A Curvature Primer. With Applications to Cosmology. Physics , General Relativity

Transcription

1 With Applications to Cosmology Michael Dine Department of Physics University of California, Santa Cruz November/December, 2009

2 We have barely three lectures to cover about five chapters in your text. To become real GR experts, it is important to eventually master all of this material. Your text, in particular, gives a good deal of geometrical insight into concepts like covariant derivative and curvature. But the basic ingredients needed to build the Einstein equations can be summarized in a few pages.

3 Vectors and Tensors First, we need to define vectors and tensors. These are defined by their transformation properties under general coordinate transformations (changes of coordinates). Contravariant vectors transform like dx µ (dx α x α = x dx µ ): µ V α = x α x µ V µ. (1) An example is provided by the four velocity, u α. Covariant vectors transform like µ : x α f (x ) = x µ x α f (x). (2) x µ

4 More generally, tensors are objects with several indices, upstairs (contravariant) or downstairs (covariant), each transforming like the indices α, µ above. For example, the metric tensor is a second rank tensor with two covariant indices: and its inverse transforms as: g (x ) αβ = x µ x α x ν x β g µν (3) g (x ) αβ = x α x µ x β x ν gµν. (4) Exercise 1: Verify the transformation laws of α ; verify that g αβ g βα is invariant, by explicitly working out the transformation laws of each term.

5 Covariant Differentiation One crucial element is the notion of a covariant derivative. If one takes the derivative of a vector, V α (x), the result does not transform as a tensor: x β V α (x ) = x µ x β ( ) x α x µ x ν V ν = x µ x α x β x ν x µ V ν + x µ 2 x α x β x ν x µ V ν. The first term looks fine, but the second term is not appropriate to a tensor. (5)

6 We can try to add something to the derivative so that it transforms covariantly, i.e. like a tensor. A hint comes from the geodesic equation: du α dτ + Γα µνu µ u ν = 0. (6) This equation, we know, is true for any choice of coordinates. Suppose that u α = u α (x), e.g. it might be a velocity field, describing the velocities of a gas of particles at different points. Then we can rewrite the derivative with respect to τ: u β x β uα + u β Γ α βγ uγ = 0. (7)

7 We can write this as: Here β u α = u α β u α = 0. (8) x β uα + Γ α βγ uγ. (9) This is, we might expect, a covariant equation. To really verify this, we should check that Γ transforms in just a way as to cancel the annoying terms in V. Let s do this explicitly. I will outline the basic steps; you work out the details as part of your homework.

8 Exercise 2: Understanding the Covariant Derivative Show, first (you ll need to use the chain rule and the transformation law for g) that g βγ ( x ρ x α = x σ x δ ) x α x β x γ x ρ g σδ (10) + 2 x σ x δ x α x β x γ g σδ + 2 x δ x σ x α x γ x β g σδ.

9 Continuation of exercise 2: Using this, show that: x α g βγ + x β g αγ (11) x γ g αβ = ( x ρ x α x σ x β x δ x γ )( ρg σδ + σ g ρδ δ g ρσ ) x σ x δ x α x β x γ g σδ

10 Continuation of exercise 2: Show that the left hand side of eqn. 11 is 2g γµγ µ αβ so we have the transformation law for Γ: ( Γ µ x ρ αβ = x σ x µ ) x α x β x δ Γ δ ρσ (12) + 2 x δ x α x β x µ x δ. From this, show that the covariant derivative of a vector with a downstairs index is: µ V ν = µ V ν Γ ρ µνv ρ. (13) Showing that the covariant derivative of a contravariant vector is as we defined above is similar.

11 Exercise 3: In polar coordinates, use the explicit form of the Christofel symbols to construct the divergence of a vector, and the Laplacian of a scalar (this is largely done for you in your textbook).

12 Curvature Consider the object µ ν V λ ν µ V λ (14) This is certainly a tensor. It turns out that it does not contain any derivatives of V λ, so we can write: µ ν V λ ν µ V λ = R λ µνσv σ. (15)

13 Exercise 4: You can evaluate R by first showing: µ ν V λ = µ ( ν V λ + Γ λ νσv σ ) (16) = µ ν V λ + µ Γ λ νσv σ + Γ λ νσ µ V σ + Γ λ µα( ν V α + Γ α νσv σ ) Γ σ µν( σ V λ + Γ λ σρv ρ ).

