Use non-italic font in math equations where appropriate: sine, cosine, sign and names

Author: Alexander Gauggel

docs/_posts/2023-10-21-study-notes-on-the-marschner-hair-shading-model.md

@@ -38,10 +38,10 @@ These functions are defined once for each path:
 $$ M_R $$, $$ M_{TT} $$ and $$ M_{TRT} $$ and their counterparts $$ N_R $$, $$ N_{TT} $$ and $$ N_{TRT} $$. The entire scattering function is defined as:
 
 \begin{equation}
-S = M_R N_R / cos^2(\theta_d) + M_{TT} N_{TT} / cos^2(\theta_d) + M_{TRT} N_{TRT} / cos^2(\theta_d)
+S = M_R N_R / \cos^2(\theta_d) + M_{TT} N_{TT} / \cos^2(\theta_d) + M_{TRT} N_{TRT} / \cos^2(\theta_d)
 \end{equation}
 
-Note that I omitted the parameters from the Marschner paper for brevity. The division by $$cos^2(\theta_d)$$ is a normalization factor that accounts for the projected solid 
+Note that I omitted the parameters from the Marschner paper for brevity. The division by $$\cos^2(\theta_d)$$ is a normalization factor that accounts for the projected solid 
 angle of the specular cone. However, this assumes $$\theta_i=\theta_r$$. This does not hold when taking surface roughness into 
 account and is therefore only an approximation, as noted in **[3]**.
 
@@ -115,22 +115,22 @@ We want to calculate the change of $$\phi$$ in regard to $$h$$, so we have to re
 The relationship between the two is directly taken from the definition of the unit circle:
 
 \begin{equation}
-sin(\gamma_i) = h \label{eq:gamma_i_to_h}
+\sin(\gamma_i) = h \label{eq:gamma_i_to_h}
 \end{equation}
 
 It follows:
 
 \begin{equation}
-\gamma_i = arcsin(h) \label{eq:gamma_i}
+\gamma_i = \arcsin(h) \label{eq:gamma_i}
 \end{equation}
 
 Putting this back into equation \eqref{eq:phi_gamma}:
 
 \begin{equation}
-\phi = -2 arcsin(h)
+\phi = -2 \arcsin(h)
 \end{equation}
 
-Taking the derivative in regard to $$h$$, keeping in mind that $$\frac d{dx} arcsin(x) = \frac1{\sqrt{1-x^2}}$$:
+Taking the derivative in regard to $$h$$, keeping in mind that $$\frac d{dx} \arcsin(x) = \frac1{\sqrt{1-x^2}}$$:
 
 \begin{equation}
 \frac{d\phi}{dh} = -2 \frac1{\sqrt{1-h^2}}
@@ -139,49 +139,49 @@ Taking the derivative in regard to $$h$$, keeping in mind that $$\frac d{dx} arc
 The next step is to express this in regards to our shading input $$\phi$$. We will use our previous definition \eqref{eq:gamma_i_to_h} to replace $$h$$ by $$\gamma_i$$ again:
 
 \begin{equation}
-\frac{d\phi}{dh} = -2 \frac1{\sqrt{1-sin^2(\gamma_i)}}
+\frac{d\phi}{dh} = -2 \frac1{\sqrt{1-\sin^2(\gamma_i)}}
 \end{equation}
 
-We can simplify this using the trigonometric identity $$cos(x) = \sqrt{1-sin^2(x)}$$:
+We can simplify this using the trigonometric identity $$\cos(x) = \sqrt{1-\sin^2(x)}$$:
 
 \begin{equation}
-\frac{d\phi}{dh} = -2 \frac1{cos(\gamma_i)}
+\frac{d\phi}{dh} = -2 \frac1{\cos(\gamma_i)}
 \end{equation}
 
 Now we solve our initial function \eqref{eq:phi_gamma} for $$\phi$$, which is simply $$\gamma_i = - \frac \phi 2$$, and plug this in:
 
 \begin{equation}
-\frac{d\phi}{dh} = -2 \frac1{cos(-\frac \phi 2)}
+\frac{d\phi}{dh} = -2 \frac1{\cos(-\frac \phi 2)}
 \end{equation}
 
-Since $$cos(x) = cos(-x)$$ we drop the minus sign inside the cosine:
+Since $$\cos(x) = \cos(-x)$$ we drop the minus sign inside the cosine:
 
 \begin{equation}
-\frac{d\phi}{dh} = -2 \frac1{cos(\frac \phi 2)}
+\frac{d\phi}{dh} = -2 \frac1{\cos(\frac \phi 2)}
 \end{equation}
 
 We can plug this into our initial definition of $$N$$ \eqref{eq:azimuthal_scattering}:
 
 \begin{equation}
-N_R = A_R |2 (-2 \frac1{cos(\frac \phi 2)}) |^{-1}
+N_R = A_R |2 (-2 \frac1{\cos(\frac \phi 2)}) |^{-1}
 \end{equation}
 
