Riemann normal coordinates

EY : 20141214 I tried posting this on expii – spearheaded by a fellow Caltech classmate and alumni, Po-Shen Loh, I strongly believe we have intersecting goals of making online learning collaborative, bringing the discussion part, the TA session part of education, online. He’s trying to do it with expii; I’m trying to do it here with my pdfs that I’m opening up in the spirit of open-source.

Riemann normal coordinates

Peter Michor’s Topics on Differential Geometry, Ch. 22 Pseudo-Riemann Metrics and Covariant Derivatives, Section 22.7 The geodesic exponential mapping, pp. 281, has the theorem, defining the \exp map, and proving its properties. I still need to work out how it fits into the picture of a connection on a vector bundle (horizontal vs. vertical), and what a spray is; I don’t understand that clearly. But the properties outlined are crystal clear.

S. Sternberg’s Semi-Riemannian Geometry and General Relativity in Ch. 6 on Gauss’s lemma works out Riemannian normal coordinates in the context of showing Gauss’s lemma, which gives a nice geometric picture (imagine radial lines emanating outward from the origin on a plane, piercing through concentric circles). I worked through 6.1-6.5 nicely.

I compared all of that with Carroll’s treatment in Spacetime and Geometry Ch. 3 Curvature Section 3.4 Properties of Geodesics around pp. 111-113. I didn’t see Wald’s General Relativity say much about it, but mention it on pp.42, Ch.3 on Curvature, Sec. 3.3. on Geodesics.

Suppose (M,g) is a semi-Riemannian manifold with metric g, and Levi-Civita connection \nabla (recall, it’s torsion-free and isometric or compatible, people call it different things).

Let x\in M, X \in \mathfrak{X}_xM = T_xM, the tangent space.

Then \exists \, ! (maximal) geodesic \gamma_X s.t. \begin{aligned}    & \quad     & \gamma_X(0) = x     & \dot{\gamma}(0) = X \end{aligned}. The geodesic equation is a system of 2nd-order ODEs. Thus, by theorem, existence and uniqueness is guaranteed. EY : 20141213 I’m not clear what maximal means in this context at this point. Equivalence relation?

From Michor, there is a exponential mapping, a smooth mapping
\exp : TM \supset V \to M
with open neighborhood V of zero section of TM, denoted \Gamma(TM)_0
s.t. (such that)
T_{0_x}( \left. \exp \right|_{T_xM} ) = 1_{T_xM} since
T_{0_x}(T_xM) = T_xM
(Sternberg works this out in a friendly manner in Sec. 6.1, The exponential map)
By the inverse function theorem, \exp_x \equiv \left. \exp \right|_{T_xM} : V_x \to W_x and it’s a diffeomorphism from the open neighborhood V_x of 0_x \in TM onto open neighborhood W_x in M.

Carroll mentions to be worried about the geodesic, over the entire M, of hitting an edge, but for V_x, it’s ok.

Then (W_x, \exp_x^{-1}) is a Riemann normal coordinate system at $x$

Note that \forall \, q \in W_x, \exists \, ! geodesic joining x to q entirely in W_x

\gamma_X(t) = \exp{(tX)}
Suppose we’re given a basis e=(e_1 \dots e_n) of T_xM
\phantom{Suppose we're given a } dual basis l^1 \dots l^n

Then on W_x,
\exp^{-1}(q) = x^i(q)e_i
\boxed{ x^i = l^i\exp^{-1} }
which is the normal coordinates or inertial coordinates, explicitly.

Let X = a^i e_i \in V_x \subset T_xM. Then

\begin{gathered}    \exp{ (tX) } = \gamma_X(t) \\   x^i( \gamma_X(t)) = l^i \exp^{-1}( \exp{(tX)} ) = l^i(tX) = ta^i   \end{gathered}
Clearly \ddot{x}^i =0. \gamma_X(t) is a geodesic obeying the geodesic equation. Then
\Gamma^k_{ij}( \gamma_X(t)) a^i a^j = 0 \quad \quad \, \forall \, k

At t=0, \gamma_X(0) = x. So
\Gamma_{ij}^k(x) a^i a^j = 0
This was true \forall \, X in the zero section of TM, \Gamma(TM)_0. So \forall \, sufficient small a^i, \\
if connection \nabla has zero torsion, \Gamma symmetric in i,j. Thus, we can conclude, \forall \, i,j,
\Gamma_{ij}^k(x) =0

Recall that a connection \nabla is isometric for \langle \, , \, \rangle (i.e. metric. Also, isometric is called compatible with the metric by some) if
X\langle Y, Z \rangle = \langle \nabla_X Y,Z \rangle + \langle Y, \nabla_X Z \rangle
Let X = \partial_i, Y = \partial_j, Z= \partial_k, in this normal coordinates.

\begin{gathered}    X \langle Y,Z \rangle = \langle \nabla_X Y, Z \rangle + \langle Y, \nabla_X Z \rangle = \langle \nabla_{\partial_i} \partial_j, \partial_k \rangle + \langle \partial_j, \nabla_{\partial_i} \partial_k \rangle = \\    = \langle \Gamma_{ij}^l \partial_l, \partial_k \rangle + \langle \partial_j , \Gamma^l_{ik} \partial_l \rangle = 0 + 0 = 0   \end{gathered}
\boxed{ \partial_i g_{jk} = 0 }
This answers Homework 3, Problem 1 of General Relativity (Ph236a) taught by Prof. Kamionkowski at Caltech on October 17, 2006, if you are working through those problem sets through self-learning and are stuck or want more (mathematical rigor).

So truly, x^i(\gamma_X(t)) = ta^i look like straight lines in this open neighborhood V_x.

Sternberg pointed out the satisfying conclusion in Sec. 6.2. Normal Coordinates, pp. 127: Mach’s principle arose from this problem:

You stand in a field looking at the stars. You let your arms rest freely at your side. You look up and the stars are not moving.

Now start spinning. The stars spin around you. You feel your arms get pulled away from your body (centrifugal force).

Why should your arms be pulled away from you when the stars are whirling? Why are your arms freely dangling when the stars don’t move?

What’s so special about you, with your arms at rest, and when the distant stars are not moving?

cf. https://en.wikipedia.org/wiki/Mach\%27s_principle

So how can the laws of physics, in the absence of forces, of stars, involve rectilinear, straight line motion, as this would involve a particular coordinate system?

According to Einstein, the metric is determined by the distribution of matter in the universe. The metric determines the connection. The isometric and torsion free connection picks out the inertial frame, with Riemannian normal coordinates x^i(\gamma_X(t)) = ta^i.


2 thoughts on “Riemann normal coordinates

  1. I beleive maximal geodesic means defined on the largest possible interval. The uniqueness and existence theorem guarantees a solution that can be extended to a maximal interval. I guess thats what maximal means here. Hope I am right!!

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s