Gravitational wave equations? I am looking for a set of equations, one to calculate GW amplitude in watts and one to calculate frequency... I believe I have located the correct frequency equation yet I cannot find a source for power??  Any ideas??
To clarify, the equation below I got from Scott Hughes, in my correspondence with him he says:
"Dear Mr Vogeler —
I’m afraid I’ve also lost the correspondence we shared a few years ago.  As I recall, I outlined the power produced by two masses orbiting each other.  It’s not too hard to reconstruct that formula: If the masses are m1 and m2, and they move on a circular trajectory of radius R, and they complete an orbit in a time T, then the amount of power they put out is
P_{gw} = (32/5)(G/c^5) m1^2 m2^2/(m1 + m2)^2 R^4 (2 pi/T)^6
Here, G is Newton’s gravitation constant, and c is the speed of light.
That factor of G/c^5 is a killer: G is a small quantity; c is quite large; c^5 is huge.  The lesson of this formula is that you need m1 and m2 to be large, and the period to be short, in order for the system to produce appreciable waves.  2 pi R/T is the speed that the bodies whirl around each other, and that indeed needs to be a substantial fraction of the speed of light for this work."
When I run these numbers, it appears I do in fact get watts?  Is this really an accurate way to express the GW power?  When I see other amplitude or power numbers given it seems to be a dimensionless unit?  For example I saw that one detector can detect an amplitude of 10^-21, but its doesn't say what unit this number is.  I assume its calculated the same, but I am not sure if its really the same as watts.
What I am doing is imagining I have a particle accellerator that I can place two 1mg masses in and spin up to 299,000 km/s.  My radius I had at 50cm, I ended up with GW=10^-18 m^2 kg/s^3 which is about one attowatt if indeed this calculates watts??  I know that if someone tried to do that the G forces would be insane, but its just a thought experiment.  I wanted to see how much velocity is needed to hit the detectable range.
 A: The power (energy over time) that is radiated away can be approximated with
$$\rm P = \dot E = \frac{d E}{d t} = -\frac{32 G^4 M_1^2 M_2^2 (M_1+M_2)}{5 c^5 r^5}$$
and the rate at which the angular momentum is lost
$$  \rm \dot L = \frac{d L}{d t} = -\frac{32 G^{7/2} M_1^2 M_2^2 \sqrt{M_1+M_2}}{5 c^5 r^{7/2}} $$
The frequency is
$$ \rm f = \sqrt{\frac{G (M_1+M_2)}{\pi r^3}} $$
and the radial relative velocity
$$ \rm \dot r = \frac{d r}{d t} = -\frac{64 G^3 (M_1 M_2) (M_1+M_2)}{5 c^5 r^3} $$
That gives a merging time of
$$ \rm t= \frac{5 c^5 r^4}{256 G^3 (M_1 M_2)(M_1+M_2)} $$
If the merger is so far away that the cosmological redshift z is much larger than 0 you have to multiply the time and divide the frequency by $\rm z+1$. The amplitude would in any case be proportional to $\rm P/f$.
Note that this are only first order approximations where the final ringdown is not accounted for, the whole fullrelativistic story can only be computed in large computer clusters.
A: Ok, I got this from Scott Hughes MIT.  This is what I was searching for and I quote:
"....if the masses are m1 and m2, and they move on a circular trajectory of radius R, and they complete an orbit in a time T, then the amount of power they put out is:
P=(32/5)(G/c^5)m1^2m2^2/(m1+m2)^2R^4(2pi/T)^6"
This is very similar to the equation above I see.
I hope I copied that correctly.  I probably didn't explain well what I needed, thank God I got lucky with a final saving email from Mr. Hughes!!  I'm going to make sure I do not lose this email!!!
Thank you all for attempting to help me.  I typically don't answer my own questions but I didn't really this did not come from my brain!!
