SciForums.com > Science > Physics & Math > How to do basic GR maths in a Schwarzschild geometry? PDA View Full Version : How to do basic GR maths in a Schwarzschild geometry? Post ReplyCreate New Thread Pete04-11-07, 02:28 AMI want to learn how to apply the formalism of GR. To get started, I want to know if and how I can use the Schwarzschild metric to determine the escape velocity at a particular location: A freely falling reference frame is approaching the event horizon of a black hole. The singularity is in the direction of the negative x-axis. At t=0, the event horizon is at x=0. At t=0, a test particle is at x=r, with velocity along the x-axis such that the particle will escape the hole, coming to rest at infinity. Can I use the Schwarzschild metric to find the velocity of the particle? If so, how? Please keep it simple for me. I guess it would be easiest to use units so that c=1 and 2GM=1 zanket04-11-07, 06:36 PMCan I use the Schwarzschild metric to find the velocity of the particle? Yes. If so, how? In GR, there is a one-to-one mathematical relationship between the (spacetime) curvature factor incorporated into the Schwarzschild metric, and escape velocity. That is, if you know the curvature factor at some r-coordinate r, then you know the escape velocity without ambiguity, and vice versa. Pg. 2-21 in this PDF ( http://www.eftaylor.com/pub/chapter2.pdf) by Taylor and Wheeler shows the curvature factor. I like to take the square root of it and call it the gravitational distortion factor: gravitational distortion factor = sqrt(1 - (R / r)) where c = 1, G = 1, R = 2M, and r is the r-coordinate, a Euclidian radius. The gravitational distortion factor incorporated into the Schwarzschild metric is the sole difference between the Schwarzschild metric and the metric for flat spacetime, also called the Minkowski metric, for SR. The gravitational distortion factor is the GR’s factor for gravitational time dilation and gravitational length contraction. Note that: gravitational distortion factor = special relativistic distortion factor(escape velocity at r) special relativistic distortion factor = reciprocal of the Lorentz factor = sqrt(1 - v^2) where v is a fraction of c (e.g. v = 0.5 = 0.5c). Then: sqrt(1 - (R / r)) = sqrt(1 - (escape velocity at r)^2) Solving for (escape velocity at r), v = sqrt(1 - (gravitational distortion factor at r)^2) Simplifying, v = sqrt(R / r) That is GR’s equation for escape velocity for Schwarzschild geometry. I derive this relationship using logic in section 4 here (http://zanket.home.att.net/). Let a clock that is fixed at some r-coordinate run at 50% of the rate of a clock fixed at a great distance. Then: gravitational distortion factor = 0.500 And so: v = sqrt(1 - 0.500^2) = 0.866 That is, where a fixed clock runs at 50% of the rate of a clock fixed at a great distance, the escape velocity is 0.866c. andrewgray04-12-07, 02:08 AMPete, A more straightforward method might be more intuitive. You must go through the brute force geodesic equations (free-fall equations). Basically, you must do a curvilinear second derivative and set it = 0: \frac{D^2x^{\alpha}}{d\tau^2} = 0 This is very messy but straightforward. Similar to proving that great circles are the geodesics on the globe. With the Schwarzchild metric, the geodesic equations yield some constants of motion whose equations are: \frac{d\phi}{d\tau} = \frac{L^2}{\mu^2 r^2} \frac{dt}{d\tau} = \frac{E/\mu}{(1-2M/r)} \left ( \frac{dr}{d\tau} \right )^2 = \frac{E^2}{\mu^2} - (1-2M/r)(1+\frac{L^2}{ \mu^2r^2}) To find the escape velocity, let L=0 and notice that this implies that the particle's velocity at infinity is zero. Since the energy E is a constant, this means that just E= μ at infinity and everywhere else along the geodesic (no kinetic energy, just rest energy at infinity, so E=γμcē=μ in geometric units where c=1). Plug E= μ and L=0 into the long equation above and get \left ( \frac{dr}{d \tau} \right )^2 = \frac{2M}{r} We see that this equation describes both a particle starting at r and getting to infinity, as well as a particle just dropped from infinity. Using the dt/dτ equation to change to dt: dt = \pm \frac{\sqrt{r/2M}}{(1-2M/r)} dr Finally, just like tangential velocity in polar coordinates is V_{\theta} = \frac{rd\theta}{dt} The radial velocity in the Schwarzchild coordinates is V_r = \frac{1}{(1-2M/r)} \left ( \frac{dr}{dt} \right ) = \pm \sqrt{2M/r} The minus sign gives the velocity that a "dropped from infinity" particle attains, and the plus sign gives the initial velocity that a particle needs in order to just barely make it to infinity (escape velocity). Andrew A. Gray Pete04-12-07, 07:03 PMThanks to you both. I have some questions but no time now. Will post later. Post ReplyCreate New Thread