Tutorial “General Relativity” Winter Term 2016/2017

Tutorial “General Relativity” Winter term 2016/2017

Lecturer: Prof. Dr. C. Greiner Tutor: Hendrik van Hees Sheet No. 2 – Solutions

will be discussed on Nov/15/16

1. Line Element Consider the two-dimensional line element given by

ds2 = x2dx2 + 2dxdy − dy2.

ab T a T . Write down gab, g and then raise and lower indices on Va = (1, −1) and W = (0, 1) . Solution: The covariant metric components can be read oﬀ the expression for the line element as x2 1 (g ) =g ˆ = . (1) ab 1 −1 The contravariant components are given by the inverse of this matrix, which is given by Kramer’s rule, using detg ˆ = −(1 + x2),

1 1 1 (gab) =g ˆ−1 = . (2) 1 + x2 1 −x2

The contravariant components of the vector V , Va = (1, −1), are given by

b ba −1 (V ) = (Vbg ) = (1, −1)ˆg = (0, 1). (3) This implies 0 1 (W ) = (g W b) =g ˆ = . (4) a ab 1 −1

2. Coordinate Transformations b In a coordinate transformation, the components of the transformation matrix Λ a are formed by taking the partial derivative of one coordinate with respect to the other ∂xb Λb = , a ∂x0a whereas basis vectors transform as 0 b ea = Λ aeb. Plane polar coordinates are related to cartesian coordinates by

x = r cos θ, y = r sin θ.

Describe the transformation matrix that maps cartesian coordinates to (holonomous) polar coordinates, and write down the polar-coordinate basis vectors in terms of the basis vectors of cartesian coordinates. Solution: Contravariant vector components transform as the coordinate diﬀerentials, ∂xa dxa = dx0b = Λa dx0b, (5) ∂x0b b and vectors are invariant objects, i.e.,

a 0b 0 a 0b 0 a V = V ea = V eb = Λ bV ea ⇒ eb = Λ bea. (6)

That deﬁnes the transformations rules for contravariant (upper indices) and covariant (lower indices) objects like tensor components and basis vectors. For the above example of polar coordinates (r, θ) we have

dx cos θ −r sin θ dr dr = = Λˆ . (7) dy sin θ r cos θ dθ dθ

From this we ﬁnd ˆ (er, eθ) = (ex, ey)Λ = (cos θex + sin θey, −r sin θex + r cos θey). (8)

Note: Here we consider the socalled holonomous coordinates and basis vectors of the curvi- linear coordinates, not the orthonormal basis vectors as usually used in three-dimensional vector calculus.

3. General Coordinate Transformations and Metric components 1 A A µ Under a coordinate transformation x = x (q ), the Minkowski-metric components ηAB transform to new metric components gµν in such a way that proper distances are invariant. 2 A B 2 µ ν In other words, the line element ds = ηABdx dx is invariant, i.e., ds = gµνdq dq .

(a) Show, that this implies that gµν is related to ηAB by

∂xA ∂xB g = η . µν ∂qµ ∂qν AB Solution: We have ∂xA ∂xB ds2 = η dxAdxB = η dqµdqν =: g dqµdqν. (9) AB AB ∂qµ ∂qν µν Since this should hold true for all dqµ, we must have

∂xA ∂xB g = η . (10) µν AB ∂qµ ∂qν

µν µν µ (b) Show, that the inverse-metric components g , i.e., g gνλ = δλ , are given by ∂qµ ∂qν gµν = ηAB . ∂xA ∂xB 1Here we write capital roman letters to indicate components with respect to an inertial Minkowksi basis. As greek indices A ∈ {0, 1, 2, 3}, and the usual Einstein summation convention is used for these indices too. Solution: We use (10) and the given equation to show that indeed (gµν) is inverse to (gµν):

∂qµ ∂qν ∂xC ∂xD gµνg = ηAB η νλ ∂xA ∂xB CD ∂qν ∂qλ ∂xC ∂qν ∂xD ∂qµ = ηABη ∂qν ∂xB ∂qλ ∂xA CD ∂xC ∂xD ∂qµ = ηABη ∂xB ∂qλ ∂xA CD (11) ∂xD ∂qµ = δC ηABη B ∂qλ ∂xA CD ∂xD ∂qµ ∂xD ∂qµ = ηABη = δA BD ∂qλ ∂xA D ∂qλ ∂xA ∂xA ∂qµ ∂qλ = = = δµ. ∂qλ ∂xA ∂qµ λ

4. Rotating frame in Special Relativity A rotating frame can be described by

t = t0, x = x0 cos(ωt0) − y0 sin(ωt0), y = x0 sin(ωt)0 + y0 cos(ωt0), z = z0.

