# Deriving Doppler effect in observer's frame

I am attempting to find the Doppler factor from the frame of the observer. In the frame of the observer, the source is moving at velocity $$v$$ towards the observer. Since the source emits at time period $$T$$ in its frame, it essentially emits at an interval of $$\gamma T$$ in observer's frame.

So $$\lambda_{\mathrm{effective}}=\lambda-v\gamma T$$ $$v$$ is the velocity.

This gives an incorrect result for frequency.

One can use the Lorentz transformations for energy and momentum to derive the Doppler effect.

For photons, $$E' = p'c$$. Assuming our inertial frames coincide at time $$t=0$$, and the source $$S'$$ is approaching $$S$$ at a constant velocity $$v$$ along x-axis, it is reasonable to write $$p' = p'_{x}$$.

Now, Lorentz transformation for energy tells us,

\begin{align} E &= \gamma(E'+vp_{x^{'}}) \\ &= \gamma \left(E' + \frac{v}{c}E'\right)\\ &= \gamma E'(1+\frac{v}{c})\\ \end{align}

Remember that $$\gamma = \displaystyle\frac{1}{\sqrt{1-\displaystyle\frac{v^2}{c^2}}}$$.

$$E= E'\frac{\big(1+\frac{v}{c}\big)^\frac{1}{2}}{\big(1-\frac{v}{c}\big)^\frac{1}{2}}$$

Also notice that, $$E \ \alpha \ \nu$$.

Hence, $$\nu = \nu'\frac{\big(1+\frac{v}{c}\big)^\frac{1}{2}}{\big(1-\frac{v}{c}\big)^\frac{1}{2}} > \nu \text{ for } v>0$$.

Since the source is approaching the observer, we have observed a higher frequency than the original.