Lecture Notes for General Relativity Fall 2010

Transcription

1 Lecture Notes for General Relativity Fall 2010 Lecturer: Professor Lam Hui Transcriber: Alexander Chen December 9, 2010 Contents 1 Lecture Basic Information Newtonian Gravity General Relativity Applications of General Relativity Special Relativity Some Notations Lecture Transformations the Preserves Distance A Note about Notation Lorentz Transformations Integration of One-form Lecture Review of Last Lecture Tensor Notations Lecture Some note about Λ Matrices More Notations Particle Mechanics Electrodynamics Lecture Continue Electrodynamics Fluid Dynamics

2 6 Lecture Einstein s Happiest Thought Equivalence Principles Some Remarks on Metric Lecture The Geodesic Equation Altenative Derivation Using Action Principle Lecture An Example Lecture Newtonian Limit Gravitational Redshift More Formalism Lecture Another Look at the Geodesic Equation Digression on Parallel Transport Lecture Principle of General Covariance Riemann Curvature Tensor Lecture Alternative Way of Looking at the Riemann Tensor The Einstein s Equations Lecture Return to Einstein s Equation Lecture Linear Perturbation Theory Newtonian Limit Lecture Summary of Last Lecture Study of the Newtonian Metric Lecture Gravitational Waves Lecture Particles in Gravitational Waves Solution to the Nonlinear Einstein Equations

3 18 Lecture The Schwartzschild Solution The Orbit of Mercury Lecture Perihelion Precession of Mercury Gravitational Redshift Lecture Going Across the Horizon Lecture Penrose Diagrams Lecture Metric for a Homogeneous, Isotropic and Expanding Universe Lecture Lecture Lecture Lecture Time Travel

4 General Relativity Lecture 1 1 Lecture Basic Information Course Instructor: Lam Hui Office Hour: Tue 4:30-5:30 pm Textbook: Gravitation and Cosmology by Steven Weiberg & Spacetime and Geometry by Sean Carroll 1.2 Newtonian Gravity The general theory of relativity is a theory of gravity. We want to contrast it to Newtonian gravity, which is characterized by the following two aspects: 1. The equation for the gravitational potential: 2 ϕ = 4πGρ 1.1) where ϕ is the gravitational potential, G the Newton constant, and ρ the matter density. 2. The equation of motion for a test particle in the gravitational field: F = m ϕ = ma 1.2) where an important assumption is made that the gravitational mass is equal to the inertial mass. 1.3 General Relativity The structure of the theory of general relativity is similar to the Newtonian one. We also have two aspects: 1. The field equation: G µν = 8πGT µν 1.3) which is the relation between the metric describing the space-time geometry) and the energy and momentum content in the space-time. 2. The equation of motion of a test particle, which is the geodesic equation. 1.4 Applications of General Relativity There are many applications of general relativity, for example: 1. Gravitational lensing 2. Cosmology 3. Black holes 4. Gravitational waves 5. Hawking Radiation 6. Time-machines 4

5 General Relativity Lecture Special Relativity We shall begin our discussion of general relativity by reviewing special relativity, and develope the language we use in discussing the general theory. First we consider 2D Euclidean space. Let us denote the distance between two arbitrary points as s. The distance can be calculated by Pythagoras theorem: s 2 = x 2 + y 2 1.4) If we rotate the coordinates, the same expression holds for distance. Distance between two points is invariant under rotation. Now we come to 3+1 dimensional space-time. Here the analog of a coordinate transformation like rotation is a change of reference frame. The points in the space-time are not physical points but rather events. The space-time distance between two arbitrary points is: s 2 = c t) 2 + x 2 + y 2 + z 2 1.5) which is unchanged under a rotation in the 3+1 dimensional space-time. From now on we will take c = 1 for notational brevity. Let us look for the physical meaning of the space-time distance s. Let us define τ 2 = s 2. Suppose we can find a frame of reference such that x 2 = 0, i.e. the observer finds the events A and B to happen at the same place, then τ is just the proper time measured in this rest frame. Such a frame can be found when τ 2 > 0, or s 2 < 0. Such separation in space-time is called time-like. Suppose the opposite. If we have τ 2 < 0 and s 2 > 0 then we can find a frame in which A and B happens at the same time. Such separation is called space-like because now s is the physical distance measured in this frame. What if τ 2 = s 2 = 0? Then we say that the separation is null or light-like, because it can only be along the trajectory of a light beam. 1.6 Some Notations We will denote a vector in 3+1 space-time as x µ where index µ runs from 0 to 3. We set x 0 = t. We will denote a vector in 3D Euclidean space as x i where the index runs from 1 to 3. We require x i = x, y, z). We write the space-time distance in Einstein s notation: s 2 = η µν x µ x ν 1.6) where summation is assumed at repeated upper and lower indices. The quantity η µν is a 4 4 matrix with 16 entries in general. But here η = diag 1, 1, 1, 1). This is called the Minkowski metric. To Be Continued... 5

6 General Relativity Lecture 2 2 Lecture Transformations the Preserves Distance Now we ask the question: what coordinate transformation will preserve the distance between two arbitrary points? First we consider the case of R 3, the 3 dimensional Euclidean space. Remember we defined the distance in this space to be δs 2 = δ ij δx i δx j. If we have a coordinate transformation x x, we want to preserve this distance: so what transformations are allowed? δs 2 = δ ij δx i δx j = δ ij δx i δx j 2.1) Translation. Obviously translation will not modify the relative coordinates of any two points. x i = x i Rotation. To see this, let s imagine a transformation x i = R i j xj where R is a 3 3 matrix. We demand that: δ ij x i x j = δ ij R i jr j l xk x l = δ kl x k x l 2.2) i.e. we require that R T R = I, which means that R O3). Here we just need the part which is connected to identity, therefore we take R SO3). Now we pass to Minkowski space, R 3,1. Similarly we demand: δs 2 = η µν δx µ δx ν = η µν δx µ δx ν 2.3) We write the transformation as x µ = Λ µ νx ν, and ask questions about Λ. Translations are still allowed. We now need the transformation to satisfy η µν Λ µ σλ ν ρ = η σρ, i.e. Λ T ηλ = η. This is the generalization of the defining relation for orthogonal group in Euclidean space. We say that Λ SO3, 1). Again we have specialized to the part connected to identity. Note that in our argument we have assumed that the coordinate transformation is linear. This is because we require the distance function to remain second order, which is in general not the case if the transformation is nonlinear. 2.2 A Note about Notation When we write A µν = B α µ C αν then in matrix form it is just A = BC. However if we write A µν = B α µc αν, then the corresponding matrix form is A = B T C. We contract neighboring indices in matrix multiplication and a transpose is needed if non-neighboring indices need to be contracted. 6

