In the paper: gravitational radiation and the motion of two point masses, in the last But one page and the last page the author derives a formula for the lifetime of a binary whose orbit is decaying due to gravitational wave radiation.

The formula for circular binaries is as follows: $$T(a_{0})=\frac{{a_{0}}^{4}}{4\beta}$$ Where $a_{0}$ is the initial semi-major axis and $\beta=\frac{64G^{3}m_{1}m_{2}(m_{1}+m_{2})}{5c^{5}}$

Now he explains that for a circular-binary with both the masses equal to one solar mass and the semi-major axis to be $\approx{10}$ solar radius($=10×695700km$), the life time turns out to be $\approx{3×10^{12}}$ but upon calculating the following occurs: $$\beta=\frac{64G^{3}m_{1}m_{2}(m_{1}+m_{2})}{5c^{5}}\approx{6.25×10^{-42}}$$ Thus, $$T\approx{\frac{(10×695700×10^{3})^{4}}{6.25×10^{-42}×4}}\approx{9.368×10^{79}}seconds$$ which is of the order of $10^{72}years$

I'm facing a similar problem within life time of elliptical binaries. I have derived a formula as follows: $$T(a_{0},e_{0})=\frac{12}{19}\frac{{\alpha_{0}}^{4}}{\beta}\frac{2e_{0}^{\frac{48}{19}}}{\frac{96}{19}}\epsilon_{F_1}$$ Where $\alpha_{0}=\frac{a(1-e^{2})}{e^{12/19}}$ and $$\epsilon_{F_1}=F_{1}\left(\frac{24}{19};1.5;-\frac{1181}{2299};\frac{43}{19},e_{0}^{2},-\frac{121}{304}e_{0}^{2}\right)$$ is Apple's hypergeometric function. Using this I tried to compute the life time of the binary PSR J1740-3052, with eccentricity $0.57$, $a=0.55solar-radius$, masses to be(in solar mass): $1.35$ and $12.5$. I got the value close to $2.2×10^{72}seconds\approx{7×10^{64}}$, but the data value is just $3.54813×10^{5}years$

Where am i going wrong? Any help is appreciated.


I cannot reproduce your numbers. I find $\beta = 2.5\times 10^{19}$ m$^4$/s (note units) and hence $T(a_0) = 7.5 \times 10^{11}$ years.

It looks to me that the author forgot to divide by the factor of 4. How you have arrived at your results I am not sure, but your $\beta$ value disagrees with mine by a factor of $M_{\odot}^2$ in SI units.


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