A Quadrupole radiation formula for gravity

Kostik
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Some help needed converting the formula for quadrupole radiation from geometrized units (##c=G=1##) to conventional units.
I derived the formula for the quadrupole radiation power emitted by a system of masses:
$$P=\frac{1}{45}\dddot{Q}_{kl}\dddot{Q}_{kl} .\quad\quad (*)$$ Note here that: (1) I am using geometrized units, so ##c=G=1##; (2) ##Q_{kl}## is the quadrupole tensor $$Q_{kl} = \int{(3x_k x_l - δ_{kl}\cdot r^2 )ρ} \, dV \quad (r^2\equiv x_k x_k)$$ and (3) repeated indices are summed over. (In Euclidean space ##R^3##, all indices are subscripts.)

This is essentially the same formula given by Landau & Lifshitz "Classical Theory of Fields, 4th Ed." (Eq. (110.16), p. 355) and Ohanian & Ruffini "Gravitation and Spacetime" (Eq. (5.75), p. 195): $$P=\frac{G}{45c^5}\dddot{Q}_{kl}\dddot{Q}_{kl} .$$ Landau & Lifshitz keep all factors of ##c## and ##G##, while Ohanian uses ##c=1## but keeps ##G## separate.

What I want to do is simply convert from geometrized units to conventional units, which means ##t## should be replaced by ##ct## and mass ##m## (or mass density ##\rho##) should be replaced by ##Gm/c^2## (or ##G\rho/c^2##).

Where I'm having trouble is that the power formula has a product ##\dddot{Q}_{kl}\dddot{Q}_{kl}##. So it seems to me that the correct way to convert ##(*)## to conventional units would be
$$P=\frac{1}{45} \left( \frac{G}{c^5} \right)^2 \dddot{Q}_{kl}\dddot{Q}_{kl} .$$ Obviously, this is wrong -- only one factor of ##G/c^5## should be used, which is also consistent with dimensional analysis. (In geometrized units, both sides of ##(*)## are dimensionless.)

I'm just not seeing why I should only add one factor of ##G/c^5##, when surely each factor of ##\dddot{Q}_{kl}## carries its own factor of ##G/c^5##? What am I missing here?
 
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You just have to insert the factors of ##G## and ##c## so the dimensions match correctly. You know that ##[G/c^5] = M^{-1} L^{-2} T^3##. You also know that ##[\dddot{Q}] = [P] = ML^2 T^{-3}##, so ##[P]/[\dddot{Q}]^2 = M^{-1} L^{-2} T^3 = [G/c^5]##
 
Kostik said:
only one factor of ##G/c^5## should be used
But you have to also convert the LHS of the equation to conventional units, so one factor of ##G/c^5## goes on the LHS, and cancels one of the two factors of ##G/c^5## on the RHS, so the end result is one factor of ##G/c^5## on the RHS, as the references you give say.
 
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PeterDonis said:
But you have to also convert the LHS of the equation to conventional units, so one factor of ##G/c^5## goes on the LHS, and cancels one of the two factors of ##G/c^5## on the RHS, so the end result is one factor of ##G/c^5## on the RHS, as the references you give say.
Thank you!
 
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