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Distance to pulsar with plasma dispersion relation

  • Thread starter Robsta
  • Start date
  • #1
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Homework Statement


Pulsars are stars that have suffered gravitational collapse. They rotate rapidly and emit a narrow
beam of radiation. The pulse lengths, at the earth, are ∼1ms and the periods are ∼1s.
Within a few months of the discovery of pulsars distance estimates were obtained by exploiting
the dispersion of the pulses in the interstellar medium, which is ionised hydrogen with an electron
density of about 105m−3
.
(a) Show that for ω2>>ωp2 where ωp is the plasma frequency of the interstellar medium, the time
delay ∆t as a function of f−2 − (f + ∆f)−2, where f is the pulse frequency, is a straight line
whose slope is a measure of the distance to the pulsar.
(b) For the pulsar CP 0328 the delay between signals at 610 and 408MHz was 0.367s; that
between signals at 408 and 151MHz was 4.18s. Find the distance to CP 0328.


Homework Equations


I know the dispersion relation for the plasma. n2= 1 - (ω2p2)
Obviously speed * time = distance to pulsar
ω=2(pi)f
n*vp=c

The Attempt at a Solution


So the pulsar emits a pulse at t=0 of two frequencies. Because of dispersion in the interstellar plasma, the waves travel at different speeds (because of their different frequencies).
So if one wave of frequency f arrives at time t, the one travelling with frequency (f + ∆f) arrives at time (t + ∆t)
I'm a bit confused because if ω2>>ωp2 then as per the formula for n above, n=1 and there would be no dispersion?

I'm really looking for some help formalising part 1
 

Answers and Replies

  • #2
34,051
9,910
Check the fraction in your dispersion relation.
I'm a bit confused because if ω2>>ωp2 then as per the formula for n above, n=1 and there would be no dispersion?
n is close to 1, but not exactly 1. The small difference leads to the small time differences: seconds, where the pulsar can be thousands of light years away so t is thousands of years.
 

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