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)

$$\delta^{2}\geq\frac{\hbar D}{M_{s}}$$
$$\frac{GM}{c^{2}}\geq\left(\frac{c}{s}D\right)>D$$

Which 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.

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 MathJAX and this post on how to use MathJax in Blogger!

N.B. There’s a good followup post on Cosmic Variance, along with an earlier entry giving some good background information.