Lectures on General Relativity and Cosmology

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General Relativity and Cosmology 1 Lectures on General Relativity and Cosmology Pedro G. Ferreira Astrophysics University of Oxford, DW Building, Keble Road Oxford OX1 3RH, UK last modified May 22, 2015 Preamble The following set of lectures was designed for the Oxford Undergraduate course and are given rd during the 3 year of the BA or MPhys course. I have tried to keep the mathematical jargon to a minimum and ground most of the explanations with physical examples and applications. Yet you will see that, although there is very little emphasis on differential geometry you still have to learn what tensors or covariant derivatives are. So be it. These notes are not very original and are based on a number of books and lectures. In particular I have used • Gravitation and Cosmology, Steven Weinberg (Wiley, 1972) • Gravity: An Introduction to Einstein’s General Relativity, James B. Hartle (Addison Wesley, 2003) • Part II General Relativity, G W Gibbons (can be found online on http://www.damtp.cam.ac.uk/research/gr/members/gibbons/partiipublic-2006.pdf) • General Relativity: An introduction for Physicists, M.P Hobson, G.P.Efstathiou and A.N.Lasenby (CUP, 2006) • General Relativity, Alan Heavens (not publicly available). • Theories of Gravity and Cosmology T. Clifton, P.G. Ferreira and C. Skordis (Physics Reports, 2012) • Cosmological Physics John Peacock (CUP, 1998) But there are many texts out there that you can consult. For a start, I will assume that the time coordinate, t and the spatial coordinate, x = 1 2 3 0 1 2 3 (x ,x ,x ) can be organized into a 4-vector (x ,x ,x ,x ) = (ct,x) where c is the speed of light. Throughout these lecture notes I will use the (−,+,+,+) convention for the metric. This means that the Minkowski metric is a matrix of the form   −1 0 0 0   0 1 0 0   η =  (1) µν  0 0 1 0  0 0 0 1General Relativity and Cosmology 2 You will note that this is the opposite convention to the one used when you were first learning special relativity. In fact you will find that, in general (but not always), books on General Relativity will use the convention we use here, while books on particle physics or quantum field theory will use the opposite convention. We will be using the convention that Roman labels (like i, j, etc) span 1 to 3 and label spatial vectors while Greek labels (such as α, β, etc) span 0 to 3 and label space-time vectors. We will also be using the Einstein summation convention. This means that, whenever we have a pair of indices which are the same, we must add over them. So for example, when we write 2 α β ds =η dx dx αβ we mean 3 3 XX 2 α β ds = η dx dx αβ α=0 β=0 Note that the paired indices always appear with one as a superscript (i.e. ”up”) and the other one as a subscript (i.e. ”down”). Finally, I am extremely grateful to Tessa Baker, Tim Clifton, Rosanna Hardwick, Alan Heavens, Anthony Lewis, Ed Macaulay and Patrick Timoney for their help in putting together the lecture notes, exercises and spotting errors. 1 Why General Relativity? Throughout your degree, you have learnt how almost all the laws of physics are invariant if you transform between inertial reference frames. With Special Relativity you can now write nd Newton’s2 law,theconservationofenergyandmomentum,Maxwell’sequations,etcinaway that they are unchanged under the Lorentz transformation. One force stands apart: gravity. Newton’s law of gravity, the inverse square law, is manifestly not invariant under the Lorentz transformation. You can now ask yourself the question: can we write down the laws of physics so they are invariant under any transformation? Not only between reference frames with constant velocity but also between accelerating reference frames. It turns out that we can and in doing so, we incorporate gravity into the mix. That is what the General Theory of Relativity is about. 