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What time of year observations of crab pulsar were made

  1. Dec 30, 2011 #1
    Hi There

    I left a related post to the problem I am having with the crab pulsar a little while ago but would like to repost in a much more coherent manner than was done the last time - I would like first to state the problem and then state my approach to solving it. hopefully then any incorrect assumptions I am making will be noticed and pointed out to me. here goes:

    The problem is, given data of the crab pulsar recorded (from which the observed frequency of the crab can be extracted) in the Canary islands, find out what time of year the data was recorded..

    My attempt at solution: First thing is to take a fourier transform of the recorded data of the crab (this data contains several thousand periods, I think). Then from this FFT, can then figure out the observed frequency of the crab, it will simply be equal to the fundamental frequency as revealed by the FFT...
    Note, the FFT was done using a Matlab script that was kindly provided to me...

    Anyway, the FFT showed that the observed frequency f_o was 29.8501253128Hz.. Also provided to me was the intrinsic frequency of the crab, f_i=29.8480959009Hz ... .

    The reason for the difference between the observed and intrinsic frequency is that there is a radial velocity between the observer and crab ( predominantly due to motion of the Earth). So using values stated for f_i and f_o, can now find the radial velocity between the observer and crab by simply using the electromagnetic doppler shift equation:

    [tex]f_{o}=\sqrt{\frac{c-v}{c+v}}f_{i}[/tex]

    where c is the speed of light. By plugging in the values of f_o and f_i into the doppler equation, I found that the radial velocity between observer and crab was equal to approximately 20.4km/s. I have made the simplifying assumption that this radial velocity is due purely to the Earth's orbital motion (ie. the Earth's rotational motion was not considered). Also note, I assumed that the Earth's orbit is a perfect circle of radius 1AU and so this implies that its tangential velocity at any given instant is given by (2*pi)/(365*24*60)*1AU which is approximately equal to 30km/s. By doing a similar calculation of the tangential velocity due to the Earth's rotation at the latitude of the Canary islands, it is found that this velocity component is approximately 400m/s. So in consideration that 400m/s<<30km/s, I think that I am justified for a first rough calculation at least to neglect the velocity due to the Earth's rotation.

    To summarise the story so far: I have calculated that the radial velocity between the observer and Crab pulsar at the time this data was recorded was approximately 20.4km/s. Further, I am considering that this velocity is due only to Earth's orbital motion (ie. have rejected rotational motion). So the next step is to figure out what time of year it would have to have been in order for the radial component of the velocity to be equal to 20.4km/s.. . And I will explain how I approached this part of the problem now:

    This is where a decent understanding of RA and declination comes in (unfortunately I lack this understanding!). The first thing I reasoned (somewhat dubiously) is that the crab pulsar lies on the same plane as the ecliptic lies at all times during the year. I reasoned this becuase the ecliptic makes an angle of 23.5 degrees to celestial equator and because the crab is at a declination of 22degrees, then the ecliptic and the crab lie on approximately the same plane (ie they are only out by 1.5degrees). Can someone please confirm that this reasoning is correct? and if it is not correct please explain clearly how to think about this problem... .

    The next thing I reasoned was that at the Vernal Equinox, when the sun is at 0hrs of RA, then the Earth is moving with its full tangential velocity towards objects that are at an RA of 6hrs. More generally, if the sun lies at Xhrs of RA (where X is an arbitrary number), then the Earth moves towards objects at an angle of (X+6)hrs RA with its full tangential velocity. Is this reasoning correct? and if it is not correct, why not?

    I then applied this reasoning to the specific problem of the Crab, which has an RA of 5hrs 34mins and with the radial velocity of 20.4km/s that I stated above - so I used all this info to calculate the time of year that the data must have been taken by the following line of math shown below:

    [tex]30\sin\theta=20.4\implies\theta=\sin^{-1}(\frac{20.4}{30})=42.8[/tex]

    (Note, the value 20.4 corresponds to radial velocity between observer and crab and the value 30 corresponds to tangential velocity of the Earth as it orbits the sun)

    so this angle of theta=42.8degrees corresponds to the difference in angle between the crab and the sun.. So convert this from degrees to hrs and mins by 6*(42.8/90)=2hrs50mins and so this implies (if my reasoning is correct of course!) that the sun was 2hrs50mins of right ascension behind the crab, that is, it was at an RA of 5hrs34mins-2hrs50mins=2hrs44mins when this data was taken. If the sun were indeed at this RA at the time of these observations, then the time of year must have been sometime in early May...

    I would be very very grateful if anyone could read carefully through my post and either confirm that my reasonings are correct or explain why they are not.... . I am happy that I have done everything correctly as far as calculating the radial velocity is 20.4km/s.
    So it is, my assumption that the Crab lies on the same plane as the eccliptic and also my assumption that the Earth moves with full tangential velocity towards objects of RA (X+6)hrs when the sun is at an RA of Xhrs that I am doubtful of... .

    I have been struggling with this problem for quite a long time now (fully understanding the coordinates of RA and declination are what I am having trouble with really) and so it would be great if anyone would like to provide advice on my approach to solving it!

    Thanks
     
    Last edited: Dec 30, 2011
  2. jcsd
  3. Dec 31, 2011 #2
    just an update.. I have decided to add a correction to what I said in my last post, quote below:

    In fact, now I think the opposite, I think that when the sun as at 0hrs RA (vernal equinox), then the Earth must be moving towards objects with its full tangential velocity that have an RA of 18hrs...
    Again can any1 confirm this?!!

    ........ I must say I rather surprised - my post has recieved over 170 views and apparantly not one person has known the answer to any of my questions.... .
     
  4. Jan 10, 2012 #3

    davenn

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    I havent really figured out the purpose of all this, maybe thats why others also havent responded.
    OK so you have come to the conclusion that it was observed some time early May ( maybe late April). Have you consulted a star almanac to see if the Crab Nebula is even above the horizon during that period of time ?
    I'm at work and dont have my planetarium software on this puter but having so it wouold be very easy to check and see if the nebula was above the horizon at that time

    Dave
     
  5. Jan 10, 2012 #4
    I'm not really sure what the purpose of this is, but if you look at the raw data, each datum should have a date/timestamp.
     
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