# Relativistic kinetic energy has no upper bound, so why is there a Schwarzschild radius?

Schwartzschild radius is the distance from the center of a body at which, the escape velocity will be equal to the speed of light, i.e. when $$\frac{2GM}{R} = c^2.$$ However, here it is assumed that the form of kinetic energy is $$\frac{1}{2} mv^2$$, which is bounded above by $$\frac{mc^2}{2}$$.

However, when the speed is high, we should use the $$(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} -1) m {c^2} = \frac{GMm}{r},$$ which will always have solutions for $$v_{escape} for any radius and mass.

So, why is there an event horizon? It seems that a particle can have enough kinetic energy to overcome the gravitational potential of a black hole.

I have not studied general relativity yet, and here I have assumed that the classical formulation of finding escape velocity (kinetic energy > potential energy) is still valid in general relativity, and the classical formula $$GMm/r$$ still holds.

• Are you using the classical $GMm/r$ as the potential energy of a small mass $m$ near a black hole with mass $M$? A good answer to your question would address this misconception. – rob Nov 27 '18 at 14:10
• Yes, I am using the classical formula for potential energy – Archisman Panigrahi Nov 27 '18 at 14:11
• Long story short, the Schwarzschild radius is not actually derived in the way that you present here. It's merely a coincidence that it matches up with the Newtonian-gravity expression for the radius with escape velocity $c$ (and if I remember correctly, the Newtonian expression actually has a slightly wrong numerical factor in front). – probably_someone Nov 27 '18 at 14:24
• @probably_someone: That should be an answer. – Ben Crowell Nov 27 '18 at 14:35
• Relativistically, if you take a material particle and boost it to a very high kinetic energy, all you're doing is giving it a very high ratio $E/m$ of energy to mass. You can make $E/m$ approach infinity, but $E/m$ is infinity for light, so all you're accomplishing is to make your particle behave just like a ray of light. – Ben Crowell Nov 27 '18 at 14:37

If we apply classical mechanics to a point particle with a large mass $$M$$ and another point particle with small mass $$m$$ such that $$m << M$$, we find (by using Newton's law for gravity) that the potential energy $$U$$ of the system is given by $$U = -\frac{GMm}{R}$$. The particle escapes when the total energy (Kinetic + Potential) equals zero, therefore the escape velocity $$v$$ must be given by $$v = \sqrt{\frac{2GM}{R}}$$ And for this we assumed that the kinetic energy is given by $$\frac{mv^2}{2}$$. If we want the escape velocity to be $$c$$, we'll have $$c = \sqrt{\frac{2GM}{R}}$$, therefore: $$R = \frac{2GM}{c^2}$$ This implies that the event horizon has radius $$R = \frac{2GM}{c^2}$$.

I suppose that your question is based on the argument above. What you meant to do was correct the argument by introducing relativistic kinetic energy. Then you found that kinetic energy has no upper bound and can have any positive value, and this appears to violate our result.

Where lies the problem? Well, the above argument is flawed, and fundamentally incorrect. A few points:

• We used a classical potential energy formula. This means that our field is newtonian, and newtonian gravitational fields violate Relativity. Relativistic gravity must be studied using General Relativity.

• There is no upper bound to potential energy in classical mechanics, since Newton's laws allow a particle with nonzero mass $$m$$ to have any velocity. This also violates Relativity, but it should be noted that both Relativity and Classical Mechanics predict that kinetic energy may have values as large as we want. We can't just say "oh, Relativity says $$v < c$$, then classical kinetic energy has an upper bound of $$\frac{mc^2}{2}$$", this makes no sense.

• Since kinetic energy is not bounded by an upper limit, classical or otherwise, the entire argument falls apart.

But if the whole argument is wrong, then why does it give the correct formula for the Schwarzschild radius?

Well, this is just a coincidence. The derivation is completely wrong, but it does indeed, by coincidence, give the right answer.

Then how is the Schwarzschild radius actually derived? How should I study this gravitational system relativistically?

Well, that's where General Relativity comes in. Your attempt to create relativistic corrections to the gravitational field fails, and the right way to do it is using General Relativity. In this case, considering total vacuum, no rotation and electric charge, you'd look for the Schwarzschild solution to Einstein's equations of General Relativity. Look up the Schwarzschild metric. (https://en.wikipedia.org/wiki/Schwarzschild_metric)

Using General Relativity and the Schwarzschild metric we find the same result for the Schwarzschild radius as the classical derivation shown above, but this is just a coincidence.

For example, we could change the scenario: make the mass $$M$$ have some size and let it be rotating. This time, classical mechanics gives the same event horizon radius, but the actual answer is different. (https://en.wikipedia.org/wiki/Kerr_metric)

As mentioned in the comments , the motivation for defining the Schwartzschild radius had nothing to do with escape velocity. It is just the distance from a central body where the Schwartzschild metric, expressed in Schwartzschild coordinates, becomes singular. For normal celestial bodies this did not matter since the radius lays inside the body(for the Sun it is 3.0 Km) , so the Schwartzschild solution did not apply there (it was supposed to apply to a spherically symmetric region of empty space ), but later people started speculating the existence of bodies so dense that the Schwartzschild radius actually ends up being in empty space and so it started to acquire importance. By chance it coincides with the distance found by John Michell in the 18th century where the escape velocity coincides with the speed of light.

You can write the energy, including the kinetic and potential part in general relativity as:

$$E=mc^2\left(\frac{\sqrt{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{v^2}{c^2\left((1-\frac{2GM}{rc^2})^2(\hat{r}\cdot\hat{v})^2+(1-\frac{2GM}{rc^2})|\hat{r}\times\hat{v}|^2\right)}}}\right).$$

This can also be written as:

$$E=mc^2\left(\frac{{1-\frac{2GM}{rc^2}}}{\sqrt{1-\frac{2GM}{rc^2}-\frac{v^2}{c^2\left((1-\frac{2GM}{rc^2})(\hat{r}\cdot\hat{v})^2+|\hat{r}\times\hat{v}|^2\right)}}}\right).$$

($$\hat{r}=\bar{r}/r,\hat{v}=\bar{v}/v$$)

As it is written here you can actually see that you need infinite energy to move at all at the Schwarzschild radius of $$r=2GM/c^2$$. You can also see that the energy of a body at rest at the Schwarzschild radius becomes zero. This is all for the case of a spherically symmetric gravitational field.

There is a Schwarzshild radius because the expression for the energy looks as above is one way to look at it.