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Fall 2019 Problem sheet 1

Problems 2 to 7 are meant to help you review , but they do not cover everything. Make sure you understand sections 1.1 to 1.4 in Carroll’s book and the more detailed discussions in chapters 4 and 5 of Hartle’s book. Problem 8 is new material, to be discussed in class

1. Astrophysical units: when mass becomes distance. 2 A of mass M has a size Rs = 2GM/c (G = Newton’s constant, c= ) called its “ Schwarzschild radius” (what is its significance?). Show that the combination 2 GM/c has units of distance. Calculate RS for a) the earth; b) the sun; c) a supermassive black hole, such as Sgr A* (Saggitarius A*, at the centre of our galaxy). Sgr A* has a mass of about four million solar masses and is r = 26, 000 light- away. What is r/Rs?

2. If you need to practice raising and lowering indices, and the Einstein summation convention, do problem 7 of chapter 1 in Carroll’s book.

3. Carroll’s problem 3 of chapter 1.

µ0 4. a) Write down the 4x4 matrix Λ ν (V ) describing a Lorentz boost with V in the positive x-direction, that is, V~ = V ˆı. This takes you from a reference frame S to a reference frame S0. A second boost that is collinear with this one will take you from S0 to S00. If S00 is moving with velocity V~ 0 = V 0ˆı. How fast is S00 moving with respect to S? Find out by multiplying the matrices and reading the velocity. 00 ~ 0 00 µ b) Now write down the boost with velocity −V from S to S :Λ ν0 (−V ) and show that their product is the identity, so S00 is the same as S. (The boosts are inverses of each other). µ00 c) Instead, consider boosts in different directions. Write down the matrix Λ ν0 (W ) for a boost in the positive y-direction with velocity W~ = W ˆ. [Note: If you boost back, you should get the identity again Λ(−W )Λ(−V )Λ(V )Λ(W ) = 1. However, in this case the order of the boosts is important: Λ(−V )Λ(−W )Λ(V )Λ(W ) is not the identity. It has a around the z axis. In other words, Lorentz boosts don’t commute. This rotation is what leads to the precession of electronic spin in the hydrogen atom, Thomas’ precession. You don’t need to show any of this, but it is good to keep it in mind.]

5. Draw a diagram showing Lorentz contraction of lengths: draw the worldlines of the two ends of a ruler in its rest frame. To make a measurement of lengths we consider the position of the two ends simultaneously, and assign the spacetime interval ∆s2 = L2. Show that an observer moving with respect to the ruler will not pick the same points in spacetime, it will pick points that are simultaneous in its reference frame – and will get a different answer for the length. [Notice the length looks shortest in the rest frame, but this is wrong! what is going on?] 6. Draw the spacetime (Minkowski) diagram for the train paradox: a “train” of 100 m is heading towards a tunnel of length 90 m. The “train” is travelling at 0.6 c (what a train!) and therefore it is Lorentz contracted to 80 m (do you agree?). The tunnel has a trap door at each end that can come down and up very fast (how fast do you need? this is not a very realistic problem... never mind). When the train is inside the tunnel both doors are simultaneously closed and open again, so fast that the train passes safely and nobody gets hurt. Now the same but from the point of view of a passenger on the train. A Lorentz contracted tunnel of length 72 m (do you agree?) is rushing towards a 100 m train. The doors close and open and still nobody gets hurt. Draw a Minkowski diagram and explain what happens.

