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Relativistic Doppler Shift and a Star breaking up

  1. May 25, 2008 #1

    TFM

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    1. The problem statement, all variables and given/known data

    A distant star at rest with respect to an observer on Earth emits light of frequency 6.690 x 10^14 Hz. The star breaks up into two remnants of equal mass, which are observed to emit light of frequency 7.135 x 10^14 Hz and 4.282 x 10^14 Hz

    Find the velocities of the remnant and the angle between line of sight and their direction of motion.


    2. Relevant equations

    [tex] v^{prime} = v \gamma [1 - \beta cos \theta] [/tex]

    [tex] v^{prime} = v \sqrt{\frac{1 + \beta}{1 - \beta}} [/tex]

    3. The attempt at a solution

    I have solved the first part, finding the speed, using:

    [tex] v^{prime} = v \sqrt{\frac{1 + \beta}{1 - \beta}} [/tex]

    assuming that since the star was initally at rest, teh centre of momentum will remain in the same polace, hence the two parts will fly away from each other.

    I got [tex] \beta [/tex] to be 0.47, and the speed of the star remnants to be 0.235c

    I am having a small problem with the second part, finding the angle. I am sure I need to use:

    [tex] v^{prime} = v \gamma [1 - \beta cos \theta] [/tex]

    with the v primke being the orginal stars frequency. but I am nto sure which of the othe frequencies to use as v, and I am not sure how to get a value of beta to use in this part, since I am sure it will be different to the value in part A.

    Any ideas will be greatly appreciated,

    TFM
     
  2. jcsd
  3. May 25, 2008 #2

    Doc Al

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    Staff: Mentor

    That's the formula for the longitudinal Doppler shift when the source is approaching. It assumes an angle of 180 degrees, so it's not relevant here.

    You can't solve the two parts separately. Hint: How are the speed and angle of the two remnants related?
     
  4. May 25, 2008 #3

    TFM

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    Would you need to use the equation [tex] c = \lambda * frequency [/tex]

    so:

    [tex] frequecy = c/ \lamda [/tex]

    so:

    [tex] \frac{c}{\lambda}^{prime} = \frac{c}{\lambda} \gamma [1 - \beta cos \theta] [/tex]

    Or I am going in the worng direction?

    TFM
     
  5. May 25, 2008 #4

    Doc Al

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    The equation to use is the first one you listed (I'll use f for frequency):

    [tex] f' = f \gamma [1 - \beta cos \theta] [/tex]
     
  6. May 25, 2008 #5

    TFM

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    So:

    [tex] f' = f \frac{[1 - \beta cos \theta]}{1 - \beta^2} [/tex]

    but we don't know beta or theta?

    TFM
     
  7. May 25, 2008 #6

    Doc Al

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    Of course you don't--that's what you're trying to find! Set up two equations (using the two observed frequencies) and you can solve for the two unknowns.

    Don't neglect to answer my question: How are the speed and angle of the two remnants related?
     
  8. May 25, 2008 #7

    TFM

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    Would the relationship between speed and angle be:

    [tex] v_{obseerved} = v*sin \theta [/tex]

    ???

    TFM
     
  9. May 25, 2008 #8

    Doc Al

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    I meant this as two questions:
    (1) How are the speeds of each remnant related?
    (2) How are their angles related?

    These should be easy to answer. Drawing a diagram will help.
     
  10. May 25, 2008 #9

    TFM

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    Considereing coinservation of momentum, then the momentum before in the Earth Observer frame should be zero, since the star has zero speed. this means that the two stars remnants must have equal and opposite speeds, to keep the center of momentum frame the same. the angle between them must be 180 degrees (Pi)?

    TFM
     
  11. May 25, 2008 #10

    Doc Al

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    Good! So how will the angles that you use in the Doppler formula relate to each other? How will their cosines relate to each other?
     
  12. May 25, 2008 #11

    TFM

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    Would one of the remnants hav an angle 0f 0, thus the cos would be 1, whilst the other would have 180, with a cos of -1?

    TFM
     
  13. May 25, 2008 #12

    Doc Al

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    No reason to think so. (But if one did have an angle of 0, then the other must certainly have an angle of 180.) Again, this is a more general case than the simple longitudinal Doppler shift.

    Call the angle that one remnant makes [itex]\theta[/itex]. (Draw that diagram!)
     
  14. May 25, 2008 #13

    TFM

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    Is the daigram required?

    TFM
     

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  15. May 25, 2008 #14

    Doc Al

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    That's a good start. Now mark the angle that each remnant makes with the line of sight. (Assume for all practical purposes that both remnants are located just where the original star was.)
     
  16. May 25, 2008 #15

    TFM

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    Would that be like this?

    TFM
     

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  17. May 25, 2008 #16

    Astronuc

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    Yes - so now find the velocity components of both stars with respect (parallel) to the line of sight, and relate them back to the velocities of the stars. Note that the star moving away is at the same angle [itex]\theta[/itex] as the star moving toward the observer.
     
  18. May 25, 2008 #17

    TFM

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    This would be breaking down into components, so, for star Remnant A:

    Horizontal speed would be:

    [tex] v*cos(\pi - \theta)[/tex]

    Vertical speed would be:

    [tex] v*sin(\pi - \theta) [/tex]

    and for Remnant B:

    Horizontal Speed would be:

    [tex] v*cos(\theta) [/tex]

    Vertical speed would be:

    [tex] v*sin(\theta) [/tex]

    ???

    TFM
     
  19. May 25, 2008 #18

    Doc Al

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    Almost. The angle should be measured from the position vector that extends from earth to the star.

    In the general formula for Doppler shift you should recognize that [itex]\theta = 0[/itex] corresponds to the ordinary longitudinal formula when the source is receding.

    There's no need to find parallel or perpendicular components of the velocity. The [itex]\beta[/itex] in the Doppler formula represents the full velocity, not a component. (Any needed component is taken care of by the [itex]\cos\theta[/itex] factor.)

    Fix your diagram (marking the correct angle), then set up two equations, one for each remnant.
     
  20. May 25, 2008 #19

    TFM

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    Should it be like this then, since I believe you measure angles clockwise

    ???

    TFM
     

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  21. May 25, 2008 #20

    TFM

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    opps, wrong diagram:

    Sorry,

    TFM
     

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