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Lorentz and Doppler and signal pick up

  1. Jun 6, 2015 #1
    Dear PF Forum,
    After so many questions about Twins Paradox and Universe Frame of Reference, I'd like to know about Lorentz and Doppler. First.
    Here V is ##\sqrt{0.75}## ≈ 86.60%. If we put V in Lorentz Transformation formula as speed, we'll have Gamma = 2.

    Okay here's the question.
    Two probes A and B.
    Clocks are synchorized.
    Distance is 100 Ly.
    Each year they're sending digital signal, containing their respective time.
    A will send A0, A1, A2, A3 for each respective year.
    B also send B0, B1, B2,... each year.

    Signal from B is always late for 100 years
    So each year A will receive:

    Code (Text):
    Signal   A Time      Interval
    B0       100.00    
    B1       101.00      1.00 year
    B2       102.00      1.00 year
    B98      198.00
    B99      199.00      1.00 year
    B100     200.00      1.00 year

    Is this true?

    When B clock (and A clock) shows 100 year, B ignites its rocket RIGHT AFTER B sents B100. Catapulted at 86% c then the rocket stops, leaving B moves steadily at 86% c to the 'west', heading toward A
    B will reach A at 115.4701 years.
    Because ##\gamma = 2##, then B will reach A according to B clock for 57.7350 years.
    When B arrive, A clock will read 115.4701
    Is this true?

    How will B PICK UP signal from A?

    From where B stays and A1 there's 1 ly
    B travels at Vb
    Vb.t + ct = 1 ly, if Vb is relative to c, then c = 1
    Vb.t + t = 1
    ##t = \frac{1}{1+V_b}/\gamma##
    t = 0.27
    Is this true?

    Code (Text):
    Signal B Time  Interval
    A0     100.00  
    A1     100.27  0.27
    A2     100.54  0.27
    A212   156.81  0.27
    A213   157.07  0.27
    A214   157.34  0.27
    A215   157.61  0.27 B meets A

    How will A receive signal from B?

    B is still sending signal.
    Signal B101 is not sent 100 lys from A because B is moving.
    Signal B101 is sent at Vb.t distance
    Signal B101 should be sent at (100-0.86) ly from A. Because ##\gamma = 2##, so B101 is sent at 98.27 ly from A. Is this true?
    B01 will be picked by A after 102 years, not 101 years, because 1 year for B is 2 years for A
    So B101 will be received by A when Ta is 102 + 98.27 = 200.27. Is this true?

    Code (Text):
    Signal   A Time      Interval  
    B0       100.00      when A received B0, A still doesn't know that B has
                         already traveled.      
    B1       101.00      1.00 year A still doesn't know that B has moved
    B2       102.00      1.00
    B98      198.00
    B99      199.00      1.00   Still doesn't know
    B100     200.00      1.00   Stil doesn't know
    B101     200.27      0.27   Suddenly A begins receiving signals at 0.27
                                years interval
    B102     200.54      0.27   Is this true?
    B103     200.80      0.27

    B155     214.74      0.27
    B156     215.01      0.27
    B157     215.27      0.27 B meets A

    Is the table above is true?

    The difference with A scenario is that the instance B ignites its rocket, the time interval for signal receiving is 0.27. Whlie A has to wait for 200 years to have 0.27 interval.
    Is this true?
    Last edited: Jun 6, 2015
  2. jcsd
  3. Jun 6, 2015 #2


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    Gold Member

    You've certainly put some effort into that but I don't see the point.

    Did you read this http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_doppler.html ?

    I attach a spacetime diagram showing that approaching receivers experience an increase in frequency, receding receivers experience a lower frequency.

    The formula for frequency shift is ##\frac{f_1}{f_2}=\sqrt{\frac{1\pm v}{1\mp v}}=\gamma ( 1\pm v)## where ##v## is the relative velocity between the receiver and sender.
    The distance between sender and receiver is not in the formula.

    Attached Files:

  4. Jun 6, 2015 #3
    Thanks for the link. I'll contemplate it
  5. Jun 6, 2015 #4
    Ah, yes the distance is not in the formula. But WHEN they see the doppler effect?
    Does B experience doppler effect immediately? While A has to wait for another 100 years to experience Doppler effect?
  6. Jun 6, 2015 #5


    Staff: Mentor

    You asked basically this same question in another thread. The answer is still the same: an observer (such as A) can't see any effects from an event (such as B firing his rocket) until the light from that event reaches the observer.
  7. Jun 6, 2015 #6
    Thanks PeterDonis,
    I think I'm getting close to my answer.
    What I find interesting is that, altough motion is relative and both observer receive signal in 0.27 years interval. B experience 0.27 years before A does?
    Does it look like an asymmetry?
    I'm still reading this: http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_doppler.html
  8. Jun 6, 2015 #7


