Question

The Crab pulsar has a very steep radio spectral index of approximately -3 (i.e. v−3) over a frequency range from 10MHz to 10GHz. If the distance to the Crab pulsar is about 2 kpc, the measured flux density at 400MHz is 650mJy, and the spin-down luminosity (i.e. ˙E ) as derived in class is 4×1038 erg s−1, what fraction of ˙E does the radio emission account for?

Answer 1

Answer on Question 55064, Physics / Astronomy | Astrophysics

Question:

The Crab pulsar has a very steep radio spectral index of approximately -3 (i.e. v-3) over a frequency range from 10MHz to 10GHz. If the distance to the Crab pulsar is about 2 kpc, the measured flux density at 400MHz is 650mJy, and the spin-down luminosity (i.e. 'E') as derived in class is $4\times 10^{38}$ erg s$^{-1}$, what fraction of 'E' does the radio emission account for?

Solution:

The distance to the Crab nebula is $D = 2\,\mathrm{kpc} = 6.2\times 10^{21}\,\mathrm{cm}$.

$$\begin{aligned} L_{\text{radio}} &= 4\pi D^2 \int_{\text{radio}} S_{\text{radio}} \, dv = 4\pi (6.2 \times 10^{21})^2 \int_{10^7}^{10^{10}} 0.65 \left(\frac{v}{4 \times 10^8}\right)^{-3} dv \\ &= 2 \times 10^{47} \left[ \frac{v^{-2}}{-2} \right]_{10^7}^{10^{10}} = 1 \times 10^{33} \text{ ergs} \times s^{-1} \end{aligned}$$

Therefore the fraction is:

$$\frac{L_{\text{radio}}}{E} = \frac{1 \times 10^{33}}{4 \times 10^{36}} = 2.5 \times 10^{-6}$$

Answer: $2.5\times 10^{-6}$ – very small

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