Announcements•Homework: Chapter 2 handout # 1, 2, 3, 4 & 7Will 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

124

8 GT R g R

c

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

12

g ggg

x x x

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

Rx x

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 RL 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!

And if you order today, we’ll throw in one more for free!!!

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.

2 2 2 2 2d dt dr r d 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 spacetimeFortunately, 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 Metric2

2 2 2 221

21

M drd dt r d

Mrr

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 2

2 2 2 221

21

M drd dt r d

Mrr

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

2

GM in meters M in kg

c

You may have heard of the Schwarzschild radius of a black hole. It is simply

Then

2

2S

GMR

c

21 1 sRM

becomesr r

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

21

drd

Mr

Here, dr is the difference in reduced circumference between two concentric shells centered on the point mass

Example ProblemA 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 1What we want to use is the space-like Schwarzschild metric. Thus

22 2 2 22

12

1

M drd dt r d

Mrr

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

122

1

drd

Mr

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

5

4 21

km

km

dr

Mr

To solve the integral, make a substitution

The substitution is r = z2 so dr = 2zdz 1 22

1

5 5 5 2

24 4 4

2

2 2 2 21 1

zkm km km

km km km z

dr rdr r dr z dz

M M r M z Mrr r

The solution to the final integral can be looked up

2 2

11

22 2

2

22 2 ln 2

2

z z

zz

z dzz z M M z z M

z M

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

22 2 2 2 22 2

1 12

1

M dr Md dt r d dt

Mr rr

12

1 11

21

Md dt

r

SoThe subscripts indicate shell 1.

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

12

2 22

21

Md dt

r

To find the fractional change in period just take the ratio of the spacetime intervals

12

12

222

1

11

21

21

Mdt

rd

d Mdt

r

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 numbersr1 = 4M and r2 = 8M

1122

1 12 2

222

1

11

2 21 18

1.22522 114

M Mdtrd M

d MMdt

Mr

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.