2 Newtonian Gravity In these lectures we will be studying the modern view of gravity. Einstein’s theory of space- time is one of the crowning achievements of modern physics and transformed the way we think aboutthefundamentallawsofnature. Itsupersededaspectacularlysuccessfultheory,Newton’s theoryofgravity. IfwearetounderstandtheimportanceandconsequencesofEinstein’stheory, we need to learn (or revise) the main characteristics of Newton’s theory.General Relativity and Cosmology 3 Newton’s Law of Universal Gravitation for two bodies A and B with masses m and m A B at a separation r can be stated in a simplified form as: m m A B F =−G 2 r −11 3 −1 −2 where G = 6.672×10 m kg s . The resulting gravitational acceleration felt by mass m A is m B g =−G 2 r 1 2 3 We can rewrite the expression in terms of vectors r = (x ,x ,x ). r −r B A F =Gm m A B 3 r is the force exerted on the mass m and the gravitational acceleration A F =m g A Consider now a few simple applications. We can use the above expression to work out the gravitational acceleration on the surface of the Earth. If we Taylor expand around the radius of the Earth, R we find ⊕ M M M h ⊕ ⊕ ⊕ g =−G =−G ≃−G (1−2 ) 2 2 2 r (R +h) R R ⊕ ⊕ ⊕ where h is the height above the surface of the Earth and M is the mass of the Earth. With ⊕ 27 8 −2 M = 5.974×10 g and R = 6.378×10 cm we find a familiar value: g≃ 9.8ms . ⊕ ⊕ The power of Newton’s theory is that it allows us to describe, with tremendous accuracy, the evolution of the Solar System. To do so, we need to solve the two body problem. We have that the equations of motion are r −r B A m¨r =Gm m i A A B 3 r with r =r =r −r . We are interested in how r evolves and we can find its evolution by A B simplifying this system. Let us first define the total mass M =m +m A B and the reduced mass m m A B µ = M Ignoring the motion of the centre of mass, we have that the Lagrangian for this system will be 1 µM L = µ r˙·r˙ +G 2 rGeneral Relativity and Cosmology 4 It is more convenient to transform to spherical coordinates so 1 x = rcos(φ)sin(θ) 2 x = rsin(φ)sin(θ) 3 x = rcos(θ) then the Lagrangian becomes 1 µM 2 2 2 2 2 2 ˙ ˙ L = µ (r˙ +r θ +r sin (θ)φ )+G 2 r The angular parts of the Euler-Lagrange equations are d 2 2 2 ˙ ˙ µ (r θ) = µr sin(θ)cos(θ)φ dt h i d 2 2 ˙ µr sin (θ)φ = 0 dt We can integrate the second equation to give 2 2 ˙ µr sin (θ)φ = J φ 2 ˙ and, multiplying the first equation by 2r θ, we can integrate to find 2 J φ 2 4 2 2 ˙ µ r θ = J − θ 2 sin (θ) Wecannowchooseacoordinatesystemsuchthattheorbitliesontheequatorialplane(θ =π/2) ˙ and J =J =J which corresponds to θ = 0. We are left with two coordinates, r and φ. θ φ Replacing the the angular momentum conservation equation in the radial equation of mo- tion, we have 2 J µM µ r¨− +G = 0 3 2 µr r which we can integrate to give us an expression for the conserved energy 2 E 1 1 J M 2 = r˙ + −G (2) 2 2 µ 2 2µ r r Note that we can define an effective potential energy which is given by 2 1 J µM V (r) = −G eff 2 2µr r Again, using conservation of energy, we can use 2 Jdt =µr dφ to reexpress time derivatives in terms of angular derivatives 2 d J d d J d J d = and = ( ) 2 2 2 2 dt µr dφ dt µr dφ µr dφGeneral Relativity and Cosmology 5 The Euler Lagrange equation then becomes 2 J d J dr J µM − =−G 2 2 3 2 r dφ µr dφ µr r If we change to u = 1/r we have 2 d u µ +u = GµM (3) 2 2 dφ J This is the simple harmonic oscillator equation with a constant driving force. We can solve this equation to find v   u 2 u 1 Gµ (µM ) 2EJ t   u = = 1+ 1+ cos(φ−φ ) 0 2 2 r J µ (GµM ) where E is the conserved energy of the system (and an integration constant) and φ is the 0 turning point of the orbit (and the second integration constant). You should recognize this as the equation for a conic with ellipticity e given by v u 2 u 2EJ t e = 1+ 2 µ (GµM ) In other words, the trajectories of the two body problem correspond to closed orbits in the form of ellipses. The Solar System is almost perfectly described in terms of these trajectories. In fact almost all the planets have quasi-circular orbits with eccentricities e≃ 1−10%. Mercury stands out, with e≃ 20%. Furthermore, Mercury’s orbit does not actually close in on itself but precesses at a rate of about 5600 arc seconds per century. This primarily due to the effect of the other planets slowly nudging it around and, again, can be explained with Newtonian mechanics. But th since the mid 19 century, it has been known that there is still an un accounted amount of precession, about 43 arc seconds per century. We shall find its origin in later lectures. Finally, it is convenient to define the Newtonian potential or gravitational potential in terms of the potential energy, V, of the system: V ≡mΦ The gravitational