 We simplify this by multiplying out the constant factors and dropping the minus sign inside the absolute value:
 
 \begin{equation}
-N_R = A_R |4 \frac1{cos(\frac \phi 2)}) |^{-1}
+N_R = A_R |4 \frac1{\cos(\frac \phi 2)}) |^{-1}
 \end{equation}
 
 Then we apply the power of minus one:
 
 \begin{equation}
-N_R = A_R |\frac1 4 cos(\frac \phi 2))|
+N_R = A_R |\frac1 4 \cos(\frac \phi 2))|
 \end{equation}
 
 Finally, pulling out the fraction outside the absolute value gives:
 
 \begin{equation}
-N_R = A_R \frac1 4 |cos(\frac \phi 2))|
+N_R = A_R \frac1 4 |\cos(\frac \phi 2))|
 \end{equation}
 
 This matches equation *(6)* from the Weta paper **[3]**. Note that they leave $$A_R$$ out of their equation, since they list it in the context of analyzing energy conservation
@@ -193,39 +193,39 @@ The Marschner paper includes a fairly detailed derivation of the Bravais Index t
 some minor errors. Luckily, there is an extended version of the appendix **[6]**. The paper appendix incorrectly states:
 
 \begin{equation}
-\eta cos \gamma = \eta' cos \delta \label{eq:gamma_delta_incorrect}
+\eta \cos \gamma = \eta' \cos \delta \label{eq:gamma_delta_incorrect}
 \end{equation}
 
 $$\gamma$$ is the inclination of the incoming vector with regard to the normal plane. $$\delta$$ is the corresponding inclination of the refracted vector.
 Solving equation \eqref{eq:gamma_delta_incorrect} for $$\eta'$$, we get:
 
 \begin{equation}
-\eta' = \frac{\eta cos \gamma}{cos \delta} \label{eq:eta_prime_incorrect}
+\eta' = \frac{\eta \cos \gamma}{\cos \delta} \label{eq:eta_prime_incorrect}
 \end{equation}
 
 Compare this to the extended appendix **[6]** that states:
 
 \begin{equation}
-\eta' = \frac{\eta cos \delta}{cos \gamma} \label{eq:eta_prime_correct}
+\eta' = \frac{\eta \cos \delta}{\cos \gamma} \label{eq:eta_prime_correct}
 \end{equation}
 
 $$ \gamma $$ and $$ \delta $$ are swapped. Initially, I was unsure which version was correct. I wrote a small Octave script to manually compute
 some refraction vectors and their projections to confirm that the extended appendix is the correct one. 
 Using the script to check the paper equations used for deriving equation \eqref{eq:eta_prime_incorrect}, I found the following two statements to be incorrect as well:
 
 \begin{equation}
-\sin \theta_i' = sin \theta_i cos \gamma
+\sin \theta_i' = \sin \theta_i \cos \gamma
 \end{equation}
 
 \begin{equation}
-\sin \theta_t' = sin \theta_t cos \delta
+\sin \theta_t' = \sin \theta_t \cos \delta
 \end{equation}
 
 The extended appendix does not go into great detail how it arrives at equation \eqref{eq:eta_prime_correct}. Let's look into this next.
 As stated in the paper appendix, Snell's law applies in the vertical normal plane:
 
 \begin{equation}
-sin \theta_i' = \eta' sin \theta_t' \label{eq:snells_law_projected}
+\sin \theta_i' = \eta' \sin \theta_t' \label{eq:snells_law_projected}
 \end{equation}
 
 $$\theta_i'$$ and $$\theta_t'$$ are the inclinations of the incoming and refracted vectors after projecting them onto the normal plane. 
@@ -247,73 +247,73 @@ Looking at the geometry of *Figure 16 (c)* in the paper appendix **[4]**, we can
 is a direction and therefore a unit vector, so $$||v_i|| = 1$$. We can use this to compute the length of the projection of $$v_i$$ on the normal plane, denoted $$v_i'$$:
 
 \begin{equation}
-cos\gamma = \frac{||v_i'||}{||v_i||}
+\cos\gamma = \frac{||v_i'||}{||v_i||}
 \end{equation}
 
 \begin{equation}
-cos\gamma = ||v_i'||
+\cos\gamma = ||v_i'||
 \end{equation}
 
 The top of the triangle between $$v_i'$$ and it's projection on the normal $$n$$ is equal to $$s_i'$$. This gives us:
 
 \begin{equation}
-sin \theta_i' = \frac{s_i'}{||v_i'||}
+\sin \theta_i' = \frac{s_i'}{||v_i'||}
 \end{equation}
 
 \begin{equation}
-sin \theta_i' = \frac{s_i'}{cos \gamma} \label{eq:incoming_projection_and_inclination}
+\sin \theta_i' = \frac{s_i'}{\cos \gamma} \label{eq:incoming_projection_and_inclination}
 \end{equation}
 
