General Relativity Final Exam Due Thursday May 11th at 3pm Circus Twins Consider the twin paradox from special relativity: Two twins, Calvin and Hobbes, begin at the same position at the same age. Calvin then flies away in a rocket at a large speed, turns around and returns to where he left Hobbes. Roughly, we can argue that since Calvin moved and Hobbes remained at rest, time‐dilation implies that the clock for Calvin passed more slowly and hence he is now younger than Hobbes when they reunite. There is nothing problematic about the argument so far. The paradox comes in when you think about how this looks from the rest frame of Calvin. From Calvin's perspective he remained at rest and it was Hobbes who moved away, and so by time‐dilation it should be Hobbes who is younger. They can't both be correct! This paradox is resolved by realizing that Calvin and Hobbes are not equivalent observers. Indeed if one of them is truly an inertial observer, say Hobbes, then Calvin's motion is necessarily accelerated (he has to speed up, slow down, turn around, speed up and slow down again). This means that the reference frame in which Calvin is at rest is a non‐inertial frame. However special relativity only equates the physics seen by different inertial observers. Since they can't both be inertial observers, Calvin and Hobbes' reference frames need not be interchangeable and their ages when they reunite will correctly reflect who did and did not remain inertial. Another way to state this is that in special relativity, there is no preferred frame among all inertial frames (all inertial observers are on the same footing), however an inertial observer is definitely distinct from an accelerated one. Note that this strangely implies that if Calvin and Hobbes are actually moving relative to each other and are both inertial observers, then they really do see the other age more slowly, however their comparison can only be done (locally) once since they cannot be at the same place and time more than once. Think about this carefully before proceeding! If however the flat spacetime of special relativity is augmented by making one of the spatial dimensions periodic, e.g. a circle, then the twin paradox gets a new twist. In this case the twins can begin at the same point at the same age and then one, say Hobbes, can remain at rest while the other, Calvin, moves at a constant velocity along the periodic dimension. After some time, they will be reunited at the original position and during this "flyby" can make a local measure of each other's age. However in this case they have both undergone constant velocity motion and hence both are inertial observers. The original twin paradox reasserts itself without any obvious resolution. a) First calculate the proper time elapsed for each observer in the periodic scenario after Calvin makes one full trip around the circle. Call the length of the periodic dimension and assume that Calvin has a velocity along the periodic dimension. This should convince you that they are definitely aging differently. b) The absolute interchangeability of inertial reference frames in Minkowski space is guaranteed by Lorentz invariance. Does this flat spacetime with one periodic dimension exhibit Lorentz invariance? Be very careful about the level at which you identify the invariance or lack thereof. c) Resolve the twins "paradox" for the case of a flat spacetime with one periodic dimension. Pac‐Man 0 out of Consider with polar coordinates , . Imagine cutting a wedge of opening angle ∆
(making an infinite "Pac‐Man") and then identifying the two edges of the cut (similar to identifying the two edges of a strip to form a cylinder). Note: In the images below, the dashed lines are infinite radius. This is called "quotienting" the original space and the result is called a quotient space. When we make this identification, the origin at 0 is the only part of the cuts which is not "moved" by the identification, i.e. it is identified with itself. We call this a "fixed point" of the identification. Generically, fixed points of this type represent curvature singularities in the resulting quotient space. In this problem we will explore this type of singularity. a) Curvature singularities are intractable in GR, so we often proceed by removing them from the space. If we remove the origin in this case (keeping all points arbitrarily close to it), demonstrate that the remaining quotient space is a smooth manifold. That is, it should satisfy all of the properties that we used to define a smooth manifold. This means that you should provide an atlas of charts with infinitely differentiable transition functions (if needed). b) Argue that the quotient space with the origin removed is geodesically incomplete. That is, find a geodesic path in this space which terminates after a finite path length. Recall that geodesic incompleteness is one way of identifying singularities. c) Now to explore the curvature of this space, consider embedding it into , ,
, ,
,
as the set of points 0. Calculate the metric induced on this surface from the embedding. d) Find the Riemann curvature associated with this metric. In words, describe the result and what it tells you about the curvature of the space. e) Now argue, based on parallel transport, that if we do include 0, then all of the curvature of the resulting space is completely concentrated at the origin. Hint: You can do this pictorially. GPS Corrections One naive fallacy that I like to dispel is the idea that when improving on Newtonian results, one must first do the special relativistic correction (for high speeds) and then follow up with the GR correction (for curved spaces). The truth is that a GR calculation does all of this at once. To that end let's consider one situation where both of these effects (velocity and curvature) play an important role. The GPS system utilizes a network of orbiting satellites carrying atomic clocks (with ~nanosecond accuracy) that send timing information to receivers. These receivers (or "GPS units") use the timing and known satellite positions to calculate their own location. The timing information sent by the satellites corresponds to intervals of proper time since the clock is at rest with respect to the satellite which is doing the transmitting. On the receiver's end these time intervals will be observed differently due to the effects of time‐dilation resulting from the relative speed of the satellite and receiver, as well as due to the differing heights in the gravitational field, i.e. the gravitational redshift. a) Consider a satellite, armed with an atomic clock, that is in a circular orbit at radius from the center of a spherically symmetric gravitational source. Consider a second clock sitting at rest at a radius . Using the appropriate metric calculate the proper time that elapses on clock 1 for a given interval of proper time elapsed on clock 2. Hint: Make sure you use the appropriate metric. b) Now assume that the source has the same mass and radius as Earth and use a satellite orbit of radius 26600 km (typical for the GPS system). Assuming the receiver is at rest on the surface of the source, evaluate the shift you calculated above in terms of a shift per day and compare to known GPS corrections. Does your result agree with the known corrections, and if not why? You will need to restore all factors of c in order to get a numerical result. Hint: Think carefully about whether the system you are studying is actually the GPS system on Earth! c) Now assume that the source is in fact a black hole and one clock is in a circular orbit just above the horizon, let's say 3.0001
. For the second clock, we will take it's position to be fixed (so you can 20001
. Calculate the time shift per day of use your earlier result) and far from the horizon, e.g. clock 1 compared to clock 2. X‐treme Physics The geometry of a spherically symmetric source of mass coordinates) by the Reissner‐Nordstrom metric: 1
2
and charge is described (in Schwarzchild 2
1
Ω The electromagnetic potential that accompanies this source can be written as: √4
, 0,0,0 Even though this is the external geometry outside of the source, it still does not satisfy the vacuum form of Einstein’s equations, i.e. 0, since the electromagnetic field provides a non‐vanishing energy momentum tensor outside of the source given by with . a) Using Mathematica (or doing it by hand), verify that this metric satisfies Einstein’s equations with the energy momentum tensor given above. b) Verify that the electromagnetic field satisfies (half of) Maxwell’s equations in vacuum, i.e. 0. in the metric above, we obtain the “extremal” Reissner‐Nordstrom geometry. One of If we set the fascinating aspects of the extremal case is that it is one of the rare scenarios where we can get exact solutions to Einstein’s equations for multiple sources. Consider the following metric which describes two extremal black holes separated along the ‐axis: 1
1
where . This metric, accompanied by an electric potential given by 1
√4
1
1
1
, 0,0,0 10
satisfies both Einstein’s and Maxwell’s equations. c) Now I want you to consider the form of this metric and observe any obvious coordinate symmetry. If you think about the fact that this represents two sources in general relativity, then the presence of this symmetry should be somewhat surprising. I want you to identify what I am alluding to and then explain this (perhaps) surprising result.