force will then be F =−∇V and the gravitational acceleration is given by g =−∇Φ The gravitational potential satisfies the Newton Poisson equation 2 ∇ Φ = 4πGρGeneral Relativity and Cosmology 6 3 The Equivalence Principle In the last section, we worked with two equations: Newton’s Law of Universal Attraction and nd Newton’s 2 Law. In both of these equations a ”mass” appears and we assumed that they are one and the same. But let us now rewrite them and be explicit about the different types of mass coming in, the inertial mass m and the gravitational mass, m : I G m a = F I m M G G V = −G r We assumed that m =m I G but are they? Their equivalence has been tested to surprising precision with Lunar Laser Ranging observations. This involves bouncing a laser pulse off reflecting mirrors sitting on the surface of the moon and measuring its orbit as it sits in the joint gravitational field of both the Earth and the Sun. The distance between the Earth and the Moon is 384,401 km and has been measured with a precision of under a centimetre TheLunarRangingexperimentcanbeseenasatypeofEo¨tvos ExperimentwheretheEarth and Moon are the test masses sitting in the gravitational field of the Sun. A laboratory based Eo¨tvos Experiment can be constructed in the following way. Consider two masses made of different material attached to either end of a rod. The rod is suspended from a string on the surface of the Earth. Each of the masses will be subjected to two forces: the gravitational pull to the centre and the centrifugal force. Hence the rod will hang at an angle relative to the vertical direction. The rod is free to rotate if there is a difference in the gravitational acceleration between the masses. This can only happen if there is a difference between m and G m . I Wecanlookatthenumbershere. Considerthetwomasses,whichhavegravitationalmasses m and m and inertial masses m and m . Denote the component of the gravitational G1 G2 I1 I2 t force in the direction that makes the rod twist to be g and the acceleration of each of the t t masses to be a and a . We then have that 1 2 t t m a = m g I1 G1 1 t t m a = m g I2 G2 2 Iftheratios ofinertial togravitational massarethesamefor both bodies, thentheacceleration will be the same for both. Any difference in the gravitational and inertial masses will lead to a ”twist” in the pendulum which can be characterized in terms of a dimensionless paramater:   m m G1 G2 t t − a −a m m 1 2 I1 I2   η = = t t m m G1 G2 a +a + 1 2 m m I1 I2 Using beryllium and titanium, the current best constraint on η is t t a −a 1 2 −13 η = = (0.3±1.8)×10 t t a +a 1 2General Relativity and Cosmology 7 Earth’s rotation Fiber T m a I Twisting direction m g G Figure 1: The pendulum for the Eotvos Experiment which is a factor of 4 better than the original Eotvos experiment of 1922. To date, one of the most successful tests is to use the Earth-Moon system in the gravitational field of the Sun as a giant Eotvos experiment. The difference, with regards to the lab based Eotvos experiments, is that the masses of the test bodies (i.e. the Earth and Moon) are not negligible any more. The test can be done by using lasers reflected off mirrors left on the Moon by the Apollo 11 mission in 1969 and, as mentioned earlier, the constraint is −13 η = (−1.0±1.4)×10 This is one of the most accurately tested principles of physics and we expect this constraint to improve by 5 orders of magnitude when space based tests can be performed. If the gravitational and inertial masses are the same, then we have that a particle in a gravitational field will obey m G ¨r = g =g m I and this will be true of any particle. This has a profound consequence-it means that I can always find a time dependent coordinate transformation (from r to R), r =R+b(t) such that ¨ ¨ R =g−b(t) = 0 Inotherwords,itisalwayspossibletopickanacceleratedreferenceframesuchthattheobserver doesn’tfeelthegravitationalfieldatall. Forexample,consideraparticleatrestnearthesurface −2 of the Earth. It will feel a gravitational pull of g = −9.8ms . Now place it in a reference 1 2 frame such that b = gt (such as a freely falling elevator). Then the particle at rest in this 2 reference frame won’t feel the gravitational pull.General Relativity and Cosmology 8 z A A A A B B B B t t = 0 t = t t = " t = t +" 1 A 1 B first pulse first pulse second