7 General Relativity Lecture Lorentz Transformations Let us do a little counting first. Because the matrix Λ is a 4 4 matrix, it has 16 entries. However we have a matrix equation Λ T ηλ = η which constrains the value of Λ. Because η is a symmetric matrix, there are 10 independent constraint equations. Therefore there are = 6 degrees of freedom. Among those 6 degrees of freedom, 3 are just rotations in 3D space. Now we want to show that the rest 3 are Lorentz boosts in the 3 directions. Consider the transformation in x and t direction discard the other 2 coordinates first): Λ 0 0 Λ 0 ) T ) Λ 0 0 Λ 0 ) Λ 1 0 Λ Λ 1 0 Λ 1 = 1 ) Writing out the equations for individual components, we get the equations: Λ 0 0 )2 + Λ 1 0 )2 = 1 Λ 0 0 Λ0 1 + Λ1 0 Λ1 1 = 0 Λ 0 1 )2 + Λ 1 1 )2 = 1 Solving these equations, we can get: 2.4) γ vγ 0 0 cosh ξ sinh ξ 0 0 Λ = vγ γ = sinh ξ cosh ξ ) where γ = is the well-known γ factor in special relativity, and the ξ, the rapidity, is defined by 1 v 2 cosh ξ = γ and tanh ξ = v. 2.4 Integration of One-form Suppose we have a particle propagating in the spacetime and traces out a time-like worldline from A to B. We want to calculate the total s from A to B that was traversed by the particle, i.e. we should write the differential form of the distance ds 2 = η µν dx µ dx ν and integrate it with respect to a parameter, say t: s = B A η µν dx µ dt dx ν B dt dt = 1 + v 2 dt 2.6) A However for a time-like worldline the above integration usually yields an imaginary number, because the speed is less than 1. In this case we should instead integrate the proper time dτ: τ = B A η µν dx µ dt which makes a lot more sense for a usual particle. dx ν dt = B A 1 v 2 dt 2.7) 7

8 General Relativity Lecture 3 3 Lecture Review of Last Lecture Recall from last time, we denote a Lorentz transformation as x µ = Λ µ αx α. We showed last time that the Lorentz transformations are those that preserve length in the Minkowski space R 3,1. Which means: ds 2 = η µν dx µ dx ν = η αβ dx α dx β 3.1) And the equality obeyed by the Lorentz transformation matrices is: the former is in component form and the latter is in matrix form. 3.2 Tensor Notations η αβ = η µν Λ µ αλ ν β, η = ΛT ηλ 3.2) Some definitions first: Vector: is an object which transforms like V µ = Λ µ αv α. For example, dx µ is a vector, the 4-velocity u µ = dxµ dτ is also a vector because dτ is a scalar under Lorentz transformations. 1-Form: Recall ds 2 = η µν dx µ dx ν. We can group the first two terms on RHS and call it dx µ = η µν x ν. This is called a 1-form. More accurately, a 1-form is an object that contracts with a vector to form a scalar. Now we can write ds 2 = dx µ dx µ. As an example, let s consider 1-forms in R 3,1. The expression is x µ = η µν x ν, and we can evaluate the components: dx 0 = dx 0, dx 1 = dx 1, dx 2 = dx 2, dx 3 = dx 3 3.3) In the previous definiton we say that the index of x ν is lowered by the metrix η µν. We can similarly get a vector from a 1-form by applying the inverse of the metric: dx ν = η νµ dx µ 3.4) note that η µν are components of η 1, instead of η itself. Somewhat confusing notation, but it makes sense because η µν η να = δ µ α. The rule of thumb is: use metric to lower indices and use its inverse to raise indices. Let s consider the object Λµ ν = η νβ η µα Λ α β, obtained by raising and lowering appropriate indices of the Lorentz transformation matrix. Written in matrix form, we call this object Λ, and it is the inverse of the transposed Lorentz transformation, Λ T Λ = 1. Written in components form, this means Λ µ α Λµ ν = δα ν. Now we are finally in a position to define a 1-form like we defined the vector. Let s consider how a 1-form transforms under Lorentz transformations. Recall that dx µ = η µν dx ν, and we know how dx ν transforms. We require that dx µ = η µν dx ν in the new frame. Working out the mathematics: dx µ = η µν Λ ν αdx α = η µν Λ ν αη αβ dx β = Λ β µ dx β 3.5) 1-Form: is an object which transforms like dx µ = Λ ν µ dx ν. This definition is similar to our definition of a vector. This definition is good because if we take any vector v µ and any 1-form u ν and contract them, the result s = v µ u µ is a scalar: s = v µ u µ = Λ µ νv ν Λ ρ µ u ρ = δ ρ ν v ν u ρ = v ν u ν = s 3.6) 8

9 General Relativity Lecture 3 This is the main reason to introduce these cumbersome notations. Question: Is ϕ a vector or 1-form? We need to look at its transformation properties: x µ ϕ xα ϕ = x µ x µ x α 3.7) Let s consider the coordinate transformation x µ = Λ µ νx ν, so we know xν directional derivative is a 1-form, so we write it as µ ϕ. x µ = Λ ν µ. Therefore the 9

