of $2\times 10^{12} {\rm G}$. For these values, Eq. \[eq:tau0\] shows that $T\gg \tau_0$ and therefore $\dot{P}/P=1/2T$ for these mature pulsars. This simple estimate for $Q_0$ agrees with the Monte Carlo results of CZ01, who find $Q(E) = 1.7 \times 10^{39} E^{-1.6}
\exp(-E/80) / {\rm GeV} /{\rm s}$, if we assume $f_e \simeq 0.03$, $\dot{N}_{100} = 1$, ${E_{\rm cut}}=80 {\rm GeV}$.

It is important to note that $f_e$ is the fraction of spin down [ *power*]{} that is injected into pairs [*after*]{} the assumed maturity age of $10^5~ {\rm yr}$. This efficiency $f_e$ is expected to be large since $e^+e^-$ are the lightest electromagnetically coupled fermions. The fraction of total initial energy injected into the ISM in pairs is very small, $\sim f_e \tau_0/T$. This argument shows that if a significant amount of the spin-down energy released before the assumed maturity age of $10^5 {\rm yr}$ were to be available in the form of $e^+e^-$ pairs injected into the ISM, then the required efficiency would be very small. We see this by noting that in the approximation that some fraction of all of the spin-down energy is injected into the ISM instantaneously in the form of $e^+e^-$ with spectrum $E^{-1.6}$ 