<p>My comment notwithstanding, <a href="http://www.astro.ucla.edu/~ianc/courses/08SA082/week6_homework4_pulsars.pdf" rel="nofollow">here's a solution to a homework problem that does the calculation</a>.  It doesn't specify the exact mechanism that converts angular momentum (aka rotational energy) into electromagnetic radiation.  In this case, it is just an assertion of the problem (partially justified with what I said in my comment: the energy must come from somewhere, and if there doesn't seem to be any sources other than angular momentum, then it must be coming from the angular momentum).</p>

<p>Slightly rephrasing that link's content for the sake of future accessibility:</p>

<blockquote>
  <p>The pulsar is radiating energy (which we observe as radio waves). Since the total
  energy in the universe must be conserved, this radio energy must come from
  somewhere. In this case, it is taken out of the rotational kinetic energy of the pulsar:
  thus, it gradually slows down.
  We're interested in a relation between the pulsar luminosity and its rotational period.
  In general, the kinetic energy of a rotating body is given by
  $$E=\frac{1}{2} I \omega^2 = 2\pi^2 I P^2.$$
  Since Luminosity is the time derivative of energy, we are
  now in a position to relate the quantities we are interested in:
  $$L= \frac{\partial E}{\partial t} = -4\pi^2 I P^{-3} \frac{\partial P}{\partial t}.$$
  Rearranging this in terms of the quantity we want – the rate of
  change of the period – gives:
  $$\frac{\partial P}{\partial t} = \frac{-L P^3}{4\pi^2 I}.$$
  If we assume that this neutron star is a homogeneous sphere (not really true, but a
  simple approximation), then its moment of inertia is just:
  $$I_{\text{sphere}}=\frac{2}{5}M R^2,$$
  and so the final rate of change of period we get is:
  $$\frac{\partial P}{\partial t} = -\frac{5}{8\pi ^2} \frac{L P^3}{MR^2}.$$</p>
</blockquote>

<p>So as long as we have measurements of the quantities on the right-hand side, we can just plug them in to get a value for the rate of change in the period.  The hardest to measure is usually the moment of inertia (where the mass, $M$, and radius, $R$, terms come from).  These are easier to get from eclipsing binary systems, since then there are nice relations between their orbital paths and masses.</p>

CC BY-SA 3.0
2015-06-16T04:34:58.093Z
Going by [this](https://sites.google.com/a/pulsarsearchcollaboratory.com/pulsar-search-collaboratory/Home/pulsar-faq#TOC-Why-do-pulsars-slow-down-over-time-) and [this](http://www.sciencedaily.com/releases/2012/10/121008091556.htm), the mechanism comes for a magnetic dipole radiation.  Which, if I understand correctly, means that the magnetic field accelerates surface (or nearby) electrons, and applies torque to them.  Acceleration means they must radiate away energy, torque means angular momentum is transferred.  It's hard to get an explanation that really spells it out in full, it seems.
zibadawa timmy
CC BY-SA 3.0
2015-06-16T09:12:23.797Z
But we don't have the quantities on the RHS. The mass, radius and moment of intertia are essentially unknowns. The pulsar "luminosity" is also rather hard to estimate in the context of this calculation.
ProfRob
CC BY-SA 3.0
2015-06-16T10:08:51.820Z
@RobJeffries Thanks for pointing that out.  I was fairly sure that most of those quantities were going to give a lot of compounding sources of measurement problems, and so there had to be another way around the issue.  Glad to see you posted about such a way.
zibadawa timmy
CC BY-SA 3.0
2015-06-15T23:43:21.720Z
2015-06-15T23:43:21.720Z
zibadawa timmy
<p>Let us suppose that the pulsar is spinning down at a uniform rate. So it has a period $P$ and a rate of change of period $dP/dt$ that is positive and constant (in practice there are also second, third, fourth etc. derivatives to worry about, but this doesn't change the principle of my answer).</p>