7. This problem is adapted from Hartle’s book. It describes a toy GPS system in flat space (here we only look at special relativity effects related to delay). Consider a two-dimensional toy model for a Satellite location system like GPS (Global Positioning System). We consider only two dimensions: the height y at which the satellites are flying overhead and a horizontal line on the ground (the x axis). You want to locate your position on the x axis based on the time signals from the satellites. The satellites fly above you with velocity V (let’s say in the positive x direction, so V~ = V ˆı) at an altitude h. They are separated from each other by a uniform distance L∗ in their rest frame. They carry clocks that are synchronized in their rest frame, and at regular intervals they broadcast signals with their time and their position in x. a) Simultaneously you receive signals from two neighbouring clocks located on either side of you, they both report the same time at broadcast. Does this mean you are midway between the clocks? (it doesn’t). Calculate the time difference between the emission of the two signals (in your rest frame). Which of the two clocks is closer to you? b) Suppose you had 24 such satellites flying in 12 hour orbits. Estimate the height and the speed (using Newtonian ) and the distance between the satellites (in what frame?). In this case, the time offset is about 3 · 10−7s. If you did not take this into account, what error would you make in your location? (in order to achieve a location accuracy of 10 m or less you need a time accuracy better than 3 × 10−8 s, the time it takes for light to cover that distance.)

8. A variational principle to calculate geodesics in a curved spacetime. Geodesics are lines that minimize (or, rather, maximize, if the signature is Lorentzian) the distance between two nearby points. Timelike geodesics can be obtained by extremizing the action

Z Z  dxµ dxν 1/2 S = |ds| = dτ −g (x) µν dτ dτ The Euler-Lagrange equations of this Lagrangian with respect to each of the coordinates xµ give the geodesic in parametric form, with parameter τ. The square root is messy, but for timelike geodesics we can use a trick that we will justify later: roughly speaking, since ds is non-zero everywhere on the geodesic, if the geodesic extremizes |ds| it also extremizes ds2 so we can also calculate it from

Z ds Z dxµ dxν S = dτ( )2 = dτ(g (x) ) , dτ µν dτ dτ which is much easier. τ is a parameter along the trajectory (in fact, it is the measured by clocks following that trajectory - more on this later), so we get the equation of the geodesics in parametric form, from the Euler Lagrange equations for each of the “fields” xµ:

d ∂L ∂L ) − = 0 . dτ ∂x˙ µ ∂xµ This approach has two advantages: there is no need for any formulas to compute the Christoffel symbols that appear in the geodesic equations (more on this later), and also (most importantly) any constants of motion are much more easily found this way. Let’s calculate an interesting example that will be useful later: the orbits around a black hole. The spacetime around a spherical mass M is described in spherical coordinates (t, r, θ, φ) by the :

 2GM  dr2 ds2 = − 1 − dt2 + + r2[dθ2 + sin2 θdφ2] r 2GM (1 − r ) so we can calculate the geodesics by extremizing

" # Z  2GM  r˙2 S = dτ − 1 − t˙2 + + r2θ˙2 + r2 sin2 θφ˙2 r 2GM (1 − r )

˙ d where = dτ .

a) Find the four Euler-Lagrange equations corresponding to t(τ), r(τ), θ(τ) and φ(τ). You can check your answers against eq (5.53) in Carroll’s book but your equations are in a more useful form for what follows. b) By looking at the equations for t(τ) and φ(τ) show that there is a conserved quantity “L”along the trajectory that we can associate with the angular momentum of the trajectory and another one coming from the time coordinate, “E”, that relates the time coordinate t to the proper time τ measured by clocks.

c) Use these constants of motion to eliminate the t˙ and φ˙ terms in the r and θ equations. Notice that equatorial trajectories with θ(τ) = π/2 (constant) are always a solution of the θ equation, and we can study those without loss of generality, other trajectories are obtained from these by rotating the axes. Derive an equation for r(τ) which only involves r(τ) and its derivatives. Answer:

" # " #2 d r˙ GM E GM r˙2 L2 − − − + = 0 dτ 1 − (2GM/r) r2 1 − (2GM/r) r2  2 r3 1 − (2GM/r)

[To be continued...]

9. Read section 2.2 of Carroll’s book carefully and make a list of all the concepts that are not familiar to you. Bring the list to class on Tuesday 24th Sept. [Do this at home].