10 General Relativity Lecture 4 4 Lecture Some note about Λ Matrices Remember Λ µ ν is the components of the Lorentz transformation matrix Λ, and we have x µ = Λ µ νx ν. The transformation for 1-forms is Λµ ν which we denote Λ in matrix form. Remember we said Λ T Λ = 1. But the transpose is very confusing. It is better just use index notation throughout. The transpose is only necessary because we are contracting the first indices from both matrices. 4.2 More Notations Recall: A vector transforms as V µ = Λ µ νv ν A 1-form transforms as W µ = Λ ν µ W ν Last time we found that ϕ/ x µ is a 1-form, which we denote µ ϕ. Now let s define some more interesting objects. Given two positive integers, we can form a m, n) tensor m upper indices) whose components are T n lower indices). We will be interested, just like the case in vectors and 1-forms, in the transformation properties of these tensors. The metric tensor g µν is a 0, 2) tensor. The transformation of a tensor, say, of the form 2, 1), then it looks like T µν ρ. It transforms like: T µν ρ = Λ µ αλ ν β Λ γ ρ T αβ γ 4.1) Basically we can play around with the metric and inverse metric tensors to raise and lower indices whatever we want. 4.3 Particle Mechanics We define the four-velocity of a particle to be: ) U µ = dxµ dt dτ = dτ, dxi = γ, γv i) 4.2) dτ where v i is the actual velocity in the 3-space. The 4-velocity rotates nicer in Lorentz transformations, whereas the 3-velocity will get mixed up with the time component. We can similarly define 4-momentum of a particle as: P µ = mu µ = γm, γmv i) nonrelativistic limit m + 12 mv2, mv i ) where m is the rest mass of the particle. Note that the 0-component of the 4-momentum is the energy of the particle E = P 0 = γm. Let s construct a Lorentz invariant out of the 4-momentum by contracting it with itself: P µ P µ = m 2 U µ U µ = m 2 dx µ If we translate the equation a little bit, and expand the summation, we can see that: dτ dx µ dτ 4.3) ) = m 2 4.4) 10

11 General Relativity Lecture 4 E 2 = p 2 + m 2 4.5) which is the famous formula from special relativity. Note that the rest mass is frame-independent, therefore a Lorentz invariant. This is ok for a massive particle, like the electron, but for a massless particle the 4- velocity is not well defined this way. For a massless particle, τ is no longer a good parametrization of the worldline, as it is zero everywhere. We then use a parameter λ called the affine parameter, and define U µ = dx µ /dλ. The magnitude of the 4-velocity is just zero because: U µ U µ = dxµ dλ dx µ dλ = 0 4.6) If we define the momentum as P µ = U µ then we also have P µ P µ = 0, and E 2 = p Electrodynamics Recall the Maxwell equations: { E = ρ Sourceful B E t = J 4.7) { E + B Sourceless t = 0 4.8) B = 0 Claim: ρ and J form a 4-vector J µ we will not prove). conservation: We have an equation expressing charge µ J µ = ρ t + J = 0 4.9) which is just the continuity equation which we are familiar with in classical electrodynamics. Note that written in this form, the equation is manifestly independent of frames, i.e. Lorentz invariant. Now we want to consider electromagnetic field. Let us look at the first Maxwell equation: i E i = ρ. Because we know that ρ is part of a 1-index object, and the differential operator is a 1-form, so we expect the electrodynamic field to have 2 indices, i.e. a rank-2 tensor. However, a rank-2 tensor has 16 degrees of freedom which is way too many if we only want electromagnetic field, which has only 6 degrees of freedom. A simple solution and correct) is to require the rank-2 tensor to be antisymmetric, which reduces the degree of freedom to 6. Now we have enough motivation to define the electromagnetic tensor: 0 E 1 E 2 E 3 F µν = E 1 0 B 3 B 2 E 2 B 3 0 B ) E 3 B 2 B 1 0 Check that F µν transforms correctly as rank-2 tensor, and that the two sourceful Maxwell equations will become: µ F µν = J ν 4.11) 11

12 General Relativity Lecture 5 5 Lecture Continue Electrodynamics Remember last time we introduced the Maxwell equations: E = ρ Sourceful B E t = J 5.1) E + B Sourceless t = 0 5.2) B = 0 and we said that the sourceful equations can be greatly simplified by the introduction of the field strength tensor: µ F µν = J ν 5.3) The advantage of writing equations in this form is that we can easily know what are the values of corresponding physical variables in another frame just do Lorentz transformations), and that the form of equations in different frames remains the same. In order to write down the other two Maxwell equations we need to define another tensor called the dual of the field strength tensor: F µν = 1 2 εµναβ F αβ 5.4) The ε µναβ is called the Lavi-Civita symbol in 4 dimensions, and is defined by the following 1, if µναβ) is an even permutation of 0123) ε µναβ = 1, if µναβ) is an odd permutation of 0123) 0, otherwise 5.5) Note that ε µναβ = ε µναβ. And both objects are rank 4 tensors under Lorentz transformations checked in homework). With the aid of the dual tensor, we can simplify the last two Maxwell equations into µ F µν = 0 5.6) To get a flavor of what exactly is the dual field tensor, let s evaluate one entry, say F 01 F 01 = 1 2 ε 0123 F 23 + ε 0132 F 32 ) = F23 5.7) If we carry out all the calculations we can find the dual tensor to have the following value: 0 B 1 B 2 B 3 F µν = B 1 0 E 3 E 2 B 2 E 3 0 E 1 5.8) B 3 E 2 E

