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I have read here, that $\frac{1}{2}mv^2$ must not be applied on a photon ever.

If i want to calculate escape velocity $v_e$ i need to use $\frac{1}{2}mv^2$ because we say that kinetic energy (positive) must be same or larger than gravitational potential (which is negative) in order for an object to escape. It is done like this:

$$ \begin{split} W_k + W_p &= 0\\ \frac{1}{2}mv_e^2 + \left(-\frac{GMm}{r}\right) &= 0\\ v_e &= \sqrt{\frac{2GM}{r}} \end{split} $$

Than we say if there is a black hole inside certain radius we call Schwarzschield radius $r=R_{sch}$ not even light can escape because its escape velocity is smaller than needed to escape. At the border of the sphere with $R_{sch}$ light can barely escape, so it must hold that $c$ equals escape velocity $v_e$. So we write down the equation below and derive $R_{sch}$.

$$ \begin{split} c &= \sqrt{\frac{2GM}{R_{sch}}}\\ R_{sch} &= \frac{2GM}{c^2} \end{split} $$

This is a well known equation, but it is derived allso using $\frac{1}{2}mv^2$ for light (photons). This is in contradiction with 1st statement in this post. So is the last equation even valid???

Frederic Brünner
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71GA
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2 Answers2

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The short answer is: no, $\frac{1}{2}mv^2$ is never valid for photons. A photon's energy is given by $$ E = h f = \hbar \omega = \frac{h c}{\lambda} $$ always. The derivation of the Schwarzschild radius you mention is an incorrect one that happens to give the right answer accidentally. The correct derivation requires general relativity.

Michael
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This is one of those cases when the ends do not justify the means... Just because you get a result that is true, from laws that aren't supposed hold in that situation, doesn't mean that the laws can be used there.

If you're asking about whether there is a deeper connection between using $\frac{1}{2}mv^2$ and getting the radius, to the best of my knowledge, there is none: it just happens to be a coincidence that they match.

Like Michael Brown said, the Schwarzschild radius can only be derived from GR, in which the Schwarzschild metric \begin{align} ds^2 = - (1-2GM/r) dt^2 + (1-2GM/r)^{-1}dr^2 + r^2d\Omega^2 \end{align} is the unique maximally symmetric solution to Einstein's equations in vacuum. (I've set $c = 1$) here. Note that something interesting is happening when $r = 2GM$, and one can go on to show that $r = 2GM$ happens to be the boundary of the causal part of the future null infinity, which we identify as the event horizon.

nervxxx
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