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What is the exact change in frequency for an approaching object?

  1. Oct 10, 2010 #1
    When light is shone from an approaching object towards a stationary observer, the light is blue-shifted. What exactly is the amount of this change in frequency? So if the wave that leaves a radar gun is, say, 500 THz, when it is travelling 100km/h, what will it be when observed by the stationary observer?

    This is NOT a homework question and I am not a student trying to do homework so please don't assume I'm being 'lazy' and not doing my homework myself. What I'm trying to figure out is if the change in frequency observed is linear compared to the speed, or if it is relativistic and a small change in frequency compared to the speed.

  2. jcsd
  3. Oct 10, 2010 #2


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    Look up the relativistic Doppler shift in any textbook.
  4. Oct 11, 2010 #3
    set the object a certain distance from the gun.
    how long before it reaches the gun.
    in that time how many cycles of radiation will it reflect
  5. Oct 11, 2010 #4
    Assuming a line of sight approach and a constant velocity.

    The Doppler formula is for c=1:

    \sqrt {{\frac {1+v}{1-v}}}

    First we convert 100km/h in terms of c:
    100 * 1000 /3600 / 299792458 = 9.2656693110597791549*10^(-8)

    Thus we get a rate of frequency of:


    Look at this graph that should give you the answer if it is linear or not:

  6. Oct 11, 2010 #5
    I thought he wanted to know what a cop with a radar gun would see if a car was moving near light speed.

    I didnt know that the radar gun was in the car (and was therefore time dilated)
  7. Oct 11, 2010 #6
    My understanding is that it doesn't matter if the radar gun is moving or stationary....it is only about the relative speed between the gun and the object it's pointing at??? Just checking.
  8. Oct 11, 2010 #7
    Funny....thanks. It didn't occur to me to search for 'relativistic' doppler shift - I had only looked into 'dopper shift' so I missed all the details on it. I get it now. And I see how at insignificant speeds there isn't much of a relativistic effect anyhow. This only leaves the effect for really fast speeds. I wonder if there has been any actual observed examples of that? I guess not yet?
  9. Oct 11, 2010 #8

    Thanks!!!! You have no idea how much it helps to clear things up when someone spells it out like that :)
  10. Oct 11, 2010 #9
    Do we have yet any observational evidence of the relativistic aspect of the Doppler shift? (ie, for high velocities where 'v' is closer to 1 - are there any examples discovered yet?)
  11. Oct 12, 2010 #10
    Here is a plot showing the relativistic and non-relativistic Doppler shift and the difference between them.


    Did you know by the way that if the motion is not line of sight but instead transverse the Doppler effect is very different? In this case it is basically the relativistic time dilation, so for a 'train' approaching or departing the station it is:

    {1 \over \sqrt{1-v^2}}

    That is exactly right!

    What does matter is acceleration, if the gun or the target accelerates (or both) we need a different formula and in these cases even the distance of the radar gun matters.

    If you keep accelerating hard enough you may even outrun an outer space speed check of your relativistic rocket. :biggrin:

    And these are just the simple scenarios, for movement in three directions the problem becomes more difficult. As far as I know there is no literature on the exact Doppler formula for 3 dimensional acceleration of the target for a given velocity, or perhaps the "high priests" prefer to keep those formulas secret. Seriously though, if anyone knows please post or give a reference to the formulas, science should not be a secret.
    Last edited: Oct 12, 2010
  12. Oct 12, 2010 #11
    thanks. yes, I was kind of avoiding the traverse situation.

    3D movement? I would see that as always being just 2D (traverse) since the 3 axis are relative and you can always align one with the movment leaving only 2 of concern.
  13. Oct 12, 2010 #12
    Yes, a poor word choice from my side.
  14. Oct 12, 2010 #13
    Hey, since you're so helpful, did you notice my question about what actual observations they've made of any sort of doppler shift than contained a relativistic effect? (ie, where v is closer to 1). (since as far as I can see, any doppler shift we measure is always with low values of v and so no real time dilation in there)
  15. Oct 12, 2010 #14
    Hey, just curious, who are the "high priests?"
  16. Oct 12, 2010 #15


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    Last edited by a moderator: Apr 25, 2017
  17. Oct 12, 2010 #16

    so if a laser beam of freq f from a stationary laser gun (the source) is reflected off a mirror moving at velocity v which is very near the speed of light then what is the freq of the reflected light as measured by the stationary observer? Note that the gun is the source.
    Last edited: Oct 12, 2010
  18. Oct 12, 2010 #17
    I was just a figure of speech, it is an allusion to ancient history where the 'scientists' were actually members of the religious caste who kept their secrets from the common people.
  19. Oct 12, 2010 #18
    Curiously enough after a brief search on the web, I find many doppler formulas for velocity but none for acceleration. I guess the derivation is easy and they don't bother to show it.
    Once you have worked out the Doppler shift of frequencies if the shift is constant there is no acceleration and if varies with time you calculate this variation per unit time and convert it to an acceleration:

    [tex]a=\frac{c \Delta fd}{2tfr}[/tex]
    a= acceleration
    c=light speed constant
    t=time it takes for the Doppler shift variation
    [tex]\Delta fd[/tex]=Doppler shift variation
    fr= frequency echoed by the source
  20. Oct 12, 2010 #19


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  21. Oct 12, 2010 #20
    Relativity teaches us that when a light source approaches we can still have it being redshifted.

    Here is a 3D graph that shows the Doppler of an approaching source for different angles. It is hard to see but if you look good you can see the shift is maximum if the angle is 0 and that for very wide angles there is actually redshift instead of blueshift (e.g. the flat parts on the sides go below 1 for high velocities).


    The formula for an accelerating observer of an inertial light source in the line of sight is (c=1):

    -g \left( 1+gx \right) ^{-1} \left( \ln \left( 1-{\frac {g}{v}}
    \right) \right) ^{-1}
    Last edited: Oct 12, 2010
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