<p>If your $r_{peak}$ is not known a priori then you have two unknowns and only one equation: this cannot be solved. At least not without a second equation or a measured $(\rho,r)$-pair (say the central density $\rho_0 = \rho(r=0)$). </p>

<p>In any case, this definition of radial density looks like an ad-hoc assumption to me, and not something that came out of a proper hydrodynamic model. So without knowing why you chose this form or the context of your disk model I can't comment on what equation is best suited to determine $r_{peak}$. A well-informed guess seems like the way to go.</p>

CC BY-SA 3.0
2014-11-26T19:48:04.347Z
This is pretty much what I wanted. The equation(s) came out of an appendix of Michael M. Woolfson's *On the Origin of Planets*.
HDE 226868
CC BY-SA 3.0
2014-11-26T16:58:25.907Z
2014-11-26T16:58:25.907Z
Michael B.
<p>I'm working with some equations to model the evolution of a circumstellar disk. One of the equations is
$$\rho(r)=Ce^{-\frac{(r-r_{peak})^2}{2 \sigma ^2}}$$
where $\rho$ is density, $r$ is the distance from the center, $C$ is a constant, $\sigma$ is one standard deviation, and $r_{peak}$ is the radius at which the density is at a maximum.</p>

<p>If the function was of the form
$$\rho(r)=Ce^{f(r)}$$
where $f(r)$ is a function of $r$, I could find the maximum easily by finding
$$\rho'(r)=Cf'(r)e^{f(r)}=0$$
and solving for $r$. However, this appears to be impossible in the current case because $\rho(r)_{peak}$ is already in the equation, at $r_{peak}$.</p>

<p>How is $r_{peak}$ determined in a given scenario? Is it determined experimentally?</p>

CC BY-SA 3.0
2014-11-26T01:21:13.773Z
Note: Some more parts of the problem are in the Sandbox post I started on Meta; I've been preparing some math in case it was needed to explain the scenario, or in case I ask other future questions about the problem.
HDE 226868
CC BY-SA 3.0
2014-11-28T20:57:54.400Z
If $r$ is the spherical radius (which is the usual convention for this symbol), then this is not a disc at all... If it is cylindrical radius, it is not a disc either, but a cylinder, since there is no $z$ dependence.
Walter
CC BY-SA 3.0
2014-11-28T22:45:13.843Z
@Walter This *is* a disk. While I didn't explain properly, this equation is valid for the areal density of the mean plane. There is a separate function for the density at a point on the $z$ axis which is also expressed exponentially; there is a drop off as the distance from the mean plane increases, meaning that after a short distance, that density becomes completely negligible. It is technically a cylinder, but it is, for all intents and purposes, a disk. Besides, a disk is really a very short cylinder, right? I can explain the $z$-dependent equation of you want.
HDE 226868
<disk>
How is the maximum density in a circumstellar disk determined?
HDE 226868
# How is the maximum density in a circumstellar disk determined?

(Asked by: username_0 on 2014-11-26)

I'm working with some equations to model the evolution of a circumstellar disk. One of the equations is
$$\rho(r)=Ce^{-\frac{(r-r\_{peak})^2}{2 \sigma ^2}}$$
where $\rho$ is density, $r$ is the distance from the center, $C$ is a constant, $\sigma$ is one standard deviation, and $r\_{peak}$ is the radius at which the density is at a maximum.

If the function was of the form
$$\rho(r)=Ce^{f(r)}$$
where $f(r)$ is a function of $r$, I could find the maximum easily by finding
$$\rho'(r)=Cf'(r)e^{f(r)}=0$$
and solving for $r$. However, this appears to be impossible in the current case because $\rho(r)\_{peak}$ is already in the equation, at $r\_{peak}$.

How is $r\_{peak}$ determined in a given scenario? Is it determined experimentally?

**Tags:** disk
> **Comments:**
> **username_0**: Note: Some more parts of the problem are in the Sandbox post I started on Meta; I've been preparing some math in case it was needed to explain the scenario, or in case I ask other future questions about the problem.  
> **username_1**: If $r$ is the spherical radius (which is the usual convention for this symbol), then this is not a disc at all... If it is cylindrical radius, it is not a disc either, but a cylinder, since there is no $z$ dependence.  
> **username_0**: @Walter This *is* a disk. While I didn't explain properly, this equation is valid for the areal density of the mean plane. There is a separate function for the density at a point on the $z$ axis which is also expressed exponentially; there is a drop off as the distance from the mean plane increases, meaning that after a short distance, that density becomes completely negligible. It is technically a cylinder, but it is, for all intents and purposes, a disk. Besides, a disk is really a very short cylinder, right? I can explain the $z$-dependent equation of you want.  

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## 1 answers

### Answer by username_2 on 2014-11-26. Score: 1 | Accepted answer

If your $r\_{peak}$ is not known a priori then you have two unknowns and only one equation: this cannot be solved. At least not without a second equation or a measured $(\rho,r)$-pair (say the central density $\rho\_0 = \rho(r=0)$).

In any case, this definition of radial density looks like an ad-hoc assumption to me, and not something that came out of a proper hydrodynamic model. So without knowing why you chose this form or the context of your disk model I can't comment on what equation is best suited to determine $r\_{peak}$. A well-informed guess seems like the way to go.

> **Comments:**
> **username_0**: This is pretty much what I wanted. The equation(s) came out of an appendix of Michael M. Woolfson's *On the Origin of Planets*.  

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