Suppose the coordinate system i is Lorentz at p, and let xi be as hypothesized in (b). We proceed by invoking condition (a) of Definition 6.2 for several paths. (These paths will correspond to sending out light rays in various directions.)
Path C: x1 = ct; x2 = x3 = 0, x4 = t (a photon traveling along the x1-axis in E4).
and hence ||T||2 = 0, and hence Definition 6.2 (a) applies. Let D be the change-of-basis matrix to the (other) inertial frame i;
By property (a) of Definition 6.2,
If we reverse the direction of the photon, we similarly get
Noting that this only effects cross-terms, subtracting and dividing by 4c gives
In other words, the first and fourth columns of D are orthogonal under the Minkowskian inner product. Similarly, by sending light beams in the other directions, we see that the other columns of D are orthogonal to the fourth column.
If, instead of subtracting, we now add (*) and (**), and divide by 2, we get
So, if we write
Similarly (by choosing other photons) we can replace column 1 by either column 2 or column 3, showing that if we take
|column i, column i||=||
Let us now take another, more interesting, photon given by
Path D: x1 = (c/21/2)t; x2 = -(c/21/2)t; x3 = 0; x4 = t, with
(You can check to see that ||T||2 = 0, so that it does indeed represent a photon.) Since ||||2 = 0, we get
and, looking at a similar photon traveling in the opposite x2-direction,
Subtracting these gives
But we already know that the second term vanishes, so we are left with
showing that columns 1 and 2 are also orthogonal.
Choosing similar photons now shows us that columns 1, 2, and 3 are mutually orthogonal. Therefore, we have
|column i, column j||=||
But, what is k? Let us now invoke condition (b) of Defintion 6.2. To measure the length of a vector in the new frame, we need to transform the metric tensor using this coordinate change. Recall that, using matrix notation, the metric G transforms to = PTGP, where P is the matrix inverse of D above. In the exercise set, you will see that the columns of P have the same property (IV) above, but with k replaced by 1/k. But,
Now, since G is just a constant multiple of an elementary matrix, all it does is multiply the last row of P by c2. So, when we take PT(GP), we are really getting the funny dot product of the columns of P back again, which just gives a multiple of G. In other words, we get
Now we invoke condition (b) in Definitoin 6.2: Take the vector = (1, 0, 0, 0) in the -frame. (Recognize it? It is the vector /1.) Since its 4th coordinate is zero, condition (b) says that its norm-squared must be given by the usual length formula:
On the other hand, we can also use to compuate ||||2, and we get
showing that k = 1. Hence, = G, and also D has the desired form. This proves (b) (and also (c), by the way).
If the change of coordinate matrix has the above orthogonality property,
|Di1Dj1 + Di2 Dj2 + Di3 Dj3 - c2Di4 Dj4||=||column i, column j||=||
then the argument in (a) (b) shows that = G (since k = 1/k = 1 here).
If = diag[1, 1, 1, -c2] at the point p, then is Lorentz at p, by the remarks preceding Definition 6.2. Thus, we are done.
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