 Solution

ID: past-exam-of-the-relativity-course-of-the-university-of-cambridge/2022/section-b/6/b/i/solution

The inverse transformation is

t = (p + q) /2

x = (q - p) /2

. Thus

d s^{2} = \frac{1}{4} [(d p + d q)^{2} - (d q - d p)^{2}] = d p d q .

(g_{ab})_{(p, q)} = (0 1/2 1/2 0), (g^{ab}) = (0220) .

Both off-diagonal entries are needed, since the quadratic line element sums over both index orders.

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