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Doppler relativity question

  1. Sep 5, 2012 #1
    I'm trying to understand Doppler and specifically I'm trying to get the same answer as I look at the same situation in different frames. I'm getting stuck though and getting a slightly different answer in one frame, hopefully someone can help me understand.

    Here's my simple example. There is a transmitter and a receiver, moving toward each other at constant speed. The receiver measures the frequency of what he receives using his own clock. I believe that the answer (the measured frequency) should be the same regardless of the frame of the observer since it is an objective and co-located thing: number of cycles of received clock per cycle of local clock.

    Case 1 is that I observe from the transmitter frame. Without invoking any relativity, the wavelength is [itex]c/f[/itex] and these pass the receiver at a speed of [itex]c+v[/itex]. The observed received frequency is [itex]\frac{c+v}{c/f}[/itex] = [itex](1+v/c)f[/itex], where f is the transmit frequency. From the frame of the transmitter though, the receiver clock looks slow by [itex]\sqrt{1-(v/c)^{2}}[/itex], so in this frame I see him measuring [itex]\sqrt{\frac{1+v/c}{1-v/c}}[/itex]. This is the non-relativistic Doppler divided by his relativistic clock slowing.

    Case 2 is that I observe from the receiver frame. Now the wavelength is (c-v)/f and these pass the receiver at c. The observed (non-relativity) frequency is [itex]\frac{c}{(c-v)/f}[/itex] = [itex]\frac{f}{1-v/c}[/itex]. Due to relativity the transmit frequency looks slow by [itex]\sqrt{1-(v/c)^{2}}[/itex], so I multiply the non-relativity version by this factor and get the same answer [itex]\sqrt{\frac{1+v/c}{1-v/c}}[/itex].

    The case that messes me up is if I pick a middle frame where they are each observed to be moving toward each other at v/2. By the same rationale as each of the previous, the non relativistic version is [itex]\frac{1+v/2c}{1-v/2c}[/itex]. Special relativity would make both the transmit and receive clock look slow by [itex]\sqrt{1-(v/2c)^{2}}[/itex], but since the transmit and receive are the same (I'd multiply numerator and denominator by this factor) this effect would cancel. I'm left with the non-relativistic version which is very very close numerically to the previous two, but not exact. What am I missing?
     
  2. jcsd
  3. Sep 5, 2012 #2
    They are not moving toward each other at v/2 in the middle frame. If the transmitter is moving toward the receiver at the speed v in the receiver's rest frame, and vice versa, then the receiver and transmitter are both moving at speed [1 - sqrt(1-v^2)]/v in terms of the rest frame of the mid point between them.
     
  4. Sep 5, 2012 #3
    I don't understand this. Is this relativity or plain old geometry? If A is still and B is moving in at v m/s, then every second they are v meters closer together. If they are both moving at 0.5v m/s inward, then each is 0.5v meters closer to the middle each second, or v meters closer to each other. Right?
     
  5. Sep 5, 2012 #4

    DrGreg

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    But that's only true when the distances and times are measured in the middle frame. In the A frame or the B frame the distances and times are different.

    It turns out that the rule for "adding" velocities isn't u + w = v but[tex]
    \frac{u + w}{1 + \frac{uw}{c^2}} = v
    [/tex]
     
  6. Sep 5, 2012 #5
    Interesting. I obviously need to do more reading. How do I think about this? The two individual movers can agree on their closure rate, but an outside observer can't?

    When I slowed my clock in my initial post, did I need to slow my velocity too, or is that sort of double booking the same effect?
     
  7. Sep 5, 2012 #6
    Thanks for this handy formula. A big improvement over my previous ,artillery technique of trial and error values entered in the addition of velocities equation.
     
  8. Sep 5, 2012 #7
    For anyone who hasn't found it, FYI:

    http://en.wikipedia.org/wiki/Velocity-addition_formula

    Everything worked for me if I used 0.5V as a nominal case, then used the addition formula when I moved to the frame of either the transmit or receive. I used the new velocity also to compute the clock slowdown factor.
     
  9. Sep 5, 2012 #8
    This seems to pose a question. In the transmitter frame, if we assume the receiver clocks are slow by the gamma factor, this would then effect an increase in observed frequency in addition to the classical increase due to closing velocity.
    AFAIK the relativistic observed Doppler factor is, in all cases, red shifted from the classical values.
    So in the transmitter frame this would seem to necessitate the clocks at the receiver to be running faster , not slower than the clocks in that frame.
    AM I missing something here?
     