pulse second pulse emitted by A received by B emitted by A received by B Figure 2: Reference frames for the gravitational redshift Einsteinhadanepiphanywhenherealizedthis. Ashesaid”... for an observer falling freely from the roof of a house there exists- at least in his immediate surroundings- no gravitational field”. It led him to formulate the Equivalence Principle of which there are three versions that we will state here. • Weak Equivalence Principle (WEP): All uncharged, freely falling test particles follow the same trajectories, once an initial position and velocity have been prescribed. • Einstein Equivalence Principle (EEP): The WEP is valid, and furthermore in all freely falling frames one recovers (locally, and up to tidal gravitational forces) the same laws of special relativistic physics, independent of position or velocity. • Strong Equivalence Principle (SEP): The WEP is valid for massive gravitating objects as well as test particles, and in all freely falling frames one recovers (locally, and up to tidal gravitational forces) the same special relativistic physics, independent of position or velocity. Youwillnotethatthesethreeequivalentprincipleshavedifferentremits. Whilethefirstmakesa simple statement about the trajectories of freely falling bodies, the second one says something about the laws of physics obeyed by the freely falling bodies and the third addresses going beyond the simplified, point-like test mass approximation. We will use WEP, SEP and EEP interchangeably throughout these lectures although they can be tested in distinct ways. Let us briefly study one of the consequences of the Equivalence Principle. Consider two observers, one at position A which is at a height h above the surface of the Earth and another at a position B on the surface of the Earth. Observer A emits a pulse every Δt to be received A by observer B every Δt . What is the relation between Δt and Δt and how is it affected by B A B the gravitational field?General Relativity and Cosmology 9 Given what we have seen above, we can think of this as two observers moving upwards with an acceleration g. We have that the positions of A and B are given by 1 2 z (t) = gt +h A 2 1 2 z (t) = gt B 2 Now assume the first pulse is emitted at t = 0 by A and is received at time t by B. A 1 subsequent pulse is then emitted at time Δt by A and then received at time t +Δt . We A 1 B have that 1 2 z (0)−z (t ) = h− gt =ct A B 1 1 1 2 1 1 2 2 z (Δt )−z (t +Δt ) = h+ gΔt − g(t +Δt ) A A B 1 B 1 B A 2 2 1 2 ≃ h− gt −gt Δt =c(t +Δt −Δt ) 1 B 1 B A 1 2 wherewehavediscardedhigherordertermsinΔt. Combiningthetwoequations,andassuming t ≃h/c we have that 1 gh Δt ≃ (1+ )Δt A B 2 c In other words, there is gravitational time dilation due to the difference in the gravitational potentials at the two points Φ −Φ =−gh so B A   Φ −Φ B A Rate Received = 1− ×Rate Emitted 2 c This is known as the gravitational redshift of light. 4 Geodesics In the previous section, we saw how important accelerated reference frames could be. From the various equivalence principles, it seems that an accelerating reference frame will be indistin- guishable from a frame in a gravitational field. Let us then study how particles and observers travel through space. In flat space, a particle follows straight lines given by solutions to the kinematic equations of motion 2 α d x = 0 2 dτ α The particle will trace out a path in space-time x (τ) and the solution is of the form α α α x (τ) =x (τ )+u (τ−τ ) i i α α where x (τ ) and u are integration constants. iGeneral Relativity and Cosmology 10 In the presence of a gravitational field, the path taken by a particle will be curved. Alterna- tively, in an accelerating reference frame, the same will happen. We call Geodesics the shortest paths between two points in space-time. We have just written down the geodesic in a flat, force free space. If we wish to do so in the presence of a gravitational field, we need to solve the Geodesic equation. As Einstein argued, according to the principle of equivalence, there should α be a freely falling coordinate system y in which particles move in a straight line and therefore satisfy 2 α d y = 0 2 dτ where τ is the proper time of the particle and hence 2 2 α β c dτ =−η dy dy αβ µ Now choose a different coordinate system, x ; it can be at rest, accelerating, rotating, etc. We α µ α µ can reexpress the y s in terms of the x , y (x ). Using the chain rule we have α µ d ∂y dx 0 = µ dτ ∂x dτ α 2 µ 2 α µ ν ∂y d x ∂ y dx dx = + µ 2 µ ν ∂x dτ ∂x ∂x dτ dτ β α Multiplying through by the inverse Jacobian ∂x /∂y we end up with the geodesic equation with the form 2 β µ ν d x dx dx β +Γ = 0 (4) µν 2 dτ dτ dτ where we have defined the affine connection β 2 α ∂x ∂ y β Γ = µν α µ ν ∂y ∂x ∂x We can also express the proper time in these new (or arbitrary coordinates) as α β ∂y ∂y 2 2 µ ν µ ν c dτ =−η dx dx ≡−g dx dx αβ µν µ ν ∂x ∂x where we have defined the metric: α β ∂y ∂y g =η µν αβ µ ν ∂x ∂x The situation is slightly different for a massless particle. Neutrinos or photons follow null paths so dτ = 0. Instead of using τ we can use some other parameter σ. We then have 2 α d y = 0 2 dσ α β dy dy −η = 0 αβ dσ dσGeneral Relativity and Cosmology 11 ne One can repeat the same derivation as above to find 2 µ α β d x dx dx µ +Γ = 0 αβ 2 dσ dσ dσ α β dy dy −g = 0 αβ dσ dσ µ So, given an Γ and g , we can work out the equations in a given reference frame. αβ αβ µ It turns out that we can simplify the calculation even further: Γ can be found from g . αβ αβ Taking the partial derivative of the metric, we have that 2 α β 2 β α ∂g ∂ y ∂y ∂ y ∂y µν = η + η αβ αβ λ λ µ ν λ ν µ ∂x ∂x ∂x ∂x ∂x ∂x ∂x µ From the definitiion of Γ we can replace it in the above expression αβ α β β α ∂g ∂y ∂y ∂y ∂y µν γ γ = Γ η +Γ η αβ αβ λµ λν λ γ ν γ µ ∂x ∂x ∂x ∂x ∂x which can be written as ∂g µν γ γ = Γ g +Γ g γν γµ λµ λν λ ∂x We can now permute indices and add them to solve and find 1 ∂g ∂g ∂g µ αν νβ αβ µν Γ = g + − αβ β α ν 2 ∂x ∂x ∂x µ Hence, as advertised, given a metric, g , we can find the connection coefficents Γ and then µν αβ solve the geodesic equation. It often useful to find the geodesic equations in terms of a variational principle. In fact, it is also a convenient method for, given a metric, calculating the connection coefficients. Consider α a path in space time x (λ). We can define the proper time elapsed between two points on that curve, A and B, to be Z Z q B B α β cτ = Ldλ = dλ −η x˙ x˙ AB αβ A A The derivatives are taken with regards to λ. It is now possible to define an action for the path α x (λ): 2 S =−mc τ and we can minimize this action to find the path which takes the most amount of proper time 0 betweenpointsAandB. Thispathwillbethegeodesic. Forexample,ifwechoosex =cλ =ct we have that v u 2 Z u dx t 2 S =−mc dt c − dtGeneral Relativity and Cosmology 12 More generally, we can use the Euler Lagrange equations: d ∂L ∂L = α α dλ ∂x˙ ∂x α GiventhatLisindependentofx (anddτ =Ldλ)wehavethattheequationofmotionbecomes β β ∂L 1 1 dx 1 dx =− η =− η αβ αβ α ∂x˙ 2L dλ 2 dτ The resulting equations then become 2 α d x = 0 2 dτ Although the above action is reparametrization invariant (i.e. we can change our definition of λ and the action is unaffected), the square root is difficult to work with. It is easier to work with Z Z α β dx dx 2 ˜ S =−m dλη =m dλL αβ dλ dλ The Euler-Lagrange equation then becomes 2 2 ∂L d ∂L dL ∂L − =−2 µ µ µ ∂x dλ ∂x˙ dλ∂x˙ Again, if we choose the affine parameter λ to be linear in τ we have that the right hand side becomes dL d dτ = c = 0 dλ dλ dλ So, for a massive particle it makes sense to choose λ =τ. Finally we, can rederive the geodesic equation from the action principle as above, but in the presence of a gravitational field. We now have Z Z µ ν dx dx 2 S =m dλg =m dλL µν dλ dλ and can apply theEuler-Lagrangeequation to obtain theGeodesicequation in a general frame. 5 Coordinate Transformations and Metrics We have seen that coordinate transformations to accelerated reference frames lead to non- trivial geodesic equations. We have considered one such transformation when deriving the gravitational redshift. Let us now look at coordinate transformations in more detail. You have extensively studied coordinate transformations between inertial reference frames (in the absence of gravitational fields) in Special Relativity. For example, consider a coordinateGeneral Relativity and Cosmology 13 transformation along the z-axis from a stationary frame to one moving at a velocity v. We have that the coordinate transformation can be expressed in a matrix form as      0 0 x˜ γ 0 0 −βγ x  1    1  x˜ 0 1 0 0 x        =   2 2      x˜ 0 0 1 0 x 