[**APPENDIX**]{}

Outer Warp + Flat Extension
===========================

Adding $\alpha$ Dependence to Outer Warp
----------------------------------------

TB96 describe how to calculate the SED for an outer warp seen at various inclination angles. Their general method for calculating the SED also includes a dependence on azimuthal viewing angle, although their detailed treatment of various occultation effects (the star blocking the far side of the disk, the disk blocking the star, etc.) does not include this dependence. Since these disks are non-axissymetric the SED can depend substantially on the azimuthal viewing angle, $\alpha$, of the observer. In this section we describe how we have added this dependence into the equations of TB96. We do not include a detailed description of the derivation of the equations, but merely state most of them and some of the geometric logic behind their modification.

The first modification is to equation (6) of TB96, which describes the calculation of the flux from the disk based on the temperature of the disk. This equation assumes that the disk is viewed along $\alpha=0$ so that the upper disk from 0 to $\pi/2$ looks the same as the upper disk from $3\pi/2$ to $2\pi$. When $\alpha\neq0$ this symmetry is broken and the individual components of the disk must be considered. While the temperature of a concave, or convex, piece of the disk does not change with viewing angle, the orientation of each of the concave, or convex pieces, changes and must be treated separately. For the outer warp this becomes:

$$\begin{aligned}
\textstyle
F_{\nu,{\bf u}} = \int^{R_{disk}}_{r_{min}}\left[\int_0^{\pi/2}B(T_{concave})f_{up}+\int_0^{\pi/2}B(T_{convex})f_{low}+\int_{\pi/2}^{\pi}B(T_{concave})f_{low}+\int_{\pi/2}^{\pi}B(T_{convex})f_{up}\right]\nonumber\\
\textstyle
+\int^{R_{disk}}_{r_{min}}\left[\int_{\pi}^{3\pi/2}B(T_{convex})f_{up}+\int_{\pi}^{3\pi/2}B(T_{concave})f_{low}+\int_{3\pi/2}^{2\pi}B(T_{concave})f_{up}+\int_{3\pi/2}^{2\pi}B(T_{convex})f_{low}\right]\end{aligned}$$

The next change comes in the appendix to the functional form of the parameter $C$. The function $C$ is used to define the points that are along the line of sight with the star. If the line intersects the star then we need to worry about whether the disk blocks the star or the star blocks the disk. If this line does not intersect the star then the disk cannot block the star and the star cannot block the disk. The definition of $C$ changes from $C=r\sin\theta$ to $C=r\sin (\theta-\alpha )$. Also the radial part of the deformation used in equation (A6) of TB96 is taken to be $H(r)=gR_{disk}\left(\frac{r}{R_{disk}}\right)^n\cos\alpha$. For $\alpha=\pi/2$ the disk along the line of sight is flat and the radial part of the height will remain at zero, while along $\alpha=\pi$ the disk curves below the midplane as expected.

The final change comes when calculating the stellar flux. In equation (A12) of TB96 we take $h(r,\theta)$ to be $h(r,\alpha)$ since this represents the part of the disk that will block the star. As the azimuthal angle increases the disk blocks less of the star because the height of the disk is smaller. We make more changes to how the stellar flux is calculated, which are described below, but when it comes to the occultation of the star by the disk this is the only change.

In the end we are able to run our models from $0<i<\pi$ and $-\pi/2<\alpha<\pi/2$. Due to the symmetry of the disk this covers all possible viewing angles allowing us to accurately model the precession of the warp, as well as observe the warp from an arbitrary angle.

Flat extension of Outer Warp
----------------------------

We have taken the outer warp model and added a flat extension beyond it in order to treat disks where the warp is not at the outer edge of the disk. The warp will shadow the outer disk, changing its temperature structure. For simplicity we assume that the outer disk is a flat blackbody. The temperature can be derived using the same formula as with the warped disk (Equation 6), but with different definitions for the integration boundaries. Half of the flat extension will be shadowed while half will not be shadowed. For the side of the flat extension that is beyond the part of the warp the goes below the midplane there is no shadowing of the disk and the integration ranges over: $$\begin{aligned}
\varepsilon_{min}=0\nonumber\\
\varepsilon_{max}=\pi/2\nonumber\\
\\
\delta_{min}=0\nonumber\\
\delta_{max}=\arcsin(R_*/d)\nonumber\\
d^2=r^2+h^2\nonumber\\\end{aligned}$$

