## Lecture 9: Geodesics and Local Inertial Frames

9. Geodesics and Local Inertial Frames

Let us now apply some of this theory to curves on manifolds. If a non-null curve C on M is paramaterized by xi(t), then we can reparamaterize the curve using arc length,

s(t) = t

a
±
gij dxidu
 dxjdu
1/2

du ,

(starting at some arbitrary point) as the parameter. The reason for wanting to do this is that the tangent vector Ti = dxi/ds is then a unit vector (see the exercises) and also independent of the paramaterization.

If we were talking about a curve in E3, then the derivative of the unit tangent vector (again with respect to s to make it independent of the paramaterization) is a measure of how fast the curve is "turning," and so we call the derivative of Ti the curvature of C.

If C happens to be on a manifold, then the unit tangent vector is still

Ti =
dxi

ds
=
dxi

dt
/
ds

dt
=

dxi/dt
±gpq dxpdt dxqdt
1/2

(the last formula is there if you want to actually compute it). But, to get the curvature, we need to take the covariant derivative:

 Pi = DTids = D(dxi/ds)ds = d2xids2 + ip q dxpds dxqds

Definitions 9.1 The first curvature vector P of the curve C is

 Pi = d2xids2 + ip q dxpds dxqds

A curve on M whose first curvature is zero is called a geodesic. Thus, a geodesic is a curve that satisfies the system of second order differential equations

 d2xids2 + ip q dxpds dxqds = 0

In terms of the parameter t, this becomes (see the exercises)

 d2xidt2 dsdt - dxidt d2s dt2 + ip q dxpdt dxqdt dsdt = 0,

where

ds

dt
=
±
gij dxidu
 dxjdu
1/2

.

Note that P is a tangent vector at right angles to the curve C which measures its change relative to M.

Question Why is P at right angles to the curve C?
Answer This can be checked as follows.

d

ds
T, T =  DTds , T + T, DTds
(Exercise Set 8 #9)
=  2 DTds , T
(Symmetry of the scalar product)
=  2 P, T
(Definition of P)
so that
P, T=  12 dds T, T .
But
T, T=  1
(Refer back to the Proof of 6.5 to check this)
whence
T, T=  P, T = 12 dds (1) = 0,
(Refer back to the Proof of 6.5 to check this)
as asserted.

Local Flatness, or "Local Inertial Frames"

Notation We will be changing some of the notation to simplify things from now on.

1. First, we shall write the Christoffel symbols of the second kind as ijk rather than ijk
2. Second, we shall continue to use comma notation for ordinary (not covariant) partial derivatives:
Ti,k instead of Ti/xk
Ti,k instead of Ti/xk etc.

In "flat space" Es all the Christoffel symbols vanish, so the following question arises:

Question Can we find a chart (local coordinate system) such that the Christoffel symbols vanish -- at least in the domain of the chart?
Answer This is asking too much; we shall see later that the derivatives of the Christoffel symbols give an invariant tensor (called the curvature) which does not vanish in general. However, we do have the following.

Proposition 9.2 (Existence of a Local Inertial Frame)
If m is any point in the Riemannian manifold M, then there exists a local coordinate system xi at m such that:

(a) gij(m) =  ±1 if j = i 0 if j i
= ±ij
(b) gij,k(m) = 0for every k.

We call such a coordinate system a local inertial frame or a normal frame.

(It follows that ijk(m) = 0 in an inertial frame.)

Before proving the proposition, we need a lemma.

 Lemma 9.3 (Some Equivalent Things) Let m M. Then the following are equivalent: (a) gpq,r(m) = 0 for all p, q, r. (b) [pq, r]m = 0 for all p, q, r. (c) prq(m) = 0 for all p, q, r.

Proof of Lemma 9.3

(a) (b) follows from the definition of Christoffel symbols of the first kind.

(b) (a) follows from the identity

gpq,r = [qr, p] + [rp, q]     (Check it!)

(b) (c) follows from the definition of Christoffel symbols of the second kind.

(c) (b) follows from the inverse identity

[pq, s] = gsrprq.

In other words, the vanishing of Christoffel symbols at any point of M is equivalent to the vanishing of the partial derivatives of the metric tensor at that point.

Click here for a proof of Proposition 9.2.

Corollary 9.4 (Partial Derivatives Look Nice in Inertial Frames)
Given any point m M, there exist local coordinates such that

 Xp|k(m) = Xpxk m

 Also, the coordinates of Xpxk m in an inertial frame transform to those of Xp|k(m) in every frame.

