Curvature 13 Nov 2008

Peter Fisher

11/26/08

8.033

We talked about massive objects “warping” or “curving” space-time. We also have been talking about the curvature k in the FRW metric. Here is a little writeup on the first topic. Start by thinking about measuring the radius R=1/C (C=curvature) of the Earth by measuring distances on the surface. Suppose we are on the surface of a sphere, but can    use GPS and measure 3D coordinates x1 , x2 and x3 . We find the distance between points   1 and 2 given by d = x1 − x2 differs from the distance measured along the surface for the sphere between 1 and 2. We can use point 3 to find s, which tells us how much the surface curves: 2

R 2 = ( R − s) + L2 4 R 2 = R 2 − 2Rs + s 2 + L2 4 ⇒R≈

L2 8s

Now think of a metric that describes a constant gravitational field close to the surface of the Earth: dτ 2 = (1 − gz ) dt 2 − dx 2 − dy 2 − dz 2 . Suppose we have two points in space-time, X1=(0,0,0,0) and X2=(xo,0,0,to). The geodesic connecting these two points will be a parabola with z (t ) = vozt − gt 2 / 2 and x (t ) = vox t . Solving, we find vox = ro t o ,

voz = gt o 2 and h = z (t o 2 ) = vozt o 2 − gt o2 8 = gt o2 8 . We can approximate the parabola as an arc of a circle and find the radius of the circle in space-time. By analogy to finding the radius of the sphere in the 3D example, we take h as the deviation of the straight-line path from 1 to 2 of the geodesic. The interval between 1 and 2 is 2

cτ = c 2 t o2 − xo2 = ct o 1 − ( xo cto ) so the

“radius” is

(

2 2 c 2τ 2 c t o 1 − ( xo cto ) = 8h gt o2

2

)=c

2

(1 − ( x g

o

cto )

2

)

So, if the motion is non-relativistic, the radius is c2 just = 9 ×1015 m=1ly . Notice that xo and to do g not appear for non-relativistic motion: the radius is the same everywhere. As a side note: think about all the possible paths through space-time from event 1 to 2. Only along the geodesic would you feel no force acting upon you at any time. (Adapted from MTW, Chapter 1)