A Neural Radiance Field (NeRF) use a volumetric rendering function to render the color for a camera ray:

\[\int_0^\infty T(t) \sigma(t) c(t) \,dt \qquad \qquad T(t) = \exp\left(-\int_0^t \sigma(s) \,ds\right)\]

Here, \(\sigma(t)\) and \(c(t)\) are volume density and RGB color properties of the point \(t\) distance into the ray. \(T(t)\) is the accumulated transmittance and can be interpreted intuitively as the amount of light from point \(t\) which will reach the camera through the volume that exists in between.

Note that taking this integral numerically involves making calculations at discrete points, say \(t_1, t_2 \ldots t_n\). However, if we treat the transmittance as a discrete phenomena along our ray, we might expect something like the following as our accumulated transmittance:

\[T(t_i) = \prod_{j = 1}^{i - 1} 1 - \sigma(t_j)\]

Here, we are choosing to interpret each \(\sigma(t_j)\) as the probability a ray passes through at the discreet point \(t_j\). This expression does not have the exponential and negative terms in the continuous version of \(T\). Where do these come from?

For me, it is intuitive to start with the accumulated radiance \(T\), reparameterized as \(T(t_1, t_2)\), representing the probability that a particle of light passes through the space between \(t_1\) and \(t_2\). This should satisfy \(T(t_1, t_2) T(t_2, t_3) = T(t_1, t_3)\) for any \(t_1, t_2, t_3\) as the conjunction of two independent probabilities. We also adopt \(T(t) = T(0, t)\) as is consistent with our earlier usage of \(T\). Then, the derivative of this function can be written as follows:

\[\begin{align} dT(t) &= \lim_{\epsilon \rightarrow 0} \frac {T(0, t + \epsilon) - T(0, t)} {\epsilon} \\ &= \lim_{\epsilon \rightarrow 0} \frac {T(0, t) T(t, t + \epsilon) - T(0, t)} {\epsilon}\\ &= - T(0, t) \lim_{\epsilon \rightarrow 0} \frac {1 - T(t, t + \epsilon)} {\epsilon} \end{align}\]

Here we define \(\sigma(t) = \lim_{\epsilon \rightarrow 0}(1 - T(t, t + \epsilon)) / \epsilon\). This gives us the following initial value problem for \(T\):

\[dT(t) = - \sigma(t) T(t) \qquad T(0) = 1\]

This differential equation can be solved algebraically:

\[\begin{align} dT(t) &= -\sigma(t) T(t) \\ \frac {dT(t)} {T(t)} &= -\sigma(t) \\ \int_0^\tau\frac {dT(t)} {T(t)} \,dt &= -\int_0^\tau \sigma(t) \,dt \\ \log(T(\tau)) - \log(T(0)) &= - \int_0^\tau \sigma(t) \,dt \\ T(\tau) &= \exp\left(-\int_0^\tau \sigma(t) \,dt\right) \end{align}\]

This derives the relationship between \(T\) and \(\sigma\) and in particular defines \(\sigma\) in relation to \(T\) as \(\lim_{\epsilon \rightarrow 0} T(t, t + \epsilon) / \epsilon\). We note in particular that this limit is not bounded so \(\sigma(t)\) takes values in \(\mathbb{R}^+\) and should not be interpreted as a probability or proportion.