Solutions Exam FY3452 Gravitation and Cosmology fall 2017 Lecturer: Professor Jens O. Andersen Department of Physics, NTNU Phone: 46478747 mob) Wednesday December 13 2017 09.00-13.00 Permitted examination support material: Rottmann: Matematisk Formelsamling Rottmann: Matematische Formelsammlung Barnett & Cronin: Mathematical Formulae Angell og Lian: Fysiske størrelser og enheter: navn og symboler Problem 1 a) The nonzero components of the metric can be read off from the line element and are g tt = fr), g rr = 1 fr), g φφ = r 2. 1) The metric is diagonal. Since the metric is indenpendent of t and φ, there are at least) two Killing vectors. These are ξ = 1, 0, 0), η = 0, 0, 1). 2) 1
The associated conserved quantities are u ξ and u η e = u ξ = fr) dt, dφ l = u η = r2. 3) Time independence implies energy conservation, while independence of φ implies conservation of the z-component of the angular momentum. b) The Christoffel symbols Γ γ αβ can be calculated from the equation of motion [ ] d L ) = L dx µ x µ, 4) where L = g αβ dx ) α dx β 1 2. We first consider µ = t. Since L is independent of t, the right-hand side of equation 4) vanishes. We find L dx t ) = 1 L fr)dxt 5) Using This yields = L, we can write L dx t ) = fr) dxt L d ] [fr) dxt and the equation of motion becomes = 0. 6) or f d2 x t 2 + f dxt dx t = 0, 7) d 2 x t 2 + f dx t dx t f = 0, 8) We can then read off the nonzero Christoffel symbols with γ = t Γ t rt = Γ t tr = 1 f 2 f, 9) where the prime indicates differentiation with respect to r. The equations of motion for α = r and α = φ can be calculated in the same manner. We list them for completeness d 2 r 2 + 1 dt dt ff 2 1 f dr dr dφ dφ rf 2 f d 2 φ 2 + 2 dr dφ r = 0, 10) = 0. 11) 2
This gives Γ r tt = 1 2 ff, Γ r rr = 1 f 2 f, Γr φφ = rf Γφ rφ = Γφ φr = 1 r. 12) The other Christoffel symboles are zero. c) For α = β = φ, we find R φφ = γ Γ γ φφ φγ γ φγ + Γγ φφ Γδ γδ Γδ φγ Γγ φδ = r Γ r φφ + Γγ φφ Γδ γδ Γδ φγ Γγ φδ [ ] = r Γ r φφ + Γr φφ Γ r rr + Γ φ rφ + Γt rt Γ r φφ Γφ φr Γφ φr Γr φφ = r [ rf] + 1 2 rf + f 1 2 rf = rf. 13) The other diagonal components of R αβ can be calculated in the same manner. This yields Contracting R αβ with the metric yields R tt = 1 2 ff + 1 ff 2 r, 14) R rr = 1 f 2 f 1 2 d) The Einstein equation in vacuum reads This yields R = f 2 f f fr, 15) r. 16) R αβ 1 2 g αβr = 0. 17) ff r = 0, 18) f fr = 0, 19) f r 2 = 0. 20) Thus f is constant. We identify the line element as that of Minkowski spacetime for f = 1. 21) 3
Problem 2 a) We first calculate the differentials in the new coordinates dφ = dφ Ωdt. 22) We can therefore make the substitution dφ dφ + Ωdt in the metric. This yields ds 2 = dt 2 + dr 2 + r 2 dφ Ωdt) 2 + dz 2 = 1 Ω 2 r 2 )dt 2 + dr 2 2Ωr 2 dφdt + r 2 dφ 2 + dz 2. 23) The relations r = x 2 + y 2 and φ = arctan y x yield dr = dφ = xdx x 2 + y + ydy 2 x 2 + y, 24) 2 xdy ydx x 2 + y 2. 25) Inserting Eqs. 24) and 25) into 23) and cleaning up, we obtain ds 2 = [1 Ω 2 x 2 + y 2 )]dt 2 + 2Ωydx xdy)dt + dx 2 + dy 2 + dz 2. 26) b) In the nonrelativistic limit, we can approximate τ = t. This implies dt d 2 t 2 = 0. This yields = 1 and d 2 x dt 2 2Ωdy dt + Ω2 x = 0, 27) d 2 y dt 2 + 2Ωdx dt + Ω2 y = 0, 28) d 2 z d 2 t = 0. 29) c) Eq. 27) can be written as a x + 2Ω v) x + Ω Ω r)) x, 30) where the subscript x means the x-component. Eqs. 28) and 29) can be written as the y - and z-components of the same equation. Thus, we have a + 2A v + Ω Ω r) = 0. 31) Being in a rotating frame of reference, fictitious forces are present. The term 2Ω v is the Coriolis force, while the term Ω Ω r) is the centrifugal force. 4