13 General Relativity Lecture 5 The familiar Lorentz force law for a charged particle moving inside a electromagnetic field can be simplified with the introduction of field tensors, too: dp µ dτ = qg µν dxσ νσf dτ with a little bit of algebra we can see this gives the usual form of Lorentz force law. Note that the zeroth component of this equation gives the rate of change of energy of the charged particle. 5.2 Fluid Dynamics Suppose we have a bunch of particles, and we want to describe the physics of this collection. We have some variables of particular interest. First is the number density n, defined as the number of particles per unit volume in the rest frame of the collection. We can also introduce a collective velocity of the particles as U µ. Using these two variables we can introduce the number current or number flux, as the dimension is number per area per time): N µ = nu µ 5.10) In particular N 0 = γn gives the number density in the boosted frame. Very similar to the idea of 4-current in Electrodynamics. How about energy density? The energy density can t be a scalar, because it is not invariant under Lorentz transform. When it transforms under a boost, it gets two γ factors, therefore we expect it to be part of a rank 2 tensor. This object is usually denoted T µν and is called the energy-momentum tensor. Let s consider an example. Consider a cloud of dust, i.e. a collection of non-interacting massive particles. The energy-momentum tensor of such a system is: 5.9) T µν = P µ N ν = mnu µ U ν 5.11) The coefficient mn is nothing other than the restframe energy density if we think of mass as energy). If we consider T 00 = mnγ 2 is just the energy density in lab frame. How about other components? Consider T 0i : T 0i = P 0 N i = mnγ 2 v i 5.12) which has the meaning of energy flux in lab frame. Similarly T i0 = P i N 0 which can be interpreted as the momentum density. Because by definiton T is symmetric, the energy flux equals the momentum density. Finally let s look at T ij : T ij = P i N j 5.13) which has the dimension of momentum per unit area per unit time, which can be called pressure. Note that this is a more general form of pressure, but it is not isotropic, so we usually call it stess. Because of this, the energy-momentum tensor is also sometimes called the stress tensor. Let s now study the energy-momentum tensor of a perfect fluid. In local fluid rest frame, the tensor is T µν = diagρ, P, P, P ) 5.14) where ρ is the energy density at the local coordinate and P is the physical pressure. The dust example above is a special case of a perfect fluid, where local pressure is zero. The above energy-momentum tensor is written in the local rest frame. We want to find the expression in some other frame. Recall that dust has T µν = ρu µ U ν, but we want one which is more general than 13

14 General Relativity Lecture 5 this. Let s make a guess T µν = ρ + P )U µ U ν, but it doesn t work. To correct this we add a term to give the right P dependence: T µν = ρ + P )U µ U ν + P η µν 5.15) Because this is a tensor by definiton, and it is right in the local rest frame, so we can conclude that this is the right definition for all frames. It is instructional to work out the explicit components of this tensor: T µν ρ + P )γ = 2 P ρ + P )γ 2 v i ) ρ + P )γ 2 v j ρ + P )γ 2 v i v j + P δ ij 5.16) In general T µν is symmetric T µν = T νµ 5.17) This is true by our definition above, but there is physical reason behing it. Ref. Schutz Chapter 4.5. In electrodynamics we have charge conservation. Here we have energy-momentum conservation constraints on the energy-momentum tensor: µ T µν = ) We want to plug the above tensor for perfect fluid into this equation. First we take the nonrelativistic limit γ 1 and P ρ. The resulting equations are: ρ t + iρv j ) = ) t ρvj ) + i ρv i v j + P δ ij) = ) = vj t + vi i v j = j P ρ 5.21) 14

15 General Relativity Lecture 6 6 Lecture 6 Starting from this lecture we will be talking about General Relativity. 6.1 Einstein s Happiest Thought Galileo observed that the inertial mass is equal to the gravitational mass, i.e. different objects fall at the same acceleration under gravitation. The inertial mass is the mass which shows up in Newton s second law, which relates force with acceleration. The gravitational mass, however, is of totally different origin. It is the coefficient of the gravitational field which shows up in Newton s law of gravitation: F = G Mm r 2 6.1) These two quantities of totally different origin turns out to be the same, so all bodies fall at the same rate under gravity. Now what is the implication of this assertion? 1. Suppose somebody is trapped in a free falling elevator. He feels weightless because he falls with the elevator at the same rate, i.e. as if there is no gravity. 2. Suppose somebody is trapped in a elevator. There is no gravitational field around, but the elevator is being accelerated upwards. Because there is force on your feet, it feels like that there is gravitational force acting on you. This means that an accelerating frame acts as though this is gravity. Einstein pushed this thought to an extreme, saying that there is absolutely no difference between an accelerating frame and gravitation. Before we examine his thought, let s firstly contrast gravity with E&M. Suppose we have 2 test particles with masses m 1, m 2, and charges q 1, q 2. Say the particles are on the surface of a planet of mass M. The motion of the two particles in the gravitational field are the same: d 2 r 1 dt 2 = d2 r 2 dt 2 = GM R 2 6.2) Now if we go to a frame r = r + GMt 2 /2R 2, then in that frame we have d 2 r /dt 2 = 0 for both particles. This is the frame where gravitational force does not exist. Note that in this local frame all objects do not feel gravity. The crucial thing is that in one transformation we can eliminate gravity from all test particles. How about E&M? The motion of the test particles on a planet with charge Q suppose such planet exists), then their equations of motion are: d 2 r 1 dt 2 = Q R 2 q 1 m 1, d 2 r 2 dt 2 = Q R 2 q 2 m 2 6.3) But the charge-to-mass ratio is not a universal constant in the universe, so different things fall in different rate, and we can t find one single frame where electric forces go away. As shown in the above example, we want to consider general coordinate transformations for gravity, i.e. x = x x), which may be highly nonlinear, unlike the Lorentz transformations. This seems necessary to describe gravity in corresponding frames. Also we can only transform away gravity locally, i.e. there is in general no global transformation to eliminate gravity in all space. This local gravity-free frame is called a local inertial frame. 15

