Announcements Homework: Chapter 2 handout # 1, 2, 3, 4 & 7 Will not be collected but expect to see problems from it on the exam. Solutions are posted. Exam 3 is Friday December 4. It will cover collisions (Chapter 8) from Schaum's Outline and the material from the two chapter handouts. Project poster presentations are Thursday December 10 at 1:30pm. Poster boards will be set up in the 3rd floor B-wing hallway. Be prepared to stand by your poster for the two hours of the exam period.

Einstein’s equation of General Relativity G is the gravitational constant and c is the speed of light. But what and are L, Tmn Rmn and gmn and what do the m and n mean?

A brief introduction to tensor calculus The Christoffel Symbol gmn and gmn are the metric tensors. It is what we are trying to solve for to be able to determine the path an object will follow through spacetime.

The Christoffel symbol is used to write out Rmn Rmn is the Ricci Curvature tensor. It describes the curvature of spacetime.

Tmn is the mass-energy-stress tensor Tmn is defined in a manner similar to Rmn. It describes the mass-energy distribution in spacetime. It also includes something akin to pressure (the stress on spacetime).

The final terms: L and R L is the cosmological constant. It is only significant when dealing with the universe as a whole. For situations near a gravitational object, it can be ignored. R is a shape factor and, again, is only important in dealing with the universe as a whole.

Putting it all together we have a set of coupled non-linear DE’s

But wait, there’s more!

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So what do you get if you solve the equations for nothing? In other words, what is the metric for empty spacetime (i.e. flat spacetime) when Tmn = 0? We only need one temporal coordinate and two spatial coordinates: radius r and polar angle f. t is the spacetime interval. We saw this before as t2 = t2 – s2. All we have done here is extend it to two spatial dimensions and make the terms differentials.

We won’t even attempt to solve the equations for non-flat spacetime Fortunately, someone already has. Within month’s of Einstein’s publication of General Relativity, Karl Schwarzschild worked out the solution for the simplest situation: spacetime around a non-rotating point mass.

The Schwarzschild Metric The equation is for polar coordinates with slight modification: the coordinate “r” is the reduced circumference. t is the observers time and t is the spacetime interval. f is the angle in polar coordinate. Spherical symmetry exists so a second angle is not needed. If we want a “space-like” form of the metric, it is

A bit about units (again!) In the relativists view, mass has units of distance. I know we already equated time with distance (lightyears and years) but that was special relativity. The conversion factor is You may have heard of the Schwarzschild radius of a black hole. It is simply Then

A bit about the Schwarzschild Metric When r = 2M the time term goes to zero and the space term goes to infinity. This is a manifestation of the coordinate system used. It is not a physical barrier. That point is called the Horizon or Event Horizon or Schwarzschild radius

Consider a rod held radially toward the center. Set off two explosions at either end of the rod such that the two events occur simultaneously to you. Then dt = 0 and the space-like form of the metric the spacetime interval gives Here, dr is the difference in reduced circumference between two concentric shells centered on the point mass

Example Problem A black hole has a mass equal to that of the Sun (1.477 km). Two concentric shells are centered on the black hole, one with a reduced circumference of 5 km and the other with a reduced circumference of 4 km. If you could measure the separation between the shells directly, what would the measurement show? In Euclidian geometry (i.e. flat spacetime) it would be 1 km.

Solution 1 What we want to use is the space-like Schwarzschild metric. Thus Since we want the radial distance between the shells, df = 0. If we want to measure a distance, then dt = 0. Thus we are left with

Solution 2 To solve, integrate from the inner shell to the outer shell

To solve the integral, make a substitution The substitution is r = z2 so dr = 2zdz

The solution to the final integral can be looked up z1 = 2 and z2 = 2.236, both in units of square root of km and M = 1.477 km. Plugging in numbers gives s = 1.723 km Watch Spaghettification video

How about determining the change in the rate of time? Light is emitted from a shell at r1 = 4M and absorbed by a shell at r2 = 8M. By what fraction is the period of the light increased due to the gravitational redshift?

Think of a clock embedded on one of the shells Two ticks on the clock constitute the two events. They occur at the same location so dr = df = 0. Thus The subscripts indicate shell 1. So

Now do the same for the other shell If I set-up another clock on the other shell and find its rate of time flow (dt2). A similar result will be obtained

To find the fractional change in period just take the ratio of the spacetime intervals In both shells dt is the time between two ticks of the clock as observed by someone on that shell. Thus they are both 1 second and they cancel out.

Plugging in the numbers r1 = 4M and r2 = 8M Thus, the rate that time flows on the inner shell is slower than the rate of time flow on the outer shell by a factor of 1.225 or 22.5% slower.