According to many textbooks, pulsars have lifetime (the time it would take the pulses to cease if $$d(P)/dt$$ were constant).

Lifetime is $$P/[d(P)/dt]$$, $$P=$$period

Why do P/[d(P)/dt] means the lifetime?

• Did you try plugging in a value for $P$ and $\frac{d(P)}{dt}$ ? Jun 30, 2021 at 14:35
• If i substitute a value, i will get an appropriate value. But I don't know if the meaning of the expression is lifetime. Jun 30, 2021 at 15:12
• I have \$500 but I am spending \$20 per day. How long will my \\$500 last ? Jun 30, 2021 at 15:19
• @gandalf61 the period is increasing. You have saved 500 dollars by putting aside 20 dollars/day - how long have you been saving? Jun 30, 2021 at 16:03

You should be able to see from a dimensional basis that the time taken to reach a certain value of something is that value divided by the rate at which it changes (assuming that you started from zero). In this case, where period is measured in seconds, in dimensional terms: $$\tau \equiv \frac{{\rm seconds}}{\rm seconds\ per\ second}$$
However, there are a couple of wrinkles to this. Firstly, $$\dot{P}$$ is positive and $$P$$ is getting bigger from some initially small value, but not zero. Secondly, $$\tau = P/\dot{P}$$ also assumes that the rate of change is itself constant with time. This is not thought to be true for pulsars because the power emitted in radiation is strongly dependent on rotation rate, and there must be a small numerical factor in front of this expression. For example, if the pulsar spins down by emitting magnetic dipole radiation then $$\tau = \frac{1}{2}\left( \frac{P}{\dot{P}}\right)$$ e.g. see here, where $$\tau$$ is the time to reach the present period $$P$$ and spindown rate $$\dot{P}$$, so long as $$P\gg P_0$$, where $$P_0$$ was the initial (small) period.