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Measuring the doppler effect.

  1. Mar 25, 2006 #1
    I have a simple question about a relativistic Doppler effect problem.
    To keep the math simple, I have intentionally left out the labels, you can view the speed of light as 300 miles per seconds, or 300 meters per second, it isn’t important.

    Speed of light = 300 per second

    Wavelength of light = 4

    Frequency of light = 75 cycles per second

    Formula: Wavelength (4) * Frequency (75) = Speed (300)

    If an observer travels towards the source at a constant speed of 2 per second, he would expect the relative speed to be 302 per second. The observer’s speed has caused the frequency to increase. He now measures the frequency to be 75.5 cycles per second. If he divides the speed (302) by the new frequency (75.5) he calculates the wavelength to be 4.

    Knowing the speed of light has to remain constant, he divides the speed of light by the measured frequency and now calculates the wavelength to be 3.973509934

    Is this right?

    When I use the following formula…

    (C-Vs)T / Sq.Root of (1-(Vs Squared /C Squared))

    C = 300
    Vs = 2
    T = 1/75

    …I calculate the wavelength to be 3.973421633

    Why do I calculate 2 different answers, what have I done wrong?
    It seems to me that 3.973509934 has to be correct in order to multiply it by the frequency and get the speed of light.
  2. jcsd
  3. Mar 25, 2006 #2


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    But note how you are contradicting yourself here. You know the speed of light is constant, so the light wave will come at you at 300 Mm/s no matter what. And yet you just added velocities like 300+2=302.
    Your velocity relative to the light wave really is just 300 Mm/s. period.

    If the source is stationary, then you've calculated the frequency 75 Hz correctly. But if the source is approaching you, the shisft in wavelength occurs not because the speed of the wave changes relative to you. The sources follows the wave which causes the wavelength to be shorter. Also, the source clock is ticking more slowly than in its rest frame which actually causes an (small) effect towards a larger wavelength.
    Last edited: Mar 25, 2006
  4. Mar 25, 2006 #3

    Did you read the entire post?
    Do you know why I calculate 2 wavelengths?
  5. Mar 25, 2006 #4


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    When the observer is moving towards the source, the observed frequency is affected in two ways.

    First, the frequency tends to increase because, from the observer's point of view, the source is moving towards him and therefore emits wavefronts at different locations. Each successive wavefront has less distance to travel, so it arrives at the observer earlier than if the source were stationary (in the observer's reference frame).

    Second, the frequency tends to decrease because the source is time-dilated with respect to the observer. That is, in the moving observer's reference frame, the source emits wavefronts less frequently.

    The derivation of the relativistic Doppler equation takes both of these effects into account. You did not take the second effect into account in your first calculation.
  6. Mar 25, 2006 #5


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    Yes, I've read the whole post. Your second answer is correct. The first is wrong because you used a wrong analysis as explained in my first post. The shifted frequency is not 302/4=75.5.
    First of all because the speed of the wave is not 302 (it's 300). Second because the wavelength is not 4, but:
    [tex]\lambda'=4\cdot \sqrt{\frac{298}{302}}[/tex]
    =your second answer.

    The measured frequency is [itex]c/\lambda'[/itex] which is slightly larger than 75.5 Hz.
  7. Mar 25, 2006 #6
    If I took this second effect into account, would not it make the wavelength in my first calculation longer since the frequency is decreasing?
  8. Mar 25, 2006 #7


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    Time dilation tends to make the wavelength longer, yes, but your first calculation of the wavelength is already messed up because you used the wrong speed.

    In the following discussion, I work in the reference frame in which the observer is at rest and the source is moving towards him at a speed of 2 units/sec.

    First, find the time-dilation factor:

    [tex]\gamma = \frac {1} {\sqrt {1 - v^2 / c^2}} = \frac {1} {\sqrt {1 - 2^2 / 300^2}} = 1.00002222[/tex]

    Time dilation decreases the frequency, so to get the frequency in the observer's reference frame, [itex]75 / \gamma[/itex] = 74.99833331. If we used this frequency to get the wavelength, we would get [itex]\lambda_0[/itex] = 300 / 74.99833331 = 4.00008889 units.

    But the source is moving towards the observer. It emits wavefronts every 1 / 74.99833331 = 0.01333363 sec. During that time, the source moves a distance [itex]\Delta \lambda[/itex] = (2 units/sec)(0.01333363 sec) = 0.02666726 units. This reduces the wavelength from the value calculated in the preceding paragraph, so the actual wavelength is [itex]\lambda_0 - \Delta \lambda[/itex] = 3.97342163 units, which agrees with the relativistic Doppler formula.

    In effect, what we have above is a derivation of the relativistic Doppler formula! Just replace the numbers with appropriate symbols, combine all the steps into a single equation, and rearrange it if necessary.
    Last edited: Mar 25, 2006
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