1. Jul 2, 2011

RK1992

can anyone say why the derivation works? my teacher went through it in class and sort of said "don't question it" (which i hate) but it's still annoying me now even though it's a few weeks since i finished college.

KE = GPE
0.5mv² = GMm/r
r=2GM/v²

and then if the escape velocity is the speed of light, then the radius of the region of space where you cant escape is given by r=2GM/c²

but the equation KE=0.5mv² is just valid for small values of v and can be obtained from the expansion of m(γ-1)c²... m(γ-1)c² is not defined for v=c, though, so why does the maths turn out nicely when its so clearly wrong to use KE=0.5mv²?

thanks :)

2. Jul 2, 2011

Subductionzon

Even though you should be using relativity to figure this out, which I never learned, it just so happens that if you do use classical mechanics you do get the right answer. I derived the same answer when I was in college. I knew the methodology was wrong and I was surprised when I got the correct answer. So I see two possibilities, either your professor, like me, found the correct answer but did not know why it was correct, in other words he was not a physicist, or he did not want to go through the relativistic derivation that would have been above the class' ability to understand.

3. Jul 2, 2011

Pengwuino

The derivation of what the Schwarzschild radius is that I have seen basically compared the time-time components of the metric in a Schwarzschild background with the classical newtonian potential in a metric theory with a small perturbation from flat spacetime. So this assumed the test particle was non-relativistic and a fair distance away so the newtonian potential was the correct limit to compare to.

4. Jul 2, 2011

RK1992

yeah, this seems much too complicated to explain to a level students ^^ :tongue:

5. Jul 2, 2011

Pengwuino

Basically, you setup a relativistic theory and you look at how that relativistic theory looks in the non-relativistic limit. The small perturbation above flat spacetime just means that you're looking at something like the earth-sun where the distances are so great and the center objects mass is so small that relativistic effects can be ignored.

Another thing to realize is that mass is not so simple in general relativity. The M in $R=2GM/c^2$ is not the same mass as in $F = GmM/r^2$. I haven't gone into GR enough to actually speak of how you approximate the newtonian mass from the GR mass though.

6. Jul 2, 2011

RK1992

heh, cant wait to study it all properly. thanks :)