sign in sign up
<<
Home , Einstein tensor, Geodesic, Geodesic deviation, Tensor field, Weyl tensor

Fall 2019 Lecture 13: deviation; Einstein field

Yacine Ali-Ha¨ımoud October 11th, 2019

GEODESIC DEVIATION

The principle of equivalence states that one cannot distinguish a uniform gravitational field from being in an accelerated frame. However, tidal fields, i.e. gradients of gravitational fields, are indeed measurable. Here we will show that the Riemann encodes tidal fields.

Consider a fiducial free-falling observer, thus moving along a geodesic G. We set up Fermi in µ the vicinity of this geodesic, i.e. coordinates in which gµν = ηµν |G and Γνσ|G = 0. Events along the geodesic have coordinates (x0, xi) = (t, 0), where we denote by t the of the fiducial observer.

Now consider another free-falling observer, close enough from the fiducial observer that we can describe its position with the Fermi normal coordinates. We denote by τ the proper time of that second observer. In the Fermi normal coordinates, the spatial components of the geodesic for the second observer can be written as d2xi d  dxi  d2xi dxi d2t  dxi  dxµ dxν = (dt/dτ)−1 (dt/dτ)−1 = (dt/dτ)−2 − (dt/dτ)−3 = − Γi − Γ0 . (1) dt2 dτ dτ dτ 2 dτ dτ 2 µν µν dt dt dt The Christoffel symbols have to be evaluated along the geodesic of the second observer. If the second observer is close µ µ λ λ µ enough to the fiducial geodesic, we may Taylor-expand Γνσ around G, where they vanish: Γνσ(x ) ≈ x ∂λΓνσ|G + 2 µ 0 µ O(x ). Moreover, since Γνσ(x , 0, 0, 0) = 0, we find ∂0Γνσ|G = 0. Thus only the spatial derivatives contribute:

µ λ k µ 2 Γνσ(x ) ≈ x ∂kΓνσ|G + O(x ). (2) Thus the geodesic equation for the second observer becomes

d2xi  dxi  dxµ dxν = −xk ∂ Γi − ∂ Γ0 + O(x2). (3) dt2 k µν k µν dt dt dt

Let us solve this equation perturbatively in xk – we will discuss later relative to what xk needs to be small. To zero-th order, we find d2xi/dt2 = 0, which is the geodesic equation for the fiducial observer. Since the fiducial i µ µ observer moreover has constant x = 0 in the Fermi normal coordinates, we thus have, to zero-th order, dx /dt = δ0 . Substituting this on the right-hand-side, we then get, at linear order,

d2xi = −xk∂ Γi + O(x2). (4) dt2 k 00 In a LICS (which the Fermi normal coordinates provideat each event along G), the components of the Riemann tensor σ σ σ i i i i are R λµν = ∂µΓνλ − ∂ν Γµλ. In particular, along G, R 0k0 = ∂kΓ00 − ∂0Γ0k = ∂kΓ00. We thus finally arrive at

d2xi ≈ −xkRi . (5) dt2 0k0

So we see that the Riemann tensor plays the role of a tidal field. To see this, consider a Newtonian potential φ such k that ∇φ = 0 at the origin, and ∂iφ ≈ x ∂i∂kφ near the origin. The Newtonian are then d2xi ≈ −xk∂ ∂ φ. (6) dt2 i k

i Thus we see that we can identify R 0k0 = ∂i∂kφ. Since φ is dimensionless (in units where c = 1), the Riemann tensor has of inverse length squared. The relevant lengthscale corresponds to the radii of of : Riemann ∼ 1/ (radius of curvature)2. To see this explictly, compute the Ricci of a 2- with radius r, you will find R = 2/r2. 2

Let us recall the different concepts that the Riemann tensor quantifies:

• It quantifies the non-commutation of covariant derivatives of vector fields • It quantifies the difference of transport√along different paths. • It has dimensions of 1/length2, and is such that 1/ Riemann ∼ radius of curvature of spacetime. • It plays the role of a tidal field, and quantifies the deviation of neighboring .