 Corollary 9.5 (Geodesics are Locally Straight in Inertial Frames) If C is a geodesic passing through m M, then, in any inertial frame, it has zero classical curvature at m (that is, d2xi/ds2 = 0).

Question Is there a local coordinate system such that all geodesics are in fact straight lines?
Answer Not in general; if you make some geodesics straight, then others wind up curved. It is the curvature tensor that is responsible for this. This involves the derivatives of the Christoffel symbols, and we can't make it vanish.

Question If I throw a ball in the air, then the path is curved and also a geodesic. Does this mean that our earthly coordinates are not inertial?
Answer Yes. At each instant in time, we can construct a local inertial frame corresponding to that event. But this frame varies from point to point along our world line if our world line is not a geodesic (more about this below), and the only way our world line can be a geodesic is if we were freely falling (and therefore felt no gravity). Technically speaking, the "earthly" coordinates we use constitute a momentary comoving reference frame; it is inertial at each point along our world line, but the direction of the axes are constantly changing in space-time.

 Proposition 9.6 (Changing Inertial Frames) If x and are inertial frames at m M, then, recalling that D is the matrix whose ij th entry is (xi/j), one has det D = det = ±1.

Proof By definition of inertial frames,

gij(m) = ±ij,

and similarly for ij, so that ij = ±gij, whence det(g**) = ± det(**) = ±1. On the other hand,

 ij = xki xlj gkl,

which, in matrix form, becomes

** = DTg**D.

Taking determinants gives

det(**) = det(DT) det(g**) det(D) = det(D)2 det(g**),

giving

±1 = ±det(D)2,

which must mean that det(D)2 = +1, so that det(D) = ±1 as claimed.

Note that the above theorem also workds if we use units in which det g = -c2 as in Lorentz frames.

 Definition 9.7 Two (not necessarily inertial) frames x and have the same parity if det > 0. An orientation of M is an atlas of M such that all the charts have the same parity. M is called orientable if it has such an atlas, and oriented if it is equipped with one.

Notes
1.
Reversing the direction of any one of the axes reverses the orientation.
2. It follows that every orientable manifold has two orientations; one corresponding to each choice of equivalence class of orientations.
3. If M is an oriented manifold and m M, then we can choose an oriented inertial frame at m, so that the change-of-coordinates matrix D has positive determinant. Further, if D happens to be the change-of-coordinates from one oriented inertial frame to another, then det(D) = +1.
4.E3 has two orientations: one given by any left-handed system, and the other given by any right-handed system.
5. In the homework, you will see that spheres are orientable, whereas Klein bottles are not.

We now show how we can use inertial frames to construct a tensor field.

Definition 9.8 Let M be an oriented n-dimensional Riemannian manifold. The Levi-Civita tensor of type (0, n) is defined as follows. If is any coordinate system and m M, then define

 i1i2...in(m) = det (Di1Di2 ... Din) = determinant of D with columns permuted according to the indices,
where Dj is the j th column of the change-of-coordinates matrix xk/l, and where x is any oriented inertial frame at m.

Notes
1. is a completely antisymmetric tensor. If is itself an inertial frame, then, since det(D) = +1 (see Note 2 above) the coordinates of (m) are given by

 i1i2...in(m) = 1, if (i1, i2, ... , in) is an even permutation of (1, 2, ... , n) -1, if (i1, i2, ... , in) is an odd permutation of (1, 2, ... , n)

2. I have not seen this tensor defined in this generality in any of the sources I consulted. Note that this tensor cannot be defined without a metric being present. In the absence of a metric, the best you can do is define a "relative tensor," which is not quite the same, and what Rund calls the "Levi-Civita symbols" in his book. Wheeler, et al. just define it for Minkowski space.

(Compare this with the metric tensor, which is also "nice" in inertial frames.)

 Proposition 9.9 (Levi-Civita Tensor) The Levi-Civita tensor is a well-defined, smooth tensor field.