For the part of the flat extension that lies behind the warp that stretches above the midplane the definition of $\delta_{min}$ changes. In this case $\delta_{min}$ is set by the angle between the warp and the point $P(r,\theta)$ in the disk.

$$\delta_{min}=\arctan\left(\frac{gR_{warp}\cos(\theta)}{(r-R_{warp})}\right)$$

This takes into account shadowing of the flat extension to the disk due to the warp. Once the temperature structure has been determined the flux can be derived using the equation for the flux from a disk (Equation 5).

Inner Warp
==========

Temperature Profile of the Inner Warp
-------------------------------------

In this section we describe the method for calculating the SED of a disk with an inner warp. In the text we laid out the basic equations from TB96 that are needed to calculate the temperature structure. As mentioned in the text the essential difference between the inner warp and outer warp comes in calculating $\delta_{max},\delta_{min},\varepsilon_{min},\varepsilon_{max}$ for each point $P(r,\theta)$, which are used in equation 6. From here the disk is split into two sides that are treated separately. The convex side is the side of the disk that faces the star on the inner edge and receives the most direct heating from the star. For this side, the integration ranges over:

$$\begin{aligned}
\varepsilon_{min}=0\nonumber\\
\varepsilon_{max}=\pi/2\nonumber\\
\\
\delta_{min}=-\arctan(\partial h/\partial r)\nonumber\\
\delta_{max}=\arcsin(R_*/d)\nonumber\\
d^2=r^2+h^2\nonumber\\\end{aligned}$$

Figure \[del\_vex\] demonstrates the limits on $\delta$ for the convex side of the disk. The definition of $\delta_{min}$, demonstrated in figure \[delblock\_vex\], comes from the inner disk blocking light from the top of the star. The inner disk will limit the field of view of the point $P(r,\theta)$ as it looks toward the star. Traveling out from the star, less of the star will be seen by the disk because the shallower slope of the disk will cause more of the star to be blocked. In the limit of a flat disk far from the star $\delta_{min}$ approaches zero and the disk can only see half of the star. If the point $P(r,\theta)$ on the disk is close enough to the star then the disk can see all of the star and $\delta_{min}=-\delta_{max}$. The limits on $\varepsilon$ assume that the scale height of the disk does not change across the face of the disk, which will be an accurate approximation far from the star.

The concave side of the disk is the side that does not directly face the star. Since it does not face the star much of the inner parts of the disk will be blocked by the warp and will only be heated by viscous dissipation. The condition for the point $P(r,\theta)$ on the concave part of the disk to see any of the star is:

$$\frac{h(r_{min},\theta)}{r-r_{min}}<\frac{R_*}{r}$$

If the point $P(r,\theta)$ meets this condition then this point can see some of the star and $\delta_{min}$ becomes (fig \[delmin\_cave\])

$$\delta_{min}=\arctan(\frac{h(r_{min},\theta)}{r-r_{min}})$$

The rest of the limits stay the same. In the limit of a perfectly flat disk $\delta_{min}$ approaches 0 and the point $P(r,\theta)$ is irradiated by only half of the star. For a large warp the only heating by this side of the disk will be from viscous dissipation because the warp will block the star over most of the disk.

Calculating the SED for an Inner Warp
-------------------------------------

In this section we describe the procedure for converting the temperature structure into a spectral energy distribution (SED). From TB96, the flux emitted by the disk is

$$F_{\nu,{\bf u}}= \int\int_{disk surface} B_{\nu}(T(r,\theta))dS{\bf n_d}\cdot{\bf u}$$

