- #1
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I finally found a result I believe for the the asymptotic metric (valid for large r) of a pair of bodies in a circular orbit emitting gravitational waves. I use spherical coordinates, ##[t, r, \theta, \phi]##.
If we let the linearized metric ##g_{\mu\nu}## be equal to the sum of a flat metric ##\eta_{\mu\nu}## and a pertubation metric ##h_{\mu\nu}##, then the approximate solution is:
$$\eta_{\mu\nu} = \begin{bmatrix} -c^2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & r^2 & 0 \\ 0 & 0 & 0 & r^2 \sin^2 \theta \end{bmatrix} \quad h_{\mu\nu} = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & k r \cos \psi & k r \sin \theta \sin \psi \\ 0 & 0 & k r \sin \theta \sin \psi & -k r \sin^2 \theta \cos \psi \end{bmatrix}$$
Here k is an arbitrary constant which depends on the quadrupole moment of the pair of bodies, and ##\psi = 2 \omega (t - r/c)## is a function of retarded time t-r/c. The wave propagates radially outward from the origin. In my actual calculations I simplified things by setting c=1 and ##\omega## = 1/2.
This metric only satisfies the Einstein field equations for large r. Because of this, rather than using the Lorentz gauge (which the solution only approximately satisfies), I used the arbitrary non-gauge formula to compute the Ricci tensor to check the solution. The specifics were:
$$\Gamma^a{}_{bc} = \frac{1}{2}( h_a{}^u{}_{,b} + h_b{}^u{}_{,a} - h_{ab}{}^{,u}) \quad R^a{}_{bcd} = \Gamma^a{}_{bd,c} - \Gamma^a{}_{bc,d} \quad R_{bc} = R^a{}_{bac}$$
Here a comma represents taking the partial derivative. Note that only partial derivatives are needed, something that surprised me enough that I rechecked my text (MTW) on the topic.
I found it convenient to re-write the metric and the Ricci tensor I computed from the above in an orthonormal basis of one-forms with
$$e^\hat{t} = dt \quad e^\hat{r} = dr \quad e^\hat{\theta} = r d\theta \quad e^\hat{\phi} = r \sin \theta \, d \phi$$
Representing the tensors in this basis by putting a "hat" over the basis symbols, I found:
$$h_{\hat{a}\hat{b}} = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & k \frac{\cos \psi}{r} & k \frac{\sin \psi}{r} \\ 0 & 0 & k \frac{\sin \psi}{r} & - k \frac{ \cos \psi}{r} \end{bmatrix} R_{\hat{a}\hat{b}} \approx \begin{bmatrix} 0 & 0 & 0 & O(\frac{1}{r^2}) \\ 0 & 0 & 0 & O(\frac{1}{r^3}) \\ 0 & 0 & O(\frac{1}{r^2}) & O(\frac{1}{r^2}) \\ O(\frac{1}{r^2}) & O(\frac{1}{r^3}) & O(\frac{1}{r^2}) & O(\frac{1}{r^3}) \end{bmatrix} $$
Note the solution isn't quite completely correct to order 1/r, because it is linearized to a flat background metric, when we really expect the static part of the metric for a pair of orbiting bodies to be not quite flat, but to have time dilation terms in ##g_{tt}## of order 1/r.
If we let the linearized metric ##g_{\mu\nu}## be equal to the sum of a flat metric ##\eta_{\mu\nu}## and a pertubation metric ##h_{\mu\nu}##, then the approximate solution is:
$$\eta_{\mu\nu} = \begin{bmatrix} -c^2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & r^2 & 0 \\ 0 & 0 & 0 & r^2 \sin^2 \theta \end{bmatrix} \quad h_{\mu\nu} = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & k r \cos \psi & k r \sin \theta \sin \psi \\ 0 & 0 & k r \sin \theta \sin \psi & -k r \sin^2 \theta \cos \psi \end{bmatrix}$$
Here k is an arbitrary constant which depends on the quadrupole moment of the pair of bodies, and ##\psi = 2 \omega (t - r/c)## is a function of retarded time t-r/c. The wave propagates radially outward from the origin. In my actual calculations I simplified things by setting c=1 and ##\omega## = 1/2.
This metric only satisfies the Einstein field equations for large r. Because of this, rather than using the Lorentz gauge (which the solution only approximately satisfies), I used the arbitrary non-gauge formula to compute the Ricci tensor to check the solution. The specifics were:
$$\Gamma^a{}_{bc} = \frac{1}{2}( h_a{}^u{}_{,b} + h_b{}^u{}_{,a} - h_{ab}{}^{,u}) \quad R^a{}_{bcd} = \Gamma^a{}_{bd,c} - \Gamma^a{}_{bc,d} \quad R_{bc} = R^a{}_{bac}$$
Here a comma represents taking the partial derivative. Note that only partial derivatives are needed, something that surprised me enough that I rechecked my text (MTW) on the topic.
I found it convenient to re-write the metric and the Ricci tensor I computed from the above in an orthonormal basis of one-forms with
$$e^\hat{t} = dt \quad e^\hat{r} = dr \quad e^\hat{\theta} = r d\theta \quad e^\hat{\phi} = r \sin \theta \, d \phi$$
Representing the tensors in this basis by putting a "hat" over the basis symbols, I found:
$$h_{\hat{a}\hat{b}} = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & k \frac{\cos \psi}{r} & k \frac{\sin \psi}{r} \\ 0 & 0 & k \frac{\sin \psi}{r} & - k \frac{ \cos \psi}{r} \end{bmatrix} R_{\hat{a}\hat{b}} \approx \begin{bmatrix} 0 & 0 & 0 & O(\frac{1}{r^2}) \\ 0 & 0 & 0 & O(\frac{1}{r^3}) \\ 0 & 0 & O(\frac{1}{r^2}) & O(\frac{1}{r^2}) \\ O(\frac{1}{r^2}) & O(\frac{1}{r^3}) & O(\frac{1}{r^2}) & O(\frac{1}{r^3}) \end{bmatrix} $$
Note the solution isn't quite completely correct to order 1/r, because it is linearized to a flat background metric, when we really expect the static part of the metric for a pair of orbiting bodies to be not quite flat, but to have time dilation terms in ##g_{tt}## of order 1/r.
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