Notes on Relativity and Cosmology - Physics Department, UCSB

9.4. STRETCHING AND SQUISHING 257
In addition, there is clearly an analogue of equation (9.20) for the second static
observer (remembering that the second one does not start at x = 0, but instead
starts at x = L):
x 2 = L + 1 2 (a s2 + a FF2 )t 2 2 + O(T 4 ). (9.24)
9.4.2 The solution
So, let’s try using these equations to solve the problem at hand. The way the I
will proceed is to substitute equation (9.22) for t 2 in equation (9.24). That way
we express both positions in terms of just t 1 . The result is
x 2 = L + 1 (
2 (a s2 + a FF2 ) 1 + a ) 2
s1
c 2 (x 2 − x 1 ) t
2
1 + O(T 4 ) + O(L 3 T 2 ). (9.25)
OK, now we will want to use the condition (9.23) that the proper distance between
the static observers does not change. This equation involves the difference
x 2 − x 1 . Subtracting equation (9.20) from equation (9.25), we get:
x 2 − x 1
= L + 1 2 (a s2 + a FF2 − a s1 )t 2 1 + (a s2 + a FF2 ) a s1
c 2 (x 2 − x 1 )t 2 1
+ 1 2 (a s2 + a FF2 ) a2 s1
c 4 (x 2 − x 1 ) 2 t 2 1 + O(T 4 ) + O(L 3 T 2 ). (9.26)
And, actually, we won’t need to keep the (x 2 − x 1 ) 2 term, so we can write this
as:
x 2 −x 1 = L+ 1 2 (a s2+a FF2 −a s1 )t 2 1+(a s2 +a FF2 ) a s1
c 2 (x 2−x 1 )t 2 1+O(T 4 )+O(L 2 T 2 ).
(9.27)
Now, this equation involves x 2 − x 1 on both the left and right sides, so let’s
solve it for x 2 − x 1 . As you can check, the result is:
x 2 −x 1 =
(
L + 1 ) (
2 (a s2 + a FF2 − a s1 )t 2 1 1 − (a s2 + a FF2 ) a )
s1 −1+O(T
c 2 t2 1
4 )+O(T 2 L 2 ).
(9.28)
But there is a standard ‘expansion’ (1 − x) −1 = 1 + x + O(x 2 ) that we can use
to simplify this. We find:
x 2 − x 1 = L+ 1 2 (a s2 +a FF2 − a s1 )t 2 1 + L(a s2 +a FF2 ) a s1
c 2 t2 1 + O(T 4 )+O(T 2 L 2 ).
(9.29)