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02941 Physically Based Rendering Reflection and Transmission Jeppe - - PowerPoint PPT Presentation
02941 Physically Based Rendering Reflection and Transmission Jeppe - - PowerPoint PPT Presentation
02941 Physically Based Rendering Reflection and Transmission Jeppe Revall Frisvad June 2016 Multiscale light modelling Light at different scales: Quantum electrodynamics (photons) Electromagnetic radiation (waves) Geometrical
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Time-harmonic Maxwell equations for isotropic materials
∇ × Hc = (σ − iωε)Ec ∇ × Ec = iωµHc ∇ · (εEc) = ∇ · (µHc) = 0 , where ◮ ω = 2πc/λ is the angular frequency of the light ◮ Re(Ec) is the electric field vector ◮ Re(Hc) is the magnetic vector ◮ and the following are the isotropic material properties:
◮ σ is the conductivity ◮ ε is the permittivity ◮ µ is the permeability
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The plane wave solution
◮ Reflection and refraction occurs at a surface marking the boundary between two (at least locally) homogeneous materials. ◮ Plane waves satisfy the time-harmonic Maxwell equations for isotropic and homogeneous materials. ◮ The plane wave equations are: Ec(x, t) = E0e−i(ωt−k·x) Hc(x, t) = H0e−i(ωt−k·x) . where
◮ x is a position in space ◮ t is time ◮ E0 and H0 are wave amplitudes ◮ k is the wave vector.
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Plane wave Maxwell equations
◮ The plane wave solution inserted in the time-harmonic Maxwell equations for isotropic, homogeneous materials: k × H0 = −ω(ε + iσ/ω)E0 k × E0 = ωµH0 k · E0 = k · H0 = 0 . ◮ This boils down to the following conditions that plane waves must satisfy to be physically realisable: k · E0 = k · H0 = E0 · H0 = 0 k · k = ω2µ(ε + iσ/ω) . ◮ Thus k is a complex vector if the material is a conductor.
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The meaning of the wave vector
◮ Consider the exponential part of the plane wave equation: e−i(ωt−k·x) = e−iωteik·x ◮ The first part concerns temporal aspects, the second part concerns spatial aspects. ◮ The wave vector k determines the spatial aspect (the wave propagation). ◮ If the material is a conductor, k = k′ + ik′′ is complex: ei k·x = ei k′·xe−k′′·x . ◮ This reveals that the phase velocity is v = ω/k′ and
◮ k′ is normal to the surface of constant phase ◮ k′′ is normal to the surface of constant amplitude
◮ They determine the direction and damping of the wave.
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The index of refraction (IOR)
◮ Recall the only remaining condition which involves material properties: k · k = ω2µ(ε + iσ/ω) . ◮ This is a golden opportunity to reduce the number of material properties. ◮ Suppose we introduce n = n′ + in′′ = c
- µ(ε + iσ/ω) .
◮ Then the condition becomes k · k = ω2 c2 n2 , where
◮ c is the speed of light in vacuo. ◮ n is called the (complex) index of refraction.
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Phase velocity and amplitude damping
◮ A closer look at the condition which describes the relation between material and wave propagation: k · k = k′ · k′ − k′′ · k′′ + i2k′ · k′′ = ω2 c2 n2 . ◮ For materials that are not strong absorbers: k′′ · k′′ ≈ 0. ◮ Then (equating the real parts): k′ ≈ ω c n′ ⇒ v = ω k′ ≈ c n′ the phase velocity v is determined by the real part of the IOR. ◮ And (equating the imaginary parts): 2k′k′′ cos θ = ω2 c2 2n′n′′ ⇒ k′′ ≈ ω c n′′ cos θ ≈ ω c n′′ , where θ (usually close to 0) is the angle between k′ and k′′.
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The Poynting vector and absorption
◮ The Poynting vector S determines the direction and magnitude of energy propagation in an electromagnetic field. ◮ For the plane wave solution, we have: S = ε0c2µ Re(Ec) × Re(Hc) . ◮ Inserting the plane wave equations and taking the magnitude of S determines the absorption of energy by the material: |S| = ε0c2µ
- Re(E0e−i(ωt−k′·x)) × Re(H0e−i(ωt−k′·x))
- e−2k′′·x .
◮ This reveals that energy is exponentially attenuated by 2k′′, or σa = 2k′′ ≈ 2ω c n′′ = 4πn′′ λ , where σa is called the absorption coefficient and λ is the wavelength in vacuo.
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Plane waves incident on surfaces
◮ Consider a plane wave incident on a smooth surface. ◮ The wave gives rise to two other waves. Altogether we have
◮ An incident wave (subscript i) ◮ A reflected wave (subscript r) ◮ A transmitted wave (subscript t)
◮ We resolve all waves into two independent components:
◮ The wave with the electric vector perpendicular to the plane of incidence: ⊥-polarised light. ◮ The wave with the electric vector parallel to the plane of incidence: -polarised light.
