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The doppler effect and time dilation context

  1. Dec 20, 2009 #1
    I'm having a hard time understanding some things regarding the doppler effect. Please excuse my Microsoft paint diagram, I'm no artist.

    http://img686.imageshack.us/img686/3949/diagram.jpg [Broken]
    http://img686.imageshack.us/img686/3949/diagram.jpg" [Broken]

    Given this scenario, which equation/approach would yield the most accurate results:
    [A] Special Relativistic Equation for Radial Doppler Shift
    or [B] Classic / Newtonian Equation.

    I'm under the impression that the primary difference between the special relativistic equation for radial doppler shift and the newtonian equation is the time-dilation term, (please correct me if I'm under the wrong impression). Additionally, does time dilation come into play in this scenario? I'm still having a hard time understanding the exact contexts in which time dilation occurs.

    Another reason I ask this question is because of the anomalous doppler observations in pioneer 10 and 11 [ http://renshaw.teleinc.com/papers/pionee2/pionee2.stm ]
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 20, 2009 #2

    It does in every scenario involving relative motion.
    The observer is not"standing perpendicular to the radial velocity of light's propogation"
    At the point in your diagram to calculate the redshift you'd use
    z=sqrt.(1+b/1-b) -1
    where b=v/c, v is the tangential velocity of the bulb.
    See derivation http://en.wikipedia.org/wiki/Relativistic_Doppler_effect" [Broken]
    You'd use transverse shift when the sorce is moving perpendicular to the line of sight.
    Last edited by a moderator: May 4, 2017
  4. Dec 20, 2009 #3
    This is true, but another significant difference is that according to the Newtonian Doppler equation an observer would never see any Doppler shift for a light source moving exactly transverse to the line of sight (observer in line with the radial spoke connecting the light source to the centre of the cylinder in your diagram). In the relativistic Doppler equation a shift is seen even when the source is neither going towards the observer nor going away if it going across the line of sight.

    Yes it does. Time dilation changes the frequency that the light is emmited at and this is on top of the shift in frequency due the classic Newtonian Doppler shift.
    Last edited: Dec 21, 2009
  5. Dec 21, 2009 #4
    The light is moving at the observer, so don't you mean blue shift? And what angle must the observer be at in order to experience blue shifting?

    I'm confused, which one is correct then? And how is the light not moving directly at the observer in the diagram?

    I thought the velocity of the bulb in the direction of the observer caused a change in frequency and energy. How does time dilation change the frequency? Would the observer actually receive more energy than the light actually produced during the moment the observer experiences blue shifting?
    Last edited by a moderator: May 4, 2017
  6. Dec 21, 2009 #5
    The relativistic Doppler solution is always the correct solution. The Newtonian Doppler shift is just an approximation. The inaccuracy of the Newtonian Doppler shift was not noticed in Newtonian times because they did not observe many things moving at significant velocities and even today the Newtonian approximation is good enough for most calculations at normal velocities in everyday experience because the error is very small at low velocities.

    The light is moving directly at the observer in your diagram. You asked what the significant difference between Newtonian Doppler shift and relativistic Doppler shift is and I said the most significant difference shows up when measuring transverse Doppler shift even though that situation is not shown in your diagram. (I did describe how your diagram would have to modified).

    It does, but it is not the only thing that can change the frequency.

    Time dilation slows everything down. If some atomic process has a characteristic vibration frequency when stationary in a lab it will have a slower vibration frequency when moving relative to an observer even before the classic Newtonian Doppler shift is considered. For example some crystals oscillate at a fixed frequency when a certain voltage is applied to them and this is the basis of some digital watches, but if that watch is moving relative to the observer the crystal will no longer be oscillating at its "natural" frequency. A light source moving towards the observer is blue shifted due to the classic Doppler shift and time dilation causes an opposite red shift but the two effects do not necessarily cancel each other out. If the source is moving away from the observer (eg on the opposite side of your cylinder) then the red shift due to classical Doppler shift and the red shift due to time dilation combine to produce an even greater red shift. This effect can be observed when looking at stars on opposite sides of a rotating galaxy that is edge on to the observer and is in agreement with the relativistic Doppler shift (including time dilation) predicted by relativity.
    Last edited: Dec 21, 2009
  7. Dec 21, 2009 #6
    Thank you kev for being so informative and helpful.

    Additionally, I'm wondering if the observer would actually receive more energy than the light actually produced during the moment the observer experiences blue shifting?
  8. Dec 21, 2009 #7
    I am going to stick my neck out here and say yes. Imagine you are riding a bike at 10kph and throw a snowball at 15kph at a target on the side of the road, then the snowball hits the target at 25kph. If you throw another snowball backwards at 15kph at the same target after you have passed it the snowball hits the target at 5kph. The energy of the snowball depends not only on the energy imparted when throwing it, but also on the energy it had before being thrown due the relative motion of the frame it was originally at rest in. In the case of a light signal the light does not change its speed but its energy does change due to the change in frequency. This change in energy is not just a perceived change, but real in the sense that an ordinary light signal of optical wavelength could be an X-ray to an observer moving at relativistic velocities towards the light source.
  9. Dec 21, 2009 #8
    Thank you kev. (Please correct me if I am wrong), but the increase in energy that the observer would experience could be perfectly modeled (with high accuracy, even with high velocities) using the following two equations:

    Frequency expressed as a function of velocity
    Moving Frequency = [(c-v)/(c+v)] * original frequency

    Energy expressed as a function of frequency
    E = hF
    Last edited: Dec 21, 2009
  10. Dec 21, 2009 #9
    I think that should be:

    Frequency expressed as a function of velocity
    Moving Frequency = sqrt[(c-v)/(c+v)] * original frequency

    Energy expressed as a function of frequency
    E = hF[/QUOTE]
  11. Dec 21, 2009 #10
    That is the first time you said 'i think'. Are you sure?
  12. Dec 21, 2009 #11
    Ha! When I said "sticking my neck out" in an earlier post I think that indicates a certain amount of uncertainty on my part ;)

    In this case I took the equation:

    [tex]\frac{f_s}{f_o} = \sqrt{\frac{1+\beta}{1-\beta}}[/tex]

    where [itex]\beta = v/c [/itex] and [itex]f_s[/itex] is the source frequency and [itex]f_o[/itex] is the frequency measured by the observer, from Wikepedia http://en.wikipedia.org/wiki/Relativistic_Doppler_effect and did not go into to it any deeper that. I said "I think" because I was assuming the Wikepedia source is correct (which is not always the case).
  13. Dec 21, 2009 #12
    So the correct relativistic & highly accurate equation is:

    http://img12.imageshack.us/img12/3498/equationy.jpg [Broken]

    Last edited by a moderator: May 4, 2017
  14. Dec 21, 2009 #13
    Which is the same as:

    Moving Frequency = sqrt[(c-v)/(c+v)] * original frequency
    Last edited by a moderator: May 4, 2017
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