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Relativistic wavelength shift for a stationary source and an receding detector is given by

$\frac{\lambda_s}{\lambda_d}=\sqrt\frac{{1-\beta}}{{1+\beta}}=\frac{{1-\beta}}{\sqrt{1-\beta^2}}=\gamma(1-\beta)$

Now, the $\gamma$ can be explained as a result of Lorentz length contraction. In other words, distance between two points in the source frame will be Lorentz contracted when seen by a moving detector.

But how is the $(1-\beta)$ factor explained by the source?

In other words, the source sees the detector detect a higher frequency due to time dilation as well as due to its motion. It also sees the wavelength to be Lorentz contracted in the frame of the detector. How does the source explain the additional factor $(1-\beta)$ needed to explain the same speed of light measured by the detector?

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  • $\begingroup$ Related : Deriving relativistic Doppler shift in terms of wavelength. $\endgroup$
    – Frobenius
    Dec 1 '19 at 12:46
  • $\begingroup$ @Frobenius Thank you sir! I have seen that question. The proof is correct, however, it derives the expression for wavelength indirectly by deriving the expression for frequency (using time dilation) and the constancy of speed of light. I am asking something different. I understand the expression can be derived that way and a part of it can be explained by length contraction. However, what is not clear to me is how is the remaining factor $(1-\beta)$ explained. $\endgroup$ Dec 1 '19 at 12:52
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    $\begingroup$ What are you expecting in terms of “explained”? Clearly you agree with the maths, so what more do you think there is here? $\endgroup$
    – Kyle Kanos
    Dec 1 '19 at 13:15
  • $\begingroup$ @KyleKanos Thank you for helping me make the question clearer. I have edited the question to address your comment. $\endgroup$ Dec 1 '19 at 13:20
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    $\begingroup$ Make use of the Lorentz transformation to solve safely problems like this one. It's dangerous to use length contraction and/or time dilation to solve problems. Make use of them to interpret the results provided by Lorentz transformations. $\endgroup$
    – Frobenius
    Dec 1 '19 at 13:41
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How does the source explain the additional factor (1−β) needed to explain the same speed of light measured by the detector?

Let us say that the detector is a physicist with a clock, who measures the time it takes for one wave to pass her, then calculates the length of the wave to be: time * speed of light.

The source is an observer with the light source. This observer observes what the physicist does. The question was what does that observer observe, right?

So the observer notes that the clock is slow, which decreases the measured time by gamma.

Then he notes that the passing speed is speed of light - speed of physicist. Passing time is proportional to passing speed, and the length of the passing thing.

So the passing time t according to the observer is: length of the passer / passing speed = L / (1 - Beta).

The physicist measures passing time: t'= t / gamma =

(L / (1 - Beta)) / gamma

The length of the passing thing according to the physicist is the speed of that thing multiplied by t, and we have set c to 1.

So according to this derivation the correct formula is:

L' = (L / (1 - Beta)) / gamma

"Passing speed" is the same thing as "closing speed", see Wikipedia if not familiar with that latter term.

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