3 3 x˜ −βγ 0 0 γ x √ 2 where β =v/c and γ = 1/ 1−β . This coordinate transformation is of the form µ µ ν x˜ =x˜ (x ) and given that it is linear, we have that the Jacobian is simply:   γ 0 0 −βγ µ   ∂x˜ 0 1 0 0   =  ν   0 0 1 0 ∂x −βγ 0 0 γ Now, in special relativity we have that the space time interval 2 2 2 1 2 2 2 3 2 ds =−c dt +(dx ) +(dx ) +(dx ) is invariant under coordinate transformations. We can restate this as 2 α β α β ds =η dx dx =η dx˜ dx˜ αβ αβ in other words, the transformation leaves the actual form of the space time interval invariant. What happens if we now consider a coordinate transformation to an accelerating refer- ence frame? Let us consider the simplest case, an accelerating reference frame, with accelera- 3 tion g, along the x direction. We have that the transformation between the old coordinates, 0 1 2 3 0 1 2 3 (x ,x ,x ,x ) and the new coordinates, (x˜ ,x˜ ,x˜ ,x˜ ), is given by 2 3 gx˜ c g 0 0 2 c x = e sinh( x˜ ) 2 g c 1 1 x = x˜ 2 2 x = x˜ 2 3 gx˜ c g 3 0 2 c x = e cosh( x˜ ) 2 g c We can transform the expression for the space-time interval to the new coordinate system: 2 α β 0 2 1 2 2 2 3 2 ds =η dx dx = −(dx ) +(dx ) +(dx ) +(dx ) αβ 3 3 gx˜ gx˜ 2 0 2 1 2 2 2 2 3 2 2 2 c c = −e (dx˜ ) +(dx˜ ) +(dx˜ ) +e (dx˜ ) 3 Note that the equivalent gravitational potential has the form Φ =gx˜ and so the interval takes the form Φ Φ 2 2 0 2 1 2 2 2 2 3 2 2 2 c c ds =−e (dx˜ ) +(dx˜ ) +(dx˜ ) +e (dx˜ ) (5)General Relativity and Cosmology 14 In fact, this expression is valid in a more general setting than just a constant gravitational acceleration. A weak, static gravitational field, Φ is equivalent to a metric of this form. For the remainder of this section let us familiarize ourselves a bit more with coordinate transformations and how they affect the metric. With a general coordinate transformation, α α β x˜ =x˜ (x ) we find that the space time interval changes as α β ∂x ∂x 2 α β µ ν µ ν ds =η dx dx =η dx˜ dx˜ ≡g dx˜ dx˜ αβ αβ µν µ ν ∂x˜ ∂x˜ In other words, under a general coordinate transformation, we have that η → g . We call µν µν the object g the metric. µν The metric contains information about the geometry of space (and space-time) we are considering. It is instructive to work through a few examples. For example, consider a 2-D sheet plane, with coordinates (x,y). The interval on that plane is given by 2 2 2 ds =dx +dy and hence the metric is very simple: it is a diagonal matrix with entries 1 0 g = ij 0 1 We can transform to polar coordinates x =rcosθ y =rsinθ to find 2 2 2 2 ds =dr +r dθ Note that now the metric is more complicated: 1 0 g = ij 2 0 r yet it still describes a plane. We could have considered a different surface, a sphere with unit radius. Itisatwodimensionalsurfaceandhenceneedstwocoordinates,(θ,φ). Theinfinitesimal interval is defined as 2 2 2 2 ds =dθ +sin (θ)dφ with metric 1 0 g = ij 2 0 sin (θ) The geometry of the surface of a sphere is obviously very different to the geometry of a plane and it is in the metric that this information is encoded.General Relativity and Cosmology 15 We are of course, interested in the geometry of 3-D space and of 4-D space time. So, for example, the interval (and metric) for Euclidean (or flat) 3-D space in Cartesian coordinates are 2 1 2 2 2 3 2 ds = (dx ) +(dx ) +(dx ) and   1 0 0   0 1 0 g =  ij 0 0 1 In spherical coordinates 1 x = rsin(θ)cos(φ) 2 x = rsin(θ)sin(φ) 3 x = rcos(θ) we have that the interval (and metric) are 2 2 2 2 2 2 2 ds =dr +r dθ +r sin (θ)dφ and   1 0 0   2 g = 0 r 0  ij 2 2 0 0 r sin (θ) Again, these two metrics (in Cartesian and spherical coordinates) describe exactly the same space. We have already seen examples of space-time metrics above. The Minkowski metric and the metric of an accelerated observer (known as a Rindler metric). Letusnowconsidertwoimportantmetric. ThefirstoneisthatofaEuclidean,homogeneous and isotropic spacetime. We have that h i 2 2 2 2 1 2 2 2 3 2 ds =−c dt +a (t) (dx ) +(dx ) +(dx ) (6) which has a metric   −1 0 0 0   2 0 a (t) 0 0   g =  µν 2  0 0 a (t) 0  2 0 0 0 a (t) A particularly important metric is a static (i.e. time independent), spherically symmetric Schwarzschild metric:     −1 2GM 2GM 2 2 2 2 2 2 2 2 2 ds =− 1− c dt + 1− dr +r dθ +r sin (θ)dφ (7) 2 2 c r c rGeneral Relativity and Cosmology 16 which has a metric     2GM − 1− 0 0 0 2 c