In this case [**u**]{} is the vector along the line of sight to the observer from the center of the star and ${\bf n_d}$ is the normal to the disk at the point $P(r,\theta)$. The vector ${\bf u}$ can be defined in terms of the azimuthal and polar angles to the line of sight, $\alpha$ and $i$ respectively. The area of the disk along the line of sight is given by

$$\textstyle
dS {\bf n_d}\cdot{\bf u} = r\left[\left(\frac1{r}\frac{\partial h}{\partial \theta}\sin\theta-\frac{\partial h}{\partial r}\cos\theta\right)\cos\alpha\sin i-\left(\frac1{r}\frac{\partial h}{\partial\theta}\cos\theta+\frac{\partial h}{\partial r}\sin\theta\right)\sin\alpha\sin i+\cos i\right]drd\theta$$

The angle $\alpha$ ranges from $-\pi$/2 to $\pi$/2 while the inclination $i$ ranges from 0 to $\pi$. This covers all possible viewing angles of the disk, since the symmetry of the disk makes some viewing angles redundant.

Splitting up the equation for the flux from the disk helps to make the problem simpler to understand and more tractable. It also fits with the fact that we do not need to calculate the temperature structure of the entire disk. The symmetry of the disk allows us the calculate the temperature of the convex and concave side from $0<\theta<\pi/2$ and then apply this temperature profile to the rest of the disk. The integral is split into eight parts:

$$\begin{aligned}
\textstyle
F_{\nu,{\bf u}} = \int^{R_{disk}}_{r_{min}}\left[\int_0^{\pi/2}B(T_{concave})f_{up}+\int_0^{\pi/2}B(T_{convex})f_{low}+\int_{\pi/2}^{\pi}B(T_{concave})f_{low}+\int_{\pi/2}^{\pi}B(T_{convex})f_{up}\right]\nonumber\\
\textstyle
+\int^{R_{disk}}_{r_{min}}\left[\int_{\pi}^{3\pi/2}B(T_{convex})f_{up}+\int_{\pi}^{3\pi/2}B(T_{concave})f_{low}+\int_{3\pi/2}^{2\pi}B(T_{concave})f_{up}+\int_{3\pi/2}^{2\pi}B(T_{convex})f_{low}\right]\label{eqn_flux}\end{aligned}$$

where $f(r,\theta)=dS {\bf n_d}\cdot{\bf u}p(r,\theta)$, and $p(r,\theta)$ is a binary function used to determine if the point $P(r,\theta)$ is visible to the observer. The integration is done over both the upper and lower sides of the disk in order to account for inclination angles greater than $90^{\circ}$ where the lower half of the disk is visible. If the inclination is 0 then the observer is face on to the upper half of the disk, which has both a concave and convex side. If the inclination is $180^{\circ}$ then the observer is face on to the lower half of the disk, which includes both a concave and a convex side. Treating each quarter of the disk separately allows us to use the symmetry of the temperature profile but still treat general azimuthal viewing angles.

### Calculating the value of p

The above description sets out the basics for how to calculate the temperature structure and SED for a warped inner disk. Most of this is derived from TB96, which treated these situations generally enough to apply to any type of warp. The main differences between this inner warp and the outer warp from TB96 comes from the calculation of $p(r,\theta)$. This section describes the conditions used to calculate $p(r,\theta)$ for the particular warp used here.

The first condition is that the observer is facing the point $P(r,\theta)$. For inclinations less than $90^{\circ}$ the observer will see mostly the upper half of the disk, while at inclinations greater than $90^{\circ}$ the observer will see mostly the lower half of the disk. There are select inclinations close to edge on where at inclinations less than $90^{\circ}$ some of the lower disk can be seen. For example, if figure \[delblock\_vex\] had an observer in the upper left viewing the disk close to edge on they would be able to see some of the lower convex side that is illustrated in the figure. In general it can be determined if the observer is facing the point $P(r,\theta)$ based on the dot product ${\bf n_d}\cdot{\bf u}$. The normal is defined as extending on the upper side of the disk and the dot product will be greater than zero if ${\bf n_d}$ and ${\bf u}$ lie along the same direction. Therefore, the upper part of the disk can be seen if the dot product is greater than 0 while the lower part of the disk can be seen when the dot product is negative.

