# Derivation of formulas in Doppler‘s effect in light [duplicate]

I was wondering how was the formula for Doppler effect in light i.e. $$f=\sqrt{\frac{(1-v/c)}{(1+v/c)}}f$$ was derived. I understood how it is derived for sound but I am unable to understand how it is derived for light . I read a few books but they were difficult for me to comprehend.A little hint of derivation would also do.

## 1 Answer

It is very simple. It all starts from considering a situation in which a listener is moving relative to a light emitting source. If you consider the source moving relative to the listener the result is the same because of symmetry. Now there are two ways to tackle this situation. One way is using Lorrentz transformation, consider two events in which the listener receives two consecutive wave fronts. Find the distance and time between the events in both the frames, and use them to calculate the frequency and/or wavelength of the wave in both frames.

But if you are not so comfortable with how to use Lorrentz transformation, let me show you a simple derivation.

Consider a source moving towards a listener and emitting light waves. let $$T'$$ be the time period of the light wave in the source's frame. The light waves will be compressed in the listeners frame due to the source moving towards the listener in the direction which the source is emitting the light waves. The wavelength of this wave in the listeners frame is given by:

$$(c-v)T = \lambda_l$$

where $$T$$ is the time period of the between each wave front emitted by the source in the listener's frame. We can write the wavelength of the listener as the speed of light over the frequency:

$$(c-v)T = \frac{c}{f_l}$$

substitute $$T$$ from the time dilation equation i.e: $$T = \frac{T'}{\sqrt{1-\frac{v^2}{c^2}}}$$

$$\frac{(c-v)T'}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{c}{f_l}$$

since $$T'$$ was the time period of the wave in the source's frame, it is the reciprocal of the frequency in the source's frame.

$$\frac{c-v}{f_s\sqrt{1-\frac{v^2}{c^2}}} = \frac{c}{f_l}$$

divide both sides by $$c$$ and take $$f_s$$ to the right

$$\frac{1-\frac{v}{c}}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{f_s}{f_l}$$

now the denominator on the left side of the equation is a difference of squares and can be factorized $$\frac{1-\frac{v}{c}}{\sqrt{1+\frac{v}{c}}\sqrt{1-\frac{v}{c}}} = \frac{f_s}{f_l}$$

simplifying we get:

$$\sqrt{\frac{1-\frac{v}{c}}{1+\frac{v}{c}}} = \frac{f_s}{f_l}$$

and that's how you derive it. Hope I helped :)