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I understand this intuitively and can picture it in my head, but when I do it on paper, the result is a sign difference that I cannot understand

enter image description here

According to this diagram the wavelength = ct-vt = t(c-v) then the periodic time T = t-(vt/c) which should be t+(vt/c) instead of minus, because the second wavefront takes a longer time to pass the observer, if the error was in the direction, why should the velocity v be negative if the velocity of light in the ---> direction is positive and the observer is moving in the same direction as the light waves? Thanks

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For a stationary source the wavelength is $\lambda=cT$. The wavelength does not depend on the motion of the observer. The position of wavefront n is $x=ct-n\lambda=ct-ncT$.

The position of the observer is $x=vt $. Setting those two equal, the observer receives wavefront n at $vt=ct-ncT$ which gives $t_n=\frac{c}{c-v}nT$. Wave front 0 is received at $t_0=0$ and wavefront 1 is received at $t_1=\frac{c}{c-v}T$.

The frequency is the inverse of that so $f_o=\frac{c-v}{c}f_s$ which has the correct sign for an observer moving away from the source.

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  • $\begingroup$ but if rearrange $t_n=\frac{c}{c-v}nT$ then $T=\frac{c-v}{c}$ $\frac{t_n}{n}$ = (1-(v/c)) $\frac{t_n}{n}$ which says that the periodic time relative to the observer is t(1-(v/c)) so it takes less time for a wavefront to pass by the observer which should be the opposite? Sorry if I'm misunderstanding but I don't get it :/ $\endgroup$ – khaled014z Oct 13 '18 at 12:49
  • $\begingroup$ T is the period of the source, so that equation doesn’t tell you about the period of the observer. The period of the observer is $t_1$ $\endgroup$ – Dale Oct 13 '18 at 12:54

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