# Physics 616

• Prof. Andrew W. Steiner
(or Andrew or "Dr. Steiner")
• Office hour: 103 South College, Thursday 11am
• Email: awsteiner@utk.edu
• Homework: Electronically as .pdf
• You may work with each other on the homework, but you must write the solution in your own words

Use down and up arrows to proceed to the next or previous slide.

## Group Work

• Complete 8.17 and 8.18 in Schutz. For 8.17, just start with Kepler's laws, and for 8.18, check out arXiv:gr-qc/0004037 .

## Generation by a Slowly Moving Source

• Now solve $$\left( - \frac{\partial^2}{\partial t^2} + \nabla^2 \right) \bar{h}_{\mu \nu} = -16 \pi T_{\mu \nu}$$ by assuming $$T_{\mu \nu} = \mathrm{Re} \left[ S_{\mu \nu}(x^{i}) e^{-i \Omega t}\right]$$ and look for a solution of the form $$\bar{h}_{\mu \nu} = \mathrm{Re} \left[ B_{\mu \nu}(x^{i}) e^{-i \Omega t}\right]$$ giving $$\left( \nabla^2+\Omega^2 \right) B_{\mu \nu} = -16 \pi S_{\mu \nu}$$
• Far away where $S=0$, we want outgoing radiation, and presume spherical symmetry. One can show $$B_{\mu \nu} = \frac{A_{\mu \nu}}{r} e^{i \Omega r} +\frac{Z_{\mu \nu}}{r} e^{-i \Omega r}$$ where $A_{\mu \nu}$ and $Z_{\mu \nu}$ (chosen to be zero) are constants.

## Generation by a Slowly Moving Source II

• Presume source nonzero inside sphere of radius $\varepsilon \ll 2\pi/\Omega$, then integrate our problem inside.
• LHS leads to $$\int \Omega^2 B_{\mu \nu} d^3 x \leq \Omega^2 |B_{\mu \nu}|_{\mathrm{max}} 4 \pi \varepsilon^3/3$$ and the other term, from which we can use the divergence theorem $$\int \nabla^2 B_{\mu \nu} d^3 x = \oint \mathbf{n} \cdot \nabla B_{\mu \nu}~dS = 4 \pi \varepsilon^2 \left( \frac{d}{dr} B_{\mu \nu} \right)_{r=\varepsilon} \approx - 4 \pi A_{\mu \nu}$$
• RHS leads to $$J_{\mu \nu} \equiv \int S_{\mu \nu} d^3 x$$
• Schutz gets $A_{\mu \nu} = 4 J_{\mu \nu}$, how?, thus $\bar{h}_{\mu \nu} = 4 J_{\mu \nu} e^{i \Omega (r-t)}/r$

## Generation by a Slowly Moving Source III

• Using our definitions $$J_{\mu \nu} e^{-i \Omega t} = \int T_{\mu \nu} d^3 x$$ Now Schutz gets (how?) $$-i \Omega J^{\mu 0} e^{-i \Omega t} = \int {T^{\mu 0}}_{,0} d^3 x$$ but energy and momentum is conserved, thus $$i\Omega J^{\mu 0} e^{-i \Omega t} = \int {T^{\mu k}}_{,k}~d^3 x = \oint {T^{\mu k}} n_k~dS$$ but the RHS vanishes so $J^{\mu 0}=0$.

## Generation by a Slowly Moving Source IV

• We can write the spatial components of J using the tensor virial theorem $$\frac{d^2}{dt^2} \int T^{00} x^{\ell} x^{m} d^3 x = 2 \int T^{\ell m}~d^3 x$$ thus defining $I$, the quadrupole moment tensor, $$I^{\ell m} \equiv D^{\ell m} e^{-i \Omega t} \equiv \int T^{00} x^{\ell} x^{m}~d^3 x$$ we get $$\bar{h}_{jk} = - 2 \Omega^2 D_{jk} e^{i \Omega(r-t)}/r$$
• There is no wave generation from monopole or dipole sources, monopole moment is just the mass, and dipole moment is the center of mass

## Choosing Coordinates

• Again choose the wave to move in the z direction and choose the TT gauge, then $$\bar{h}_{xx}^{TT}= - \bar{h}_{yy}^{TT}= - \Omega^2 \left( {I\!\!\!-}_{xx} - {I\!\!\!-}_{yy} \right) e^{i \Omega r}/r$$ and $$\bar{h}_{xy}^{TT}=- 2 \Omega^2 {I\!\!\!-}_{xy} e^{i \Omega r}/r$$ where $${I\!\!\!-}_{jk} \equiv I_{jk} - \frac{1}{3} \delta_{jk} I^{\ell}_{\ell}$$ is the trace-free quadrupole moment tensor

## Two Masses Separated by a Spring

• Remember quadrupole tensor is $$I^{\ell m} \equiv \int T^{00} x^{\ell} x^{m}~d^3x$$
• For our two masses oscillating with angular frequency $\omega$ and amplitude $A$ separated by a distance $\ell_0$: $$\begin{eqnarray} T_{xx} &=& m x_1^2 + m x_2^2 \\ &=& \mathrm{constant} + mA^2 \cos 2 \omega t + 2 m \ell_0 A \cos \omega t \end{eqnarray}$$
• Wave equation is linear (even though second derivatives appear, the metric does not appear with any exponent)
• The $\cos \omega t$ part gives $$\begin{eqnarray} {I\!\!\!-}_{xx} &=& \frac{4}{3} m \ell_0 A \mathrm{Re}(e^{-i \omega t}) \\ {I\!\!\!-}_{yy} &=& {I\!\!\!-}_{zz} = -\frac{2}{3} m \ell_0 A \mathrm{Re}(e^{-i \omega t}) \end{eqnarray}$$ and off-diagonal parts are zero

## Two Masses Separated by a Spring II

• For radiation traveling in the z-direction remember $$\bar{h}_{xx}^{TT}= - \bar{h}_{yy}^{TT}= - \Omega^2 \left( {I\!\!\!-}_{xx} - {I\!\!\!-}_{yy} \right) e^{i \Omega r}/r$$ and $$\bar{h}_{xy}^{TT}=- 2 \Omega^2 {I\!\!\!-}_{xy} e^{i \Omega r}/r$$ thus $$\bar{h}_{xx}^{TT}= - \bar{h}_{yy}^{TT}= - 2 m \omega^2 \ell_0 A e^{i \omega (r-t)}/r$$ and $$\bar{h}_{xy}^{TT}= 0$$
• There is no radiation traveling in the x-direction

## Two Masses Separated by a Spring III

• For the $\cos 2 \omega t$ part thus $$\bar{h}_{xx}^{TT}= - \bar{h}_{yy}^{TT}= - 4 m \omega^2 A^2 e^{2 i \omega (r-t)}/r$$ and $$\bar{h}_{xy}^{TT}= 0$$
• Thus the total gravitational radiation is the sum $$\bar{h}_{xx}^{TT}= - 2 m \omega^2 A \left[ \ell_0 e^{i \omega (r-t)} + 2 A e^{2 i \omega (r-t)} \right]/r$$
• Some numbers: $m=10^3$ kg, $\ell_0 = 1$ m, $A=10^{-4}$ m, $\omega=10^4~\mathrm{s}^{-1}$ then the amplitude is $(10^{-34} m)/r$