14 Exercise 4: Continued Now subtract the same expression with µ and ν interchanged. Show that the result is From this read off: ( µ Γ λ νσ ν Γ λ µσ)v σ (17) +(Γ λ µαγ α νσ Γ λ ναγ α µσ)v σ. R λ σνµ = µ Γ λ νσ ν Γ λ µσ + Γ λ µαγ α νσ Γ λ ναγ α µσ. (18)

15 More on Curvature From the Riemann tensor, R λ σνµ, one can construct two other tensors by contracting indices. These are the Ricci tensor, R αβ and the Ricci scalar, R: R αβ = R γ αγβ R = g αβ R αβ. (19) As we will see shortly, it is actually these two objects which enter into the Einstein equation.

16 The Stress Tensor Our goal is to write a tensor equation which relates the curvature of space-time gravity to the source of curvature (energy, momentum, mass). We have now tensors which can describe curvature the Riemann tensor, and tensors obtained by contracting indices with g αβ, another tensor. We need a tensor which describes energy and momentum. This is the energy momentum or stress tensor, T µν. This tensor can be described in a simple situation, a fluid consisting of particles of mass m with density ρ(x), and with a four-velocity at each point, u α (x). The stress tensor is then: The components are, in flat space: T αβ = mρu α u β. (20) T 00 = mργ γ (21)

17 This looks the relativistic energy density, except for an extra factor of γ. But this extra factor is readily understood. Consider a gas at uniform density at rest; now Lorentz boost. Because of the Lorentz contraction, there is an extra γ factor in the density (and hence the energy density). Similarly, T i0 = mργv i γ (22) which is the relativistic momentum density in the i th direction, except, again, for the extra γ factor which accounts for the Lorentz contraction. The components T ij, for a gas of uniform density, are the pressure (consider the non-relativistic case).

18 For other systems, there is still a stress tensor, though its description may be different. For example, for the electromagnetic field, the stress tensor is built out of E and B fields; its 00 component is the usual expression for the energy density; it s 0i components are the components of the Poynting vector. We can understand the stress tensor more generally by considering the question of energy conservation. This is expressed, in flat space, by the equation T αβ = 0. (23) x β To understand the significance of this equation, note first that it is really four equations, one for each possible value of α. Take the case α = 0.

19 Then 0 T 00 + i T i0 = 0. (24) Integrating over a volume V this gives d T 00 d 3 x = d 3 x i T i0 (25) dt V The left hand side is the time derivative of the energy in the volume. The right hand side is the integral of the divergence of a vector (T i0 ), so we can use Gauss s theorem to write it as a surface integral. The result is d dt V T 00 d 3 x = dan i T i0 (26) So T i0, the momentum density, is also the energy flux in the i th direction.

20 Taking the j th component of the conservation equation and proceeding in the same way, one learns that T ij is the flux of i component of momentum in the j direction. This is the same as the i th component of the pressure in the j th direction. In the case of an isotropic medium, T ij = pδ ij (27) and p really acts like a normal pressure. If the medium is not isotropic, one can have shear forces (stresses), which do not act perpendicular to any particular plane. This is the origin of the term stress tensor.

21 Now we make our earlier fluid example more relativistic. We write, for a perfect fluid" T αβ = diag(ρ, p, p, p). (28) This is completely isotropic. It will be important when we discuss cosmology.

22 We can write this is a more covariant fashion, which generalizes to curved spacetime: T αβ = (ρ + p)u α u β + η αβ p. (29) (To check this, consider u = (1, 0, 0, 0)). The stress tensor in curved spacetime is then: T αβ = (ρ + p)u α u β + g αβ p. (30) It should obey the covariant version of the conservation law: α T αβ = 0. (31) This will be important in discussing curved spacetime. In that case, it will not be quite such a simple statement of energy-momentum conservation.

23 The Einstein Equations So now we are in a position to write down the Einstein equations. We want an equation of the form: R αβ arg αβ = ct αβ. (32) This is a covariant equation. The constant a is fixed by the requirement that the covariant divergence must vanish. There is an important identity, called the Bianchi identity, which can be used to fix a: β (R αβ 1 2 gαβ R) = 0. (33) We can give a rough derivation by considering the case of linearized gravity.

