Deriving Relativistic Doppler Effect through length contraction

Question

Just before I explain my problem, please note that all I am looking for is for someone to point out where I got it wrong. Don't downvote this post because my reasoning is wrong - I know it is wrong, and I want to know why.

So, we have a source and an observer, moving away from each other at velocity $v$, with the source constantly emitting light in the observer's direction. In the source's frame of reference, the distance between two light wavefronts is $λ$, and the frequency is $f=\frac{c}{λ}$. In the same reference frame, the observer will receive the light signals separated by the distance $λ+vτ$, where $τ=\frac{1}{f}$. However, due to length contraction, the observer will think the source is underestimating the length of $λ$, and so, in the observer's frame of reference, the wavefronts are separated by $\frac{λ+vτ}{γ}$, with $γ$ the Lorentz Factor. The observer's frequency of light in the observer's reference frame, then, is simply $f'=\frac{γc}{λ+vτ}$ Furthermore, $τ$ is $\frac{λ}{c}$, and $f=\frac{c}{λ}$, so $f'=\frac{γ}{τ+βτ}=\frac{γf}{1+β}$, $β$ being $v/c$. Recalling the Lorentz Factor equals to $\frac{1}{\sqrt{1-β^2}}$, this gives: $f'=\frac{1}{(1-β)^\frac{1}{2}(1+β)^\frac{3}{2}}$, which obviously isn't right. Can someone tell me what is the mistake that I have done?

Answer 1

For concreteness (and so that we can more easily compute and visualize quantities), let's analyze from a frame where the source is at rest and the receiver moves with velocity $v=(3/5)c$.
Further, let the source period be $\tau=10$
and so (in units where $c=1$) the source wavelength is $\lambda=c\tau=10$.

Let's visualize this on a spacetime diagram on drawn on rotated graph paper.

The source and receiver meet briefly at event O.
So, after O, they are receding from each other.
With velocity $v=\frac{PQ}{OP}=(3/5)c$,
the time-dilation factor is $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}=\frac{OP}{OQ}=(5/4)$
and the Bondi $k$-factor $k=\frac{OR}{OT}=\sqrt{\frac{1+(v/c)}{1-(v/c)}}=2$.

Suppose a light-signal was emitted at the meeting event O, and then again at event $T$, one period ($\tau_{source}=10$) later in the source frame.
So, the source wavelength is the distance between wavefronts in the source frame (that is, the "separation between two lightlike signal-lines" in the source frame). From the diagram, $\lambda_{source}=c\tau=10$, as expected.

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I will address specifics of your approach after the diagram. But first, I will make an important comment on wavelengths and length-contraction.

• In the source frame, imagine a ruler at rest with a marking at $x=10$. Interpret this as "where the source says the previous wavefront is located when the source emits the next signal". Note that this marking has a worldline parallel to the source.
  Although the "separation between these timelike-worldlines" is equal to $\lambda_{source}$ in the source frame, these timelike-worldlines are only indirectly related to the source-wavelength [which is the "separation between the lightlike signal-lines"].
  In the receiver-frame the "separation between these timelike-worldlines" is given by $OX=\frac{\lambda}{\gamma}=\frac{10}{5/4}=8$, in accordance with length-contraction.
  However, this is not the wavelength observed by the receiver---
  the observed-wavelength is the "separation between the lightlike signal-lines" given by $OW=20$ in the receiver frame. ($OW=k(O\lambda_{source}=(2)(10)=20.$).
  The point is: the observed-wavelength (separation between lightlike signal-lines)
  doesn't directly involve length-contraction (involving parallel timelike-lines).

Now to your approach...
I believe you next reference event $R$, when the receiver receives the second signal after meeting at $O$. To determine $R$'s coordinates in the source frame, find the intersection of the receiver's worldline through $O$ ($x=vt$) with the forward light-signal emitted at event $T$ ($x=c(t-\tau)$).
I get $t_R=\frac{1}{1-\beta}\lambda/c=25$ and $x_R=\frac{\beta}{1-\beta}\lambda=15$.
I'm not sure where your "$\lambda+v\tau$"$=(10)+(\frac{3}{5})(10)=16$ comes from.

Note that $x_R$ is the distance to the source in the source frame when the receiver observes the second signal. This is not the separation between wavefronts. In the frame of the receiver, when the reception occurs, the receiver says that she is $\displaystyle\frac{x_{R}}{\gamma}=\frac{15}{5/4}=12$ units away from the source [which, again, is not the observed wavelength $OW=20$].
So, again, length-contraction doesn't seem to help find the observed-wavelength.

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To find the observed-wavelength $OW$, use the observed-period $OR$.
By similar triangles, time-dilation factor is $\gamma=\frac{OP}{OQ}=\frac{25}{OR}$ so that $\tau_{obs}=OR=(25)/\gamma=(25)/(5/4)=20$. Then, $\lambda_{obs}=c\tau_{obs}=20$, which is $OW$.
Symbolically, $$ \begin{align*} \lambda_{obs}=c\tau_{obs}=c\frac{t_R}{\gamma} &=c\frac{\frac{1}{1-\beta}\lambda/c}{\gamma}\\ &=\frac{1}{1-\beta}(\sqrt{1-\beta^2})\lambda\\ &=\sqrt{\frac{1+\beta}{1-\beta}}\lambda_{source}=(2)(10)=20, \end{align*} $$ which, although is a "length", features the Doppler factor (the Bondi $k$-factor), not the length-contraction factor.
Again, distinguish the "separation between lightlike signal-lines" from the "separation between parallel timelike-lines".
Similarly, $$\tau_{obs}=\sqrt{\frac{1+\beta}{1-\beta}}\tau_{source}=(2)(10)=20,$$ which is longer than the source period of $\tau_{source}=10$, as expected for a receding receiver.

Answer 2

If light was a stream of particles with a small mass, moving almost at c in all normal frames, then we could use the length contraction formula without any problems.

Luckily we can never be sure that photon does not have some small mass.

So if we pick a speed close enough to c, we get answers that current experiments can't prove wrong.

So let's see how that could be done. The rest-length of a light pulse is gamma times its length when it's moving.

A very special observer moving with the light pulse would see all normal observer's measuring sticks being very short. The special observer may think the normal observers with their short measuring sticks would measure the light pulse to be even longer than it is in the special observer's frame. But that's just wrong. Length of an moving object is not measured with a measuring stick. How exactly would we measure the length of a moving object using a measuring stick?

So, first we go to the frame of the pulse, then we ask what is the length of the pulse in two normal frames. That's a simple calculation of speeds of objects and lengths of said objects. I mean first we calculate the speed of the pulse in the two normal frames, and then we calculate the Lorentz-contractions at those two speeds.