The invariant line element reads ds2 = c2dt2 − dx2 − dy2 − dz2

(a) Calculate the metric components in the rotating frame. Solution: We get (with x0 = x00 = ct = ct0), using

 dx00  µ 0 0 0 0 0 0 0 0 0 0 µ 0ν ∂x  dx cos(ωt ) − dt ωx sin(ωt ) − dy sin(ωt ) − dt ωy cos(ωt )  (dx ) = dx =   , ∂x0ν −dx0 sin(ωt0) + dt0ωx0 cos(ωt0) + dy0 cos(ωt0) − dt0ωy0 sin(ωt0) dz0 (12) after some algebra

ds2 = (dx0)2 − dx2 − dy2 − dz2 ω2 ωy0 ωx0 = (dx00)2 1 − (x02 + y02) − dx02 − dy02 − dz02 + 2dx00dx0 − 2dx00dy0 . c2 c c (13)

From this one reads oﬀ the covariant metric components in the new coordinates,

 ω2(x02+y02) ωy0 ωx0  1 − c2 c − c 0 0  ωy −1 0 0  (g0 ) =g ˆ0 =  c  . (14) µν  ωx0   − c 0 −1 0  0 0 0 −1 We also note the contravariant metric components, which are given by matrix inversion to  ωy0 −ωx0  1 c c 0 0 2 02 2 0 0  ωy −1 + ω y − ω x y 0  (g0µν) =g ˆ0−1 =  c c2 c2  . (15)  −ωx0 ω2x0y0 ω2x02   c − c2 −1 + c2 0  0 0 0 −1 (b) The aﬃne connections (Christoﬀel symbols) for the primed coordinates are given as

1 ∂g0 ∂g0 ∂g0 Γρ = g0ρσ νσ + µσ − µν . µν 2 ∂x0µ ∂x0ν ∂x0σ

Calculate the non-vanishing aﬃne connections. (c) Derive the geodesic equation in a rotating frame. Use your results from (b) to derive the relativistic centrifugal- and the Coriolis force. Hint: It is easier to ﬁrst derive the equations of motion for the geodesic from the quadratic form of the Lagrangian,

1 dx0µ dx0ν L = g0 , (16) 2 µν dλ dλ i.e., using the Euler-Lagrange equations

d ∂L ∂L g0µν − = 0, dλ ∂x˙ 0ν ∂x0ν

which then take directly the form of the geodesic equation (proof that!)

D2xµ d2x0µ dx0α dx0β := + Γµ = 0. Dλ2 dλ2 αβ dλ dλ µ From this it is easy to read off the Christoffel symbols Γ αβ. Solution: Following the hint, we first prove the claimed connection between the Christoffel symbols and the Euler-Lagrange equations with the above given Lagrangian: d ∂L 1 = g0 x¨0α + ∂ g0 x˙ 0αx˙ 0β = g0 x¨0α + (∂0 g0 + ∂0 g0 )x ˙ 0αx˙ 0β (17) dλ ∂x˙ 0ν να β να να 2 α βν β αν ∂L 1 = ∂ g0 x˙ 0αx˙ 0β. (18) ∂x0ν 2 ν αβ Writing down the Euler-Lagrange equations and contracting with gµν finally leads to 1 x¨0µ + g0µν(∂ g0 + ∂ g0 − ∂ g0 ) =x ¨0µ + Γµ x˙ 0αx˙ 0β = 0. (19) 2 α βν β αν ν αβ αβ Now for the above example of a rotating reference frame, first we calculate

d ∂L ∂L − dλ ∂x˙ 0ν ∂x0ν  2 2 2 00 0 0 0 0 2 00 0 0 0 0 2 (1 + ω ρ /c )¨x + ωy x¨ /c − ωx y¨ /c − 2ω x˙ (x x˙ + y y˙ )/c (20)  ωy0x¨00/c − x¨0 + ω2x0(x ˙ 00)2/c2 + 2ωx˙ 00y˙0/c  =   = 0.  −ωx0x¨00/c − y¨0 − 2ωx˙ 00x˙ 0/c + ω2y0(x ˙ 00)2/c2  −z¨0 Multiplying this covariant vector components with the contravariant metricg ˆ−1 we get  x¨00  x¨0 − ω2x0(x ˙ 00)2/c2 − 2ωx˙ 00y˙0/c   = 0, (21) y¨0 − ω2y0(x ˙ 00)2/c2 + 2ωx˙ 00x˙ 0/c z¨0