◮ Maxwell’s equations require that the field vectors are continuous across the surface boundary. Let us call this the continuity condition. ◮ The continuity condition must hold at all times and no matter where we place the point of incidence in space.
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Conditions imposed by Maxwell’s equations
◮ Using the continuity condition: E⊥i + E⊥r = E⊥t . ◮ This must also hold for x = 0, therefore E ⊥
0i e−iωit + E ⊥ 0re−iωrt = E ⊥ 0te−iωtt .
◮ This is true only if ωi = ωr = ωt . ◮ With the condition k · k = n2ω2/c2, this reveals ki · ki n2
i
= kr · kr n2
i
= kt · kt n2
t
which shows how the index of refraction governs the propagation of plane waves
- f light.
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The plane of incidence
ni nt Θi Θr Θt Ei Et Er z x k z,i k x,i k z,r k x,r k x,t k z,t
◮ At the boundary, z = 0, we have at the time t = 0: E ⊥
0i ei(kx,ix+ky,iy) + E ⊥ 0rei(kx,rx+ky,ry) = E ⊥ 0tei(kx,tx+ky,ty)
which must hold for all x and y.
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The plane of incidence
ni nt Θi Θr Θt Ei Et Er z x k z,i k x,i k z,r k x,r k x,t k z,t
◮ The result is kx,i = kx,r = kx,t and ky,i = ky,r = ky,t .
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The plane of incidence
◮ The result is kx,i = kx,r = kx,t and ky,i = ky,r = ky,t . ◮ Maxwell’s equations require that k is perpendicular to E0. ◮ Since E ⊥
0i (by definition) is perpendicular to the plane of incidence, ki must be
parallel to the plane of incidence. ◮ Then ky,i = 0. ◮ Meaning that ky,r = ky,t = ky,i = 0. ◮ In other words, the reflected and transmitted vectors lie in the plane of incidence.
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The law of reflection
◮ Using the following results (found previously): ki · ki n2
i
= kr · kr n2
i
kx,i = kx,r and ky,r = ky,i = 0 . ◮ We find: k2
z,i = k2 z,r
with the solutions kz,r = kz,i or kz,r = −kz,i. ◮ Only the solution kz,r = −kz,i describes a wave on the right side of the surface. ◮ The law of reflection is then: The reflected wave lies in the plane of incidence, the (complex) angle of reflection is equal to the (complex) angle of incidence.
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The law of refraction
◮ Using the following results (found previously): ki · ki n2
i
= kt · kt n2
t
kx,i = kx,t ◮ We find: ni kx,i √ki · ki = nt kx,t √kt · kt
- r (using complex angles)
ni sin Θi = nt sin Θt . ◮ This is called generalised Snell’s law. ◮ The law of refraction is then: The refracted wave lies in the plane of incidence, the (complex) angle of refraction follows the generalised Snell’s law.
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Continuity of the magnetic vector
◮ One of the plane wave Maxwell equations is: H0 = 1 ωµk × E0 . ◮ The x component of this equation is H0x = 1 ωµ(kyE0z − kzE0y) . ◮ Using the perpendicular components and the result that ky,i = ky,r = ky,t = 0, we have H0x = − kz ωµE0y and (0, E0y, 0) = E ⊥ . ◮ The continuity condition (invoked on the x component of the magnetic vector) is then − kz,i ωiµi E ⊥
0i − kz,r
ωrµi E ⊥
0r = − kz,t
ωtµt E ⊥
0t .
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The Fresnel equations for reflection
◮ The newly found continuity condition: − kz,i ωiµi E ⊥
0i − kz,r
ωrµi E ⊥
0r = − kz,t
ωtµt E ⊥
0t .
◮ Using the law of reflection kz,r = −kz,i and the result that angular frequencies are equal ωi = ωr = ωt, we have kz,iE ⊥
0i − kz,iE ⊥ 0r = kz,t
µi µt E ⊥
0t .
◮ Ignoring magnetic effects, we set µi/µt = 1. ◮ Using the other continuity condition, E ⊥
0i + E ⊥ 0r = E ⊥ 0t, we get
kz,iE ⊥
0i − kz,iE ⊥ 0r = kz,t(E ⊥ 0i + E ⊥ 0r)
⇔ E ⊥
0r = kz,i − kz,t
kz,i + kz,t E ⊥
0i .
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The Fresnel equations for reflection
◮ The result we arrived at was E ⊥
0r = kz,i − kz,t
kz,i + kz,t E ⊥
0i .
