(a) (b)

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).

This gives

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;

so that

By property (a) of Definition 6.2,

so that

If we reverse the direction of the photon, we similarly get

Noting that this only effects cross-terms, subtracting and dividing by 4c gives

that is,

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

showing that

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

we have

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

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).

(b) (c)

If the change of coordinate matrix has the above orthogonality property,

then the argument in (a) (b) shows that = G (since k = 1/k = 1 here).

(c) (a)

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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