 Solution

ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2015/paper-1/6/e/solution

The generator acts only in the

c t, z

plane, with its square equal to the identity on that two-dimensional subspace. Its exponential is

Λ = cosh κ 00 sinh κ 01000010 sinh κ 00 cosh κ .

For the vector in (c), this gives

c t^{'} = \frac{1}{2} (cosh κ + sinh κ cos θ), z^{'} = \frac{1}{2} (sinh κ + cosh κ cos θ),

while

x^{'} = x

and

y^{'} = y

. Since the vector is future-null, its spatial magnitude is

c t^{'}

. Therefore

\frac{z ^{'}}{c t ^{'}} = \frac{c o s θ + t a n h κ}{1 + t a n h κ c o s θ} = cos θ^{'}, \frac{x ^{'2} + y ^{'2}}{c t ^{'}} = \frac{s i n θ}{R} .

The transverse azimuth does not change. These are exactly the direction and scale obtained from the spinor congruence. In particular

c t^{'} = R c t

and

tan (θ^{'} /2) = e^{- κ} tan (θ /2)

. The azimuth-sign qualification in (c) is compatible with this check because a boost along

z

leaves both transverse coordinates unchanged.

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