Proof

(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

T = ( c, 0, 0, 1 ),

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;

 Dki = ixk ,

so that

i = DkiTk
=
 D11 D12 D13 D14 D21 D22 D23 D24 D31 D32 D33 D34 D41 D42 D43 D44
 c 0 0 1
.

By property (a) of Definition 6.2,

(1)2 + (1)2 + (1)2 - c2(1)2 = 0,

so that

(cD11 + D14)2 + (cD12 + D24)2 + (cD13 + D34)2 - c2(cD14 + D44)2 = 0 ... (*)

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

(-cD11 + D14)2 + (-cD12 + D24)2 + (-cD13 + D34)2 - c2(-cD14 + D44)2 = 0 ... (**)

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

D11D14 + D12D24 + D13D34 - c2D14D44 = 0;

that is,

column 1, column 4 = 0 = e1, e4 .

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

c2[D11D11 + D12 D12 + D13 D13 - c2D14 D14] + [D41D41 + D42 D42 + D43 D43 - c2D44 D44] = 0,

showing that

c2 column 1, column 1 = - column 4, column 4 .

So, if we write

column 1, column 1 = k,

then

column 4, column 4 = -c2k       ......     (***)

Similarly (by choosing other photons) we can replace column 1 by either column 2 or column 3, showing that if we take

column 1, column 1 = k,

we have

column i, column i =  k if 1 i 3 -kc2 if i = 4
.

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

T = c/21/2, -c/21/2, 0, 1 .

(You can check to see that ||T||2 = 0, so that it does indeed represent a photon.) Since ||||2 = 0, we get

(D11c /21/2- D21c /21/2+D41)2 + (D12c /21/2- D22c /21/2+D42)2 + (D13c /21/2- D23c /21/2+D43)2 - c2(D14c /21/2- D24c /21/2+D44)2= 0

and, looking at a similar photon traveling in the opposite x2-direction,

(D11c /21/2+ D21c/21/2+D41)2 + (D12c/21/2+ D22c/21/2+D42)2 + (D13c/21/2+ D23c/21/2+D43)2 - c2(D14c/21/2+ D24c/21/2+D44)2= 0

Subtracting these gives

2c2[D11D21 + D12 D22 + D13 D23 - c2D14 D24] + 4c/21/2[D21D41 + D22 D42 + D23 D43 - c2D24 D44] = 0.

But we already know that the second term vanishes, so we are left with

D11D21 + D12 D22 + D13 D23 - c2D14 D24 = 0,

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 =  0 if i j k if 1 i = j 3 -kc2 if i = j = 4
.....      (IV)

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,

= PTGP

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

= PTGP = G/k.

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:

||||2 = 1.

On the other hand, we can also use to compuate ||||2, and we get

||||2 =1/k,

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,

Di1Dj1 + Di2 Dj2 + Di3 Dj3 - c2Di4 Dj4 = column i, column j =  0 if i j 1 if 1 i = j 3 -c2 if i = j = 4
,

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.