With

u = t - r

, direct integration gives

I_{xx} = \frac{m a ^{2}}{3} cos^{2} ω u, I_{yy} = \frac{m a ^{2}}{3} sin^{2} ω u, I_{x y} = \frac{m a ^{2}}{3} sin ω u cos ω u, I_{zz} = 0.

Therefore the time-dependent spatial field is

\overset{ˉ}{h}_{ij}^{rad} = \frac{4 m a ^{2} ω ^{2}}{3 r} - cos 2 ω u - sin 2 ω u 0 - sin 2 ω u cos 2 ω u 0 000 .

2 ω

. On the positive

z

-axis this matrix is transverse and traceless. It is a rotating combination of the plus polarization and cross polarization, with amplitude proportional to

1/ z

, exactly as for a plane wave propagating in the

z

direction.

ii

Solution

G = c = 1

P = \frac{1}{5} ⟨ Q ..._{ij} Q ..._{ij} ⟩, Q_{ij} = I_{ij} - \frac{1}{3} δ_{ij} I_{kk} .

Here

I_{kk} = m a^{2} /3

is constant, so its trace subtraction has no third derivative. If

B = 4 m a^{2} ω^{3} /3

, then

I ..._{xx} = B sin 2 ω t, I ..._{yy} = - B sin 2 ω t, I ..._{x y} = - B cos 2 ω t .

Thus

I ..._{ij} I ..._{ij} = 2 B^{2}

and

⟨ P ⟩ = \frac{32}{45} m^{2} a^{4} ω^{6} .

Restoring units multiplies this by

G / c^{5}

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