$E=\left(\frac{c^{2}}{32\pi G}\right)\omega^{2}f^{2}$

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.

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.*

## 4 comments:

So.... can LIGO detect gravitons? Give a conclusion for the lay masses!

Hey. How's NoCal?

Updated, with complete lecture & conclusions.

Excellent, and this time I was paying attention, and not being a smart-ass. Is there measurable advancement in LIGO technology that something practical could eventually be brought to production? Is there an equivalent of Moore's Law for a LIGO?

Well, coincidentally, I wrote a paper summarizing the latest findings and advances for LIGO here:

http://insecure.ucdavis.edu/Members/adam/physics/ligo.pdf/view

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