 We can apply the equivalent steps to the refracted vector to obtain:
 
 \begin{equation}
-sin \theta_t' = \frac{s_t'}{cos \delta} \label{eq:refracted_projection_and_inclination}
+\sin \theta_t' = \frac{s_t'}{\cos \delta} \label{eq:refracted_projection_and_inclination}
 \end{equation}
 
 Substituting \eqref{eq:incoming_projection_and_inclination} and \eqref{eq:refracted_projection_and_inclination} into equation \eqref{eq:snells_law_projected}:
 
 \begin{equation}
-\eta' \frac{s_t'}{cos \delta} = \frac{s_i'}{cos \gamma}
+\eta' \frac{s_t'}{\cos \delta} = \frac{s_i'}{\cos \gamma}
 \end{equation}
 
 \begin{equation}
-\eta' = \frac{s_i'}{s_t'} \frac{cos \delta}{cos \gamma}
+\eta' = \frac{s_i'}{s_t'} \frac{\cos \delta}{\cos \gamma}
 \end{equation}
 
 Substituting $$\eta$$ into this using equation \eqref{eq:eta_ratio} yields equation \eqref{eq:eta_prime_correct}. As a reminder:
 
 $$
-\eta' = \eta \frac{cos \delta}{cos \gamma}
+\eta' = \eta \frac{\cos \delta}{\cos \gamma}
 $$
 
-The next step is to get rid of the $$cos\delta$$ term. The extended appendix observes:
+The next step is to get rid of the $$\cos\delta$$ term. The extended appendix observes:
 
 \begin{equation}
-\eta sin \delta = sin \gamma
+\eta \sin \delta = \sin \gamma
 \end{equation}
 
-We transform this for substitution by using the trigonometric identity $$ sin(x) = \sqrt{1 - cos^2(x)} $$:
+We transform this for substitution by using the trigonometric identity $$\sin(x) = \sqrt{1 - \cos^2(x)} $$:
 
 $$
 \begin{aligned}
-\eta sin \delta &= sin \gamma \\
-\eta \sqrt{1 - cos^2 \delta} &= sin \gamma \\
-\eta^2(1 - cos^2 \delta) &= sin^2 \gamma \\
-\eta^2 - \eta^2 cos^2 \delta &= sin^2 \gamma \\
-\eta^2 cos^2 \delta &= \eta^2 - sin^2 \gamma
+\eta \sin \delta &= \sin \gamma \\
+\eta \sqrt{1 - \cos^2 \delta} &= \sin \gamma \\
+\eta^2(1 - \cos^2 \delta) &= \sin^2 \gamma \\
+\eta^2 - \eta^2 \cos^2 \delta &= \sin^2 \gamma \\
+\eta^2 \cos^2 \delta &= \eta^2 - \sin^2 \gamma
 \end{aligned}
 $$
 
 This intermediate result is also listed in the extended appendix. When we take the square root of this, we get:
 
 \begin{equation}
-\eta cos \delta = \sqrt{\eta^2 - sin^2 \gamma}
+\eta \cos \delta = \sqrt{\eta^2 - \sin^2 \gamma}
 \end{equation}
 
 Finally,     we substitute this term into equation \eqref{eq:eta_prime_correct} to obtain:
 
 \begin{equation}
-\eta'(\gamma) = \frac{\sqrt{\eta^2 - sin^2 \gamma}}{cos \gamma}
+\eta'(\gamma) = \frac{\sqrt{\eta^2 - \sin^2 \gamma}}{\cos \gamma}
 \end{equation}
 
 # Angle for computing $$\eta'$$
@@ -340,44 +340,44 @@ As noted in **[7]**, the internal path segment length $$l$$ is incorrect in the
 There it is incorrectly stated:
 
 \begin{equation}
-l_{Marschner}(\gamma_t) = 2 + 2cos(2\gamma_t) \label{eq:path_length_marschner}
+l_{\mathrm{Marschner}}(\gamma_t) = 2 + 2\cos(2\gamma_t) \label{eq:path_length_marschner}
 \end{equation}
 
 We can derive the correct term using the law of cosines. The law of cosines relates the length of the triangle sides using the cosine of the angle:
 
 \begin{equation}
-c^2 = a^2 + b^2 - 2ab cos \gamma \label{eq:law_of_cosines}
+c^2 = a^2 + b^2 - 2ab \cos \gamma \label{eq:law_of_cosines}
 \end{equation}
 
