I'm confused here and hoping someone can give me a straightforeward answere... the derivation of the schwarzschild radius comes from the fact that at the event horizon the gravitational potential energy is equal to the kinetic energy it had went in left the singularity... i know this is a classical method and quite simple but it works and that is 1/2 mV^(2) = GMmr^(-1) r = 2GMV^(-2) in this case V = C so; r = 2GMV^(-2) which is correct... but if i try to derive it in terms of forces i'm off by a factor of 2... the force of attraction between to massive bodies due to gravity is F = GMmr^(-2) and the centripetal force of an object in orbit is F = mV^(2)r^(-1) imagining a partical orbiting a singularity at the speed of light... GMmr^(-2) = mV^(2)r^(-1) and V = c so; r = GMC^(-2) which is off by a factor of 2... my quesiton is simply, why?
Actually, your first classical calculation is what's off by a factor of 2, but the second one is valid in the classical regime. It looks like you attempted to apply the Virial Theorem but forgot to add a factor of 1/2 on the right-hand side, i.e. \(\frac{1}{2}mv^2=\frac{GMm}{2r}\), and then you'd set \(v=c\) to complete the classical derivation. So by either method, your classical value for the Schwarzschild radius should be off by a factor of 2, and this you can chalk up to the fact that Newton's law only works well in very weak gravitational fields (i.e. the fields generated by stars at large distances).
The orbital velocity is not the same as the escape velocity. If you want to launch a rocket into Earth orbit, you need to launch it at a certain speed. If you want it to escape from Earth completely, on the other hand, you need to make it go faster than the orbital speed. When it comes to the event horizon of a black hole, we're interested in the escape velocity being the speed of light, which has nothing to do with orbits.
But escape velocity refers to the velocity needed to reach \(\infty\), not necessarily the same thing as avoiding getting sucked into a black hole...
The Schwarzschild radius is defined to be the radius at which the escape velocity would be equal to the speed of light, not any other radius (such as the radius needed to stably orbit the hole at a safe distance, which is larger than the Schwarzschild distance).
Are you entirely sure about that? As far as I understand it, the Schwarzschild radius isn't defined that way at all- it's defined as the radius where the Schwarzschild metric of General Relativity becomes singular. If a clock ticks off two events at the event horizon, it registers them at the same time even if they're occuring at different times as seen from a distance.
CptBork: You started talking about the "classical" (i.e. non-general relativistic) derivation of the radius. Then, in your last post you suddenly decided to switch to the general relativistic definition. We can discuss either one, but please be consistent.
Well my point is just that there is no "classical" Schwarzschild radius. I think what writers usually mean when they refer to that, is the point beyond which anything going less than the speed of light is guaranteed to fall further inwards. In that case the "classical definition" is off by a factor of 1/2, the proper calculation uses GR and the metric singularity turns out also to be the point of no return.
Sure there is. The concept of a dark massive object that light couldn't escape from existed long before Einstein (or Schwarzschild) came along. What problem do you have with the escape velocity definition?
Escape velocity isn't what we're looking for. Like I said, escape velocity is the velocity needed to guarantee that you'll be able to drift an arbitrary distance away from the black hole without being stopped (i.e. the velocity needed to reach \(\infty\)). Not at all the same thing as the velocity needed to get past a certain radius.
No particular velocity is ever needed to get past a certain radius from a massive object, unless you specify a starting point and no propulsion in between (which is how escape velocity is defined). So, I'm not sure what you're arguing.
If you want to determine the point of no return, on the assumption nothing goes faster than \(c\), then what we want to calculate is the radius at which you can place an object such that it's guaranteed to be pulled in closer to the black hole, unless it moves at \(c\) or faster. So we want to find the radius at which a classical projectile travelling at \(c\) will just barely be able to maintain a circular orbit without being pulled in any closer. This is how you would classically determine the "point of no return", again with the added assumption that nothing goes faster than \(c\). This definition will be off by a factor of 2, hence the OP had the right idea, they just made a mistake in their first calculation which accidentally gave them the correct GR answer.
Isn't it called the photon sphere? Event horizon is the limit of points where you cannot escape from the black hole no matter which way you headed.
No, but it's not completely trapped and doomed to spiral inwards, either. That's what I defined at the end of my last post, when James asked for reference radii. The radius calculated in the OP is the smallest radius at which a classical beam of light can avoid spiralling inwards, assuming you always confine it to travel at \(c\). If you go out to double that radius, that's the smallest radius from which a classical photon could escape out to an observer arbitrarily far away, in this case assuming the photon slows down over time, which is what James was referring to. In reality that's actually the point of no escape whatsoever.
Put it this way: If I take the actual Schwartzschild radius and do a classical calculation, I can always reach other points further away from that radius even if I'm launched at less than \(c\). But I understand where the disagreement and confusion is, because that happens at the OP's radius as well. But the idea is that 1/2 the actual Schwartzschild radius is the radius at which a circular beam of light is classically trapped, and the real Schwartzschild radius is where a real beam of light can be trapped, in both cases with circular orbits. If light tries to escape from the real Schwartzschild radius, it gets redshifted out of existence. 'Night Please Register or Log in to view the hidden image!
Cpt, don't you mean twice, not half. And isn't it 3 times the Schwarzchild radius? Or 3/2? I haven't got any notes to hand and it has been a long time. If it weren't 7.30am, my week off and if I were in my own house I'd go find the relevant result in a book but it is and I'm not so I won't. Suffice to say if you write down the Euler-Lagrange equations for the SC metric, slap in \(\theta = \frac{\pi}{2}\) and then set \(ds^{2} = 0\) you'll get the relevant equations for working out the path of a light which is going around and around the black hole. It'll depend on the angular momentum value L and I think that'll give you the Schwarzchild radius too (though its obvious from the metric), as that'll correspond to a photon with no angular momentum, while the circular orbit will have all its motion in angular momentum. In both cases you're solving \(\dot{r} = 0\). The circular orbit is unstable, if you perturb it inwards the photon spirals in, if you perturb it outwards the photon spirals out. The photon already pointing directly outwards on the event horizon is 'stuck' there at best, any perturbation will cause it to move onto a geodesic which takes it into the centre. As for the original post, the fact Newtonian approximations give you so similar a result as the full GR one is often noted in the opening blurb of GR books or lecture notes, where a reference is made to some guy hundreds of years ago (his name escapes me) who first conceived the notion of an object so massive nothing could get away from it. /edit http://en.wikipedia.org/wiki/Photon_sphere 3/2, my vague recollection of a 3 somewhere was right.
Damnit, I messed that one up. For some reason I had it in my head that light could still perform a circular orbit at the Schwarzschild radius, but it looks like it can only do that at 1.5x this radius. Classically yeah, I'm pretty sure it should be at 0.5x the Schwarzschild radius, as in the OP's second method of calculating. But ok, I misunderstood the point James was making, depends on how you define "escape"... classically, the radius from which light can escape to \(\infty\) is the same as the radius from which a beam of light cannot return even a millimetre in GR, no matter what direction it's pointing. I'm pretty sure my first course in modern physics treated the problem classically in terms of circular orbits, I could be wrong but that's how I recall it, and I presume it would be because anything in a lower radius travelling less than \(c\) wouldn't make it far even if it were travelling directly away from the black hole. This just shows you can't really define an analogue to GR event horizons using Newtonian gravity, as you wouldn't need something to escape to \(\infty\) in order to be able to see it leaving a black hole. The OP was, as I understand it, doing both calculations in terms of circular orbits, and they missed the factor of 1/2 in the first method when applying the Virial theorem, which is why they got two different answers.
