Optically thin clouds and the Eddington Limit

As part of my independent study I’m going through “Radiative Processes in Astrophysics” by George B. Rybicki and Alan P. Lightman as a way to inform my work. This is in tandem with the other things I’m doing like the literature review and working with HYPERION so this is a pretty light introduction. I plan on working on one problem for each chapter that seems particularly informative or relevant. This first problem is from chapter 1 Fundamentals of Radiative Transfer and it’s problem 4.

1.4a] Show that the condition that an optically thin cloud of material can be ejected by radiation pressure from a nearby luminous object is that the mass to luminosity ration (M/L) for the object be less than, \kappa/(4\pi Gc)
where G = gravitational constant, c = speed of light, k = mass absorption coefficient of the cloud material (assumed to be independent of frequency)

To begin let’s assume that the luminous object is spherically symmetric. In that case we can use the standard definition of flux,

F=\dfrac{L}{4\pi r^2}

From equation (1.34) the force per unit mass is given by,

f=\dfrac{1}{c}*\int \dfrac{\kappa L}{4pi r^2}dv

Since cloud is independent of frequency of the frequency dependencies falloff and we’re left with,

f=\dfrac{\kappa L}{c4pi r^2}dv

In order for material to get objected the radiative pressure needs to be larger than the force of gravity being exerted from the object to the cloud,

\dfrac{GM}{r^2} < \dfrac{\kappa L}{c4pi r^2}

Then simple algebra shows that,

\dfrac{M}{L} < \dfrac{k}{4 \pi c G}

1.4b] Calculate the terminal velocity v attained by such a cloud under radiation and gravitational forces alone, if starts from rest and a distance R from the object.

Terminal velocity is when the gravitational force is equal to the counter force (for skydivers it’s drag, for this example it’s the radiation pressure). So let’s make a force called Geff that’s the radiation pressure subtracted from the gravitational force. This is the force that the cloud actually feels. We can set the Geff force equal to the general velocity equation and solve for v.

1.4c] A minimum value for kappa may be estimated for pure hydrogen as that due to Thomson scatter off free electrons, when the hydrogen is completely ionized. Show that the maximum luminosity that a central mass M can have and still not spontaneously eject hydrogen by radiation pressure is,

L_{EDD}=\dfrac{4\pi Gcm_H}{\sigma_T}

This limit is called the Eddington limit! And it’s really cool! It is usually spoken about the context of the accretion disks of black holes. But do you know what else has accretion disks? Pre-main sequence stars such as RR Tau. Basically, when you have a black hole (or star or what have you) with an accretion disk there are photons coming off the disk. This is why Quasars are so bright. Let’s think about why accretion disk have energy to create photons in the first place. An accretion disk forms when an object with a deep gravity well starts sucking objects toward it. The object it has gravitational potential energy which then gets converted to kinetic energy as it gets closer to the object. This is why accretion disks spin and why they’re luminous. The Eddington Limit basically sets a limit for the rate of accretion. The reason is clear. In our everyday life we’re not knocked over by the light coming off of our lamps even though the the photons coming from our lamps have radiation pressure it just simply isn’t enough to affect us in any significant way. But the the combined pressure of photons from an accretion disk can add up to be enough to start counteracting the force of gravity. So when the radiation pressure is larger than the gravitational force the Eddington Limit has been exceeded. When radiation pressure is equal to gravity that is the Eddington Limit. The answer to this problem is simple. You just make the solution to part a an equality as sub in the mass scattering coefficient for kappa.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s