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


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.