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Doppler shift of two ships

  1. Jan 7, 2016 #1

    Buckethead

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    Imagine a mother ship traveling away from earth at .5c and a shuttle takes off from the mother ship in the same direction of travel at .5c relative to the mother ship. The shuttle fires a blue laser (400nm) to a transparent screen mounted on top of the mother ship as it is traveling away. Due to the Doppler shift of

    [itex]1-v/c[/itex] = .5 for v=.5c

    the wavelength approximately doubles at this speed causing the blue light to light the screen with a deep red color (around 800nm) . A telescope on earth views the screen and since the screen is moving away from the earth at .5c it gets red shifted again to an infrared color of around 1600nm. Between the shuttle and the mother ship the speed is fairly non-relativistic (.5c) as is the speed between the ship and the Earth so I'm just using the simpler form of the Doppler shift to calculate the wavelengths.

    My questions are:

    1) If the above assumptions about calculating the total Doppler shift is wrong, why?

    2) What is the apparent speed of the shuttle as seen from Earth and how do I calculate it?

    3) If I use just the apparent speed of the shuttle as seen from Earth in the Doppler formula will the answer differ from my calculation above (assuming my assumptions were correct of course).

    Thanks
     
  2. jcsd
  3. Jan 7, 2016 #2

    Nugatory

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    You use the formula for relativistic velocity addition (google will find it for you) and it will be .8c

    That's not "fairly non-relativistic".... Something like .05c might be.
     
  4. Jan 7, 2016 #3

    marcus

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    Also there's a relativistic doppler shift formula.
    http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/reldop2.html

    If something is approaching at .5c you take the square root of

    (1+β)/(1-β) =1.5/0.5 = 3

    so the frequency you receive is √3 times the frequency sent.

    In your case the little ship is receding at .5c so there is a minus sign on the beta
    (1+β)/(1-β) =0.5/1.5 = 1/3
    so the ratio is the square root of 1/3
     
    Last edited: Jan 7, 2016
  5. Jan 7, 2016 #4

    Mister T

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    If you want the wavelength to double you need speeds of 0.6, not 0.5. The Doppler factor is ##\sqrt{\frac{1+0.6}{1-0.6}}=2##.

    The shuttle's speed relative to Earth is ##\frac{0.6+0.6}{1+(0.6)(0.6)}=\frac{1.2}{1.36} \approx 0.88##.

    The Doppler factor at this speed is ##\sqrt{\frac{1+\frac{1.2}{1.36}}{1-\frac{1.2}{1.36}}}=4##.
     
  6. Jan 7, 2016 #5

    pervect

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    If you use the relativistic formula, https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

    the wavelength increases by sqrt(3) for ##\beta = 0.5## using the formula
    $$\sqrt{\frac{1+\beta}{1-\beta}}$$

    where ##\beta = v/c##

    I double-checked the sign of ##\beta##, according to Wiki they look right. In words, the relativistic time dilation factor makes the relativistic doppler shift have a lower magnitude (sqrt(3) ##\approx## 1.32## than the non-relativistic doppler shift.

    The doppler shifts multiply, so the factor of sqrt(3) becomes a 3:1 doppler shift, making the wavelignth 1200nm.


    To get the velocity of the ship relative to earth, use the relativistic velocity addition formula:

    $${\beta}_{tot} = \frac{\beta_1 + \beta_2}{1 + \beta_1 \beta_2} = \frac{1}{1.25} = .8$$

    Substituting ##\beta=.8## into the relativistic doppler formula gives a doppler shift of 3, as expected.
     
  7. Jan 7, 2016 #6

    Buckethead

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    Fantastic! Thank you all very much for your help. These were exactly the answers I was looking for. This site (and the people in it) is amazing!
     
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