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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???

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$\frac{1}{2}mv^2 = \frac{p^2}{2m}$ isn't even generally correct for massive particles: in only applies where $v \ll c$. For high velocities it is mandatory to use the correct relativistic expression $\sqrt{m^2c^4 + p^2c^2} - mc^2$. Obviously $c \not\ll c$ so you get $\sqrt{0 + p^2c^2} - 0 = pc$. –  dmckee Jan 18 '13 at 15:38
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It's a random coincidence. In particular, your derivation doesn't change if the "black hole" is spinning, but in full general relativity, you'll have $R = \frac{G}{c^{2}}\left(M + \sqrt{M^{2} - \frac{J^{2}c^{2}}{G^{2}M^{2}}}\right)$, where $J$ is the angular momentum of the black hole. –  Jerry Schirmer Jan 18 '13 at 16:23
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Indeed it is coincidence. See also my answer to this question –  Chris White Jan 18 '13 at 17:13
    
Our professor derived it like this. Furthermore he didnt even warn us the derivation is WRONG! But i am amazed! What a coincidence indeed :) –  71GA Jan 18 '13 at 19:21
    
@71GA Congrats. I never managed to catch my professors on any errors more serious than flipped signs or the like. :) –  Dan Neely Jan 18 '13 at 20:49

2 Answers 2

up vote 5 down vote accepted

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.

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Missing those factors of 2 perhaps? –  Chris White Jan 18 '13 at 17:10
    
ahha yes. i'll edit my post, thanks –  nervxxx Jan 18 '13 at 17:12

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.

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