General Relativity Fall 2017 Lecture 2: the Tangent Space, Coordinate Bases

General Relativity Fall 2017 Lecture 2: the Tangent Space, Coordinate Bases

General Relativity Fall 2017 Lecture 2: The tangent space, coordinate bases Yacine Ali-Ha¨ımoud September 7th 2017 This mostly mathematical lecture introduces concepts which wil be central to differential geometry and general relativity. The goal of this lecture is to formalize what is the underlying vector space which we use in Newtonian physics as well as special relativity, and to start thinking about vectors as living in a different space than the space(time) where material points (or events) live. A. The Levi-Civita tensor [carry-over from Lecture 1] The Levi-Civita tensor , which will be used to convey the notion of volume. This is a tensor of rank (0; n) that is completely antisymmetric. The space of such tensors is of dimension 1: given a basis e(1); :::e(n) and the value 12:::n ≡ (e(1); :::; e(n)), one can determine the value that takes for any n-tuple. The only (possibly) non-zero components of are of the form σ(1)σ(2)...σ(n) = s(σ)12:::n; (1) where σ 2 Pn, and s is its signature. The components of change as follows for a change of basis: µ1 µ2 µn X σ(1) σ(2) σ(n) 0 0 µ ...µ = Λ 0 Λ 0 :::Λ 0 µ1...µn = Λ 0 Λ 0 :::Λ 0 s(σ)12:::n: (2) 1 n µ1 µ2 µn µ1 µ2 µn σ2Pn So in particular, 0 X σ(1) σ(2) σ(n) 1:::n = Λ 1 Λ 2 :::Λ n s(σ)12:::n; (3) σ2Pn where I have changed the notation to put primes on tensors rather than indices, as it makes things clearer here. The bottom line is, the right-hand-side is just the definition of the determinant of Λ seen as a matrix: 0 12:::n = det(Λ)12:::n: (4) This tells us how one (hence all) components of change under changes of coordinates. The components of the metric change as follow: µ ν gµ0ν0 = Λ µ0 Λ ν0 gµν : (5) Seen as a matrix equation, this is g0 = ΛTgΛ. Taking the determinant of this relation, det(g0) = det(ΛTgΛ) = det(ΛT) det(g) det(Λ) = det(Λ)2 det(g). Therefore we find that the determinant of g must be of the same sign in any basis, and that s det(g0) det(Λ) = ± : (6) det(g) So we conclude that 0 12:::n = sign [det(Λ)] 12:::n ; (7) pj det(g0)j pj det(g)j p i.e., up to changes of sign, the ratio 12:::n= j det(g)j is invariant under changes of basis. We can now define THE Levi-Civita tensor, as the completely antisymmetric rank (0; n) tensor whose component 12:::n = 1 in an orthonormal basis. The choice of that basis is not unique, an in particular we would get a sign flip 2 for 12:::n if we chose a basis with two flipped vectors. So the Levi-Civita tensor is only defined up to that arbitrary choice of orientation. Since in such a basis det(g) = ±1 we then have, in any other basis, 0 p 0 12:::n = sign [det(Λ)] j det(g )j: (8) In summary, the Levi-Civita tensor is the completely antisymmetric tensor of rank (0; n) whose compo- nent 12:::n is 1 in a \positively oriented" orthonormal basis. Exercise: show that in 3-dimensions, (A; B; C) ≡ A · (B × C), which is the volume of the parallelepiped with sides (A; B; C). B. The spacetime interval, Lorentz transformations Consider the usual flat physical space R3. If we pick cartesian coordinates (x; y; z), we know that the square distance between two points separated by (∆x; ∆y; ∆z) is ∆`2 = ∆x2 +∆y2 +∆z2. This distance is independent of the chosen cartesian coordinates (up to rescalings of course): if two coordinate systems (x; y; z) and (x0; y0; z0) are related by a rotation, then ∆`2 = ∆x2 + ∆y2 + ∆z2 = ∆x02 + ∆y02 + ∆z02. This notion gets generalized to the relativistic spacetime interval ∆s2 = −∆t2 + ∆x2 + ∆y2 + ∆z2, where here and throughout we use units in which the speed of light c = 1. Let us review why this is the relevant interval. We recall the first fundamental postulate of special relativity: there exists coordinate systems (or reference frames) in which the speed of light jd~x=dtj = c = 1, independent of the velocity of the emitter. To this postulate we also add the assumption of homogeneity of spacetime (i.e., there are no preferred spacetime locations) and of isotropy of space (i.e., there are no preferred spatial directions). Pick cartesian coordinates (x1; x2; x3) [i.e. a system of rigid, orthogonal metersticks] on space, and a time t ≡ x0, defined by synchronization through light signals, i.e. defined such that the time interval ∆t it takes light