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In the last lecture, we saw how we can use a Riemannian metric to measure distance. Here, we look at a very special metric.
First a general comment: We said in the last section that, at any point p in a Riemannian manifold M, we can find a local chart at p with the property that the metric tensor g_{**} is diagonal, with diagonal terms 1. In particular, we said that Minkowski space comes with a such a metric tensor having signature (1, 1, 1, 1). Now there is nothing special about the number 1 in the discussion: we can also find a local chart at any point p with the property that the metric tensor g_{**} is diagonal, with diagonal terms any nonzero numbers we like (although we cannot choose the signs).
In relativity, we take deal with 4dimensional manifolds, and take the first three coordinates x^{1}, x^{2}, x^{3} to be spatial (measuring distance), and the fourth one, x^{4}, to be temporal (measuring time). Let us postulate that we are living in some kind of 4dimensional manifold M (since we want to include time as a coordinate. By the way, we refer to a chart x at the point p as a frame of reference, or just frame). Suppose now we have a particle  perhaps moving, perhaps not  in M. Assuming it persists for a period of time, we can give it spatial coordinates (x^{1}, x^{2}, x^{3}) at every instant of time (x^{4}). Since the first three coordinates are then functions of the fourth, it follows that the particle determines a path in M given by
so that x^{4} is the parameter. This path is called the world line of the particle. Mathematically, there is no need to use x^{4} as the parameter, and so we can describe the world line as a path of the form
where t is some parameter. (Note: t is not time; it's just a parameter. x^{4} is time). Conversely, if t is any parameter, and x^{i} = x^{i}(t) is a path in M, then, if x^{4} is an invertible function of t, that is, dx^{4}/dt 0 (so that, at each time x^{4}, we can solve for the other coordinates uniquely) then we can solve for x^{1}, x^{2}, x^{3} as smooth functions of x^{4}, and hence picture the situation as a particle moving through space.
Now, let's assume our particle is moving through M with world line x^{i} = x^{i}(t) as seen in our frame (local coordinate system). The velocity and speed of this particle (as measured in our frame) are given by
v  =  dx^{4} 
,  dx^{4} 
,  dx^{4} 
speed^{2}  =  dx^{4} 
2 
+  dx^{4} 
2 
+  dx^{4} 
2 
. 
The problem is, we cannot expect v to be a vector  that is, satisfy the correct transformation laws. But we do have a contravariant 4vector
T^{i}  =  dt 
(T stands for tangent vector. Also, remember that t is not time). If the particle is moving at the speed of light c, then
dx^{4} 
2 
+  dx^{4} 
2 
+  dx^{4} 
2 
=  c^{2}  ...... (I) 
dt 
2 
+  dt 
2 
+  dt 
2 
=  c^{2}  dt 
2 
(using the chain rule) 
dt 
2 
+  dt 
2 
+  dt 
2 
  c^{2}  dt 
2 
=  0. 
Now this looks like the normsquared, T^{2}, of the vector T under the metric whose matrix is
g_{**}  =  diag[1, 1, 1, c^{2}]  = 

In other words, the particle is moving at lightspeed T^{2} = 0 T is null under this rather interesting local metric. So, to check whether a particle is moving at light speed, just check whether T is null.
Question What's the c^{2} doing in place of 1 in the metric?
Answer Since physical units of time are (usually) not the same as physical units of space, we would like to convert the units of x_{4} (the units of time) to match the units of the other axes. Now, to convert units of time to units of distance, we need to multiply by something with units of distance/time; that is, by a nonzero speed. Since relativity holds that the speed of light c is a universal constant, it seems logical to use c as this conversion factor.
Now, if we happen to be living in a Riemannian 4manifold whose metric diagonalizes to something with signature (1, 1, 1, c^{2}), then the physical property of traveling at the speed of light is measured by T^{2}, which is a scalar, and thus independent of the frame of reference. In other words, we have discovered a metric signature that is consistent with the requirement that the speed of light is constant in all frames in which g_{**} has the above diagoal form (so that ita makes sense to say what the speed if light is.)
Definition 7.1 A Riemannian 4manifold M is called locally Minkowskian if its metric has signature (1, 1, 1, c^{2}). 
For the rest of this section, we will be in a locally Minkowskian manifold M.
Note If we now choose a chart x in locally Minkowskian space where the metric has the diagonal form diag[1, 1, 1, c^{2}] shown above at a given point p, then we have, at the point p:
(a) If any path C has T^{2} = 0, then
dt 
2 
+  dt 
2 
+  dt 
2 
  c^{2}  dt 
2 
=  0 
(b) If V is any contravariant vector with zero x^{4}coordinate, then
(a) says that we measure the world line C as representing a particle traveling with light speed, and (b) says that we measure ordinary length in the usual way. This motivates the following definition.
Definition 7.2 A Lorentz frame at the point p M is any coordinate system ^{i} with the following properties:
(a) If any path C has the scalar T^{2} = 0, then, at p,
(Note: In general, , is not of this form, since _{ij} may not be be diagonal) (b) If V is a contravariant vector at p with zero ^{4}coordinate, then
(Again, this need not be ^{2}.) 
It follows from the remark preceding the defintion that if x is any chart such that, at the point p, the metric has the nice form diag[1, 1, 1, c^{2}], then x is a Lorentz frame at the point p. Note that in general, the coordinates of T in the system ^{i} are given by matrix multiplication with some possibly complicated changeofcoordinates matrix, and to further complicate things, the metric may look messy in the new coordinate system. Thus, very few frames are going to be Lorentz.
Physical Interpretation of a Lorentz Frame What the definition means physically is that an observer in the frame who measures a particle traveling at light speed in the xframe will also reach the conclusion that its speed is c, because he makes the decision based on (I), which is equivalent to (II). In other words: 
Question Do all Lorentz frames at p have the property that metric has the nice form diag[1, 1, 1, c^{2}]?
Answer Yes, as we shall see below.
Question OK. But if x and are two Lorentz frames at the point p, how are they related?
Answer Here is an answer. First, continue to denote a specific Lorentz frame at the point p by x.
Theorem 7.3 (Criterion for Lorentz Frames)
The following are equivalent for a locally Minkowskian manfifold M: (a) A coordinate system ^{i} in Minkowski space M is Lorentz at the point p (b) If x is any frame such that, at p, G = diag[1, 1, 1, c^{2}], then the columns of the changeofcoordinate matrix
satisfy
where the inner product is defined by the matrix G. (c) = diag[1, 1, 1, c^{2}] 
We will call the transformation from one Lorentz frame to another a generalized Lorentz transformation.
An Example of a Lorentz Transformation We would like to give a simple example of such a transformation matrix D, so we look for a matrix D whose first column has the general form a, 0, 0, b , with a and b nonzero constants. (Why? If we take b = 0, we will wind up with a less interesting transformation: a rotation in 3space.) There is no loss of generality in taking a = 1, so let us use 1, 0, 0, _{}/c . Here, c is the speed of light, and _{} is a certain constant. (The meaning of _{} will emerge in due course). Its normsquared is (1  _{}^{2}), and we want this to be 1, so we replace the vector by
(  1 (1  _{}^{2})^{1/2} 
,  0  ,  0  ,  (1  _{}^{2})^{1/2} 
)  . 
This is the first column of D. To keep things simple, let us take the next two columns to be the corresponding basis vectors e_{2}, e_{3}. Now we might be tempted to take the forth vector to be e_{4}, but that would not be orthogonal to the above first vector. By symmetry (to get a zero inner product) we are forced to take the last vector to be
(  (1  _{}^{2})^{1/2} 
,  0  ,  0  ,  (1  _{}^{2})^{1/2} 
)  . 
This gives the transformation matrix as
D  = 