<p>Now let's assume you can measure the period very accurately - say you look at the pulsar today and measure its radio signals for a few hours, do a Fourier transform of the signal and get a nice big peak with a period of 0.1 seconds (for example).</p>

<p>With that period, you can "fold" the data to create an average pulse profile. This pulse profile can then be cross-correlated with subsequent measurements of the pulse to determine an offset between the predicted time of "phase zero" in the profile, calculated using the 0.1 s period, and the actual time of phase zero. This is often called an "O-C" curve or a residuals curve.</p>

<p>If you have the correct period and $dP/dt=0$, then the residuals will scatter randomly around zero with no trend as you perform later and later observations (see plot (a) from Lorimer &amp; Kramer 2005, The Handbook of Pulsar Astronomy). If the initial period was in error, then the residuals would immediately begin to depart from zero on a linear trend.</p>

<p>If however, you have the period correct, but $dP/dt$ is positive, then the residuals curve will be in the form of a parabola (see plot (b)).</p>

<p>If you have second, third etc. derivatives in the period, then this will affect the shape of the residuals curve correspondingly.</p>

<p>The residuals curve is modelled to estimate the size of the derivatives of $P$. The reason that $dP/dt$ can be measured so precisely is that pulsars spin fast and have repeatable pulse shapes, so changes in the phase of the pulse quickly become apparent and can be tracked over many years. </p>

<p><img src="https://i.sstatic.net/0BMzj.png" alt="Pulsar residual timing curves"></p>

<p>Mathematically it works something like this. The phase $\phi(t)$ is given by
$$\phi(t) \simeq \phi_0 + 2\pi \frac{\Delta t}{nP} - \frac{2\pi}{2}\frac{(\Delta t)^2}{nP^2} \frac{dP}{dt} + ...,$$
where $\phi_0$ is an arbitrary phase zero, $\Delta t$ is the time between the first and last observation and $n$ is the integer number of full turns the pulsar has made during that time. If the period is approximately correct, then $n = int(\Delta t/P)$.</p>

<p>The "residual curve" would be given by
$$\phi_0 - 2\pi\frac{\Delta t}{nP} -\phi(t) \simeq  \frac{2\pi}{2}\frac{(\Delta t)^2}{nP^2} \frac{dP}{dt} + ...,$$</p>

<p>For example, if the period of a $P \sim 0.01$ second pulsar changed by a picosecond in a year, then there would be an accumulated residual of almost $10^{-4}$ seconds after 1 year of observation. Depending on how "sharp" the pulse is, then this shift of about 1% in the phase of the pulse might be detectable. </p>

<p>Perhaps needless to say, but there are a host of small effects and corrections to make in order to get this very high precision timing. You need to know exactly how the Earth is moving in its orbit. The proper motion of the pulsar on the sky also has an effect. These and more can be found in Lorimer and Kramer's book, but there is also a summary <a href="http://www.cv.nrao.edu/course/astr534/PulsarTiming.html" rel="noreferrer">here</a>.  </p>

CC BY-SA 3.0
2015-06-16T09:54:47.517Z
2015-06-16T10:00:10.997Z
2015-06-16T10:00:10.997Z
ProfRob
<p>I've been to some talks that mention how stable the period of a millisecond pulsar is over long periods of time. Recently, it was mentioned that astronomers have calculated the change in period over time to be less than 10^-12 seconds per year for several pulsars. No one I've talked to seems to know any details of this calculation.
How do we calculate such small differences in period? How much data must be collected and what are the exposure times for imaging such rapid phenomena?
A source/paper would be excellent. I apologize that I don't have a citation for the 10^-12s figure, but the lack of citation is mostly my reason for posting this question.</p>

CC BY-SA 3.0
2015-06-15T23:25:43.453Z
I'm not sure of the details of the calculation, either, and would be interested to see some details.  I know the basic idea: the pulsar is emitting a lot of electromagnetic radiation, and that constitutes an energy loss.  The energy has to have come from something. If it doesn't have an accretion disk or the intense magnetic field of a magnetar, then the most likely source is the conversion of angular momentum.  As its angular momentum is lost, the rotation speed goes down.  I don't know the conversion mechanism or calculations right now, though.
zibadawa timmy
<astrophysics><pulsar><observational-astronomy>
Pulsars: How do astronomers measure minute changes in period (~picoseconds per year)?
Sam
# Pulsars: How do astronomers measure minute changes in period (~picoseconds per year)?

