1. Energy Density ($u$):
For a stationary source, energy spreads spherically. At distance $r$: $$ u = \frac{I_{static}}{c} = \frac{P}{4\pi r^2 c} $$ 2. Relative Speed ($v_{rel}$):
The detector moves towards the source against the wave direction. $$ v_{rel} = c + v_d $$ 3. Intensity ($I$):
$$ I_a = u \cdot (c + v_d) = \frac{P}{4\pi r^2 c}(c + v_d) $$ $$ I_a = \frac{P}{4\pi r^2} \left( 1 + \frac{v_d}{c} \right) $$

1. Geometry of Emission:
Let $t$ be the time taken for the wave to travel from emission point to detector.
• Radius of wavefront: $R = c t$.
• Distance source moved: $d = v_s t$.
• Current distance (given): $r = R – d = (c – v_s)t$.
From this, we find the “Emission Radius” $R$ in terms of current distance $r$: $$ t = \frac{r}{c – v_s} \implies R = c \left( \frac{r}{c – v_s} \right) = \frac{r}{1 – v_s/c} $$ 2. Energy Density ($u$):
The energy $P$ is spread over a sphere of radius $R$ (not $r$). Additionally, the “shell” of energy is compressed by the Doppler factor $(1 – v_s/c)$. The derivation leads to: $$ u = \frac{P (1 – v_s/c)}{4\pi r^2 c} $$ *(Why? Because $u \propto \frac{1}{R^2} \times \text{Compression}$. The $1/R^2$ term dominates the increase from compression.)* 3. Intensity ($I$):
Detector is stationary, so $v_{rel} = c$. $$ I_b = u \cdot c = \frac{P (1 – v_s/c)}{4\pi r^2 c} \cdot c $$ $$ I_b = \frac{P}{4\pi r^2} \left( 1 – \frac{v_s}{c} \right) $$

We combine the effects found in (a) and (b).
1. The moving source establishes an energy density field $u$ described in part (b).
2. The moving detector intercepts this field with relative velocity $v_{rel}$ described in part (a). $$ I_c = u_{source\_moving} \cdot (c + v_d) $$ $$ I_c = \left[ \frac{P}{4\pi r^2 c} \left( 1 – \frac{v_s}{c} \right) \right] \cdot (c + v_d) $$ $$ I_c = \frac{P}{4\pi r^2} \left( 1 – \frac{v_s}{c} \right) \left( 1 + \frac{v_d}{c} \right) $$