24 Exercise 5: Show that the Bianchi equation is satisfied for weak gravitational fields ( linearized gravity"). First, show that in this case R αβγδ = 1 2 ( 2 g αβ x β x γ 2 g αγ x β x δ 2 g βδ x α x γ (34) + 2 g βγ x α x δ ) Then contract with g αγ η αγ to form the Ricci tensor. (Think of g αβ = η αβ + h αβ where h αβ is small). Now just take the ordinary divergence, β R βδ. Compare with δ R.

25 So a = 1/2. The constant c is fixed by the requirement that for weak sources and fields, one obtains the Newtonian limit. For non-relativistic systems, and weak gravity (g 00 = (1 + 2φ/c 2 )): So we obtain the full Einstein equation: R 00 = 1 c 2 2 φ. (35) R αβ 1 2 g αβr = 8πGT αβ. (36)

26 Cosmology As an application of the Einstein equation, and especially the stress tensor, we can consider the universe as a whole. Assume homogeneous, isotropy (sounds crazy at first; see lecture and discussion in your book for evidence). What sort of metric? ds 2 = dt 2 + a(t)d x d x (37) is both homogeneous and isotropic. We will see that a(t) grows with time, when we plug into Einstein s equations. So distance between points grows with time (consider a balloon with marked points as a two dimensional model). This is not the most general homogeneous, isotropic cosmology. It describes a flat universe, in the sense that the universe at any fixed time is flat. There are two other possibilities: closed (positive curvature) and open (negative curvature). These are described in your text and will be described in lecture, time permitting.

27 Hubble: observed an empirical rule for the velocity of nearby galaxies as compared with their distance: V = H 0 d. (38) He obtained V from the redshift; d is more complicated (see your books discussion of the cosmic distance ladder). H 0 has only been reliably obtained in recent years (not least from data from HST): H 0 = 72 ± 7km/s/Mpc. (39)

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30 Let s understand the connection to a(t) in our model. We need to understand the cosmological red shift. Radial light rays in the Friedman metric behave as R = t1 t 0 dt = R. (40) a(t) The ratio of frequencies observed at times t 0 and t 1 is ω(t 1 ) ω(t 0 ) = a(t 0) a(t 1 ). (41) This can be seen, for example, by studying: t1 +δt 1 t 0 +δt 0 dt t1 a(t) = R t 0 From this one has δt 1 /a(t 1 ) = δt 0 /a(t 0 ). dt a(t) + δt 1 a(t 1 ) δt 0 a(t 0 ). (42)

31 Now we can understand Hubble s law. For nearby galaxies, a(t + δt) a(t 0 ) 1 + δt ȧ(t 0) a(t 0 ) 1 + dh 0. (43) So So 1 + z = λ 0 λ 1 = ω 1 ω 0 = a(t 0) a(t 1 ). (44) z = λ λ z is called the red shift by astronomers. = Hd. (45)

32 Robertson-Walker We want to solve Einstein s equations for our metric. We need to discuss, first, the energy-momentum tensor. Energy conservation: de = pdv. (46) Any fixed coordinate volume changes with a: d(ρa 3 V c ) = pd(a 3 V c ) (47) or d dt (ρa3 ) = p d dt a3. (48)

33 This can be seen more directly from the divergence of the stress tensor. The only non-vanishing components of the Christofel symbol are: are: So Γ j 0i = δ ij ȧ a Γ0 ij = δ ij ȧ a. (49) µ T µν = µ T µν + Γ µ µαt αν + Γ ν µαt αµ = 0. (50) Taking the µ = 0 component of this equation, and noting that all spatial derivatives vanish: 0 T 00 + Γ i i0 T 00 = 0. (51) This is the energy conservation equation above.

34 Epochs in the History of the Universe If we have an equation of state then, plugging in, p = wρ (52) dρ dt a3 + ρ(1 + w)3a 2 ȧ = 0. (53)

35 Matter Domination Matter (astronomers mean "non-relativistic matter" by this): w 0. So and d dt (ρa3 ) = 0. (54) ρ = ρ(t 0 )( a(t 0) a(t) )3. (55)

36 Radiation Domination Radiation: p = 1 3 ρ: ρ(t) = ρ(t 0 ) [ ] a(t0 ) 4. (56) a(t) For a blackbody, ρ r = g π3 (k B T ) 4 30 ( c) 3 (57) For photons, g = 2, corresponding to two possible polarization states. Other massless particles (more precisely, k B T m, one counts similarly. E.g. for T > o K, g = 2 + ( ) 7/8. (58)

