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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.

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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.

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