<p>For reference, the relevant bit from the paper is:</p>

<blockquote>
  <p>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.</p>
</blockquote>

<p>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.</p>

<blockquote>
  <p>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 ... - How much energy is emitted (per second) from the source in the beam?</p>
</blockquote>

<p>A spherical cap has surface area of $A = 2\pi Rh$, and here $R = 21\,\mathrm{kly}$. Now, I'm unclear what <em>cross-sectional radius</em> 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.</p>

CC BY-SA 3.0
2014-11-13T11:39:05.560Z
Many thanks. Sorry, yes I should have said half angle = 5 degrees which is a made-up equivalent of their detailed beam intensity map. I haven't found any reported power spectra for the beam. Maybe we could assume it is flat across the 4 bands (which would quadruple estimated P ). Maybe also should double P because there are two antipodial beams? But P is still much much lower than the gravitational wave energy loss.
steveOw
CC BY-SA 3.0
2014-11-13T02:31:11.860Z
2014-11-13T02:31:11.860Z
Stan Liou
<p>Following on from <a href="https://astronomy.stackexchange.com/questions/6051/how-did-the-hulse-taylor-pulsar-provide-confirming-evidence-of-general-relativit">an earlier question</a> about the very interesting <a href="https://en.wikipedia.org/wiki/Hulse%E2%80%93Taylor_binary" rel="nofollow noreferrer">Hulse-Taylor binary pulsar</a>.  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).</p>

<p><a href="http://arxiv.org/pdf/astro-ph/0407149v1.pdf" rel="nofollow noreferrer">Weisberg &amp; Taylor, 2004</a> 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?  </p>

CC BY-SA 3.0
2014-11-12T22:29:07.810Z
I don't know about the question being "recent" . . . But I'm assuming that [the Wikipedia page on gravitational waves (and associated energy loss)](https://en.wikipedia.org/wiki/Gravitational_wave#Power_radiated_by_orbiting_bodies) was unsatisfying?
HDE 226868
CC BY-SA 3.0
2014-11-12T22:33:10.903Z
@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
CC BY-SA 3.0
2014-11-12T22:36:56.733Z
@HDE226868 I rewrote my previous comment having previously misread your question.
steveOw
<general-relativity><pulsar>
Hulse-Taylor binary pulsar - what is the rate of mass/energy loss from the source?
steveOw
# Hulse-Taylor binary pulsar - what is the rate of mass/energy loss from the source?

(Asked by: username_0 on 2014-11-12)

Following on from [an earlier question](https://astronomy.stackexchange.com/questions/6051/how-did-the-hulse-taylor-pulsar-provide-confirming-evidence-of-general-relativit) about the very interesting [Hulse-Taylor binary pulsar](https://en.wikipedia.org/wiki/Hulse%E2%80%93Taylor_binary). 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](http://arxiv.org/pdf/astro-ph/0407149v1.pdf) report that the beam has "a flux density of about 1 mJy at 1400 MHz." (mJy = milliJansky, thanks username_2). 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?

**Tags:** general-relativity, pulsar

> **Comments:**
> **username_1**: I don't know about the question being "recent" . . . But I'm assuming that [the Wikipedia page on gravitational waves (and associated energy loss)](https://en.wikipedia.org/wiki/Gravitational_wave#Power_radiated_by_orbiting_bodies) was unsatisfying?  
> **username_0**: @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.  
> **username_0**: @HDE226868 I rewrote my previous comment having previously misread your question.  

---

## 1 answers

### Answer by username_2 on 2014-11-13. Score: 2 | Accepted answer

For reference, the relevant bit from the paper is:

> 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.

> 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 ... - How much energy is emitted (per second) from the source in the beam?

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.

> **Comments:**
> **username_0**: Many thanks. Sorry, yes I should have said half angle = 5 degrees which is a made-up equivalent of their detailed beam intensity map. I haven't found any reported power spectra for the beam. Maybe we could assume it is flat across the 4 bands (which would quadruple estimated P ). Maybe also should double P because there are two antipodial beams? But P is still much much lower than the gravitational wave energy loss.  

---

