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Considering that $S$ (fixed) and $S'$ are inertial references which coincides when $t=t'=0$ and their coordinates $(ct,x,y,z)$ and $(ct',x',y',z')$ are related by Lorentz's transformations, calculate what are the frequences $f$ and $f'$ of a moving monocromatic light source (velocity $\vec{v}=v\hat z$ ) in $S$ and $S'$, respectively ( this light is at rest in relation with $S'$). To do this calculation, assume that, from someone's point of view who is in $S'$, a wave crest of this light is sent at the instant $t'_1$ while the next is sent at the instant $t'_2=t'_1+\delta t'$.

What I have tried was replacing the position $z=vt$ in Lorentz's transformation and obtain a relation between $t'$ and $t$ as a function of $\gamma$. But I don't really know how to link this with $t'_2=t'_1+\delta t'$.

If anyone could help me with that!

Qmechanic
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Aaron
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1 Answers1

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The way I should solve this is by direct use of the Lorentz transformation: $$ct = \gamma (ct' + z' v/c), \quad x = x', \quad y = y', \quad z = \gamma (z' + t'v),$$ inversely, $$ct' = \gamma (ct - z v/c), \quad x' = x, \quad y' = y, \quad z' = \gamma (z - tv),$$

In $S'$ the light wave is described by $$\phi = \sin 2\pi f' (t' - z'/c).$$

Substituting $t'$ and $z'$ with the expressions above gives $$\phi = \sin 2\pi f' (\gamma(t-zv/c^2) - \gamma(z-tv)/c) = \sin 2\pi f' \gamma (1+v/c) (t-z/c).$$

We see that the frequency in $S$ is given by $f = f' \gamma (1+v/c) = \sqrt{\frac{1+v/c}{1-v/c}} f'$.

md2perpe
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  • Nice simple answer. Would you consider editing to define phi for S and phi dash for S' together with the usual connotations just to make it even clearer? – m4r35n357 Dec 02 '18 at 09:58
  • @m4r35n357. I'm not sure how you mean. – md2perpe Dec 02 '18 at 18:45
  • Phi is defined against dashed quantities, S' f' z'. You ould call that phi dash. The define an undashed phi with S, f, z. No big deal, I still like the answer! – m4r35n357 Dec 02 '18 at 23:17
  • @m4r35n357. But then I need to explain why $\phi' = \phi.$ As of now, I don't use different symbols since the value is assumed to be the same; it's just the functional form that differs between systems. – md2perpe Dec 03 '18 at 06:38