which is indeed in the form (19). We can immediately read off the non-vanishing Christoffel symbols, ω2x0 ω ω2y0 ω Γ1 = − , Γ1 = Γ1 = − , Γ2 = − , Γ2 = Γ2 = . (22) 00 c2 02 20 c 00 c2 01 10 c The geodesic equations are given by (21). Since the Lagrangian (16) does not explicitly depend on the affine parameter λ the “Hamiltonian” ∂L H = p0 x˙ 0µ − L, p0 = = g0 x˙ 0ν (23) µ µ ∂x˙ 0µ µν is conserved. Now in our case H = L, and this implies that by choosing H = c2/2 we define λ = τ to be the proper time. According to the first equation in (21) we have

x00 = ct = Acτ, (24)

where we have choosen the origin of the coordinate time to coincide with the origin of proper time, and A is an integration constant to be determined. Then the spatial part of the equations of motion can be rewritten as 0 0 0 2 0 ~x¨ + 2A~ω × ~x˙ + A ~ω × (~ω × ~x ) = 0 with ~ω = 0 . (25) ω

Multiplying with the invariant mass of the particle m and solving for m~x¨0 leads to the spatial components of the inertial Minkowski forces

m~x¨0 = K~ 0 = −2mA~ω × ~x˙ 0 − mA2~ω × (~ω × ~x0). (26)

It is clear that in this case the temporal component of the Minkowski force K0 = 0. (d) Solve the equations of motion with the choice λ = τ for the world-line parameter. Hint: The only non-trivial equations are that for x0 and y0. Here the task is tremen- dously simpliﬁed by introducing the complex auxilliary variable ξ0 = x0 +iy0 and derive an equation of motion for it. Then the solution for x0 and y0 is given by x0 = Re ξ0 and y0 = Im ξ0. Solution: Written out in components the equations of motion (25)

x¨0 − ω2A2x0 − 2ωAy˙0 = 0, y¨0 − ω2A2y0 + 2ωAx˙ 0 = 0, z¨0 = 0. (27)

To solve the equations for x0 and y0 we introduce the complex variable

ξ0 = x0 + iy0. (28)

Then the ﬁrst two equations (27) are obviously the real and imaginary parts of the equation ξ¨0 + 2iωAξ˙0 − ω2A2ξ0 = 0. (29) As a homogeneous linear diﬀerential equation of motion we try to solve it by making the ansatz ξ0(τ) = B exp(−iΩτ). (30) Plugging this in (29) leads to the characteristic equation

(Ω − Aω)2 = 0 ⇒ Ω = Aω. (31)

Since we ﬁnd only one solution for Ω, we need to ﬁnd another independent solution of (29). To that end we insert the ansatz

ξ0(τ) = B(τ) exp(−iAωτ), (32) leading to ¨ 0 0 0 0 B(τ) = 0 ⇒ B(τ) = B1 + B2τ, B1,B2 ∈ C. (33) 0 0 Writing B1 = B1 exp(−iϕ1), B2 = B2 exp(−iϕ2) with B1,B2 ≥ 0 the general solution of (29) is 0 ξ (τ) = B1 exp(−iAωτ + iϕ1) + B2τ exp(iAωτ + iϕ2), (34) i.e.,

0 0 x (τ) = Re ξ (τ) = B1 cos(Aωτ + ϕ1) + B2τ cos(Aωτ + ϕ2), 0 0 (35) y (τ) = Im ξ (τ) = −B1 sin(Aωτ + ϕ1) − B2τ sin(Aωτ + ϕ2).

Of course, there are only six independent integration constants determined by the 0 0 ˙ 0 ˙ 0 initial values ~x0 = ~x (τ = 0) and ~x0 = ~x (τ = 0). It is clear that in this case dt A = = γ = const, (36) dτ and it is determined by

0µ 0ν 2 2 2 2 2 gµνx˙ x˙ = c A − B2 − C1 = c , (37) i.e., r B2 + C2 A = 1 + 2 2 . (38) c2