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A laser, when stationary for observer $S$, emits light of frequency $f_o$ and wavelength $\lambda_o$. Another observer $L$ is sitting on the laser and also reports the same frequency and wavelength. $L$ is also carrying a stick of length $l_o$ in his hand.

Let's say the laser is now set in motion, moving with a velocity $v$ towards the stationary observer $S$. $S$ now sees that the harmonic oscillator generating the light wave moves in a frequency $f_o\gamma$ which thus should also be the frequency of the light wave observed by him.

The wavelength of light is now observed by $S$ to:

$$\lambda= \frac{ c-v}{f} = \frac{ c-v}{f_o\gamma} = \frac{ c-v}{c\gamma}\lambda_o$$

The length of the stick carried by $L$ is now observed by $S$ to be: $$l=\frac {l_o}{\gamma}$$

Now, imagine that both the wavelength of the light and the length of the stick observed by $S$ when $L$ is moving are actually the same.

$$l=\lambda$$

$$\frac {l_o}{\gamma} = \frac{ c-v}{c\gamma}\lambda_o$$ $$ l_o = \frac{ c-v}{c}\lambda_o$$ $$l_o < \lambda_o$$ This would mean that $S$ observed that the length of the stick was less than the wavelength of the light when $L$ was stationary but when $L$ is in motion, they both contract to the same length.

$$l=\lambda $$ $$l_o < \lambda_o$$

Why would such two different lengths contract to be exactly the same? Or in other words, why would two different lengths face different degrees of contraction when both move with the same velocity?

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2 Answers 2

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While $\lambda= \frac{ c-v}{c}\gamma\lambda_o$ is correct [where $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}$],
$\lambda(laser)= \frac{ c-v}{c}\gamma^2\lambda_o$ is incorrect since $\lambda(laser)\neq \gamma \lambda$.
The correct expression is $\lambda(laser)= k \lambda$, where $k=\sqrt{\frac{1+(v/c)}{1-(v/c)}}=\frac{c+v}{c}\gamma$="the Doppler factor".
Then $\lambda(laser)= k \lambda= k( \frac{ c-v}{c}\gamma\lambda_o )=\frac{c^2-v^2}{c^2}\gamma^2\lambda_o=\lambda_o$, as expected.

The length contraction formula (using $\gamma$) involves the proper-length between two parallel timelike worldlines (like the ends of a ruler).
However, the wavelength $\lambda$ is the apparent-length between two parallel lightlike "wavefront-worldlines"... and this involves the Doppler factor $k$.

Consider the following spacetime diagram on rotated graph paper.
This diagram is a little involved... so I will lead you through it.

Doppler effect - spacetime diagram on rotated graph paper

Alice is the receiver at rest.
Bob is the moving source, with velocity $v=(3/5)c$. [See triangle OPQ. $v=\frac{PQ}{OP}$ and $\gamma=\frac{OP}{OQ}$ and $k=(1+\frac{v}{c})\gamma$.] So, $\gamma=(5/4)$ and $k=2$.

Suppose Bob carries a ruler of proper-length $L=10$ (along OY, between worldlines OQ and its parallel XY) and happens to emit light with source-wavelength $\lambda_0=10$ (along OY, between the forward light rays emitted by Bob at events E and O).

Alice observes

  • Bob's ruler along OX to be $L_{Alice}=\frac{L}{\gamma}=\frac{(10)}{(5/4)}=8$ (length contraction).
  • the distance between wavefronts along OW as $\lambda_{rec, Alice}=\frac{1}{k}\lambda_o=\frac{1}{(2)}(10)=5$ (Doppler effect).

For completeness, I use the diagram to derive your first equation.

  • The source wavelength in the source frame is $\lambda_o=OY=10$
  • The source period in the source frame is $T_o=EO=10$
  • The source period in the receiver frame is $T_{o,Alice}=FO=EG=\gamma EO=(\frac{5}{4})10=12.5$.
  • In elapsed time $T_{o,Alice}$, Alice observes
    • the light ray from event E move a distance $GW=c\ EG=(1)(12.5)=12.5$.
    • the source move a distance $GO=EF=v\ EG=(\frac{3}{5})(12.5)=7.5$
      (to event O, when it sends out the next light ray)
    • the wavelength as $\lambda=\lambda_{rec}=OW=(GW-GO)=(c-v)\gamma T_0=(1-\frac{3}{5})(\frac{5}{4})(10)=5$
    • the receiver period $T_{rec,Alice}=NO=\frac{OW}{c}=5$.

Thus, $\lambda=\lambda_{rec}=OW=(c-v)\gamma T_0=(c-v)\gamma \frac{\lambda_0}{c}=\frac{c-v}{c}\gamma OY=\frac{1}{k} OY$

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The person not moving relative to the laser will observe a wavelength $\lambda_{\rm o}$ whatever wavelength the observer moving relative to the laser measures.

You cannot then transform from the frame which is moving relative to the laser to the frame of the laser as there is no source of light in the frame moving relative to the laser.

The HyperPhysics article Relativistic Doppler Shift shows which formulae you should be using or the low speed approximation.

A formula like $\lambda(laser)= \frac{ c-v}{c}\lambda_o$ which is derived for mechanical waves should not be used and "So, the person on the laser sees the wavelength to decrease." is not true.

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