BIANCHI IDENTITY

In a LICS, the fully-covariant components of the Riemann tensor are 1 R = (∂ ∂ g − ∂ ∂ g + ∂ ∂ g − ∂ ∂ g ) . [LICS] (7) δλµν 2 λ µ νδ λ ν µδ δ ν µλ δ µ νλ

In a LICS, we moreover have ∇γ Rρσµν = ∂γ Rρσµν . We then find that the sum of cyclic permutations of the three first indices of this tensor vanishes:

∇γ Rρσµν + ∇ρRσγµν + ∇σRγρµν = 0 . Bianchi identity (8)

This was derived in a LICS, but, being a tensorial identity, it holds in any .

Let us contract this identity on the 3rd and 5th index, to get

σ σ σ σ 0 = ∇γ Rρσµ + ∇ρRσγµ + ∇σRγρµ = ∇γ Rρµ − ∇ρRγµ + ∇ Rγρµσ, (9) where Rαβ is the Ricci tensor. Let us now contract on γ and µ, and obtain  1  0 = ∇γ R − ∇ R + ∇σR = 2∇γ R − g R = 2∇γ G , (10) ργ ρ ρσ ργ 2 ργ ργ where R is the Ricci scalar and

1 G = R − R g is the . (11) αβ αβ 2 αβ

α α 1 α 1 The Einstein tensor is the -reversed Ricci tensor. Indeed, G α = R α − 2 Rδ α = R − 2 R × 4 = −R. Thus we 1 γ also have Rαβ = Gαβ − 2 G γ gαβ. To summarize, we found that the Einstein tensor is -free:

αβ ∇αG = 0 . Contracted Bianchi identity (12)

EINSTEIN EQUATIONS

2 ij Poisson’s equation relates the Laplacian of the Newtonian potential to the density: ∇ φ = δ ∂i∂jφ = 4πGρ, i where G is Newton’s constant. Now, since we identified R 0k0 = ∂i∂kφ, in Fermi normal coordinates, we thus have 2 i µ 0 ∇ φ = R 0i0 = R 0µ0 = R00, since R 000 = 0 from the antisymmetry of Riemann in its last two indices. In the 00 Newtonian limit, the density is the 00 component of the -energy tensor, so we have R00 ≈ 4πGT = 4πGT00, using the Minkowski metric to lower and raise indices.

Now we want to write a tensorial equation, not just an equality between sepcific components of two in a particular coordinate system. Naively, one may think that Rαβ = 4πGTαβ is the way to go, but, if Tαβ is to be the total stress-energy tensor, it is divergence-free, and the Ricci tensor Rαβ is not in general. So we’ll have better chances with the Einstein tensor, which is divergence-free. In the non-relativistic limit, T 0i viT 00 and T ij ∼ vivjT 00, where vi is the characteristic velocity of matter. Thus the trace of the stress-energy tensor is γ 2 2 1 γ 1 1 γ T γ = −T00 +O(v ) = −ρ+O(v ). Hence T00 − 2 T γ g00 ≈ 2 ρ. So the Poisson equation is R00 ≈ 8πG(T00 − 2 T γ g00). 1 γ If we now try out Rαβ = 8πG(Tαβ − 2 T γ gαβ), and trace-reverse, we get the Einstein field equations

Gαβ = 8πG Tαβ . (13) 3

Because of the contracted Bianchi identity, this equation is consistent with the conservation of the total stress- energy tensor. This equation allows to directly determine the 10 components of the Ricci tensor, given the stress-energy tensor. It does not, a priori, fully determine the Riemann tensor, which contains 10 more independent components in the . Now, the Weyl tensor is defined as 1 C = R + g R − g R + g g R. (14) αβγδ αβγδ β[γ δ]α α[γ δ]β 3 α[γ δ]β As you will show in the homework, the Bianchi identity implies the following differential equation for the Weyl tensor: 1 ∇ρC = ∇ R + g ∇ R. (15) ρσµν [µ ν]σ 6 σ[µ ν] Plugging in Einstein’s field equation, we see that the divergence of the Weyl tensor is sourced by the gradients of the stress-energy tensor:

 1  ∇ρC = 8πG ∇ T + g ∇ T γ . (16) ρσµν [µ ν]σ 3 σ[µ ν] γ

νµ ν This equation is qualitatively similar to Maxwell’s equations, ∇µF = J . As we will see in more detail later on, this can be seen as an equation for gravitational waves.