16 General Relativity Lecture Equivalence Principles We are now ready to introduce the equivalence principle. There are several versions of this principle: 1. Weak Equivalence Principle) At every space-time point in some gravitational field, it is possible to choose a locally inertial coordinate system where test particles move as if there is no gravity. 2. Einstein Equivalence Principle) At every space-time point in some gravitational field, it is possible to choose a locally inertial coordinate system where the laws of physics takes the same form as in an unaccelerated frame in the absence of gravity i.e. special relativity). This is much more general than the above, because it covers all laws of physics, including E&M, QM, etc.) 3. Strong Equivalence Principle The statement that inertial mass equals gravitational mass includes all possible contribution to mass, including gravitational binding energy itself. The last statement is very hard to test, as in usual objects the gravitational binding energy is extremely weak. The only objects in which gravitational contribution to mass is significant are black holes and neutron stars. In order to consider general coordinate transformations, we need to go beyond Minkowski metric, and consider ds 2 = g µν x)dx µ dx ν 6.4) where g µν is in general dependent on spacetime coordinate and g µν η µν. g µν will be our gravitational field, which is defined to be symmetric, as the antisymmetric part does not have any physical meaning. As we talked about in the first lecture, we will consider two kinds of questions: 1. How does mass source g µν? 2. Given g µν, how do test particles move? The second question is easier to answer, since we can always use the equivalence principle to locally eliminate the gravitational force, so that particles will only move in a straight line. But the question becomes how to cleverly choose such a local inertial frame. 6.3 Some Remarks on Metric Remember we defined ds 2 = g µν dx µ dx ν to be the invariant interval under any coordinate transformation. That is to say, the meric transforms as: g µν = xα x µ x β x ν g αβ 6.5) In special relativity the partial derivatives are just Λ matrices, because the transformation was linear. Now we consider more general transformations. Note that in special relativity the Minkowski metric is invariant under Lorentz transformations, this is just a special case of the above transformation law. It is useful to remember some simple examples which builds our intruition. For example we have the cartisian coordinates in R 2 : ds 2 = dx 2 + dy 2, g µν = diag1, 1) 6.6) We can also consider R 2 in a totally different frame, i.e. the polar coordinates, where ds 2 = dr 2 + r 2 dθ 2, g µν = diag1, r 2 ) 6.7) 16

17 General Relativity Lecture 6 We have the transformation law from cartesian coordinates to polar coordinates: g mn = g ij x i x m x j x n 6.8) Check that this transformation works, yielding the explicit coordinates we wrote down above. Another example is S 2, which is embedded in R 3. Consider first the inherited metric from R 3 which is flat. By the relation x 2 + y 2 + z 2 = R 2, we have z = ± ) R 2 x 2 y 2 xdx, dz = R 2 x 2 y + ydy 6.9) 2 R 2 x 2 y 2 so we can get the metric to be: ds 2 = R2 y 2 R 2 x 2 y 2 dx2 + R2 x 2 2xy R 2 x 2 y 2 dy2 + R 2 x 2 dxdy 6.10) y2 We can also use spherical coordinates x = R sin r R cos θ y = R sin r R sin θ z = R cos r R 6.11) and we can obtain ds 2 = dr 2 + R 2 sin 2 r R dθ2 6.12) with the range of variables 0 θ < 2π and 0 r πr. This metric looks very like the polar coordinates for R 2, but with different topology. 17

18 General Relativity Lecture 7 7 Lecture The Geodesic Equation The this lecture and next, we will attempt to replace the Newton s second law by a generalized equation which describes the motion of a particle under gravitational field, which is just the geodesic equation. The essence of the geodesic equation is that we don t think of gravity as a force on the particle, but the particle just falls freely in a curved space-time. There are two ways that we can derive the geodesic equation, first is from the equivalence principle, and the next is from the principle of least action. We will do the former in this lecture. Remember the equivalence principle which says that there always exists a local inertial coordinate system in which there is no gravity, i.e. space-time is flat ds 2 = g µν dx µ dx ν = η µν dξ µ dξ ν 7.1) where ξ µ is the local coordinates where the space-time is flat. The statement that the particle feels no force translates into d 2 ξ dτ 2 = 0 7.2) where τ is the affine parameter and dτ 2 = ds 2. This equation is written in the local inertial frame, so it is natural to ask what does it look like in another frame? Let s do a coordinate transformation: d dξ α dτ dτ = d dτ [ ξ α dx µ x µ dτ ] = 0 7.3) ξ α d 2 x µ x µ dτ ξ α dx ν dx ν x ν x µ dτ dτ = 0 7.4) We multiply the second equation by a term x λ / ξ α, we can write it in a new and illuminating way d 2 x λ dτ 2 + dx µ dx ν Γλ µν dτ dτ = 0 7.5) where Γ λ µν = xλ 2 ξ α ξ α x µ x ν 7.6) is called the Christoffel symbol. From the definition it is evident that the µν indices are symmetric because partial derivatives are symmetric. However it is not a tensor although it has three indices, and we will return to this issue later. The equation 7.5) is called the geodesic equation, and it explicitly involves the geometry of spacetime. Now we want to write the Christoffel symbol in terms of the metric g µν. Remember we changed the coordinates from the original space-time to a flat inertial one, so ξ α ξ β g µν = η αβ x µ x ν 7.7) we want to evaluate the derivative of metric, and identify terms in the Christoffel symbol 2 ξ α ξ β λ g µν = η αβ ξ λ x µ x ν + ξα 2 ξ β ) x µ x λ x ν ξ α ξ β = η αβ x γ Γγ λµ x ν + ξα ξ β ) 7.8) x µ x γ Γγ λν = g γν Γ γ λµ + g µγγ γ λν 18

19 General Relativity Lecture 7 Now we will massage the equation a little bit and create some different copies of it by changing around indicies, and add them up to isolate one copy of Γ: Γ λ µν = 1 2 gλρ µ g ρν + ν g ρµ ρ g µν ) 7.9) 7.2 Altenative Derivation Using Action Principle Remember the proper time defined by d 2 τ = d 2 s is a manifestly invariant quantity under coordinate transformation. If we want to construct an invariant action it is a good starting point. So let us define the action to be dx S = dτ = g µ dx ν µν dλ 7.10) dλ dλ where λ is an affine parameter of the motion of the particle. This action makes sense because a particle should minimize the proper time it travels when it feels no force. We carry out the variation as follows we use dot for derivatives with respect to λ) δs = = = = dλδ g µν ẋ µ ẋ ν = dτ δg µνẋ µ ẋ ν ) 2 g µν ẋ µ ẋ ν 1 dλ 2 g µν ẋ µ ẋ σg ν µν ẋ µ ẋ ν δx σ + g µν δẋ µ ẋ ν + g µν ẋ µ δẋ ν ) 1 dλ 2 g µν ẋ µ ẋ σg ν µν ẋ µ ẋ ν δx σ σ g µν ẋ σ ẋ ν δx µ σ g µν ẋ σ ẋ µ δx ν 2g µν ẍ µ δx ν ) 1 dλ 2 g µν ẋ µ ẋ σg ν µν ẋ µ ẋ ν µ g σν ẋ µ ẋ ν ν g µσ ẋ ν ẋ µ 2g µσ ẍ µ )δx σ 7.11) Because the variation is arbitrary, the term in the bracket must vanish, and we have our equation of motion: d 2 x ρ dλ 2 + dx µ dx ν Γρ µν dλ dλ = ) where Γ is the same symbol defined above. It is the exactly the same equation we got above. 19

20 General Relativity Lecture 8 8 Lecture 8 Note that in our definition of a local inertial frame, we left out one condition, which is that the derivative of the metric should also vanish, i.e. g µν = η µν, Γ α µν = 0 8.1) 8.1 An Example Let us we write down a metric which describes a worm hole: ds 2 = dt 2 + dr 2 + b 2 + r 2 )dθ 2 + sin 2 θdφ 2 ) 8.2) This looks like a connected hole in the embeding diagram. b is the radius of the throat of the hole. We need to calculate the Christoffel symbols for this metric. The metric is just g µν = b 2 + r 2 ) 0 8.3) b 2 + r 2 ) sin 2 θ ) So the inverse metric is just g µν 1 = diag 1, 1, b 2 + r 2, 1 b 2 + r 2 ) sin 2. Let s calculate all the nonzero θ Christoffel symbols and they are: Γ r θθ = r, Γr φφ = r sin2 θ, Γ θ rθ = r b 2 + r 2 8.4) Γ θ φφ = sin θ cos θ, Γφ rφ = r b 2 + r 2, Γφ φθ = cot θ 8.5) We can plug in these Christoffel symbols into the geodesic equation and get the equation for components: ẍ t = ẗ = 0 8.6) r + Γ r θ θθ 2 + Γ r φ φφ 2 = r r θ 2 r sin 2 θ φ 2 = 0 8.7) θ + 2Γ θ θṙ rθ + Γ θ φφ = θ + 2r b 2 + r 2 ṙ θ sin θ cos θ φ 2 = 0 8.8) φ + 2r b 2 + r 2 ṙ φ + 2 cot θ θ φ = 0 8.9) Consider an initial value problem where initially r = R, ṙ = v, θ = φ = 0, and let the particle have a radial free-fall into the wormhole. The dot on the quantities are derivatives with respect to τ. The initial four-velocity is just U µ = dxµ ) dτ = 1 + v 2, v, 0, ) In our geodesic equations if θ and φ start at zero, then θ = 0 and φ = 0 and these two quantities will not change at all. Therefore the radial equation is simple, too: r = 0. The particle essentially moves with constant speed on the r direction, and the proper) time it takes to travel from r = R to r = R is just τ = 2R v 8.11) 20

21 General Relativity Lecture 8 Note we have some conserved quantities for our metric. Obviously the metric does not explicitly depend on time t and angle φ, however there is another symmetry. We will discuss this later by introducing Killing vectors. Let s look at the Lagrangian again with the affine parameter equal to proper time, and L = dτ dτ = 1. Writing out the Euler-Lagrangian equation we have: d dτ L ẋ µ ) L x µ = ) We can look at the µ = 0 component, i.e. time component, first. Working out the equation we get immediately ẗ = 0. Carrying out all the calculations we can get all the equations of motion shown above. Note that L/ φ = 0, therefore if we define L/ φ = π φ, then by equation of motion we have π φ = 0. So we have a conserved quantity. This is a special case of the Noether Theorem. 21

22 General Relativity Lecture 9 9 Lecture Newtonian Limit Remember we had the geodesic equation d 2 x µ dτ 2 We want to recover the Newton s equation in the limit of weak field: + dx α dx β Γµ αβ dτ dτ = 0 9.1) g µν = η µν + h µν, h µν 1, 0 h µν 0 9.2) Another assumption we make is that the motion is nonrelativistic, i.e. dx/dτ 1. Under this assumption the only contribution to the acceleration is just We need to evaluate the Christoffel symbol d 2 x µ dτ 2 + dt dt Γµ 00 dτ dτ = 0 9.3) Γ µ 00 = 1 2 gµλ [2 0 g λ0 λ g 00 ] = 1 2 ηµλ λ h ) We ignored the time derivatives because everything moves slowly and these terms are subdominate. Now we can write out the µ = i components of the geodesic equation, discarding the terms with v which are subdominate and taking dt/dτ 1 This looks like the Newton s law of motion. We identify d 2 x i dτ 2 = 1 h 00 2 x i 9.5) h 00 = 2φ, g 00 = 1 + 2φ) 9.6) where φ is identified as the gravitational potential. Note that φ 1 in most circumstances: On Earth s surface φ 10 9 On Sun s surface φ 10 6 On the surface of a White dwarf φ 10 4 On the surface of a proton φ Near black holes we have φ O1) 22

