Eccentric Disk

Eccentric Disk Model

Following the approach of Hughes, Factor, Chiang et al., we model an eccentric circumstellar disk as a series of apse-aligned ellipses. We model the mass distribution within each ellipse using a mass-on-a-wire approach. In this document, we derive the mass distribution for a single ellipse as a function of orbital position.

To start, we assume that the \(i\)th ellipse we have chosen to model has a total mass of \(m_i\). The linear density of this wire, \(\lambda_i\), with units [g/cm], is proportional to the amount of time spent in that section of the orbit. This means that the density will be largest at apoapse, when a test particle in the orbit is traveling the slowest. Conversely, the density will be smallest at periapse, when the test particle is moving fastest. The linear density is uniquely defined as a function of orbital position. We can either measure the orbital position in terms of path length along the perimeter of the ellipse, \(s\), where \(s=0\) denotes periapse, or, we can measure orbital position as a function of true anomaly1 \(\theta\), which is the same as the angle \(f\) in Meredith’s notes.

Once we have defined \(\lambda_i\), we can check that we have the right definition by integrating the linear density over the perimeter of the elliptical orbit–we should recover \(m_i\).

Derivation of linear density

After making the assumption that the linear density should be proportional to the amount of time spent in the orbit, we can use the equations of orbital motion to solve for \(\lambda_i\). The amount of time \(dt\) spent in a section of the wire \(ds\) is the inverse of the velocity at that orbital position

\[\frac{dt}{ds} = \frac{1}{v}\]

This same section of the wire \(ds\) also contains a small chunk of mass \(dm\), which provides the definition of linear density

\[\lambda_i = \frac{dm}{ds}\]

Because the elliptical orbit is closed, we know the normalization terms for each of these quantities. The total mass in the ring is \(m_i\) and the total amount of time spent in the orbit is the period

\[T = \frac{2 \pi}{\mu^2} \frac{h^3}{(1 - e^2)^{3/2}}\]

where \(e\) is the eccentricity of the wire, \(\mu = G M_\ast\) is gravitational parameter, and \(h\) is the angular momentum. Therefore, the fractional amount of time spent in the section of the wire should be equal to the fractional amount of mass contained in that section \[\frac{dt/ds}{T} = \frac{dm/ds}{m_i} = \frac{\lambda_i}{m_i}\]

Solving for \(\lambda_i\) (units of [g/cm]), we have

\[\lambda_i = m_i \frac{dt/ds}{T} = \frac{m_i}{T} \left(\frac{1}{v} \right)\]

We can find the velocity using orbital equations for conservation of mechanical energy for an ellipse

\[\frac{v^2}{2} - \frac{\mu}{r} = \frac{-\mu}{2 a}\]

where \(a\) is the semi-major axis of the ellipse,

\[a = \frac{h^2}{\mu} \frac{1}{1 - e^2}\]

we can also find the radial distance \(r\) of the orbital position \[r = a \frac{1 - e^2}{1 + e \cos \theta}\]

Using these equations and the equation for the period \(T\), we can solve for \(\lambda_i\). First, we rearrange the mechanical energy equation for velocity \[v = \sqrt{\mu \left ( \frac{2}{r} - \frac{1}{a} \right )}\]

and plug in for \(r\) \[v = \sqrt{\mu \left ( \frac{2 ( 1 + e \cos \theta)}{a (1 -e^2)} - \frac{1}{a} \right )} = \sqrt{\frac{\mu}{a}} \sqrt{\frac{1 + 2 e \cos \theta + e^2}{1 - e^2}}\]

Before we plug in for the period \(T\), let’s first simplify by using the equation for semi-major axis \(a\) to eliminate \(h\) \[h = \sqrt{ a \mu (1 - e^2)}\] and so we have \[T = \frac{2 \pi}{\mu^2} \frac{a^{3/2} \mu^{3/2} (1 - e^2)^{3/2}}{(1 -e^2)^{3/2}} = \frac{2 \pi a^{3/2}}{\sqrt{\mu}}\] Finally, solving for \(\lambda\) \[\lambda_i = \frac{m_i}{T} \left(\frac{1}{v} \right) = m_i \frac{\sqrt{\mu}}{2 \pi a^{3/2}} \frac{\sqrt{a} \sqrt{1 - e^2}}{\sqrt{\mu} \sqrt{1 + 2 e \cos \theta + e^2}}\] \[\boxed{ \lambda_i = \frac{m_i}{2 \pi a} \frac{\sqrt{1 - e^2}}{\sqrt{1 + 2 e \cos \theta + e^2}} \quad \quad [\textrm{g} / \textrm{cm}] }\]