  10. Sep 5, 2012 #9
    That pulse, presented as ticking of time in both frames; is taken as an "instant", so it's really a sample of length (distance) from that "transmitter" frame according to your proper time. The "complication" of events that happened all in the same location(determined by defining the FoR, in this case the transmitter in comparative motion) is a sample of the proper time from that "transmitter" frame, while maybe obsolete after a long & speedy trip, the "Transmitter" is not as old as the "Receiver" :smile:.

    Consider the EM that's within the Transmitter FoR. Specifically the EM one would find in between the atoms that make up the transmitter itself, how do you interpret the "Doppler Effect" there? As a length contraction or an issue of time dilation? (RoS tee hee hee)

    So I guess I'm safe to say, using "Doppler effect" as a measure, "flips" the "metric" (particular measure either Length or distance) when making measurement comparisons. How symmetrical:rolleyes:

    To look at it different, your "clock" is broken, the two "mirrors" (transmitter/receiver) are moving towards each other...

    oh and blue shifted happens too, or is it less red shifted...
     
    Last edited: Sep 5, 2012
  11. Sep 5, 2012 #10
    I think it's true that the velocity thing always causes a red shift from the perspective of the receiver. In this case I'm not the receiver, but someone in a different frame looking at the receiver and determining what his measurement would be. To me, his local clock (used to make the measurement) looks slow. It doesn't look slow to him.

    Now this notion of velocity not being velocity adds confusion since you have to decide in which frame v is defined. The answer to my original question was that there could be some v that both transmitter and receiver agree on, but then someone in another frame would see something different. If I make it so that they are each going 0.5v toward a middle point frame, each of the individual guys would see a closure of something other than v. Confusing.
     
  12. Sep 5, 2012 #11
    In this instance, if they are traveling 0.5c and -0.5c respectively , in the middle frame ,then the transmitter and receiver are traveling 0.8c relative to each other in their own frames. They of course would both agree on this ;-)
     
  13. Sep 6, 2012 #12
    I've got another example that's hanging me up. Same transmit/receive scenario as before, except now the transmit and receive people are relatively stationary.

    If I sit in the obvious stationary frame, Doppler is zero, fobserved = ftransmit.

    I'm trying to get the same answer if I switch to a fast moving frame where the transmit and receive are moving such that the direction of travel is perpendicular to the line between them. In this case, the transmitter will appear to "lead" the receiver. I can draw a right triangle with one leg being the line between transmit and receive, another being the distance traveled by the receiver during the light time delay, and the hypotenuse being the light path. If I did my math right, sin(θ) = v/c, where θ is the angle of the lead from the transmitter. The velocity component along the light path would then be [itex]v^{2}/c[/itex] and the received frequency factor would be [itex]\frac{1-v^{2}/c^{2}}{1+v^{2}/c^{2}}[/itex]. I think...

    But I need something to cancel all that stuff out. The clocks will look slower in this frame, but it will be a common factor between transmit and receive and will cancel out. How do I make this consistent? Thanks again.
     
  14. Sep 7, 2012 #13
    Anyone have an idea about my last example?
     
  15. Sep 7, 2012 #14

    Mentz114

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  16. Sep 7, 2012 #15
  17. Sep 7, 2012 #16

    pervect

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    It sounds like a "light clock" to me - and the point is that light clocks (as do all other clocks) run slower when moving due to time dilation.
     
  18. Sep 7, 2012 #17
    It kind of is a light clock, but I'm thinking about Doppler as observed by the receiver, not the time it takes to make the trip.
     
  19. Sep 7, 2012 #18

    Mentz114

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    I don't know if I understand your setup, but if there is no relative motion between the sender and receiver then there's no Doppler effect.

    Is this what you mean by 'moving together' ? Or do you mean they are approaching each other ?
     
  20. Sep 7, 2012 #19
    Yeah, the Doppler should come out zero. What I'm trying to do is just an exercise where I want to observe the situation from a moving frame and arrive at the same answer. It's similar to the scenarios in my first message in this thread; I had a transmitter and receiver moving toward each other and proved to myself that I could determine the frequency measured by the receiver in the transmit frame, receive frame, or a "stationary" frame in which they were both moving - and get the same answer.

    In this case I don't know how to get the same answer (zero doppler) if I analyze from a frame where two relatively stationary guys are moving together perpendicular to the line between them. Like a light clock where the transmitter is at the top and the receiver is at the bottom. The non-zero angle of incidence at the receiver induces Doppler that I don't know how to account for.

    I hope that makes sense.
     
  21. Sep 7, 2012 #20
    If I am understanding you I think Mentz114 had the right solution.
    The transverse Doppler blueshift related to theta at the transmitter, is exactly countered by the transverse redshift at the receiver. The angle being of course the same at both ends
    the end being zero net shift which you already knew.
     
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