23 General Relativity Lecture Gravitational Redshift We have seen that the metric in the weak field limit is ds 2 = g 00 dt 2 + 2g 0i dx i dt + g ij dx i dx j 9.7) and we have g 00 = 1+2φ) and 0 g 00 = 0. Under assumption of nonrelativistic speed, the only important contribution is from the first term. Now let s consider a wave propagating from an observer at point 1 to another observer at point 2. Suppose two successive wave peaks passes through point 1 and then passes through point 2. We choose the coordinate separation to be t 1 = t 2, because the metric is static. The time differences recorded by the two observers are determined by the proper time they record in their frame. We can calculate the proper time from observer 1 as under the nonrelativistic assumption) dt dt τ 1 = g 00 dλ dλ dλ = g 00 1) t 1 9.8) Similarly we have τ 2 = g 00 2) t 2. So we can calculate the ratio of the proper times τ 2 τ 1 = 1 + 2φ φ φ 2 φ 1 ) + oφ 2 ) 9.9) Because we know that the frequency of the photon ν 1/τ, so we know that when φ 1 > φ 2 then ν 1 > ν 2 so we have redshift φ 1 < φ 2 then ν 1 < ν 2 so we have blueshift 9.3 More Formalism Remember we could raise and lower indices in Special Relativity using the inverse metric and the metric. But now in GR we need to replace η µν with g µν and the inverse metric no longer has the same components as the metric. Remember also that in SR we defined the transformation property for vectors and 1-forms as V µ = Λ µ νv ν, W µ = Λ ν µ W ν 9.10) where Λ are Lorentz transformations. Now in GR we need to replace Λ µ ν = x µ / x ν. The tensors transform the same way, just replace the Lorentz transforms with derivatives. Note that in SR V µ / x ν is a tensor, but this in general is not true in GR: V µ x ν = xγ x ν x γ x µ x β V β = xγ x µ V β x ν x β x γ ) + xγ x ν 2 x µ x β x γ V β 9.11) It is the second term that destroys the tensorial transformation property. We can rewrite the second term as x γ 2 x µ x ν x λ x γ = xγ x µ 2 x α x β xγ x µ x ν x β x λ x γ = x α x ν x β Γβ λγ 9.12) 23

24 General Relativity Lecture 9 This motivates the definition of a covariant derivative, which converts a tensor into another tensor α V β = α V β + Γ β αγ V γ 9.13) Let s try to work out the covariant derivative of various objects: The covariant derivative of a scalar is just the derivative of that scalar α φ = α φ. α ω β = α ω β Γ µ αβ ω µ. ρ T µν λ ) = ρt µν λ + Γν ραt µα λ + Γµ ραt αν λ Γα ρλ T µν α The way we defined the Christoffel symbol satisfies the metric compatibility condition which says that α g µβ = ) If we define the connection as not metric compatible, then there is extra degrees of freedom, and we can actually convert them into scalar fields. Let s check some consistency. First we need that α V β W β ) = α V β W β ). This can be checked directly: α V β W β ) = α V β )W β + V β α W β ) 9.15) We can check that the plus and minus terms with Christoffel symbols exactly cancel to yield only the derivative term. Another thing to check is that 2 yields the correct Laplacian in different coordinate systems. For example in 2D we have 2 φ = 2 φ x φ y 2 = 1 r φ ) + 1r 2 r r r 2 φ ) This is checked in the homework. Finally we calculate the divergence of V µ µ V µ = µ V µ + Γ µ µαv α = µ V µ + 1 g α g V α = 1 g µ g V µ ) 9.17) The derivation involve using determinant trick. 24

25 General Relativity Lecture Lecture 10 Remember the definition of a geodesic is the curve on which the integral dτ is extremized. Also remember that the principle of equivalence tells us around any space-time point, we can always choose a coordinate such that a particle under no other force moves in a straight line with constant velocity. This frame is called local inertial frame, and is characterized by the condition g µν x) = η µν, α g µν x) = ) at that point x. Note that the second order derivatives of g µν do not vanish in the local inertial frame, so we have curvature even in the inertial frame, but as long as we don t go too far then the curvature doesn t really matter as it is a second order effect. These two principles both lead to the geodesic equation d 2 x µ dτ 2 + dx α dx β Γµ αβ dτ dτ = ) which is true in general coordinates. In locally coordinates this translates into d 2 x µ dτ 2 = ) which is just a motion at constant velocity. Note that the Christoffel symbol is not a tensor, although it has three indices and looks like one. We worked out in the homework the transformation property of this connection. However the geodesic equation is a tensor equation because the second derivative of x µ is not a tensor either. However, the parts which do not behave nicely under transformation from these quantities cancel so that the sum transforms nicely Another Look at the Geodesic Equation We want to write the geodesic equation in a new way to show manifestly that it is a good tensor equation. Let s first consider the 4-velocity U µ = dx µ /dt which is a good tensor under coordinate transformations. It obeys the transformation law dx µ dτ = x µ dx ν x ν 10.4) dτ However x µ is not a good tensor because it doesn t transform as above, whereas dx µ is a good tensor. Now recall that we defined the covariant derivative which is a good tensor regardless of general coordinates, i.e. α U µ = α U µ + Γ µ αβ U β 10.5) αu µ = xβ x µ x α x ν βu ν 10.6) The geodesic equation can be readily written as: U α α U µ = dxα dx µ α dτ dτ + dx β ) Γµ αβ = d2 x µ dτ dτ 2 + dx α dx β Γµ αβ dτ dτ = ) So in this tensor form we know manifestly that the geodesic equation is a good tensor equation, and it has the same form in all general coordinates. 25

26 General Relativity Lecture Digression on Parallel Transport Firstly let us define the concept of parallel transport. Consider a path xλ) parametrized by some parameter λ. We want to define the transport of a vetor V µ along this path, and define a derivative D Dλ V µ such that D Dλ V µ = dv µ dλ D Dλ V µ is a good tensor. in a local inertial frame. First note that dv µ /dλ is not a good tensor because dv µ dλ = d dλ x µ x ν V ν ) x µ x ν d dλ V ν 10.8) Now we claim that the definition DV µ Dλ = dxα dλ αv µ 10.9) satisfies the two conditions above. This is easy to check because for the first requirement, in local inertial frame the covariant derivative just reduces to ordinary derivative. And for second requirement the covariant derivative and the velocity vector are good tensors, so the product is a tensor. Now that we have defined a derivative, and want to know what does it mean. We claim that parallel transport means that DV µ /Dλ = 0 along the path of interest xλ). The intuitive picture is a vector being transported along the curve xλ) while keeping its length and orientation with respect to the path. Consider the 2-sphere S 2. The path we are interested is part of the equator, i.e. part of a great circle. Then parallel transport of a vector along this path is just moving it while keeping the angle with the tangent vector of the path. Now consider the geodesic equation U α α U µ = 0 again, it can be written as DU µ Dλ = ) along the geodesic xλ). The geodesic is the path on which the 4-velocity is parallel-transported. Let s look at the 2-sphere again. If it is an Euclidean space then two parallel lines do not intersect. But on the sphere two parallel straight lines can be defined as two geodesics from two points at which their tangent vectors are both perpendicular to the geodesic connecting theme. But these kind of parallel lines eventually meet, so this space is a curved space. Another relation between curvature and parallel transport is that in flat space if you parallel transport a vector along a loop, it will come back to itself, while in a curved space after parallel transport along a closed loop the vector comes back at an angle. 26