(Asked by: username_0 on 2015-06-15)

I've been to some talks that mention how stable the period of a millisecond pulsar is over long periods of time. Recently, it was mentioned that astronomers have calculated the change in period over time to be less than 10^-12 seconds per year for several pulsars. No one I've talked to seems to know any details of this calculation.
How do we calculate such small differences in period? How much data must be collected and what are the exposure times for imaging such rapid phenomena?
A source/paper would be excellent. I apologize that I don't have a citation for the 10^-12s figure, but the lack of citation is mostly my reason for posting this question.

**Tags:** astrophysics, pulsar, observational-astronomy

> **Comments:**
> **username_1**: I'm not sure of the details of the calculation, either, and would be interested to see some details.  I know the basic idea: the pulsar is emitting a lot of electromagnetic radiation, and that constitutes an energy loss.  The energy has to have come from something. If it doesn't have an accretion disk or the intense magnetic field of a magnetar, then the most likely source is the conversion of angular momentum.  As its angular momentum is lost, the rotation speed goes down.  I don't know the conversion mechanism or calculations right now, though.  

---

## 2 answers

### Answer by username_2 on 2015-06-16. Score: 6 | Accepted answer

Let us suppose that the pulsar is spinning down at a uniform rate. So it has a period $P$ and a rate of change of period $dP/dt$ that is positive and constant (in practice there are also second, third, fourth etc. derivatives to worry about, but this doesn't change the principle of my answer).

Now let's assume you can measure the period very accurately - say you look at the pulsar today and measure its radio signals for a few hours, do a Fourier transform of the signal and get a nice big peak with a period of 0.1 seconds (for example).

With that period, you can "fold" the data to create an average pulse profile. This pulse profile can then be cross-correlated with subsequent measurements of the pulse to determine an offset between the predicted time of "phase zero" in the profile, calculated using the 0.1 s period, and the actual time of phase zero. This is often called an "O-C" curve or a residuals curve.

If you have the correct period and $dP/dt=0$, then the residuals will scatter randomly around zero with no trend as you perform later and later observations (see plot (a) from Lorimer & Kramer 2005, The Handbook of Pulsar Astronomy). If the initial period was in error, then the residuals would immediately begin to depart from zero on a linear trend.

If however, you have the period correct, but $dP/dt$ is positive, then the residuals curve will be in the form of a parabola (see plot (b)).

If you have second, third etc. derivatives in the period, then this will affect the shape of the residuals curve correspondingly.

The residuals curve is modelled to estimate the size of the derivatives of $P$. The reason that $dP/dt$ can be measured so precisely is that pulsars spin fast and have repeatable pulse shapes, so changes in the phase of the pulse quickly become apparent and can be tracked over many years.

![Pulsar residual timing curves](https://i.sstatic.net/0BMzj.png)

Mathematically it works something like this. The phase $\phi(t)$ is given by
$$\phi(t) \simeq \phi\_0 + 2\pi \frac{\Delta t}{nP} - \frac{2\pi}{2}\frac{(\Delta t)^2}{nP^2} \frac{dP}{dt} + ...,$$
where $\phi\_0$ is an arbitrary phase zero, $\Delta t$ is the time between the first and last observation and $n$ is the integer number of full turns the pulsar has made during that time. If the period is approximately correct, then $n = int(\Delta t/P)$.