r     −1   2GM   0 1− 0 0 2 g = c r  µν   2  0 0 r 0  2 2 0 0 0 r sin (θ) This metric is of particular importance. It corresponds to the space time of a point like mass and can be used to describe the space time around stars, planets and black holes. To finish, let us calculate the geodesics for the homogeneous metric to find out what the connection coefficients are. The action is: X 2 2 2 2 i 2 ˙ L =−c t +a (t) (x˙ ) df ˙ where f = . The Euler-Lagrange equations are: dλ X ada 0 i 2 x¨ + (x˙ ) = 0 c dt 1 da i 0 i x¨ +2 x˙ x˙ = 0 acdt We can now read off the connection coefficients (and be careful not to over count with the factor of 2 in the second expression): 1 da 0 Γ = a δ ij ij c dt 1 da i i Γ = δ 0j j acdt 6 The Newtonian Limit and the Gravitational Redshift Revisited Interestingly enough, with what we have done we can already start relating the geodesic equa- tion with the Newtonian regime of gravity. Let us look at the case where the gravitational field is extremely weak and stationary and particles are moving at non-relativistic speeds so v≪c. Let us start with the geodesic equation: 2 µ α β d x dx dx µ +Γ = 0 αβ 2 dτ dτ dτ i The non-relativistic approximation means that dx/dτ ≪ d(ct)/dτ so the geodesic equation simplifies to 2 2 µ 0 d x dx µ +Γ ≃ 0 00 2 dτ dτ If the gravitational field is stationary we have that ∂g /∂t = 0 and we have µν 1 ∂g µ 00 µλ Γ =− g 00 λ 2 ∂xGeneral Relativity and Cosmology 17 Now consider a weak field g =η +h µν µν µν µ where we assume thath ≪ 1. We can expand Γ to first order in h to find µν 00 µν 1 ∂h 00 µ µν Γ =− η 00 ν 2 ∂x Only the spatial parts of η survive (which are 1). Hence we have µν 1∂h 00 µ Γ =− 00 µ 2 ∂x The geodesic equation then becomes 2 2 µ 0 d x 1 dx ∂h 00 = 2 ν dτ 2 dτ ∂x which in vectorial notation becomes 2 2 0 d r 1 dx = ∇h 00 2 dτ 2 dτ 2 d r For small speeds dt/dτ≃ 1 and comparing with the Newtonian result =−∇Φ we see that 2 dt 2Φ h =− 00 2 c Hence in the weak-field limit   2Φ g =− 1+ 00 2 c Note that, if we take the metric that we found in equation 5 and expand to linear order (i.e. take the weak field limit), we get exactly the same expression. We can rederive our expression for the gravitational redshift directly from the metric. Con- sider the proper time again 2 2 2 µ ν −ds =c dτ =−g dx dx µν i and pick a stationary system so dx = 0. We then have √ dτ = −g dt. 00 In the weak field case we have   Φ dτ≃ 1+ dt 2 cGeneral Relativity and Cosmology 18 Note that t and τ only coincide if Φ = 0 so clocks run slowly in potential wells. We can now compare the rate of change at two points, A and B, to get v u u dτ g (A) A 00 t = dτ g (B) B 00 which in the weak field limit gives the ratio of frequencies ν dτ Φ −Φ B A B A = ≃ 1− 2 ν dτ c A B which is equivalent to the expression we found in Section 3. We can define the gravitational redshift ν −ν Φ −Φ A B B A z ≡ = grav 2 ν c A −6 This is a very small effect- for the Sun it is∼ 10 . One way to look for this effect is by studying the spectral lines emitted from atoms which are very close to a massive body and hence deep into a gravitational potential. This effect has been observed in the Sun and white dwarfs but these observations are not very accurate. A classic test of the gravitational redshift was undertaken by Pound and Rebka in 1960 at 57 Harvard. They used a 22.5 metre high tower where they placed an unstable nucleus, Fe at the top and the bottom. The nucleus (at the top) would emit gamma-rays with a certain frequency related to their energy. These rays would fall to the bottom and interact with the 57 57 Fe there. If the gamma rays of the observer were the same as the emitter, the Fe at the bottom of the shaft would react. But because of the gravitational redshift, the frequency was shifted and the absorption was less efficient. By changing the velocity of the source at the top of the tower, the experimenters could compensate for the gravitational effect and measure it to within 1%. Bettermeasurementsofthegravitationalredshiftingoflightcanbeobtainedon(ornear)the Earth where, even though the gravitational field is much, much weaker, there is the possibility to make very precise measurements. One way to do this is to send a rocket up into orbit with 4 a hydrogen-maser clock and emitting pulses to a ground station. At an altitude of 10 Km, the 2 −10 change in gravitational potential will be gh/c ≃ 10 . Note that this effect is minute, almost 5 orders of magnitude smaller than the simple Doppler effect due to the motion of the rocket. Yet it is still possible to constrain the effect to within 0.002%. 