Now we determine if the point $P(r,\theta)$ is blocked by either the star or the disk. First we consider whether the star blocks the far side of the disk. This applies for inclinations less than $90^{\circ}$ where the star may block part of the upper convex side, as is demonstrated in figure \[inclim\]. The limit at which this condition applies is given by

$$\tan i_{lim}=\frac{r_{min}-R_*\cos i_{lim}}{gr_{min}\cos\alpha+R_*\sin i_{lim}}$$

[cccc]{} 0.005& 59.7 & 78.2 & 83.9\
0.01& 59.4 & 77.9 & 83.7\
0.03& 59.3 & 76.7 & 82.5\
0.05& 57.2 & 75.6 & 81.4\
0.07& 56.1 & 74.5 & 80.3\
0.1& 54.5 & 72.8 & 78.6\
0.3& 44.7 & 62.3 & 67.8\
0.5& 36.7 & 53.1 & 58.3\

If the inclination is greater than $i_{lim}$ then part of the disk is blocked by the star. In this case we can use the discussion of TB96 to determine if the point $P(r,\theta)$ is blocked by the star (figure \[disk\_block\] and table \[ilim\]). Defining $C=r\sin(\theta-\alpha)$ the only time the star can block the disk is when $C<R_*$ otherwise $p(r,\theta)=1$. If $C<R_*$ then $p(r,\theta)=1$ only if $r\cos\theta<r_D\cos\theta_D$, where $\theta_D=\pi-\arcsin(C/r_D)$ and $r_D$ is the positive root of the following equation:

$$\left(-H(r_D)\sqrt{1-\left(\frac{C}{r_D}\right)^2}-z_M\right)\sin i+\left(r_D\sqrt{1-\left(\frac{C}{r_D}\right)^2}+x_M\right)\cos i=0$$

where $H(r)=gr_{min}\left(\frac r{r_{min}}\right)^{-n}\cos\alpha$ and M is the point on the northern hemisphere of the star in the plane (P,[**u**]{},z) such that [**u**]{} is tangent to the star at this point:

$$\begin{aligned}
x_M=-\sqrt{R_*^2-C^2}\cos i\\
y_M=C\\
z_M=\sqrt{R_*^2-C^2}\sin i\end{aligned}$$

Now consider inclinations greater than $90^{\circ}$. In this case the disk can still be blocked by the star if the warp is small enough and the inclination angle is close enough to 90 degrees (figure \[disk\_block2\]). This affects both the upper and lower part of the disk, but only from $\theta\in[\alpha+\pi/2,\alpha+3\pi/2]$. The condition for the point P to be hidden by the star is:

$$|h(r,\theta)|<\frac{R_*\cos(i-\pi/2)\tan(\pi-i)-r}{tan(\pi-i)}$$

As above this only applies when $C=R_*\sin(\theta-\alpha)<R_*$.

Another possibility to consider when the inclination is greater than $90^{\circ}$ is that the warp on the lower concave side is steep enough that is blocks part of this side of the disk (figure \[disk\_block3\]). This condition only applies to parts of the lower concave side, from $\theta\in[\alpha+\pi/2,\alpha+3\pi/2]$. The condition is that $\gamma_1<\gamma_2$ where $\gamma_1=i-\pi/2$ and

$$\gamma_2=\arcsin\left(\frac{h_m-h(r,\theta)}{PM}\right)$$

In this case M is the highest point on the disk on the line of sight to point P. For $r\sin(\pi-\theta)>r_{min}$ the disk does block itself, but for $r\sin(\pi-\theta)<r_{min}$ we have $r_m=r_{min}$ and $\theta_m=\arcsin(r\sin(\theta-\alpha)/r_m)$. The quantity PM is the distance between the point P and the point M ($\sqrt{(h_m-h)^2+(r_m-r)^2}$).

Once all of the conditions have been considered the flux from the disk can be calculated using equation \[eqn\_flux\]. These conditions would apply to any type of warp whose maximum height above the midplane occurs at the inner edge of the disk, as opposed to at the outer edge of the disk, regardless of the exact functional form of the warp (ie. power law vs. exponential).

