 Solution

ID: past-exam-of-the-general-relativity-ii-course-of-the-university-of-oxford/2021/2/d/solution

A, B, C, F

u

by integration so that

A^{''} = \frac{c}{2}, B^{''} = \frac{d}{2}, C^{''} = \frac{e}{2}, F^{'} = - \frac{f}{2} .

For example, for any fixed

u_{0}

one may take

A (u) = \frac{1}{2} \int_{u_{0}}^{u} (u - s) c (s) d s, B (u) = \frac{1}{2} \int_{u_{0}}^{u} (u - s) d (s) d s, C (u) = \frac{1}{2} \int_{u_{0}}^{u} (u - s) e (s) d s, F (u) = - \frac{1}{2} \int_{u_{0}}^{u} f (s) d s .

A generator, specified as a covector with components in the null coordinates

(u, v, x, y)

, is

χ = A d v + B d x + C d y + (- A^{'} v - B^{'} x - C^{'} y + F) d u .

Indeed, with

δ h_{A B} = \partial_{A} χ_{B} + \partial_{B} χ_{A}

, the mixed components cancel:

δ h_{uv} = A^{'} - A^{'} = 0, δ h_{ux} = B^{'} - B^{'} = 0, δ h_{u y} = C^{'} - C^{'} = 0.

All components without a

u

index vanish as well, and

δ h_{uu} = - 2 A^{''} v - 2 B^{''} x - 2 C^{''} y + 2 F^{'} = - c v - d x - ey - f .

Since

n^{♭} = d u

, only the

uu

component needs changing. Thus

\overset{˚}{h} = ψ d u^{2}, ψ = a (u) (x^{2} - y^{2}) + b (u) x y .

Yes, $\overset{˚}{h}$ satisfies both equations (3). Its trace is zero,

\partial_{v} ψ = 0

\partial^{A} \overline{\overset{˚}{h}}_{A B} = 2 \partial_{v} ψ δ_{B}^{u} = 0, □ ψ = 4 \partial_{u} \partial_{v} ψ + \partial_{x}^{2} ψ + \partial_{y}^{2} ψ = 2 a - 2 a = 0.

The two surviving functions encode the two gravitational wave polarizations; constant and linear transverse terms have been removed by the explicit linearized coordinate gauge transformation.

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