    Staff: Mentor

    I'm not sure what you mean by this.
  9. Jun 6, 2015 #8
    I mean A has to wait 100 years before receiving signal from B in 0.27 year interval.
    B immediately receiving signal from A at 0.27 years interval once he starts his engine.
  10. Jun 7, 2015 #9


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    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, B sees a change in the Doppler shift in the signal from A the moment he changes his velocity, because he is changing his velocity. A has to wait 100 yrs to see the change in Doppler shift from B, because he did not change his velocity, B did.

    Try looking at it this way: Add two more observers C and D. C stays alongside A while D initially stays along B. Now at some point, B fires it engines to slow down and reverse its relative motion with respect to A and C. D does not. Now B and D have a relative motion with respect to each other, so you would not expect them to see the same Doppler shift from A, and since D did nothing, it shouldn't see any change, so B should. In fact, you can even imagine a 5th observer E, who is moving relative to A at the same velocity as B is after he changes velocity. B and D meet E coming from the other direction and B changes velocity to match that of E when they meet so that aftarward the are flying side by side. Again, it is obvious Once B and E match speeds they should see the same Doppler shift from A, after all they are flying side by side. And since E did nothing, it should be B that sees a change in the Doppler shift.

    During all of this A and C are doing nothing and neither see any change in the Doppler shift in B until they see B math speeds with E, and it takes 100 yrs after the fact for this information to reach them.
  11. Jun 7, 2015 #10
    Thank you very much Janus for your help.
    Somehow I get the idea.
    Actually I want to know about how twins paradox occures while the universe has no frame of reference.
    Should motion is relative in space. Acceleration is not relative, this is right, isn't it.
    Shouldn't A and B see the other participant is moving while each of them see themselves respectively at rest.
    And what medium does light travel in space.
    Somehow there's asymmetry here! It might, it might not be the answer to Twins Paradox or the universe frame of reference. But there is a proof, at least for me, that even though motion is relative, both observer experience a different situation!

    Supposed A and C are in the "west".
    B and D are in the "east"
    A: Supposed B and D travels at 0.8c to the west.
    B: Supposed now B travels at 0.5c while D still travels at 0.8c to the west, can I say that?
    C: Does E travels from the east to the west at 0.8c? You said who is moving relative to A at the same velocity as B
    D: Or does E travels from the west to the east? B and D meet E coming from the other direction
    If E travels from the east to the west at 0.8c, how can D meet E?
    F: Will A and C notice that the Doppler shift for B somehow changes for a while, because between "B" and "E", B travels at 0.5c?
    Thanks for any help
  12. Jun 7, 2015 #11


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    Your questions have been answered many times but you seem to believe there is some mystery here that you can solve with an absolute reference frame.

    There is no mystery, the twins'paradox' is well understood.
  13. Jun 7, 2015 #12
    Come on...
    "The theory is not good if it can't be explained it to a six year old",
    Albert himself say that, or because I'm not six years old :smile:
  14. Jun 7, 2015 #13


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    Science Advisor

    It doesn't have to be good, just right.
  15. Jun 7, 2015 #14
    It's not mystery, I think I type asymmetry.
    But you could say that the twins paradox is a mystery, at least for me.
    Still hard to figure it out. But, I'm getting closer to my answer with Doppler effect. (or not closer?)
  16. Jun 7, 2015 #15


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    Maybe I misinterpreted what you mean. I don't mean to discourage you.

    You can calculate the twins ages using Doppler.

    I do know of any physical reason why clocks ( time itself) behaves this way.
  17. Jun 7, 2015 #16
    Just Doppler, not Doppler and Lorentz?
  18. Jun 7, 2015 #17


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    The Doppler shift formula for the ratio of frequencies is ##\gamma (c-v)## where ##v## is the relative velocity of source and receiver. There is a ##\gamma## there which is not in the pre-relativistic expression.

    In one of your many threads there is a spacetime diagram ( thank ghwells) showing the clock ticks ( which are also subject to Doppler shift) of the traveller received by the home base. You should find it and try to understand it.
  19. Jun 7, 2015 #18
    Perhaps I should understand spacetime diagram first.
    Thanks for giving me the right direction.
  20. Jun 7, 2015 #19


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    OK, here is something that could help. Look at the diagram and then play the movie. Can you work out how the movie was generated from that diagram ?

    Attached Files:

  21. Jun 7, 2015 #20
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