37 Cosmological Constant (Dark Energy) Vacuum: T µν should be Lorentz covariant (if g µν η µν ). T µν = ρη µν. (59) This is the infamous cosmological constant. Now Then ρ is constant. d dt (ρa3 ) ρ d dt a3 = 0. (60)

38 The Friedman Equation Plugging the metric into Einstein s equations yields the Friedman equation: Note ȧ 2 8πρ 3 a2 = 0. (61) H 2 (t) = 8πρ 3. (62) With H 0 the Hubble constant today, we can naturally define ρ crit = 3H2 0 8π = h 2 g/cm 3. (63) Our universe, as we will see, is described by the simple, flat (fixed time slice) Friedman models (this fact has been known reliably only for about 10 years, but it was predicted by models of inflation").

39 Calling Ω m = ρ m ρ crit ; Ω r = ρ r ρ crit ; Ω V = ρ V ρ crit (64) Ω m + Ω r + Ω V = 1. (65) From a variety of sources, we know: Ω m.30 (baryons + darkmatter); Ω V 0.7 Ω r is negligible.

40 Different Epochs in the History of the Universe With these results, we can run the clock backwards. Important eras: 1 Matter dominated era: kt < 10eV. (ȧ ) 2 = 8πρ a 3. (66) Since ρ m = 1 a 3, a t 2/3. (67) 2 Radiation dominated era: kt > 10 ev. (ȧ ) 2 = 8πρ a 3. (68) Since ρ rad = 1 a 4, 3 Vacuum dominated era: starts now. a t 1/2. (69)

41 Exercise 6: Show that, when vacuum energy dominates, the universe expands exponentially. Determine, roughly the age of the universe when: a. Matter and radiation recombine (assume radiation domination, and that the temperature is approximately kt = 10 ev). b. Atomic nuclei form, kt 1 MeV.

42 More General FRW Metrics Closed FRW Models: Spatial geometry is that of three sphere: Analog of spherical coordinates: W 2 + X 2 + Y 2 + Z 2 = 1. (70) X = sin χ sin θ cos φ Z = sin χ cos θ (71) Y = sin χ sin θ sin φ W = cos χ (note constraint above is satisfied). Now plug in to find metric: ds 2 = dt 2 + a(t) 2 sin 2 χ(dθ 2 + sin 2 θdφ 2 ). (72)

43 Open FRW Universes: W 2 + X 2 + Y 2 + Z 2 = 1. (73) Analog of spherical coordinates: X = sinh χ sin θ cos φ Z = sinh χ cos θ (74) Y = sinh χ sin θ sin φ W = cosh χ (Again, note constraint above is satisfied. Now ds 2 = dt 2 + a(t) 2 sinh 2 χ(dθ 2 + sin 2 θdφ 2 ). (75) Now by a little relabeling, this can be brought to a standard form r = sin χ closed r = χ flat r = sinh χ open (76) [ ] dr ds 2 = dt 2 + a(t) kr 2 + r 2 (dθ 2 + sin 2 θdφ 2 ). (77) Here k = 0 for an open universe, k = 1 for closed, and k = 1 for open.

44 Cosmology Summary 1 The Universe is well-described by a FRW metric: ds 2 = dt 2 + a 2 (t)(dr 2 + r 2 dω 2 ). 2 In the radiation dominated era (t < 10 5 yrs or so), a(t) t 3 In the matter dominated era, a(t) t 2/3. 4 At the present time, the universe is becoming dark energy dominated, a(t) e t/t 0

45 At all times, T 1/a(t) Can think of temperature as a clock". When radiation dominated, ρ g T 4 where g counts light particles (all spins counted separately). Here k B = 1; T = 1 ev 10 4o K. H(t) ρ/m 2 p = c T 2 M. H(t) = ȧ a = 1 2t so we can related the temperature to the time.

46 Important times: T = 1 MeV t 100 sec T = 1 ev t sec 1 MeV: nucleosynthesis. 1 ev: recombination Dark matter comes to dominate the energy density about that time. Temperature today: [t(ev )/t(3 o )] 2/3 = (10 5 / ) 2/3 = 1/3000 (i.e. temperature, age of universe correlated properly).