Problem 3 a) We know that the covariant derivative of a scalar is the the usual partial derivative. If B β is a covariant vector s = A β B β is a scalar and we can write α s = s x α. 32) Using the Lebniz rule we can also write α A β B β = α A β) B β + A β α B β ) = s x α = A β B β x α + Aβ x α B β. 33) Substituting the expression for the covariant derivative a contravariant vector in Eq. 33), we find A β α B β ) + Γ β αγa γ B β = A β B β x α. 34) Swapping dummy indices β and γ, this reads Since A β is arbitrary, we must have A β α B β ) + Γ γ αβ Aβ B γ = A β B β x α. 35) α B β = B β x α Γβ αγb β. 36) b) As a photon propagates in a gravitational field, its frequency ω changes. For example, if a photon propagates radially outwards in a Schwarzschild spacetime being emitted at r A and being detected at r B, the frequencies are related as ω B = 1 2M r A 1 2M, 37) r B where M is the mass of the planet. Since r B > r B, we find ω B < ω A, i.e. gravitational redshift. c) If an observer sees the same universe in all directions, it is isotropic around the point in space of the observer. If it is isotropic for all observers in the universe, it is globally isotropic. If all observers see the same universe, it is homogeneous. These concepts are not equivalent. A uniform magnetic field in one direction, clearly breaks isotropy, but the universe can still be homogeneous. 5
d) The term F µν F µµ is gauge invariant as it is constructed out of the field tensor, which we know is invariant. The second term transforms as j µ A µ j µ A µ where χ is a well-behaved function. The change is = j µ A µ + µ χ). 38) L = j µ µ χ. 39) The action also changes S = = d 4 x L d 4 xj µ µ χ 40) This can be written as S = = d 4 x [ µ χj µ ) χ µ j µ ] d 4 x [ µ χj µ )], 41) where we have used current conservation, µ j µ = 0. The Lagrangian changes by a total derivative, which is allowed. The action does not change. Problem 4 We denote the ejected four-momentum by p e and the remaining four-momentum by p f. The initial four-momentum is denoted by p. Conservation of four-momentum gives This yields p = p e + p f. 42) p 2 e = m 2 m 2 f 2p f p 43) Since the ejected material has zero rest mass, we have p 2 e = 0. The initial four-momentum p is spaceship at rest) p = m 1, 0, 0, 0. 44) 6
To evaluate the product p f p, we only need the zeroth component of the four momentum p f. This denoted by p t f r). Conservation of energy gives p t f r) The spaceship must be at rest at r =, whence e = 1. ) = m f e. 45) p t f r) = m f. 46) Writing m f = mf, where f is the fraction, and using the expressions for the four-momentum p, Eq. 43) can be written as The solution for f is m 2 1 + f 2 ) 2m2 f 2M R f = 1 ± 1 2M R = 0. 47). 48) The positive solution yields f > 1, which must be rejected on physical grounds. Hence, the fraction is The limit is 2M R f = 1 1 2M R f horizon = lim R 2M f Thus, nothing can escape if the spaceship starts at the horizon.. 49) = 0. 50) 7