to travel 0 2 2 µ ν a distance j∆~xj is such that 0 = −(∆x ) + (∆~x) ≡ ηµν ∆x ∆x , where ηµν ≡ diag[−1; 1; 1; 1]. Such a coordinate system is said to be inertial (we'll say more on this later on). Now let us pick another inertial coordinate system (t0; x0; y0; z0), i.e. by definition, a coordinate system for which µ0 ν0 ηµ0ν0 ∆x ∆x = 0 along light paths, where again ηµ0ν0 ≡ diag[−1; 1; 1; 1]. By homogeneity of spacetime, the primed µ0 µ0 µ inertial coordinates must be linearly related to the unprimed inertial coordinates: ∆x = Λ µ∆x and vice-versa, µ µ µ0 µ0 ∆x = Λ µ0 ∆x . So we are looking for transformations Λ µ such that µ ν µ0 ν0 µ0 ν0 µ ν 0 = ηµν ∆x ∆x , 0 = ηµ0ν0 ∆x ∆x = ηµ0ν0 Λ µΛ ν ∆x ∆x : (9) where we emphasize that ηµ0ν0 has the same numerical values as ηµν . In matrix form, we are looking for transformat- sions Λ such that ∆xTη∆x = 0 , ∆xT(ΛTηΛ)∆x = 0: (10) This implies (though this is not completely trivial to prove) that ΛTηΛ = κη, where κ is a constant, i.e. −∆t02 + ∆x02 + ∆y02 + ∆z02 = κ −∆t2 + ∆x2 + ∆y2 + ∆z2 : (11) Now we know that spatial rotations (with t0 = t) satisfy this equation with κ = 1. Let us consider a boost, i.e. a change of coordinates such that the primed system has velocity v with respect to the unprimed system. By isotropy of space, the constant κ(v) in Eq. (11) must only depend on the magnitude of v. Now take three cordinate systems S1;S2;S3, such that S2 has velocity v2 wrt S1, S3 has velocity v3 wrt S1, and S3 has velocity v23 wrt S2. Then we must have κ(v2)κ(v23) = κ(v3). Clearly, v23 must depend on the angle between v2 and v3 (it clearly does in the limit of small boosts). For the last equation to hold regardless of the angle, we therefore must have κ = 1, independently of the velocity. Hence we find that the transformations that preserve the condition −dt2 + d~x2 = 0 along light paths are such that µ0 ν0 Λ µΛ ν ηµ0ν0 = ηµν : (12) This equation defines Lorentz transformations. A fortiori, this implies that the infinitesimal spacetime interval ds2 is invariant under Lorentz transformations, i.e. independent of the inertial frame. So the generalization of the Euclidean infinitesimal length squared ∆`2 = ∆x2 +∆y2 +∆z2, which holds in cartesian coordinates and is invariant under rotations, is the Lorentzian spacetime interval ∆s2 = −∆t2 + ∆x2 + ∆y2 + ∆z2, which holds for inertial coordinate systems, and is invariant under Lorentz transformations. 3 C. Tangent space, coordinate bases in flat spacetime We are used to thinking of the flat spacetime as a vector space, since it is just R4. In fact however, we don't use the vector structure of spacetime itself: given two spacetime points P and Q, there is no meaning in P + Q or 2P . Spacetime (which we shall refer to as M for now on) is merely a set that \looks like" R4. We will formalize this when we talk about manifolds. In fact the relevant vector space that we use for expressing physical laws is that of infinitesimal displacements near a spacetime event. To convince yourself, think about the following: how does one calculate the length of a curve in R3, or equivalently the total spacetime interval along a curve in spacetime? We do so by adding the lengths of infinitesimal straight segments. Similarly, the velocity of a particle is an infinitesimal diplacement divided by the infinitesimal time it takes to travel across it. We formalize the notion of “infinitesimal displacement along a direction X" with derivative operators. Define F as the set of differentiable functions f : M! R. Given a spacetime point P and spacetime coordinates xµ, we define the partial derivative operators ( F! R @(µ) : @f : (13) f 7! @(µ)f ≡ @xµ P Suppose we are given a curve Q(λ) on spacetime, where λ is some parameter, and Q(0) = P . The coordinates take the values xµ(λ). Then the rate of change of a function f along this curve is obtained by the chain rule: µ µ df dx @f dx = = @(µ)f: (14) dλ P dλ @xµ P dλ dxµ This defines the operator d/dλ ≡ dλ @(µ). More generally, we define the tangent space at P , VP , as the space µ spanned by the @(µ), i.e. for any vector V 2 VP , V = V @(µ). A vector V acting on a function simply gives its derivative along V . Even more generally, the tangent space is the space of linear operators on F satisfying Leibniz's rule, V (fg) = f(P )V (g)+g(P )V (f), i.e.

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