 As seen in *Figure 9* in **[4]**, $$a=1$$ and $$b=1$$, since this is the radius of the unit circle. We require the angle opposite of $$l$$.
 Since we know that the sum of all angles in a triangle is $$\pi$$, we know $$\gamma= \pi - 2 \gamma_t$$. Inserting these values into \eqref{eq:law_of_cosines} we get:
 
 $$
 \begin{aligned}
-l^2 &= 1^2 + 1^2 - 2 cos(\pi - 2 \gamma_t) \\
-l^2 &= 2 - 2 cos(\pi - 2 \gamma_t)
+l^2 &= 1^2 + 1^2 - 2 \cos(\pi - 2 \gamma_t) \\
+l^2 &= 2 - 2 \cos(\pi - 2 \gamma_t)
 \end{aligned}
 $$
 
-Then we simplify this using $$cos(x) = cos(-x)$$ and $$cos(x) = -cos(x - \pi)$$:
+Then we simplify this using $$\cos(x) = \cos(-x)$$ and $$\cos(x) = -\cos(x - \pi)$$:
 
 $$
 \begin{aligned}
-l^2 &= 2 - 2 cos( 2 \gamma - \pi) \\
-l^2 &= 2 + 2 cos(2 \gamma_t)
+l^2 &= 2 - 2 \cos( 2 \gamma - \pi) \\
+l^2 &= 2 + 2 \cos(2 \gamma_t)
 \end{aligned}
 $$
 
 Taking the square root, the final path segment length is:
 
 \begin{equation}
-l = \sqrt{2 + 2 cos(2 \gamma_t)} \label{eq:path_length_derived}
+l = \sqrt{2 + 2 \cos(2 \gamma_t)} \label{eq:path_length_derived}
 \end{equation}
 
 Comparing this to equation \eqref{eq:path_length_marschner}, it is clear that the latter is missing the square root. In **[7]**, the path length is given as:
 
 \begin{equation}
-l_{Zinke} = 2 cos(\gamma_t) \label{eq:path_length_zinke}
+l_{\mathrm{Zinke}} = 2 \cos(\gamma_t) \label{eq:path_length_zinke}
 \end{equation}
 
 Note that my notation differs slightly from **[7]**. For one, I assume that the fiber radius is one, so I omitted $$r=1$$ . Furthermore, they relate the 
@@ -390,35 +390,35 @@ or when taking their absolute value. We can show that they are equal. First, we
 
 $$
 \begin{aligned}
-l &= \sqrt{2 + 2 cos(2 \gamma_t)} \\
-l &= \sqrt{2} \sqrt{1 + cos(2 \gamma_t)} \\
-l &= \sqrt{2} \sqrt{2} \sqrt{\frac{1 + cos(2 \gamma_t)}{2}} \\
-l &= 2 \sqrt{\frac{1 + cos(2 \gamma_t)}{2}}
+l &= \sqrt{2 + 2 \cos(2 \gamma_t)} \\
+l &= \sqrt{2} \sqrt{1 + \cos(2 \gamma_t)} \\
+l &= \sqrt{2} \sqrt{2} \sqrt{\frac{1 + \cos(2 \gamma_t)}{2}} \\
+l &= 2 \sqrt{\frac{1 + \cos(2 \gamma_t)}{2}}
 \end{aligned}
 $$
 
-Then we use the trigonometric identity $$cos(\frac{x}{2}) = sign(cos\frac{x}{2})\sqrt{\frac{1+cos(x)}{2}}$$ to arrive at:
+Then we use the trigonometric identity $$\cos(\frac{x}{2}) = \mathrm{sign}(\cos\frac{x}{2})\sqrt{\frac{1+\cos(x)}{2}}$$ to arrive at:
 
 \begin{equation}
-sign(cos(\gamma_t)) l = 2 cos(\gamma_t)
+\mathrm{sign}(\cos(\gamma_t)) l = 2 \cos(\gamma_t)
 \end{equation}
 
 Taking the absolute value of this equation gives:
 
 \begin{equation}
-|sign(cos(\gamma_t)) l| = |2 cos(\gamma_t)|
+|\mathrm{sign}(\cos(\gamma_t)) l| = |2 \cos(\gamma_t)|
 \end{equation}
 
 From equation \eqref{eq:path_length_derived}, we know that $$l \geq 0$$ and $$l$$ is zero when $$cos(\gamma_t)$$ is zero, so we can drop the sign:
 
 \begin{equation}
-|l| = |2 cos(\gamma_t)|
+|l| = |2 \cos(\gamma_t)|
 \end{equation}
 
 Since we know that $$l$$ is never negative, we can drop the absolute value on the left side of the equation as well:
 
 \begin{equation}
-l = |2 cos(\gamma_t)|
+l = |2 \cos(\gamma_t)|
 \end{equation}
 
 This matches \eqref{eq:path_length_zinke}, except for the absolute value.