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

Doppler for light

  1. Jul 19, 2015 #1
    Dear PF Forum,
    I wonder about this Doppler formula.
    https://en.wikipedia.org/wiki/Doppler_effect#General
    ##f=(\frac{c+V_r}{c+V_s})f_0##
    The speed of sound is 343m/s
    Supposed S (Source) moves to the east, toward R(Receiver) at 70 m/s and R moves to the east at 40 m/s. So the formula is:
    ##f=(\frac{343-40}{343-70})f_0##
    But R will say. No, I'm at rest, it's S who moves toward me at 70-40 = 30 m/s. So the doppler factor would be:
    ##f=(\frac{343}{343-30})f_0##
    No, we can't do that. ##\frac{343-40}{343-70} \neq \frac{343}{343-30}##. Okay...
    What about blue/red shifted?
    Every where we go, the light will always travel at 300 thousands km/s according to us, right.
    What is the formula of doppler ratio, for the frequency that is received by the receiver based on the frequency sent by source?
    R is at West
    S is at East.
    If S moves to west at 0.7c and R moves to west at 0.4c?
    A: ##\frac{c-0.4}{c-0.7}## or B ##\frac{c}{c-0.3}## or even C ##\frac{c+0.3}{c}##

    If S moves to west at 0.7c and R moves to east at 0.4c?
    D: ##\frac{c-0.4}{c-0.7}## or E: ##\frac{c+1.1}{c}## or F ##\frac{c}{c-1.1}##
    I suspect F is impossible, but then I have no idea at all.
    Unless..., there is aether
     
  2. jcsd
  3. Jul 19, 2015 #2

    Mentz114

    User Avatar
    Gold Member

    The relativistic Doppler effect is reciprocal and depends on the relative velocity ##\beta=v/c##. If the receiver approaches the emitter then the freqency shift is ##f'=f\gamma(1+\beta) = f \sqrt{\frac{1+\beta}{1-\beta}}##. If they are separating then reverse the sign of ##\beta##.
     
  4. Jul 19, 2015 #3
    Ahh, again my hastily question. Of course no matter where we moves, it's the signal from the source that is counted, not from us. Why the hell that we bother with our own signal :eek:
     
  5. Jul 19, 2015 #4

    jtbell

    User Avatar

    Staff: Mentor

    Because, from the Wikipedia article that you referred to:
    This formula is for classical waves such as sound waves in air (in which case the speeds are relative to the air), ripples on a water surface (in which case the speeds are relative to the water), etc. It does not apply to light, for which you must use the relativistic Doppler effect which makes no reference to a medium, and uses only the relative speed of source and observer, and the the (invariant) speed of light:

    https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

    (Mentz beat me to it while I was struggling with the formatting)
     
  6. Jul 19, 2015 #5
    Yes, right. I should have read other answer in my previous post.
    And ##\sqrt{\frac{1+\beta}{1-\beta}}## is different from ##\frac{1+v}{1-v}##. No square root in sound doppler
     
  7. Jul 19, 2015 #6

    Mentz114

    User Avatar
    Gold Member

    Plugging in some numbers, if ##\beta=1/2## then the frequency ratio is ##\sqrt{\frac{1+1/2}{1-1/2}}=\sqrt{3}## ( better check this !) or ##1/\sqrt{3}## if ##\beta## is -1/2

    [thanks to Stephanus for pointing out my arithmetic error]
     
    Last edited: Jul 19, 2015
  8. Jul 19, 2015 #7
    I think if ##\beta = \frac{1}{2}## it's not ##\sqrt{2}##, but ##\sqrt{\frac{1.5}{0.5}} = \sqrt{3}##
    And if ##\beta = -\frac{1}{2}##, it is ##\sqrt{\frac{0.5}{1.5}} = \sqrt{\frac{1}{3}}##. Are you sure?
     
  9. Jul 19, 2015 #8

    Mentz114

    User Avatar
    Gold Member

    You're right. I have arithmetic dyslexia.
     
  10. Jul 19, 2015 #9
    One more thing.
    A moves to west at 0.4c according to rest observer
    B moves to west at 0.7c according to rest observer
    ##s = \frac{u+v}{1+uv}; 0.7 = \frac{0.4+v}{1+0.4v}; v = \frac{5}{12}##
    What is ##\beta##?
    ##0.3 = \frac{18}{60}## or ##\frac{25}{60}## There's slightly different.
    In short how to calculate ##\beta##?
    Relative to A or relative to "rest" observer that watches A moves 0.4c from him and B moves 0.7c from him.
     
  11. Jul 19, 2015 #10

    Mentz114

    User Avatar
    Gold Member

    The answer depends on whether A starts east or west of B.

    But all you need to do is find the relative velocity of A wrt B as seen in A and B coords.
     
    Last edited: Jul 19, 2015
  12. Jul 19, 2015 #11
    Thanks. It's clear for me now
     
  13. Jul 19, 2015 #12

    Mentz114

    User Avatar
    Gold Member

    I should have said that the relative velocity between two things is invariant. It will be the same in any coords.
     
  14. Jul 19, 2015 #13
    What about this:
    B in east of A
    A moves to east at 0.4c wrt Rest Observer (R)
    B moves to east at 0.3c wrt A
    R will see that A moves at (of course) 0.4c and
    But, R will see that B moves at ##\frac{0.4+0.3}{1+0.12} = 0.625c##
    Even tough we still calculate ##\beta## from "B moves to east wrt A NOT wrt R" right?
     
  15. Jul 19, 2015 #14

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    In the relativistic Doppler shift formula
    $$f_o = \sqrt{\frac{1-\beta}{1+\beta}}f_s,$$ how is ##\beta## defined? Look that up, and you'll have the answer to your question.
     
  16. Jul 19, 2015 #15

    Mentz114

    User Avatar
    Gold Member

    What is your point ?

    If two inertially moving objects have relative velocity ##\beta## then they have this in all frames. If you think you've found differently then you made a mistake.

    I can't see what the problem is.
     
  17. Jul 19, 2015 #16

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    This isn't correct. Stephanus's example clearly shows that the B's velocity relative to A as seen by A is (5/12)c whereas the observer in the rest frame would say B's velocity relative to A is (3/10)c.
     
  18. Jul 19, 2015 #17

    Mentz114

    User Avatar
    Gold Member

    OK, thank you.

    What I should have said was 'The relative velocity between A and B as measured in A and B coords is invariant'.

    (I did say that but then I crossed it out for some reason.)
     
  19. Jul 19, 2015 #18

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    To clarify, I know what you were trying to say, and you're correct that all observers would get the same value for ##\beta## to plug into the formula. It depends on what exactly one means by "relative velocity". If you take "B's velocity relative to A" to mean B's velocity in A's rest frame, you're indeed correct. On the other hand, the phrase could also mean "how much faster B is moving compared to A" as seen in a given reference frame." Now that I think about it, perhaps your interpretation is more common.
     
  20. Jul 19, 2015 #19

    Mentz114

    User Avatar
    Gold Member

    I take the definition given by Malament which is valid when the worldlines coincide and involves spatial and temporal projections (this is the first definition you give). But, as you say different interpretations are possible.
     
  21. Jul 19, 2015 #20

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    No. You are confusing separation speed versus relative speed. Separation speed is coordinate dependent, relative speed is invariant. It is defined from the dot product of two 4 velocities, which is obviously invariant. This dot product gives gamma for the relative speed. This is related to all the idiocy spouted about superluminal relative speeds. These are invariably separation speeds not relative speeds. Separation speeds have no upper limit, even in flat spacetime. See the Milne model for an example.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Doppler for light
Loading...