7 Orbits I: the Perihelion of Mercury It is now time to revisit the two body problem. We have already worked this out for the Newtoniancasebutwecannowseewhathappensifweconsiderthemoregeneralcase. Strictly speaking we will be studying the motion of mass µ in a central potential sourced by a mass M. We can use the Schwarzschild metric that we introduced in equation 7 of the previous section. The total mass is M and the reduced mass is µ . The action for the geodesic equation for (t(λ),r(λ),θ(λ),φ(λ)) in this metric is   2 2GM r˙ 2 2 2 2 2 2 2 ˙ ˙ ˙ L = 1− c t − −r (θ +sin θφ ) 2GM 2 rc 1− 2 rcGeneral Relativity and Cosmology 19 The angular Euler-Lagrange equations are exactly as in the Newtonian two body problem we previously solved d 2 2 2 ˙ ˙ (2r θ) = 2r sinθcosθφ dλ d 2 2 ˙ (2r sin θφ) = 0 dλ and we can solve them in the same way, placing the orbit on the equatorial plane, choosing ˙ integration constants such that θ = 0 so that J 2 ˙ r φ = µ The timelike component of the geodesic obeys     d 2GM 2 ˙ c 1− t = 0 2 dλ rc which can be integrated to give   2GM ˙ 1− t =k 2 rc 2 2 For a massive particle we have L =c 2 2 2 2 c k −r˙ J 2   c = − 2 2 2GM µ r 1− 2 c r which can be rewritten   2 2GM J 2GM 2 2 2 2 r˙ + 1− =c k −c + 2 2 2 c r µ r r Rearranging we find 2 2 J 2GM 2GMJ 2 r˙ + − − = constant (8) 2 2 2 3 2 µ r r µ r c We can compare this expression with the one we found in the Newtonian case in equation 2: 2 h 2GM 2 r˙ + − = constant 2 r r There is an extra term in the General Relativistic case. Furthermore, in the Newtonian case we are taking derivatives with regards to t while in the General Relativistic case we are using the affine parameter λ. We would now like to find the orbits of motion in this system. As in the Newtonian case, using conservation of angular momentum we can change the independent variable from λ to φ. Furthermore, we can transform to u = 1/r so that drdφ 1 duJ J du 2 r˙ = =− u =− 2 dφdλ u dφµ µ dφGeneral Relativity and Cosmology 20 2 2 Dividing through by J /µ , equation 8 becomes 2 2 du 2GMµ 2GM 2 3 +u − u− u =constant 2 2 dφ J c du Differentiating by φ and dividing by 2 we find dφ 2 2 d u Gmµ 3GM 2 +u = + u 2 2 2 dφ J c You will see that this equation has a very similar form to equation 3 with an extra bit. It is 2 J useful to rescaleu so as to assess how important the correction is. DefineU = u. We then 2 GMµ have 2 d U 2 +U = 1+ǫU (9) 2 dφ 2 2 2 2 2 −7 with ǫ≡ 3G M µ /J c . We will see that ǫ in the case of Mercury, is of order 10 , so very small. We therefore assume that U can be split into a Newtonian part, U and a small, general 0 relativistic correction, U . We have that U = 1+ecos(φ) which we can plug into equation 9 1 0 to find (to lowest order): 2 2 2 d U e e 1 2 2 2 +U =ǫU =ǫ1+2ecos(φ)+e cos (φ) =ǫ1+ +2ecos(φ)+ cos(2φ) 1 0 2 dφ 2 2 The complementary function is as before but the particular integral takes the form " 2 2 e e U =ǫ 1+ +eφsin(φ)− cos(2φ) 1 2 6 With time, the dominant term will be proportional to φ so that, adding the complementary function and the particular integral we have U≃ 1+ecos(φ)+ǫeφsin(φ) This corresponds to the Taylor expansion of U≃ 1+ecosφ(1−ǫ) I.e. the period of the orbit is now 2π/(1−ǫ) and not 2π. The orbit does not close in on itself. 30 We can work out what this correction is for Mercury. Taking M ≃ 2×10 kg, the orbital 10 −7 period T = 88 days and the mean orbital radius r = 5.8×10 m, we find ǫ≃ 10 so that precession rate is approximately 43 arc seconds per century. This effect, first detected by Le th Verrier in the mid 19 century is obscured by a number of other effects. The precession of the equinoxes of the coordinate systems contributes to about 5025” per century while the other planets contribute about 531” per century. The Sun also has a quadropole moment which contributes a further 0.025” per century. Taking all these effects into account still leaves a precession of△θ for which the current best estimate is △θ = 42.969”±0.0052” per century The prediction from General Relativity is △θ≃ 42.98” per century

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