The "residual curve" would be given by
$$\phi\_0 - 2\pi\frac{\Delta t}{nP} -\phi(t) \simeq \frac{2\pi}{2}\frac{(\Delta t)^2}{nP^2} \frac{dP}{dt} + ...,$$

For example, if the period of a $P \sim 0.01$ second pulsar changed by a picosecond in a year, then there would be an accumulated residual of almost $10^{-4}$ seconds after 1 year of observation. Depending on how "sharp" the pulse is, then this shift of about 1% in the phase of the pulse might be detectable.

Perhaps needless to say, but there are a host of small effects and corrections to make in order to get this very high precision timing. You need to know exactly how the Earth is moving in its orbit. The proper motion of the pulsar on the sky also has an effect. These and more can be found in Lorimer and Kramer's book, but there is also a summary [here](http://www.cv.nrao.edu/course/astr534/PulsarTiming.html).


---

### Answer by username_1 on 2015-06-15. Score: 3

My comment notwithstanding, [here's a solution to a homework problem that does the calculation](http://www.astro.ucla.edu/~ianc/courses/08SA082/week6_homework4_pulsars.pdf). It doesn't specify the exact mechanism that converts angular momentum (aka rotational energy) into electromagnetic radiation. In this case, it is just an assertion of the problem (partially justified with what I said in my comment: the energy must come from somewhere, and if there doesn't seem to be any sources other than angular momentum, then it must be coming from the angular momentum).

Slightly rephrasing that link's content for the sake of future accessibility:

> The pulsar is radiating energy (which we observe as radio waves). Since the total
> energy in the universe must be conserved, this radio energy must come from
> somewhere. In this case, it is taken out of the rotational kinetic energy of the pulsar:
> thus, it gradually slows down.
> We're interested in a relation between the pulsar luminosity and its rotational period.
> In general, the kinetic energy of a rotating body is given by
> $$E=\frac{1}{2} I \omega^2 = 2\pi^2 I P^2.$$
> Since Luminosity is the time derivative of energy, we are
> now in a position to relate the quantities we are interested in:
> $$L= \frac{\partial E}{\partial t} = -4\pi^2 I P^{-3} \frac{\partial P}{\partial t}.$$
> Rearranging this in terms of the quantity we want – the rate of
> change of the period – gives:
> $$\frac{\partial P}{\partial t} = \frac{-L P^3}{4\pi^2 I}.$$
> If we assume that this neutron star is a homogeneous sphere (not really true, but a
> simple approximation), then its moment of inertia is just:
> $$I\_{\text{sphere}}=\frac{2}{5}M R^2,$$
> and so the final rate of change of period we get is:
> $$\frac{\partial P}{\partial t} = -\frac{5}{8\pi ^2} \frac{L P^3}{MR^2}.$$

So as long as we have measurements of the quantities on the right-hand side, we can just plug them in to get a value for the rate of change in the period. The hardest to measure is usually the moment of inertia (where the mass, $M$, and radius, $R$, terms come from). These are easier to get from eclipsing binary systems, since then there are nice relations between their orbital paths and masses.

> **Comments:**
> **username_1**: Going by [this](https://sites.google.com/a/pulsarsearchcollaboratory.com/pulsar-search-collaboratory/Home/pulsar-faq#TOC-Why-do-pulsars-slow-down-over-time-) and [this](http://www.sciencedaily.com/releases/2012/10/121008091556.htm), the mechanism comes for a magnetic dipole radiation.  Which, if I understand correctly, means that the magnetic field accelerates surface (or nearby) electrons, and applies torque to them.  Acceleration means they must radiate away energy, torque means angular momentum is transferred.  It's hard to get an explanation that really spells it out in full, it seems.  
> **username_2**: But we don't have the quantities on the RHS. The mass, radius and moment of intertia are essentially unknowns. The pulsar "luminosity" is also rather hard to estimate in the context of this calculation.  
> **username_1**: @RobJeffries Thanks for pointing that out.  I was fairly sure that most of those quantities were going to give a lot of compounding sources of measurement problems, and so there had to be another way around the issue.  Glad to see you posted about such a way.  

---

