. This would imply it has infinite density. Shouldn't that make it collapse into a black hole? Why doesn't it?

# [Physics] Why don’t electrons collaspe into black holes?

black-holeselectrons

#### Related Solutions

You are asking a wrong question. Here is the problem with your reasoning.

You are assuming a Schwarzschild metric and a homogenous distribution of mass. But the Schwarzschild geometry describes a vacuum spacetime. So you can't use it for a spacetime filled with matter. For a cosmological spacetime filled with matter, like our universe, the suitable metric to use would be something else, like the FRW for example.

You could only use the Schwarzschild spacetime if you assumed a sphere of some uniform density $\rho$ and vacuum outside the radius of the sphere.

Let me illustrate how things would work out then. As you can see, a particular density corresponds to a particular $R_s$, lets call it $R_s(\rho)$. So if you had a sphere of matter with a radius $R_1$ grater than $R_s(\rho)$, then you couldn't apply the formula $R_s(\rho)=c\sqrt{\frac{3}{8\pi G \rho}}$. You would have to use the Schwarzschild metric only in the vacuum region outside of the sphere. So you would have then $R_s=\frac{8\pi G\rho R_1^3}{3c^2}$. In order to see how the $R_s$ compares with $R_s(\rho)$, you can replace the density with $\rho=\frac{3c^2}{8\pi G R_s(\rho)}$. So you would get that the Schwarzschild radius for a sphere of uniform density $\rho$ and radius $R_1>R_s(\rho)$ is $R_s=\left(\frac{R_1}{R_s(\rho)}\right)^2R_1$, which is grater than the radius of the sphere. So the sphere is inside its Schwarzschild horizon. If on the other hand, the radius $R_1$ is smaller than $R_s(\rho)$, then the corresponding horizon would have to be inside the sphere. But inside the sphere the Schwarzschild metric doesn't apply. So it isn't necessary that there should be a horizon inside the matter distribution.

If you apply these to the universe and assume for example that the radius of the visible universe is the radius $R_1$ of the sphere, then you would have a horizon radius (using your numbers) that would be almost 10 times the radius of the observable universe. So, the entire universe would have to be in a black hole of radius of 460 billion light-years. So the assumption that we should see black holes with horizons of radii of 13.9 billion light-years is not correct.

If one assumes the above point of view, one could say that the universe is a white hole that is exploding.

I hope that all these are helpful and not confusing.

## Best Answer

The angular momentum and charge of an electron are both large enough that a black hole would not form. If you believe classical general relativity all the way down to the scale of an electron (and you really shouldn't), then the electron will form a naked singularity.

More exactly, for the case of a spinning body, the horizon is at the zero of

$$r^{2} - 2Mr + a^{2}$$

Where $a$ is the angular momentum divided by the mass (in $G = c = 1$ units), and $M$ is the black hole's mass. If you put in the numbers for the electron, this equation has no real zeroes. Adding charge to the picture will only make the matter worse.