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9. Geodesics and Local Inertial Frames
Let us now apply some of this theory to curves on manifolds. If a nonnull curve C on M is paramaterized by x^{i}(t), then we can reparamaterize the curve using arc length,
s(t)  =  t a 
 du  , 
(starting at some arbitrary point) as the parameter. The reason for wanting to do this is that the tangent vector T^{i} = dx^{i}/ds is then a unit vector (see the exercises) and also independent of the paramaterization.
If we were talking about a curve in E_{3}, 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 T^{i} the curvature of C.
If C happens to be on a manifold, then the unit tangent vector is still
T^{i}  =  ds 
=  dt 
/  dt 
= 

(the last formula is there if you want to actually compute it). But, to get the curvature, we need to take the covariant derivative:
P^{i}  =  ds 

=  ds 

=  ds^{2} 
+  p q 
ds 
ds 
Definitions 9.1 The first curvature vector P of the curve C is
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

In terms of the parameter t, this becomes (see the exercises)
dt^{2} 
dt 
  dt 
dt^{2} 
+  p q 
dt 
dt 
dt 
=  0, 
where
dt 
= 

. 
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.
ds 
T, T  = 

(Exercise Set 8 #9)  
= 

(Symmetry of the scalar product)  
= 

(Definition of P) 
P, T  = 

T, T  = 

(Refer back to the Proof of 6.5 to check this) 
T, T  = 

(Refer back to the Proof of 6.5 to check this) 
Local Flatness, or "Local Inertial Frames"
Notation We will be changing some of the notation to simplify things from now on.
In "flat space" E_{s} 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 x^{i} at m such that:
We call such a coordinate system a local inertial frame or a normal frame. 
(It follows that _{i}^{j}_{k}(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) g_{pq,r}(m) = 0 for all p, q, r. (b) [pq, r]_{m} = 0 for all p, q, r. (c) _{p}^{r}_{q}(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
g_{pq,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] = g_{sr}_{p}^{r}_{q}.
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

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, d^{2}x^{i}/ds^{2} = 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 spacetime.
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 (x^{i}/^{j}), one has

Proof By definition of inertial frames,
and similarly for ^{ij}, so that ^{ij} = ±g^{ij}, whence det(g_{**}) = ± det(_{**}) = ±1. On the other hand,
_{ij}  =  ^{i} 
^{j} 
g_{kl}, 
which, in matrix form, becomes
Taking determinants gives
giving
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 = c^{2} 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 changeofcoordinates matrix D has positive determinant. Further, if D happens to be the changeofcoordinates from one oriented inertial frame to another, then det(D) = +1.
4.E_{3} has two orientations: one given by any lefthanded system, and the other given by any righthanded 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 ndimensional Riemannian manifold. The LeviCivita tensor of type (0, n) is defined as follows. If is any coordinate system and m M, then define

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
i_{1}i_{2}...i_{n}(m)  =  1, if (i_{1},_{ }i_{2}, ... , i_{n}) is an even permutation of (1, 2, ... , n) 1, if (i_{1},_{ }i_{2}, ... , i_{n}) 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 "LeviCivita 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 (LeviCivita Tensor)
The LeviCivita tensor is a welldefined, smooth tensor field. 
Proof To show that it is welldefined, 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 changeofcoordinates matrices D and E, then DE is the changeof coordinates from one inertial frame to another, and therefore has determinant 1. Now,
_{ i1i2...in}(m)  =  det (D_{i1}D_{i2} ... D_{in})  
=  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 welldefined at each point. We now show that it is a tensor. If and _{} are any two oriented coordinate systems at m and changeofcoordinate 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 (E_{r1}E_{r2} ... E_{rn}) respectively, then at the point m,
_{ k1k2...kn}  =  det (D_{k1}D_{k2} ... D_{kn})  
= 


(by definition of the determinant(!) since _{ i1i2...in} is just the sign of the permutation!)  
= 


= 

showing that the tensor transforms correctly. Finally, we assert that det (D_{k1}D_{k2} ... D_{kn}) is a smooth function of the point m. This depends on the changeofcoordinate matrices to the inertial coordinates. But we saw that we could construct inertial frames by setting
^{j} 
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 E_{3}, the LeviCivita tensor coincides with the totally antisymmetric thirdorder tensor _{ijk} in Exercise Set 4. In the Exercises, we see how to use it to generalize the crossproduct.
Exercise Set 9
1. Recall that we can define the arc length of a smooth nonnull curve by
s(t)  =  t a 
 du  . 
Assuming that this function is invertible (so that we can express x^{i} as a function of s) show that
ds 
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 g_{abc} = g^{ab}_{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 LeviCivita tensor, then, in any frame, _{ i1i2...in} = 0 whenever two of the indices are equal. Thus, the only nonzero coordinates occur when all the indices differ.
7. Use the LeviCivita 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
is a scalar.
8. The Volume 1Form (A Generalization of the Cross Product)
If we are given n1 vector fields X(2), X(3), . . . , X(n) on the nmanifold M, define a covariant vector field by
where is the LeviCivita tensor. Show that, in any inertial frame at a point m on a Riemannian 4manifold, 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 threedimensional parallelepiped spanned by these vectors by justifying the following facts.
(a) Restricting your attention to Riemannian 4manifolds, 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 A^{1} = B^{1} = C^{1} = 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 3vectors a, b and c respectively by ignoring their first (zero) coordinate, then
which we know to be ± the volume of the parallelepiped spanned by a, b and c.
(c) Defining C^{2} = C_{i}C_{j}g^{ij} (recall that g^{ij} is the inverse of g_{kl}), 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 LeviCivita tensor of type (n, 0), and show that
_{ i1i2...in}^{ j1}^{ j2} ... ^{ jn}  =  1 if (i_{1},_{ }... , i_{n}) is an even permutation of (j_{1},_{ }... , j_{n})
1 if (i_{1}, ... , i_{n}) is an odd permutation of (j_{1},_{ }... , j_{n} 
. 
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