<p>It's a version of the standard derivation of the exponential function from its differential equation, by separation of variables.</p>
<p>To review that standard derivation: suppose we want a function <span class="math-container">$y(x)$</span> obeying the differential equation <span class="math-container">$\mathrm{d}y/\mathrm{d}x = -ky$</span>, for some constant <span class="math-container">$k$</span>. Then we can solve the equation as follows:
<span class="math-container">$$
\begin{aligned}
\mathrm{d}y &amp;= -ky \, \mathrm{d}x \\
\frac{\mathrm{d}y}{y} &amp;= -k \, \mathrm{d}x \\
\int \frac{\mathrm{d}y}{y} &amp;= -\int k \, \mathrm{d}x \\
\ln y &amp;= -kx \\
y &amp;= e^{-kx}
\end{aligned}
$$</span>
(there should actually be some constants of integration in there, but I left them out since they're not important for this answer.)</p>
<p>Now, suppose we generalize and make the constant <span class="math-container">$k$</span> into a function <span class="math-container">$k(x)$</span>. Then we can repeat this derivation, but we will not be able to do the integral on the right side, since <span class="math-container">$k(x)$</span> is unspecified. The result will then be:
<span class="math-container">$$
y = e^{-\int k(x) \, \mathrm{d}x}
$$</span></p>
<p>The derivation for the transmittance along a ray is just the same, but with some variables renamed: <span class="math-container">$y$</span> becomes <span class="math-container">$L$</span>, <span class="math-container">$x$</span> is now the parameter <span class="math-container">$t$</span> along the ray, and <span class="math-container">$k(x)$</span> is called <span class="math-container">$\sigma_\mathrm{t}(\mathrm{p})$</span>.</p>

CC BY-SA 4.0
2021-11-13T21:24:23.167Z
2021-11-13T21:24:23.167Z
Nathan Reed
<p><a href="https://i.sstatic.net/6utiT.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/6utiT.png" alt="beam transmittance" /></a></p>
<p>In pbrt v3, the book gives this description of beam transmittance, but I don't know how to solve the differential equation like it says to get <strong>Tr</strong> , can someone please tell me how to solve the differential equation? Thanks a lot.</p>

<rendering><algorithm><physically-based>
How is the beam transmittance calculated in PBRT V3?
Wenjian Zhou
# How is the beam transmittance calculated in PBRT V3?

(Asked by: username_0 on 2021-11-11)

[![beam transmittance](https://i.sstatic.net/6utiT.png)](https://i.sstatic.net/6utiT.png)

In pbrt v3, the book gives this description of beam transmittance, but I don't know how to solve the differential equation like it says to get **Tr** , can someone please tell me how to solve the differential equation? Thanks a lot.

**Tags:** rendering, algorithm, physically-based


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## 1 answers

### Answer by username_1 on 2021-11-13. Score: 1 | Accepted answer

It's a version of the standard derivation of the exponential function from its differential equation, by separation of variables.

To review that standard derivation: suppose we want a function $y(x)$ obeying the differential equation $\mathrm{d}y/\mathrm{d}x = -ky$, for some constant $k$. Then we can solve the equation as follows:
$$
\begin{aligned}
\mathrm{d}y &= -ky \, \mathrm{d}x \\
\frac{\mathrm{d}y}{y} &= -k \, \mathrm{d}x \\
\int \frac{\mathrm{d}y}{y} &= -\int k \, \mathrm{d}x \\
\ln y &= -kx \\
y &= e^{-kx}
\end{aligned}
$$
(there should actually be some constants of integration in there, but I left them out since they're not important for this answer.)

Now, suppose we generalize and make the constant $k$ into a function $k(x)$. Then we can repeat this derivation, but we will not be able to do the integral on the right side, since $k(x)$ is unspecified. The result will then be:
$$
y = e^{-\int k(x) \, \mathrm{d}x}
$$

The derivation for the transmittance along a ray is just the same, but with some variables renamed: $y$ becomes $L$, $x$ is now the parameter $t$ along the ray, and $k(x)$ is called $\sigma\_\mathrm{t}(\mathrm{p})$.


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