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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.
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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?
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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.
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The Christoffel symbol is used to write out Rmn
Rx x
Rmn is the Ricci Curvature tensor. It describes the curvature of spacetime.
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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).
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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.
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Putting it all together we have a set of coupled non-linear DE’s
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But wait, there’s more!
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And if you order today, we’ll throw in one more for free!!!
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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.
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.
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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.
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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
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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
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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
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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
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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.
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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
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Solution 2To solve, integrate from the inner shell to the outer shell
5
4 21
km
km
dr
Mr
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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
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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
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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?
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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.
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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
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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.
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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.