Proof
(a) (b)
Suppose the coordinate system ^{i} is Lorentz at p, and let x^{i} 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: x^{1} = ct; x^{2} = x^{3} = 0, x^{4} = t (a photon traveling along the x^{1}axis in E_{4}).
This gives
and hence T^{2} = 0, and hence Definition 6.2 (a) applies. Let D be the changeofbasis matrix to the (other) inertial frame ^{i};
D_{k}^{i}  =  x^{k} 
, 
so that
^{i}  =  D_{k}^{i}T^{k}  
= 

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

. 
Let us now take another, more interesting, photon given by
Path D: x^{1} = (c/2^{1/2})t; x^{2} = (c/2^{1/2})t; x^{3} = 0; x^{4} = 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 x^{2}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  = 

..... (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 = P^{T}GP, 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 c^{2}. So, when we take P^{T}(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 normsquared 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,
D_{i}^{1}D_{j}^{1} + D_{i}^{2} D_{j}^{2} + D_{i}^{3} D_{j}^{3}  c^{2}D_{i}^{4} D_{j}^{4}  =  column i, column j  = 

, 
then the argument in (a) (b) shows that = G (since k = 1/k = 1 here).
(c) (a)
If = diag[1, 1, 1, c^{2}] at the point p, then is Lorentz at p, by the remarks preceding Definition 6.2. Thus, we are done.
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