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.

Outline

We found in vacuum the wave equation $$ \left( - \frac{\partial^2}{\partial t^2} + \nabla^2 \right) \bar{h}^{\alpha \beta} = 0 $$
Now we retain the time dependence and assume a solution of the form $$ \bar{h}^{\alpha \beta} = A^{\alpha \beta} \exp \left( i k_{\alpha} x^{\alpha} \right) $$ with $ k^{\alpha} $ and $A^{\alpha \beta}$ complex
Rewrite the wave equation $$ \eta^{\mu \nu} {\bar{h}^{\alpha \beta}}_{,\mu \nu} =0 $$ then using our guess for the solution $$ {\bar{h}^{\alpha \beta}}_{,\mu} = i k_{\mu} \bar{h}^{\alpha \beta} $$

Thus $$ \eta^{\mu \nu} {\bar{h}^{\alpha \beta}}_{,\mu \nu} = - \eta^{\mu \nu} k_{\mu} k_{\nu} \bar{h}^{\alpha \beta} = 0 $$ which vanishes only if $$ k^{\mu} k_{\nu}=0 $$
If we define $ \vec{k} = (\omega,\mathbf{k})$, then $$ \omega^2 = |\mathbf{k}^2| $$
We are still allowed to exploit the freedom provided by gauge transformations, thus we can require $$ {\bar{h}^{\alpha \beta}}_{,\beta}=0 $$ which implies $$ A^{\alpha \beta} k_{\beta} =0 $$ or $A$ is orthogonal to $\vec{k}$.

We can change gauge by solving $$ \left( - \frac{\partial^2}{\partial t^2} + \nabla^2 \right) \xi_{\alpha} =0 $$ so we solve this as before with $$ \xi_{\alpha} = B_{\alpha} \exp \left( i k_{\mu} x^{\mu} \right) $$ and then $B_{\alpha}$ can be chosen to ensure $$ {A^{\alpha}}_{\alpha} = 0 $$ and $$ A_{\alpha \beta} U^{\beta} = 0 $$ where $\vec{U}$ is some fixed four-velocity
This is the "transverse-traceless" gauge, and has the property that $$ \bar{h}^{TT}_{\alpha \beta} = h^{TT}_{\alpha \beta} $$

Choose $$ U^{\beta} = {\delta^{\beta}}_{0} $$ and orient the wave in the z-direction so that $$ \vec{k} = ( \omega,0,0,\omega) $$ then one can show that $$ A_{\alpha \beta}^{TT} = \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & A_{xx} & A_{xy} & 0 \\ 0 & A_{xy} & -A_{xx} & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) $$

From "An Introduction to QFT" by Sterman

The equation of motion (aka Euler-Lagrange equation) for the photon field, $ A^{\mu} $ in QED is $$ \left( \partial_{\nu} \partial^{\nu} \right) A^{\mu} = 0 $$
This looks familiar! We solve with $$ A^{\mu}(x) = a^{\mu} \exp \left(-i k_0 x_0 + i \mathbf{k} \cdot \mathbf{x} \right) $$ with $\omega=|\mathbf{k}|$.
Then use the gauge condition $$ a^{\mu} k_{\mu} = 0 $$ which takes us from four independent components to three
In this case, we define $$ a^{\mu} = \alpha k^{\mu} + \epsilon^{\mu} $$ where $\epsilon^{0} = \boldsymbol{\epsilon} \cdot \mathbf{k} =0 $

We remove the $\alpha k^{\mu}$ term with a new gauge transformation $$ \alpha(x) = i \alpha \exp \left( -i k^{\mu} x_{\mu} \right) $$ thus giving two independent components rather than four
This gives rise to the two different polarizations of the photon
Things become more complicated for fermions, and this leads to "ghosts" in QFT
This doesn't mean gravitational waves are just like photons, in fact their polarizations are different

The two polarizations are quite a bit different from photons
Panel (b) represents $h^{TT}_{xx} \neq 0, h_{xy}^{TT}=0$, change in metric referred to as $h_{+}$ vs. $h_{\times}$
Spin 2 excitations rather than spin 1
Not gravity waves (see e.g. this paper)

Wave traveling in z direction with "+" polarization $$ ds^2 = -dt^2 + \left[ 1+h_+ \left(z-t\right)\right] dx^2 + \left[ 1-h_+ \left(t-z\right)\right] dy^2 + dz^2 $$ and presume two objects, one at $x=0$ and the other at $x=L$.
A photon moving between the two objects has an effective coordinate speed $$ \left(\frac{dx}{dt}\right)^2 = \frac{1}{1+h_+} $$ so that the time taken to go to $x=L$ is $$ t_{\mathrm{far}} = t_{\mathrm{start}}+\int_0^{L} \left\{1+h_+[t(x)]\right\}^{1/2}~dx \approx t_{\mathrm{start}}+L+\frac{1}{2} \int_0^{L} h_+(x+t_{\mathrm{start}}) dx $$ and the total time is $$ t_{\mathrm{return}} \approx t_{\mathrm{start}}+ 2L+\frac{1}{2} \int_0^{L} h_+(t_{\mathrm{start}}+ x) dx +\frac{1}{2} \int_0^{L} h_+(t_{\mathrm{start}}+x+L) dx $$

This implies $$ \frac{d t_{\mathrm{return}}}{d t_{\mathrm{start}}} = 1+ \frac{1}{2} \left[h_+\left(t_{\mathrm{start}}+2 L\right)-h_+\left(t_{\mathrm{start}}\right)\right] = \frac{{\nu}_{\mathrm{return}}}{{\nu}_{\mathrm{start}}} $$
Thus all we must do is measure the change in frequency
Additionally, we send the light down and back several times before measuring the frequency change

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.

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 $\bar{h}_{\mu \nu} = 4 J_{\mu \nu} e^{i \Omega (r-t)}/r$

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 $J^{\mu 0}=0$.

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

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