User:WillowW/Gravitational radiation

On page 355 of Landau and Lifshitz's book, The Classical Theory of Fields (Volume 2 of their Course of Theoretical Physics, Pergamon Press, New York, 1975, translated by Morton Hammermesh, ISBN 0-08-025072-6), they define the mass quadrupole tensor as

$D_{ij}=\int d\mathbf {x} \rho (\mathbf {x} )\left(3x^{i}x^{j}-\delta _{ij}r^{2}\right)$

where ρ is the mass density field, x is the position vector, δ_ij is the Kronecker delta, and i and j range over the spatial coordinates 1-3. Lower down on the same page, they give the power radiated into a given solid angle dΩ

Failed to parse (unknown function "\dddot"): {\displaystyle dP = \frac{G}{36\pi c^{5}} \left[ \frac{1}{4} \left( \dddot{D}_{i}{j} n_{i} n_{j} \right)^{2} + \frac{1}{2} \dddot{D}_{ij}^{2} - \dddot{D}_{ab} \dddot{D}_{ik} n_{j} n_{k} \right] d\Omega }

which, when integrated, gives the total power radiated (the loss of energy E per unit time t)

Failed to parse (unknown function "\dddot"): {\displaystyle P = - \frac{dE}{dt} = \frac{G}{45 c^{5}} \dddot{D}_{ij}^{2} }

This quadrupole formula assumes that the source is much smaller than the wavelength of the emitted radiation. In the following Problem 2 on p. 356–357, they quote from PC Peters and J Mathews (Phys. Rev., 131, 435 (1963)) that the mean radiated power is

$\langle P\rangle ={\frac {32G^{4}m_{1}^{2}m_{2}^{2}\left(m_{1}+m_{2}\right)}{5c^{5}a^{5}}}{\frac {1}{\left(1-e^{2}\right)^{7/2}}}\left(1+{\frac {73}{24}}e^{2}+{\frac {37}{96}}e^{4}\right)$

where a is the major semiaxis and e is the ellipticity. Here again G is the gravitational constant, which results from averaging over the exact formula

$P=-{\frac {dE}{dt}}={\frac {8G^{4}m_{1}^{2}m_{2}^{2}\left(m_{1}+m_{2}\right)}{15a^{5}c^{5}\left(1-e^{2}\right)^{5}}}\left(1+ecos\psi \right)^{4}\left[12\left(1+ecos\psi \right)^{2}+e^{2}\sin ^{2}\psi \right]$

By an analogous calculation, the average loss of angular momentum is

$-{\frac {dL}{dt}}={\frac {32G^{7/2}m_{1}^{2}m_{2}^{2}{\sqrt {m_{1}+m_{2}}}}{5c^{5}a^{7/2}}}{\frac {1}{\left(1-e^{2}\right)^{2}}}\left(1+{\frac {7}{8}}e^{2}\right)$

For circular motion, e = 0 and

$P=\omega {\frac {dL}{dt}}$