Neural Radiance Fields (NeRF)

A camera ray passes through three samples: a half-transparent red, a half-transparent green, then a fully opaque blue surface behind them. Drag the red sample's opacity. Push it toward 1 and it steals almost all the color for itself — nothing behind it can show through an opaque sample. Drop it toward 0 and green, then blue, get their turn.

A NeRF is a tiny MLP: feed it a 3D point and a viewing direction, it hands back a density $\sigma$ (how "solid" that point is) and a color. Marching a ray through the scene takes samples along it, turns each one's density into an opacity via the step size $\delta$ between samples, and composites them exactly like alpha-blending layers in image editors — front-to-back:

• $\sigma_i$ — the density the MLP predicts at sample $i$; $\delta_i$ — the step size (world-space gap) to the next sample.
• $\alpha_i$ — sample $i$'s own opacity: denser samples, or longer steps through them, block more light.
• $T_i$ — transmittance: the fraction of light that survives every nearer sample before reaching sample $i$.
• $c_i$ — the RGB color the MLP predicts at sample $i$.

1. Sample 1 (red): alpha=0.5

   Full transmittance so far ($T_1 = 1$), so its contribution is $1 \times 0.5 = 0.5$ — half the pixel is already claimed before any other sample is even considered.

2. Sample 2 (green): alpha=0.5

   Only $T_2 = 1 - 0.5 = 0.5$ of the pixel is left to compete for. Contribution $= 0.5 \times 0.5 = 0.25$.

3. Sample 3 (blue, fully opaque): alpha=1

   $T_3 = 0.5 \times (1-0.5) = 0.25$ remains. Contribution $= 0.25 \times 1 = 0.25$. Total: $0.5 + 0.25 + 0.25 = 1$ — the ray terminated on something fully opaque, so every bit of the pixel got assigned.

Now drag density $\sigma$, not raw opacity — the actual quantity a NeRF's MLP predicts. Watch how quickly $\alpha = 1 - e^{-\sigma\delta}$ saturates toward 1: by $\sigma \approx 3$ the sample is already almost fully opaque, which is why NeRF densities are usually shown on a log-ish scale in papers.

Two more density values for sample 1 (step size $\delta = 1$, samples 2-3 unchanged):

1. sigma = 2 (moderately dense)

   $\alpha_1 = 1 - e^{-2} \approx 0.8647$. $T_2 = 1 - 0.8647 = 0.1353$, so sample 3's final weight drops to $T_2 \times (1-0.5) \times 1 \approx 0.0677$ — a dense first sample starves everything behind it.

2. sigma = 0.1 (nearly transparent)

   $\alpha_1 = 1 - e^{-0.1} \approx 0.0952$. $T_2 \approx 0.9048$, sample 3's weight becomes $0.9048 \times 0.5 \times 1 \approx 0.4524$ — almost the same as if sample 1 weren't there at all.