# Doppler effect in light (Observer moving away from source)

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

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

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.
• 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 :/ – khaled014z Oct 13 '18 at 12:49
• 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 $t_1$ – Dale Oct 13 '18 at 12:54