Spiral Wave
===========

Temperature Profile of Spiral Wave
----------------------------------

The third type of disk that we attempt to model contains a spiral wave. As with the warped disks we follow the derivation of TB96 to derive the temperature structure and SED for this disk. The derivation for the temperature structure is very similar to that of the inner warp, only with slightly different definitions of the boundaries. For the part of the disk inside the wave, the disk is not blocked by the wave but the amount of the star seen can change. For points far from the wave, the disk is like a flat disk and $\delta_{min}=0$. For points on the wave, as it rises above the midplane, more of the lower half of the star will become visible. How much of the lower half of the star is visible depends on the location and height of the point on the wave. In this case the lower limit on $\delta$ is:

$$\delta_{min}=-\arctan(h/(r-r_{min}))\\$$

This limit will continue to increase until the point on the wave can see the entire star and then $\delta_{min}=-\delta_{max}$. The other limits stay the same as in the previous models:

$$\begin{aligned}
\varepsilon_{min}=0\nonumber\\
\varepsilon_{max}=\pi/2\nonumber\\
\\
\delta_{max}=\arcsin(R_*/d)\nonumber\\
d^2=r^2+h^2\nonumber\\\end{aligned}$$

For the parts of the disk behind the wave, some of the star may be obscured. In this case:

$$\begin{aligned}
\delta_{min}=\arctan(h_{sw}/(r-r_{sw}))\\
h_{sw}=gr_{min-sw}(1-m\theta/2\pi)\nonumber\\
r_{sw}=r_{min-sw}(1+n\theta)\nonumber\\\end{aligned}$$

This is similar to the concave side of the inner warp, where the warp can obscure part of the star. The only difference is that the maximum height of the disk does not occur at the inner edge of the disk, but instead occurs at the location of the spiral wave. When $\delta_{min}>\delta_{max}$ then the entire star is blocked and that point on the disk is only heated by viscous dissipation. With these definitions and equation (6) we can calculate the temperature of the disk.

Calculating SED of Spiral Wave
------------------------------

The flux from the disk is given by:

$$F_{\nu,{\bf u}}=\int^{R_{disk}}_{r_{min}}\int^{2\pi}_{0}B(T_{disk})f_{up}$$

There is no symmetry in the disk that allows us to split the disk into different parts, as with the concave and convex pieces of the inner warp. We also only consider inclinations less than $90^{\circ}$, where we only see the upper disk, since the lower disk will look the same as the upper disk.

The one occultation effect we include is the blocking of the disk by the wave. For $\alpha-\pi/2<\theta<\alpha+\pi/2$, the near side of the disk, the wave can block the part of the disk that is at smaller radius than the wave. For $\alpha+\pi/2<\theta<\alpha+3\pi/2$, the side of the disk on the other side of the star from the observer, the outer disk can be blocked by the wave. This effect can become important for modest inclinations, given the typical wave heights we consider here. To determine if a point on the disk is blocked by the wave we first need to determine where the line of sight intersects the wave. This is illustrated in figure \[wave\_block1\] and is given by:

$$x = r\sin\theta = r_{min-sw}(1+n\theta_M)\sin(\theta_M)$$

Here $\theta_m$ is the azimuthal coordinate of the point where the wave intersects the line of sight. We assume that the point M lies at the peak of the spiral wave. This is only an approximation, although the narrowness of the wave make it an accurate one. The angle between the line connecting the points P and M and the midplane is $\gamma$ (Fig. \[wave\_block\]). When $\gamma>\pi/2-i$ then point $P(r,\theta)$ is blocked.

$$\tan\gamma=\left(\frac{h_M-h_P}{r_M-r_P}\right)$$

We ignore occultation effects due to the star blocking the disk, which we did consider in the inner and outer warp model. Based on our experience with the warped disks and the typical dimensions of the disk, these are negligible effects that will only play a role very close to edge on. We also do not consider situations where the wave on the near side of the disk can block the far side of the disk. The exclusion of these two effects prevents us from considering the spiral wave at inclinations very close to $90^{\circ}$.

