In one method to determine the age of a pulsar one assumes the spinning frequency $\nu$ to obey $\nu=-k\cdot\dot{\nu}^n$ where $k$ is some constant and $n$ is the braking index of the pulsar.
For the Crab and a few other pulsars $\nu$, $\dot{\nu}$ and $\ddot{\nu}$ have been determined from observations. The braking index can be calculated using $n=\frac{\nu\cdot\ddot{\nu}}{\dot{\nu}^2}$.
The characteristic age $\tau$ of the pulsar is often calculated as $\tau=\frac{P}{2\cdot\dot{P}}$ where $P$ is the spin period.
This relation follows from $\tau=\frac{P}{(n-1)\cdot\dot{P}}\cdot\big[1-\big(\frac{P_0}{P}\big)^{n-1}\big]$ assuming the initial spin period $P_0$ to be much smaller than the present period $P$ and $n=3$, the value expected if the spinning down is exclusively caused by magnetic dipole radiation.
However, the calculated braking index of the Crab and other pulsars differs from 3. As a consequence, $\tau\neq\frac{P}{2\cdot\dot{P}}$.
Is there another way to determine the age of a pulsar and the initial spin period, if $\nu$ and $\dot{\nu}$ (and $\ddot{\nu}$) have been measured many times over some decades? Assuming $n\neq3$ and no glitches.

  • $\begingroup$ Use the formula you've quoted? If $n$ is measured, then why wouldn't that be valid? I don't think there's a way to estimate the initial period unless you know $\tau$ (which of course we do for the Crab pulsar). $\endgroup$ – Rob Jeffries May 15 '15 at 21:32
  • $\begingroup$ If $n\neq3$, then $\tau=\frac{P}{2\cdot\dot{P}}$ doesn't follow from $\tau=\frac{P}{(n-1)\cdot\dot{P}}\cdot\big[1-\big(\frac{P_0}{P}\big)^{n-1}\big]$. Another approach is to expand $\nu$ in a Taylor series $\nu(t)=\nu(t_0)+\dot{\nu}\cdot(t-t_0)+\frac{1}{2}\cdot\ddot{\nu}\cdot(t-t_0)^2$, but this doesn't help if both $t_0$ and $\nu(t_0)$ are unknown for another pulsar, not the Crab. $\endgroup$ – gamma1954 May 15 '15 at 21:44
  • $\begingroup$ I guess I'm not following. If $n=2.8$ say, then why can't you plug $n=2.8$ into that equation? $P_0$ can still be ignored if $n>2$ and the pulsar is sufficiently old. $\endgroup$ – Rob Jeffries May 15 '15 at 21:53
  • $\begingroup$ $\tau=\frac{P}{(n-1)\cdot\dot{P}}\cdot\big[1-\big(\frac{P_0}{P}\big)^{n-1}\big]$ can be used, but how do we know if a certain pulsar is old enough to ignore $P_0/P$? The other way around, in the case of the Crab, $n=2.51$ and $P=0.033\text{s}$ and the known age give $P_0=0.019\text{s}$, so $\big(\frac{P_0}{P}\big)^{n-1}=0.43$, which isn't to be ignored. $\endgroup$ – gamma1954 May 15 '15 at 22:24
  • $\begingroup$ Right, but these are problems if $n=3$ too. $\endgroup$ – Rob Jeffries May 15 '15 at 22:48

The formula you have quoted is an intrinsically unreliable age estimator for a number of reasons. (i) Yes, you may have to assume $n=3$ for magnetic dipole radiation (or at least that an $n$ measured now has been constant); (ii) you may have to assume $P_0/P=0$, which will be unrealistic for young pulsars and (iii) it assumes that the geometry of the magnetic field, its level of misalignment with the rotation axis and the strength of the B-field remain unchanged.

An obvious way to make progress is direct estimates of $\tau$. This can come either by seeing the supernova (e.g. for the Crab nebula), or you can associate the pulsar with a supernova remnant and estimate its age from the expansion rate of the remnant.

However, something that is more applicable to many pulsars is to use a kinematic age. Pulsars are often born with large "kick" velocities. You can measure this velocity (proper motions and line of sight velocities are measurable) and then you can extrapolate this backwards with a model for the Galactic potential and see how long ago the pulsar would have been in the Galactic plane. This works because we expect massive stars to have been within ~100 pc of the Galactic plane (unless they themselves were runaways from a previous supernova explosion in a high mass binary system).

More details can be found in Noutsos et al. (2003), who use kinematic ages to constrain the distribution of $n$ and $P_0$.

  • $\begingroup$ Thank you, the kinematic age approach is enough for me to explore $\endgroup$ – gamma1954 May 15 '15 at 22:35

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