27 General Relativity Lecture Lecture 11 Remember our formulation of the geodesic equation, we can either write down the explicit form with the Christoffel symbol: d 2 x µ dτ 2 + dx α dx β Γµ αβ dτ dτ = ) or can write the covariant derivative of velocity equals zero: D Dτ U µ = ) where U µ is the 4-velocity. For massless particle like photon the worldline is null, so we can t parametrize the path using proper time. We need to choose some other λ as the affine parameter. Usually for massless particles, however, it is usually more convenient to use the equation g µν dx µ dx ν = 0 which is often stronger than the geodesic equation Principle of General Covariance The principle of general covariance states the following. An equation that describes some physical law in the presence of gravity satisfies the following 2 properties: The equation is generally covariant, i.e. it takes the same form in all coordinate systems. The same equation holds in the absence of gravity, i.e. it agrees with special relativity if g µν = η µν and Γ µ αβ = 0. Note that this is just a restatement of the equivalence principle. But this provides a machinery to easily write down equations that are automatically covariant and including gravity. Note that we don t have to write the equation in manifestly covariant form, just like the usual geodesic equation, but it is useful and possible to write the same equation in general tensor form, which is in general simpler and makes life easier. Let s consider some examples. Firstly we can derive the geodesic equation using this principle. Consider a massive particle and U µ = dx µ /dτ. In special relativity the equation of motion for a free particle is du µ dτ = 0 = dxα dτ du µ dx α = U α α U µ 11.3) Note in the final part the derivative is just usual derivative. Now using the above principle, we just replace the usual derivative by covariant derivative, so we get the equation U α α U µ = ) This is a good tensor equation, and it is just our geodesic equation. The principle boils down to replace ordinary derivatives with covariant derivatives. Let s try this for E&M. Remember the Maxwell equations with source was written as We just replace the ordinary derivative with covariant derivative µ F µν = J ν 11.5) µ F µν = J ν 11.6) 27

28 General Relativity Lecture 11 There is also the source-free Maxwell equations µ F µν = ) Now it is tempting to just replace the ordinary derivative with a covariant derivative. But there is one caveat. When we defined the dual of the electromagnetic field tensor as F µν = 1 2 εµναβ, we left an important part. We actually need to define it as F µν = 1 2 εµναβ, ε µναβ = 1 g ε µναβ 11.8) In this way we can write the covariant version of the equation to be Finally let s do fluid mechanics. generalized as µ F µν = ) The conservation of energy-momentum equation can be directly µ T µν = 0 µ T µν = ) Let s look at this equation in more detail. Write out the component form µ T µν = µ T µν + Γ µ µβ T βν + Γ ν µβ T µβ = ) This can be interpreted as follows. Formerly if we want to change the energy density, we need the fluid to flow out or into the volume element, which relates the time change of energy density with the divergence of fluid momentum. Now the extra terms mean that another way to change energy density is to move in a direction where the gravity is different. The charge conservation equation also gets covariantized, because the current density involves the volume element, which is related to the space-time measure Riemann Curvature Tensor We now want to introduce a tensor to characterize whether your space-time is flat or not. By the equivalence principle, at any point we can choose a coordinate frame where the metric is Minkowski and first derivative vanishes, but in general we can t touch the second derivative. So we anticipate that the curvature is related to the second derivatives of the metric. Remember if we parallelly transport a vector on the surface of a 2-sphere over a complete loop, the vector will not return to itself. We would use this property and consider an infinitesimal loop and ask under what condition does the vector return to itself. Suppose we have a family of curves of constant x 1, and another family of curves of constant x 2. The loop we choose is the loop enclosed by the curves x 1 = a and x 1 = a + δa, x 2 = b and x 2 = b + δb. Let s consider first the segment on x 2 = b. The path can be characterized as x 2 dx 2 λ) = b, dλ = 0, dx 1 dλ ) The parallel transport of vector V along this path means dx µ dλ µv α = ) 28

29 General Relativity Lecture 11 But remember that the second component of velocity is 0, so we can write the above equation as dx 1 dλ 1V α = 0, 1 V α = 0, 1 V α = Γ α 1µV µ 11.14) So we can write the change of vector component as A,B,C,D clockwise) V α B) = V α A) a+δa a Γ α 1µV µ x 2 =bdx ) Similarly we can work out the change of the component in all four segments, and get V α A after ) V α A) = Γ α 2µV µ x 1 =adx 1 Γ α 2µV µ x 1 =a+δadx 1 + Γ α 1µV µ x 2 =b+δbdx 1 Γ α 1µV µ x 2 =bdx 1 = δaδb [ 1 Γ α 2µ V µ) + 2 Γ α 1µ V µ)] We can expand the derivatives and use the relations i V α = Γ α iµ V µ to get 11.16) [ ] δv α = δaδb 1 Γ α 2δ ) + Γα 2µΓ µ 1δ + 2Γ α 1δ ) Γα 1µΓ µ 2δ V δ 11.17) This is a special case of a more general identity [ ] δv λ = µ Γ λ βν + νγ λ βµ Γλ µη Γη βν + Γλ νη Γη βµ V β δx µ δx ν 11.18) And the terms inside the square brackets are called the Riemann curvature tensor [ ] = R α δλσ. Note there is a minus sign in the convention. As promised, this quantity is related to the second derivative of the metric. 29