Notes on General Relativity Ryan Barnett Abstract Notes for: Tensor Calculus and General Relativity (MA7), Spring 2017 1 SpecialLecture 1 Relativity(13/1/14) We start by stating the Postulates of Special Relativity: 1. The speed of light in vacuum c ≈ 3:0 × 108 m/s is the same in all inertial reference frames. 2. The laws of nature are the same in all inertial reference frames. An inertial reference frame is a frame of reference in which Newton’s rst law holds (i.e. it is not accelerating). While Postulate 2 is consistent with Newtonian physics, Postulate 1 is not. Postulate 1 has been conrmed by numerous experimental tests (e.g. the Michelson- Morley experiment). 1.1 The Lorentz Transformation Let’s start by considering the relationship between spatial and temporal coordinates in dierent reference frames. We take two reference frames K and K0 as shown in the gure. For convenience we take the origins to coincide at time t = t0 = 0. K coordinates : r = (x; y; z) and t K0 coordinates : r0 = (x0; y0; z0) and t0 1 1 In Newtonian physics, the coordinates transform according to a Galilean transforma- tion: x0 = x − vt y0 = y z0 = z t0 = t (absolute time): These transformation rules are at odds with Postulate 1. For instance, consider a particle moving at the speed of light in K0: x0 = ct0. Then its speed in K will be c + v. This motivates us to seek an alternative transformation rule that is consistent with the Postulates of special relativity. Let’s assume the transformation to be linear and write ct0 AB ct = (1) x0 CD x y0 = y z0 = z: We will attempt to deduce A; B; C; D by insisting that Postulate 1 is satised and considering four types of motion. Note that since x and ct have the same dimensions (length) the elements in the above matrix will be dimensionless. Case 1: We rst consider a particle moving at the speed of light along the x-axis in the K frame: x = ct; y = 0; z = 0. Then by the rst Postulate, in the K0 frame we must have x0 = ct0; y0 = 0; z0 = 0. Inserting this into Eq. 1 gives A + B = C + D: Case 2: We next consider a particle sitting at the origin of K0: x0 = y0 = z0 = 0. Then in frame K we will have x = vt; y = 0; z = 0. Insertion into Eq. 1 gives Cc = −Dv Case 3: We now consider a particle sitting at the origin of K. Going through the same procedure for this case gives Av = −Cc Case 4: Finally, we consider a particle moving along the y-axis in K at the speed of light: 2 2 dx0 dy0 2 0 x = 0; y = ct; z = 0. Postulate 1 requires dt0 + dt0 = c , z = 0. With Eq. 1 we nd A2 = 1 + C2: 2 The equations for these four Cases can be solved to determine A; B; C; D. The solution gives 0 0 1 0 v 1 0 1 ct γ − c γ 0 0 ct B x0 C B − v γ γ 0 0 C B x C Lecture 2 (14/1/14)B C = B c C B C (2) @ y0 A @ 0 0 1 0 A @ y A z0 0 0 0 1 z where 1 γ = : p1 − (v=c)2 We have arrived at a Lorentz transformation. We will refer to γ as the relativistic factor. A rotation-free Lorentz transformation (as in the above) is a Lorentz boost. One can verify that the inverse transformation (from K0 to K) can be obtained by replac- ing v ! −v. That is, v 0 ct γ c γ ct = v 0 (3) x c γ γ x with y = y0, z = z0. Special relativity forces us to abandon the notion of absolute time, and to consider four- dimensional spacetime. A point (ct; x; y; z) in spacetime is called an event. 1.2 Some Consequences of the Lorentz Transformation 1.2.1 Simultaneity In Newtonian physics, events simultaneous in one frame are simultaneous in another. We will now illustrate how things are dierent in special relativity. Consider a passenger standing in the middle of a train car which is moving at velocity v 4 as shown in the picture. The person emits a pulse of light at time t0 = 0. We will consider the time it takes for the pulse to reach the front and back of the train in dierent reference frames. We take K0 to be the rest frame of the train. Then 4 0 0 0 0 xF = L; xB = −L; tF = tB = L=c 0 0 0 0 ∆t = tF − tB = 0 (Simultaneous in