Proof To show that it is well-defined, we must show independence of the choice of inertial frames. But, if and are defined at m M as above by using two different inertial frames, with corresponding change-of-coordinates matrices D and E, then D-E is the change-of coordinates from one inertial frame to another, and therefore has determinant 1. Now,

 i1i2...in(m) = det (Di1Di2 ... Din) = det D I i1i2...in (where I i1i2...in is the identity matrix with columns ordered as shown in the indices) = det DE I i1i2...in (since E has determinant 1; this being where we use the fact that things are oriented!) = det E I i1i2...in (since D = I) = i1i2...in,

showing it is well-defined at each point. We now show that it is a tensor. If and are any two oriented coordinate systems at m and change-of-coordinate matrices D and E with respect to some inertial frame x at m, and if the coordinates of the tensor with respect to these coordinates are
k1k2...kn and r1r2...rn = det (Er1Er2 ... Ern) respectively, then at the point m,

k1k2...kn = det (Dk1Dk2 ... Dkn)
=  i1i2...in x i1k1 xi2k2 ... xinkn
(by definition of the determinant(!) since i1i2...in is just the sign of the permutation!)
=  i1i2...in x i1r1 xi2r2 ... xinrn r1k1 r2k2 ... rnkn
=  r1r2...rn r1k1 r2k2 ... rnkn

showing that the tensor transforms correctly. Finally, we assert that det (Dk1Dk2 ... Dkn) is a smooth function of the point m. This depends on the change-of-coordinate matrices to the inertial coordinates. But we saw that we could construct inertial frames by setting

 xij m = V(j)i,

where the V(j) were an orthogonal base of the tangent space at m. Since we can vary the coordinates of this base smoothly, the smoothness follows.

Example
In E3, the Levi-Civita tensor coincides with the totally antisymmetric third-order tensor ijk in Exercise Set 4. In the Exercises, we see how to use it to generalize the cross-product.

Exercise Set 9
1.
Recall that we can define the arc length of a smooth non-null curve by

s(t) = t

a
±
gij dxidu
 dxjdu
1/2

du .

Assuming that this function is invertible (so that we can express xi as a function of s) show that

 dxids 2 = ±1.

2. Derive the equations for a geodesic with respect to the parameter t.

3. Obtain an analogue of Corollary 9.3 for the covariant partial derivatives of type (2, 0) tensors.

4. Use inertial frames argument to prove that gab|c = gab|c = 0. (Also see Exercise Set 3 #1.)

5. Show that, if the columns of a matrix D are orthonormal, then det D = ±1.

6. Prove that, if is the Levi-Civita tensor, then, in any frame, i1i2...in = 0 whenever two of the indices are equal. Thus, the only non-zero coordinates occur when all the indices differ.

7. Use the Levi-Civita tensor to show that, if x is any inertial frame at m, and if X(1), . . . , X(n) are any n contravariant vectors at m, then

det X(1)| . . . |X(n)

is a scalar.

8. The Volume 1-Form (A Generalization of the Cross Product)
If we are given n-1 vector fields X(2), X(3), . . . , X(n) on the n-manifold M, define a covariant vector field by

(X(2) X(3) ... X(n))j = j i2...inX(2)i2 X(3) i3... X(n) n2,

where is the Levi-Civita tensor. Show that, in any inertial frame at a point m on a Riemannian 4-manifold, ||X(2) X(3) X(4)||2 evaluated at the point m, coincides, up to sign, with the square of the usual volume of the three-dimensional parallelepiped spanned by these vectors by justifying the following facts.
(a) Restricting your attention to Riemannian 4-manifolds, let A, B, and C be vectors at m, and suppose -- as you may -- that you have chosen an inertial frame at m with the property that A1 = B1 = C1 = 0. (Think about why you can you do this.) Show that, in this frame, ABC has only one nonzero coordinate: the first.
(b) Show that, if we consider A, B and C as 3-vectors a, b and c respectively by ignoring their first (zero) coordinate, then

(ABC)1 = a.(bc),

which we know to be ± the volume of the parallelepiped spanned by a, b and c.
(c) Defining ||C||2 = CiCjgij (recall that gij is the inverse of gkl), deduce that the scalar ||ABC||2 is numerically equal to square of the volume of the parallelepiped spanned by the vectors a, b and c. (Note also that ||ABC||2always get the same answer, no matter what coordinate system we choose.)

9. Define the Levi-Civita tensor of type (n, 0), and show that

 i1i2...in j1 j2 ... jn = 1     if (i1, ... , in) is an even permutation of (j1, ... , jn) -1   if (i1, ... , in) is an odd permutation of (j1, ... , jn .

Last Updated: January, 2002
Copyright © Stefan Waner