Stellar Flux
============

Next we consider the flux coming from the star. We follow a similar procedure as above where the equation for the stellar flux is modulated by a binary function ($\varepsilon(\phi,\psi)$) which equals 1 when that part of the star is not blocked by the disk and it equals zero when the star is blocked by the disk. The angles $\phi,\psi$, shown in figure \[star\], are the azimuthal and polar angles of a point on the surface of the star relative to the center of the star and the z axis (the same z axis as for the disk). The x-axis of this coordinate system is in the same direction as the line of sight, and will differ from the x-axis of the disk by the angle $\alpha$. The flux from the star is:

$$F_*=B_{\nu}\int\int_{surface}\varepsilon(\phi,\psi)d{\bf A}$$

To determine the surface over which we integrate, we need to know the points of the star that are seen by the observer (ie. which side of the star is facing the observer). These points will be those that have ${\bf u}\cdot d{\bf A}\geq0$ where

$$\begin{aligned}
{\bf u}=\sin i\hat{x}+\cos i\hat{z}\\
d{\bf A}=R_*^2\sin\psi d\psi d\phi(\cos\phi\sin\psi\hat{x}+\sin\phi\cos\psi\hat{y}+\cos\psi\hat{z})\end{aligned}$$

The evaluation of $\varepsilon(\phi,\psi)$ will depend on the type of warp/wave and the orientation of the observer. First consider inclinations less than $90^{\circ}$. In this case the warp/wave streching above the midplane may block some of the star. The entire star will be blocked if the following condition is met:

$$h(r,\alpha)-r\tan(\pi/2-i)>R_*$$

where r is location of the peak of the vertical disturbance and $h$ is the maximum height of the warp or wave at the angle $\alpha$. The exact value of $r$ and $h$ will depend on whether we are considering the outer warp, inner warp, or spiral wave (ie. $r=r_{min}$, $h(r,\alpha)=gr_{min}\cos(\alpha)$ for the inner warp). This condition is illustrated in figure \[star\_block\] for the inner warp, which is the disk that is the most likely to occult the star.

None of the star will be blocked if the inclination is less than $i_{lim}$ (discussed earlier). A generic version of the equation for $i_{lim}$ that can be applied to all of the disks is:

$$\tan i_{lim}=\frac{r-R_*\cos i_{lim}}{h(r,\alpha)+R_*\sin i_{lim}}$$

When the inclination falls between these two limits only a fraction of the star is blocked. We can use the discussion of TB96 section A2.2 to determine if a point of the star’s surface is hidden by the star

The point Q is a point on the surface of the star that intersects the line of sight and the upper edge of the disk. If the point $N(\phi,\psi)$ lies above Q then the observer can see this part of the star, otherwise it is hidden and $\varepsilon(\phi,\psi)=0$. The vertical coordinate of Q, $z_Q$, is the greatest root of the following equation:

$$(1+\tan^2 i)z_Q^2-2\tan i(h\tan i-r\cos\alpha)z_Q+(h\tan i-r\cos\alpha)^2-R_*^2+R_*^2\sin^2\psi\cos^2\phi=0$$

If $z_N\geq z_Q$ then $\varepsilon(\phi,\psi)=1$ otherwise $\varepsilon(\phi,\psi)=0$ with $z_N$ being given by:

$$z_N=-R_*\sin\psi\sin\phi\sin i+R_*\cos\psi\cos i$$

For the inner and outer warp we consider inclinations greater than $90^{\circ}$ where the disk can still block part of the star, although it is less likely because the disk curves away from the observer. This is illustrated in figure \[star\_block2\] for the inner warp, but can also apply to the outer warp in the limit that $h$ goes to zero. The point at which the line of sight is perpendicular to the normal of the disk sets a limit to the distance z above the midplane that an observer can see. If this distance is less than the radius of the star then some of the star is blocked by the disk. The point D at which the the line of sight is perpendicular to the disk occurs when ${\bf u}\cdot{\bf n}=0$ (figure \[star\_block2\]). If this condition is met at a radius $r_D$ then a point $N(\phi,\psi)$ on the stellar surface will be blocked if

$$R_*\cos\psi>h(r_D,\alpha)+r_D\tan(i-\pi/2)$$

If there is no point at which ${\bf u}\cdot{\bf n}=0$ then $r_D=r_{min}$ and the same condition for being able to see the star is used. When this condition is met $\varepsilon(\phi,\psi)=0$, otherwise $\varepsilon(\phi,\psi)=1$.

All of these different occultations are combined to determine $\varepsilon(\phi,\psi)$. The flux is determined by integrating over the entire surface and added to the flux from the disk to create the observed SED.
