Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B Why do photons allow Doppler shift

  1. Dec 4, 2017 #51

    Buckethead

    User Avatar
    Gold Member

    Yes.
    Yes, I understand that. Because my examples use slow spaceship speeds, Its fine to just use the Newtonian portion of the Doppler formula. But either way the end result is very nearly the same.
     
  2. Dec 4, 2017 #52

    PeterDonis

    Staff: Mentor

    No, I'm not. The only relative velocity I've mentioned is the relative velocity between the source and the receiver. The relative velocity of the receiver and the thingie that is sent from the source to the receiver doesn't matter as long as it's the same for two successive thingies. That does not require that the speed of the thingie be frame-independent; the relative speed of receiver and thingie can be different for different receivers. It just has to be the same for the same receiver with two successive thingies from the same source.

    We are not talking about a change in energy. We are only talking about a change in frequency, i.e., how often each thingie arrives. (It is true that, for light, energy is proportional to frequency, but we haven't gotten there yet; that's a separate question from how frequency itself behaves.)

    If we are talking about SR, you have this backwards. As far as SR is concerned, there is one Doppler shift. It can be useful to split that one shift up, conceptually, into a "Newtonian" piece and a "relativistic" piece (the latter involving things like time dilation that don't appear in Newtonian physics), but that's a feature of human models, not of the actual physics.

    However, before you even try to understand the SR version with light, you need to understand the simpler Newtonian version, where we just have thingies going from source to receiver and we assume any SR effects like time dilation are negligible.
     
  3. Dec 5, 2017 #53

    Ibix

    User Avatar
    Science Advisor

    Are you perhaps not taking into account relativistic velocity addition? In a frame where the source is moving the separation rate between source and light pulse is not c, but in the frame where the source is at rest it is c. You can't ignore this by keeping the source velocity low because it's the speed of light that is being transformed. That's non-Newtonian by definition.
     
  4. Dec 5, 2017 #54

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    In fact, all you need is a diagram for the receiver's frame of reference. Try with the source moving away, as that might be easier to see.

    Note that, as @Ibix has pointed out, the separation velocity between the source and the light wave (in the receiver's frame) lies between ##0## and ##2c##. You may be confusing "relative" velocity with "separation" velocity here.

    Take a source, receiver and an light wave and 1D motion.

    The velocity of the light wave "relative" to the source/receiver means the velocity in the reference frame of the source/receiver. This is invariant (the same in both frames).

    However, the separation velocities are:

    Source frame: ##c-v## (speed at which the light wave is catching up the receiver)

    Receiver frame: ##c+v## (speed at which the light wave is moving away from the source).

    Where ##v## is the relative velocity of the source and the receiver.
     
  5. Dec 5, 2017 #55

    Dale

    Staff: Mentor

    Not one word of @Nugatory implied a medium in any way. That is simply something that you mentally added on your own. All we can do is recommend that you not do that.

    @Nugatory simply picked a frame that was convenient and analyzed the problem completely in that frame. That is a general approach that you should use. Pick a frame. It can be any frame that you think will simplify the math. Common choices are the rest frame of the emitter, the rest frame of the receiver, or the frame where they are moving at equal speeds in opposite directions. Just pick one and work it through in that frame.

    If you get stuck, just show us your work and we can get you unstuck

    Doppler isn’t directly about the energy or momentum of waves/photons/baseballs. It is about frequency. How much time is between two successive waves/photons/baseballs? Please ignore any question of energy or momentum and focus exclusively on the time. Calculate the time between two, not the energy or momentum of one. If you don’t have at least two with a given time between emissions then you are not doing a Doppler problem.
     
  6. Dec 5, 2017 #56

    Mister T

    User Avatar
    Science Advisor
    Gold Member

    But that derivation is just a heuristic device. There are other ways to do the derivation.

    In a practical way it's okay, but the newtonian viewpoint is entirely wrong. You can't patch it up by adopting it as a valid viewpoint to which relativistic "corrections" are added only when necessary. That approach works only in practice. The relativistic viewpoint is a worldview that works at all speeds.
     
  7. Dec 5, 2017 #57

    Mister T

    User Avatar
    Science Advisor
    Gold Member

    As he points out, he's not doing that. Perhaps what you mean here is the the relative speed between the source and the baseball changes when the baseball is thrown. Note that when the light pulse is created its speed becomes ##c##.
     
    Last edited: Dec 5, 2017
  8. Dec 7, 2017 #58
    I think the Doppler effect for light is easiest understood in terms of period and wavelength first.

    Say you have a receiver at rest, with a source of light approaching head-on at (normalized) speed ##\beta##.

    From the source's perspective, a time ##c \Delta t## elapses during the emission of a single cycle (crest to crest). That's the wave's period ##cT## in the source's rest frame: ##cT = c \Delta t##. But for light, the period ##cT## is equal to the wavelength ##\lambda##, so we actually have ##c \Delta t = \lambda##.

    By time dilation, that ##c \Delta t## is equivalent to ##\gamma c \Delta t## in the receiver's frame: ##c \Delta t^\prime = \gamma c \Delta t = \gamma \lambda## (where I've designated the receiver's frame the primed frame).

    So in the receiver's frame, the source emits a second wavefront a time ##c \Delta t^\prime = \gamma \lambda## after emitting the first. What we want to know is: what's the distance between the source and the first wavefront at this moment? The answer will tell us the wavelength ##\lambda^\prime## in the receiver's frame.

    Well, during this time ##\gamma \lambda##, the first wavefront moves closer to the receiver a distance ##\gamma \lambda##. But the source itself also moves toward the receiver during this time, a distance ##\gamma \lambda \beta##. Thus, the first wavefront is "ahead" of the source by a distance ##\gamma \lambda - \gamma \lambda \beta## when the second wavefront is emitted:

    ##\lambda^\prime = \gamma \lambda ( 1 - \beta )##.

    Then since wavelength and frequency ##\nu## are inversely proportional, we have:

    ##\nu = \gamma \nu^\prime (1 - \beta )##.

    For a receding source, the minus sign becomes a plus.

    For the general case, the source might not be approaching or receding head-on. Then we use only the component of the source's displacement that's parallel to the light wave's direction of propagation (as opposed to the total distance the source travels). This isn't difficult, but it involves defining an angle, and that can be done in either frame, so one must be mindful of signs and primes. IIRC, Einstein's original derivation uses the angle in the source's frame, but most other treatments I've seen use the angle in the receiver's frame. Can lead to some confusion. (Anyway, the angles are related by the aberration formula.)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted