# Hulse-Taylor binary pulsar - what is the rate of mass/energy loss from the source?

Following on from an earlier question about the very interesting Hulse-Taylor binary pulsar. The high-frequency (radio) beam from the spinning pulsar sweeps across Earth about 17 times per second. The total power of the gravitational radiation (waves) emitted by the binary system is calculated from GTR to be $7.35 × 10^{24}$ watts at present (declining as the orbital period and radius diminish).

Weisberg & Taylor, 2004 report that the beam has "a flux density of about 1 mJy at 1400 MHz." (mJy = milliJansky, thanks Stan Liou). Assuming that this flux is uniform across a conical beam with cross-sectional radius 5 arc degrees and assuming that the source is 21,000 light years from Earth and has mass = 1.44 Solar Masses:- How much energy is emitted (per second) from the source in the beam? Also (if it is possible to estimate a reasonable range of values) what might be the rate of steady mass loss from such a source?

• I don't know about the question being "recent" . . . But I'm assuming that the Wikipedia page on gravitational waves (and associated energy loss) was unsatisfying? – HDE 226868 Nov 12 '14 at 22:29
• @HDE226868. I have enough info on graviational wave energy loss. It is the other forms of mass/energy loss that I am seeking data on, e.g. thru the beam or other possible (steady, non-cataclysmic) processes. – steveOw Nov 12 '14 at 22:33
• @HDE226868 I rewrote my previous comment having previously misread your question. – steveOw Nov 12 '14 at 22:36

The observable pulsar is a weak radio source with a flux density of about $1\,\mathrm{mJy}$ at $1400\,\mathrm{MHz}$. ... Our most recent data have been gathered with the Wideband Arecibo Pulsar Processors (“WAPPs”), which for PSR B1913+16 achieve $13\,\mathrm{\mu s}$ time-of-arrival measurements in each of four $100\,\mathrm{MHz}$ bands, using $5$-minute integrations.
Those are millijanskys, aka milli flux units, so that the flux density is about $$1\,\mathrm{mJy} = 10^{-29}\,\frac{\mathrm{W}}{\mathrm{m}^2\cdot\mathrm{Hz}}\text{,}$$ and hence the detected irradiance is on the order of $10^{-27}\,\mathrm{W}/\mathrm{m}^2$. Since we're about to make rather uncertain assumptions anyway, I won't bother worrying about doing more than an order-of-magnitude calculation.
A spherical cap has surface area of $A = 2\pi Rh$, and here $R = 21\,\mathrm{kly}$. Now, I'm unclear what cross-sectional radius means if measured as an angle, but I take it to mean that the opening half-angle of the cone is $\frac{\vartheta}{2} = 5^\circ$, in which case $$A = 2\pi R^2\left(1-\cos\frac{\vartheta}{2}\right) \sim 10^{39}\,\mathrm{m}^2\text{.}$$ Thus, the power would be $P\sim 10^{12}\,\mathrm{W}$, but note that in addition to the assumptions you've just listed, we're only talking about a particular radio band.