<p>The form of the rendering equation that uses only the BRDF ($f$ in your example, often called $f_r$) and integrates over one hemisphere does not account for transmission.</p>

<p>When adding in transmission, it's fairly common to add a second integral over the opposite hemisphere, using a different BTDF function (bidirectional <em>transmission</em> distribution function). This is equivalent to an integral over the full sphere of directions with a BSDF function, but since that function would usually have to be defined as a piecewise function, writing it as two integrals can be more straightforward.</p>

CC BY-SA 3.0
2015-08-21T16:41:42.903Z
BSDF = Biderectional Scattering Distribution Function
cifz
CC BY-SA 3.0
2015-08-21T15:11:24.357Z
2015-11-03T17:19:11.853Z
2015-11-03T17:19:11.853Z
John Calsbeek
<p>In most text books that I have seen, this is how the rendering equation is written:</p>

<p>$$L_0( \omega_0)= L_e(\omega_0)+\int_{\Omega}{f(\omega_i, \omega_0)L_i(\omega_i)\,\mathrm{d}\omega_i}$$</p>

<p>Where $\Omega$ is defined to be a hemisphere (and all those functions depend on more variables, omitted here for simplicity's sake).</p>

<p>Now suppose the surface being rendered is some kind of glass, or some transparent plastic. Why would it make sense to only integrate over a hemisphere? I would imagine that there can be incoming light from any direction, and thus the integration domain should be the entire sphere. How is the light coming from behind the glass accounted for?</p>

CC BY-SA 3.0
2016-09-04T21:57:01.943Z
note that the subscript is not a 0 (zero), but an O (oh).  it reads like... "Light out at the out angle equations light emitted towards the out angle plus ... ".  o and i are complements, meaning out and in respectively (:
Alan Wolfe
<transparency><theory>
Why integrate over a hemisphere (and not a sphere) to solve the rendering equation?
Mon ouïe
# Why integrate over a hemisphere (and not a sphere) to solve the rendering equation?

(Asked by: username_0 on 2015-08-21)

In most text books that I have seen, this is how the rendering equation is written:

$$L\_0( \omega\_0)= L\_e(\omega\_0)+\int\_{\Omega}{f(\omega\_i, \omega\_0)L\_i(\omega\_i)\,\mathrm{d}\omega\_i}$$

Where $\Omega$ is defined to be a hemisphere (and all those functions depend on more variables, omitted here for simplicity's sake).

Now suppose the surface being rendered is some kind of glass, or some transparent plastic. Why would it make sense to only integrate over a hemisphere? I would imagine that there can be incoming light from any direction, and thus the integration domain should be the entire sphere. How is the light coming from behind the glass accounted for?

**Tags:** transparency, theory

> **Comments:**
> **username_1**: note that the subscript is not a 0 (zero), but an O (oh).  it reads like... "Light out at the out angle equations light emitted towards the out angle plus ... ".  o and i are complements, meaning out and in respectively (:  

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## 1 answers

### Answer by username_2 on 2015-08-21. Score: 20 | Accepted answer

The form of the rendering equation that uses only the BRDF ($f$ in your example, often called $f\_r$) and integrates over one hemisphere does not account for transmission.

When adding in transmission, it's fairly common to add a second integral over the opposite hemisphere, using a different BTDF function (bidirectional *transmission* distribution function). This is equivalent to an integral over the full sphere of directions with a BSDF function, but since that function would usually have to be defined as a piecewise function, writing it as two integrals can be more straightforward.

> **Comments:**
> **username_3**: BSDF = Biderectional Scattering Distribution Function  

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