A lecture given 10/27/08 by Professor Freeman Dyson of the Institute of Advanced Studies at Princeton, in honor of the 100th anniversary of the founding of the University of California, Davis.
E=\left(\frac{c^{2}}{32\pi G}\right)\omega^{2}f^{2}
is the energy per gravity wave, where f is the dimensionless amplitude/strain.
E_{s}=\frac{\hbar\omega^{4}}{c^{3}}
is the energy per graviton, taken from $\hbar\omega$ energy times $\frac{\omega^3}{c^3}$ density
f=\left(32\pi\right)^{\frac{1}{2}}\left(L_{p}\frac{\omega}{c}\right)
is the strain per graviton.
L_{p}=\left(\frac{G\hbar}{c^{3}}\right)^{\frac{1}{2}}=1.4\times10^{-33}cm
\delta=\left(32\pi\right)^{\frac{1}{2}}L_{p}
Gives the linear displacement per graviton.
Note that spherical objects can't radiate gravitational waves, and that binary stars produce kilohertz gravity waves.
LIGO's threshold is therefore $10^{37}$ gravitons.
M\delta^{2}\geq\hbar T
is the uncertainty in position and velocity.
D\leq\left(\frac{GM}{c^{2}}\right)
(from combining previous two equations)
\delta^{2}\geq\frac{\hbar D}{M_{s}}
\frac{GM}{c^{2}}\geq\left(\frac{c}{s}D\right)>D
exceeds the Schwarzschild radius, so impossible.
Then the Bohr-Rosenfeld argument is:
\Delta E_{x}(1)\Delta E_{x}(2)\approx\hbar\left|A(1,2)-A(2,1)\right|
where A(2,1) is the field from dipole 2 at location 1.
Then the detector is described by:
D_{ab}=m\int\Psi_{b}^{*}xy\Psi_{a}d\tau
where a is the initial state, b is the final state, and m is the detector mass.
\sigma(\omega)=\left(4\pi^{2}G\frac{\omega^{3}}{c^{3}}\right)\sum_{b}\left|D_{ab}\right|^{2}\delta(E_{b}-E_{a}-\hbar\omega)
S_{a}=\int\sigma(\omega)\frac{d\omega}{\omega}
is the logarithmic average taken over the graviton cross section.
S_{a}=4\pi^{2}L_{p}^{2}Q
Now consider the gravitophotoelectric effect, where the graviton removes an electron.
Q=\int\left|\left(x\frac{\partial}{\partial y}+y\frac{\partial}{\partial x}\right)\Psi_{a}\right|^{2}d\tau
Q=\frac{\int\bar{r}^{4}\left[f'(r)\right]^{2}d\bar{r}}{\int r^{2}\left[f(r)\right]^{2}dr}
\int r^{4}\left[f'+\left(\frac{3}{2}r\right)f\right]^{2}dr>0
Q>\frac{3}{4}
f(r)=r^{-n}e^{-\frac{r}{R}}
Q=1-\frac{n}{6}
4\pi^{2}L_{p}^{2}=4\pi^{2}G\frac{\hbar}{c^{3}}=8\times10^{-65}cm^{2}
This means that if you take a detector the mass of the Earth, squash it into a large flat sheet, and run it for the lifetime of the universe, you'll detect 4 gravitons.
From the Sun, there are $10^{8}$W of gravitons and $10^{25}$W of neutrinos, and we can detect gravitons about $10^{-35}$ less than neutrinos.
Special thanks to LaTeXMathML and yourequations.com! If you see funny symbols between dollar signs (for some of the equations), you need to pick a MathML capable browser. Most of the equations, however, are handled by yourequation's Javascript LaTex renderer.
