Quadrupole radiation formula for gravity

[Kostik]

TL;DR Summary

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?

 

[ergospherical]

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 $[P]/[\dddot{Q}]^2 = M^{-1} L^{-2} T^3 = [G/c^5]$

 

[PeterDonis]

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.

 

[Kostik]

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!