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( 3 x^{i} x^{j} - \delta_{ij} r^{2} \right) $$

where ρ is the mass density field, x is the position vector, δij 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



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)



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{32 G^{4} m_{1}^{2} m_{2}^{2} \left( m_{1} + m_{2} \right)}{5 c^{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{8 G^{4} m_{1}^{2} m_{2}^{2} \left( m_{1} + m_{2} \right)}{15 a^{5} c^{5} \left( 1 - e^{2} \right)^{5}} \left( 1 + e cos \psi \right)^{4} \left[ 12 \left( 1 + e cos \psi \right)^{2} + e^{2} \sin^{2} \psi \right] $$

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



-\frac{dL}{dt} = \frac{32 G^{7/2} m_{1}^{2} m_{2}^{2} \sqrt{m_{1} + m_{2}}}{5 c^{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} $$