, 
and hence the new coordinates (by integrating everything in sight; using the boundary conditions ^{i} = 0 when x^{i} = 0) as
^{1}  =  x^{1}  _{}cx^{4} (1  _{}^{2})^{1/2} 
;  ^{2} = x^{2}  ;  ^{3} = x^{3}  ;  ^{4}  =  x^{4}  _{}x^{1}/c (1  _{}^{2})^{1/2} 
; 
Notice that solving the first equation for x^{1} gives
Since x^{4} is just time t here, it means that the origin of the system has coordinates (_{}ct, 0, 0) in terms of the original coordinates. In other words, it is moving in the xdirection with a velocity of
so we must interpret _{} as the speed in "warp;"
This gives us the famous
Lorentz Transformations of Special Relativity
If two Lorentz frames x and have the same coordinates at (x, y, z, t) = (0, 0, 0, 0), and if the frame is moving in the xdirection with a speed of v, then the coordinates of an event are given by

Exercise Set 7
1. What can be said about the scalar dx^{i}/dt^{2} in a Lorentz frame for a particle traveling at (a) sublight speed (b) superlight speed?
2. (a) Show that, if x^{i}(t) is a timelike path in the Minkowskian manifold M so that dx^{4}/dt 0, then d^{4}/dt 0 in every Lorentz frame . In other words, if a particle is moving at sublight speed in any one Lorentz frame, then it is moving at sublight speed in all Lorentz frames.
(b) Conclude that, if a particle is traveling at superlight speed in one Lorentz frame, then it is traveling at superlight speeds in all such frames.
3. Referring to the Lorentz transformations for special relativity, consider a "photon clock" constructed by bouncing a single photon back and forth bewtwwen two parallel mirrors as shown in in the following figure.
Now place this clock in a train moving in the xdirection with velocity v. By comparing the time it takes between a tick and a tock for a stationary observer and one on the train, obtain the time contraction formula ( in terms t) from the length contraction one.
4. Prove the claim in the proof of 7.3, that if D is a 44 matrix whose columns satisfy
column i, column j  = 

, 
using the Minkowski inner product G (not the standard inner product), then D^{1} has its columns satisfying
column i, column j  = 

. 
[Hint: use the given property of D to write down the entries of its inverse P in terms of the entries of D.]
5. Invariance of the Minkowski Form
Show that, if P = x^{i}_{0} and Q = x^{i}_{0} + x^{i} are any two events in the Lorentz frame x^{i}, then, for all Lorenz frames ^{i}, one has
[Hint: Consider the path x^{i}(t) = x_{0}^{i} + x^{i}t, so that dx^{i}/dt is independent of t. Now use the transformation formula to conclude that d^{i}/dt is also independent of t. (You might have to transpose a matrix before multiplyingÉ) Deduce that ^{i}(t) = z^{i} + r^{i}t for some constants r^{i} and s^{i}. Finally, set t = 0 and t = 1 to conclude that ^{i}(t) = _{0}^{i} + ^{i}t, and apply (c) above.]
6. If the ^{i}system is moving with a velocity v in a certain direction with resepct to the x^{i}system, we call this a boost in the given direction. Show that successive boosts in two perpendicular directions do not give a "pure" boost (the spatial axes are rotatedno longer parallel to the original axes). Now do some reading to find the transformation for a pure boost in an arbitrary direction.
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