◮ Diving by E ⊥
0i
√ki · ki and using cos Θ = kz/√ki · ki, we find the amplitude ratio ˜ r⊥ = E ⊥
0r
E ⊥
0i
= ni cos Θi − nt cos Θt ni cos Θi + nt cos Θt . ◮ In a similar way, we can find ˜ r = E
0r
E
0i
= nt cos Θi − ni cos Θt nt cos Θi + ni cos Θt . ◮ These are called Fresnel’s equations for reflection.
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Computing reflectances and transmittances (energy ratios)
◮ To translate Fresnel’s amplitude ratios into energy ratios, one uses the squared absolute value: R⊥ = |˜ r⊥|2 and R = |˜ r|2 . ◮ The Fresnel reflectance for unpolarized light is then R = 1 2(R⊥ + R) . ◮ The transmittances are one minus the reflectances: T⊥ = 1 − R⊥ , T = 1 − R , T = 1 − R . ◮ Please note that the Fresnel equations work with complex indices of refraction.
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Reflection and transmission of radiance
◮ Requiring energy conservation at the surface boundary: Φi = Φr + Φt , where Φi, Φr, and Φt are the incident, reflected, and transmitted energy fluxes. ◮ Transformed to radiances: Li cos θi dA dωi = Lr cos θr dA dωr + Lt cos θt dA dωt
- Li cos θi dωi
= Lr cos θr dωr + Lt cos θt dωt . ◮ Another way to write the differential solid angles: dωi = sin θi dθi dφi dωr = sin θr dθr dφr dωt = sin θt dθt dφt .
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Using the law of reflection
◮ Using the law of reflection θi = θr and the plane of incidence φi = φr = φt, the solid angles become dωi = sin θi dθi dφi dωr = sin θi dθi dφi dωt = sin θt dθt dφi . ◮ By insertion, we can then change the energy conservation condition as follows: Li cos θi dωi = Lr cos θr dωr + Lt cos θt dωt
- Li cos θi sin θi dθi
= Lr cos θi sin θi dθi + Lt cos θt sin θt dθt .
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Compression of solid angle upon transmission
◮ Using the law of refraction (Snell’s law), we have (neglecting the imaginary part
- f the IOR)
sin θt = n′
i
n′
t
sin θi and dθt dθi = d dθi sin−1 n′
i
n′
t
sin θi
- = n′
i cos θi
n′
t cos θt
. ◮ By insertion, we can then change the energy conservation condition as follows: Li cos θi sin θi dθi = Lr cos θi sin θi dθi + Lt cos θt sin θt dθt
- Li
= Lr + n′
i
n′
t
2 Lt . ◮ Finally, using Fresnel reflectance R, we have Lo = RLi + n′
i
n′
t
2 (1 − R)Li .
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02576 Physically Based Rendering
Russian Roulette
Jeppe Revall Frisvad June 2016
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Doing efficient rendering
◮ Time vs. variance.
◮ Efficiency = (variance × rendering time)−1 .
◮ Tree vs. path. ◮ Splitting vs. Russian roulette.
◮ Splitting is a sum over all the branches. ◮ Russian roulette is sampling a path.
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Demo: The golden bunny
◮ Using 10 samples per pixel. Russian roulette Russian roulette Russian roulette No splitting Split on 1 reflection Split on 2 reflections Time: 14.4 s Time: 14.8 s Time 14.7 s
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Demo: The glass elephant
◮ What does a glass elephant on a flatly coloured background look like? ◮ Using one sample per pixel. No Russian roulette No Russian roulette Russian roulette Split on 5 bounces Split on 10 bounces No splitting Time: 4.4 s Time: 13.7 s Time 2.6 s
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Splitting
◮ Sum all the events.
◮ Example: Reflection multiplied by reflectance. ◮ Example: Sum reflection and transmission.
◮ Exact but expensive. ◮ When does the recursion stop?
◮ Often not before the combinatorial explosion. ◮ It might never.
◮ When to use splitting?
◮ Where variance is most striking. ◮ Example: The first (or first few) light bounces.
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Russian roulette
◮ Either - or. ◮ Sample an event.
◮ Example: Reflection or absorption. ◮ Example: Reflection or transmission.
◮ How? Sample a step function. ◮ What are the steps?
◮ The probabilities of an event. ◮ Example: Fresnel reflectance.
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Sampling a step function
Algorithm: sample ξ ∈ [0, 1] uniformly; if (ξ < P1)
call event 1; divide by p1;
else if (ξ < P2)
call event 2; divide by p2;
else if (ξ < P3)
. . .
else if (ξ < P4)
. . .
1 1 2 3 4
Discrete pdf
p - probability 1 1 2 3 4
Discrete cdf
P - cumulave
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