K ) 3 0 In this subscripts F; B denote the front and back of the train. For instance, tF is the time in K0 at which the light reaches the front of the train. On the other hand, in K (the platform frame) through the Lorentz transformation we have γL v γL v t = γ(t0 + x0 v=c2) = 1 + ; t = γ(t0 + x0 v=c2) = 1 − F F F c c B B B c c Lv ∆t = t − t = 2γ (Not simultaneous in K0) F B c2 1.2.2 Time Dilation Consider a clock sitting at the origin in K0. The time intervals in K and K0 are related by the Lorentz transformation: v ∆t = γ(∆t0 + ∆x0) = γ∆t0 c2 (∆x0 = 0 since the clock is motionless in K0). The time recorded by a clock in its rest frame is referred to as the proper time τ and so ∆τ = ∆t/γ Since γ > 1 (for v 6= 0) the time interval in K is longer: “Moving clocks run slowly”. This phenomenon is known as time dilation. 1.2.3 Length Contraction 0 0 Consider a rod of length L0 at rest in K and oriented along the x -axis: 0 0 0 0 0 xF = L0; xB = 0; ∆x = xF − xB = L0: Now transform to K: 6 1 x = L + vt; x = vt; ∆x = x − x = L /γ: F γ 0 B F B 0 We see that ∆x = ∆x0/γ < ∆x0. This is known as length contraction. 4 7 1.3 Relativistic Addition of Velocities Consider a particle moving at velocity w in K0: dx0=dt0 = w. What is the particle’s velocity u = dx=dt as measured in K? In Newtonian physics, the result is u = v + w: On the other hand, by taking the dierential of the Lorentz transformation, we have dx = γ(dx0 + vdt0) = γdt0(w + v) dt = γ(dt0 + vdx0=c2) = γdt0(1 + vw=c2): Dividing these equations gives dx v + w u = = vw : (4) dt 1 + c2 This gives the correct way to add velocities in special relativity. In comparing this with the Newtonian result, we see that the denominator serves to enforce the speed limit of light. Note that this formula will be modied if the particle is not moving along the x0-axis. As our derivation did not rely on w being constant, Eq. 4 is true even if the particle in K0 is accelerating. Eq. 4 can also be arrived at by considering a combination of Lorentz boosts. Take refer- ence frame K and K0 to be as before. Now consider a third frame K00 in which the particle discussed previously is at rest. That is, the origin of K00 is moving away from the origin of 0 0 K at velocity w along the x -axis. Now dene the rapidity v through v = tanh( ) c v with analogous relations for u and v. Then the matrix which transforms or “boosts” (ct; x) to (ct0; x0) is (compare to Eq. 2) v γ −γ c cosh( v) − sinh( v) −σ v λ(v) ≡ v = = e (5) −γ c γ − sinh( v) cosh( v) 0 1 where σ = (check). In this we take the exponential of a square matrix A to be 1 0 A P1 An dened through e = n=0 n! . Writing the Lorentz boost in such a way enables us to nd a simple way to evaluate combinations of boosts. The boost from K to K00 is given by λ(u) = e−σ u = λ(w)λ(v) = e−σ w e−σ v = e−σ( w+ v): Thus the rapidities combine in a simple way: u = v + w: (6) Eq. 4 can be deduced from Eq. 6 (check). 5 1.4 Relativistic Acceleration Now we move on to nd how to transform a particle’s acceleration between inertial reference frames. We start by taking the dierential of Eq. 4 (recall that u = dx=dt and w = dx0=dt0): 1 1 du = 2 vw 2 dw: γ (1 + c2 ) Next, take the dierential of an equation from our Lorentz transformation: v vw dt = γ(dt0 + dx0) = γ(1 + )dt0: c2 c2 Dividing these equations gives du d2x 1 1 d2x0 = = dt dt2 γ3 vw 3 dt02 1 + c2 d2x d2x0 which relates dt2 and dt02 . Example: The relativistic rocket. We consider a passenger aboard a rocket who always d2x0 “feels” a constant acceleration a = dt02 . What is the rocket’s velocity u(t) measured in the K frame? We take the inertial frame K0 to be moving away from the origin of K with the same speed as the rocket at a particular time (a momentarily comoving frame – more on this later). Then at this particular time, u = v and w = 0. This gives a dierential equation for u: du u23=2 = 1 − a: dt c For simplicity, let’s assume the rocket starts from rest: u(0) = 0.