LAGRANGIAN FORMULATION OF GENERAL RELATIVITY

0 0 0 0 0 – Consider a LICS {xµ }. The 4-volume element is dV = d4x0 = dx0 dx1 dx2 dx3 . In terms of general coordinates {xµ}, we have

" 0 # ∂xµ dV = d4x0 = det d4x. (17) µ ∂x Now, the metric components change as

0 0 0 0 ∂xµ ∂xν ∂xµ ∂xν g = g 0 0 = η 0 0 , (18) µν ∂xµ ∂xν µ ν ∂xµ ∂xν µ ν since gµ0ν0 = ηµ0ν0 in the LICS. Seing this as a operation and taking the , we find

" 0 #2 ∂xµ g ≡ det[g ] = − det . (19) µν ∂xµ

Hence, we find that the coordinate-invariant 4-volume element is √ dV = −g d4x . (20) √ Since −g = 1 in any LICS, this expression is independent of the initial LICS we started from.

Integration – The integral of a scalar function f is well defined: in any coordinate system, Z Z √ dV f = d4x −g f (21) can be computed by discretizing spacetime and summing up the values of f at each “cell”, and taking the limit of this process when the spacetime cell size goes to zero. On the other hand, the integral of a vector or tensor field is meaningless in curved spacetime. Think of the integral as a sum. To sum vectors, you need them to belong to the same . There is no common vector space in curved spacetime: at different points p, q of the , the vectors V |p and V |q belong to different vectors spaces. Only in flat spacetime can we define such integrals. First parallel-transport the vector field to a 4 single point of spacetime (it doesn’t matter which one). This operation is independent of the path in flat spacetime, hence well defined. Then sum these components to perform the integral.

Einstein-Hilbert – We can recover Einstein’s field equations by minimizing the following action with respect to variations of the metric: 1 Z √ Z √ S = S + S ,S ≡ d4x −g R, S ≡ d4x −g L , (22) EH M EH 16πG M M where SEH is called the Einstein-Hilbert action and LM is the Lagrangian density of matter, which is a scalar function. Indeed, a lengthy calculation (see e.g. Carroll 4.3) shows that, upon varying the metric components by δgµν , the Einstein-Hilbert action changes by 1 Z √ δS = − d4x −g Gµν δg . (23) EH 16πG µν

Given the matter Lagrangian density LM , we define the stress-energy tensor through √ 1√ δ( −gL ) = −g T µν δg . (24) M 2 µν We see that extremizing the total action, i.e. setting δS = 0 for any variation of the metric about the physical solution, implies Gµν = 8πGT µν . The Lagrangian formulation is particularly useful to define modified- theories. The Einstein-Hilbert action is the simplest action one could think of, with Lagrangian density Lgrav ∝ R, the Ricci scalar. But one could 2 2 2 µν 2 µνρσ imagine, for instance, additional terms of the form Lgrav ∼ R + L1R + L2R Rµν + L3R Rµνρσ, or even higher powers of the Riemann tensor. Since the Riemann√ tensor has dimensions of inverse length squared, the parameters Li must have dimensions of length. Now, 1/ R is the characteristic radius of curvature of spacetime. Thus the additional terms that we wrote down would only become relevant when the curvature of spacetime is comparable with or smaller than one of the lengthscales Li.

GEOMETRIC UNITS

We have already set c = 1, i.e. we use a system of units such that 1 length unit is traveled by light in one time unit. To avoid keeping Newton’s constant everywhere, we will moreover set G = 1. Remember that Newton’s constant is such that d2x/dt2 ∼ GM/r2, hence G has dimensions of length3 time−2 mass−1. So setting G = 1 means that, given length and time units, we choose the mass units such that G = 1. Having chosen c = 1, means that we can identify times and . For instance, 1 second = 300, 000 km – what this really means is that c×(1 second) = 300, 000 km.

Similarly, having chosen G = 1 means we can, in addition, identify masses with lengths or times. A number that is very useful to remember is the mass of the Sun in geometric units:

M ≈ 1.5 km ≈ 5 µs , (25)

2 3 −6 by which we really mean GM /c ≈ 1.5 km and GM /c ≈ 5 × 10 second.

What this means in practice is that we drop all factors of c and G from all calculations, though we still make sure that they are dimensionally correct – for instance, if L is a length, one could have an equation of the form M ∼ L, but not an equation of the form M ∼ L2. At the end of the calculation, if we want to express results in a particular unit (km, seconds, cm, solar masses, etc...) we simply multipy by the correct powers of G and c to convert the results.