When to chose Uniform (Isotropic) or Cos (Lambertian/Effusive) Desorption

General discussion
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Lately I've been think about this question so I started to look back at the theroetical derrivations and experiemental data and how this translates into single trajectory simulations like molflow. I'm still not satified that I understand this completely yet, but will muse this more over the weekend. I thought I would start this thread if others want to contribute to this. This would also apply to the type of reflection you choose for a facet, uniform or diffuse which I assume means isotropic or cos/lambertian respectively.

This is what I have so far. Experimentally, reflection is diffuse/cos/lambertian (The Physical Basis of UHV, Redhead) for 99.9% of most "working" surfaces used in UHV. For crystaline surfaces you can get some spectular reflection and diffraction. Also the more likely gas accommodates to a surface the more likely it will also reflect as diffuse/cos/lambertian. This is because reflection, in this case, is a two step procecess of adsorption then desorption; also from Redhead. This would imply that any desorption source should also be diffuse/cos/lambertian.

However, when I assume particles leave a surface with a 1-sided MWBD (Maxwell-Boltzmann velocity Distribution) I obtain an isotropic source. I have not worked backwards to see what type of velocity distribution you get when you assume a diffuse/cos/lambertian source. I would have guess particles would leave as a 1-sided MWBD, but the experimental data says otherwise.

So do real isotropic sources exist? I guess a sphere whose outer surface is diffuse/cos/lambertian would become isotropic if treated as a point source. Point sources don't exist in molflow so should you always choose diffuse/cos/lambertian in order for it to be physically real? Maybe I've got it backwards. Perhaps all particles leave a "flat" surface isotropically (1-sided MWBD), but a real "working" UHV surface is rough on a microscopic scale so is made up of many "flat" surfaces pointing in many directions. I wonder if this produces a diffuse/cos/lambertian distribution? If this is so, then would a highly polished SS surface behave as a "flat" surface?

If you have a volume with a MWB gas in it and with a small apeture (small compared to the rest of the surface area of the volume), then gas comming out of the apeture will be diffuse/cos/lambertian.

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Thank you Alan for the input.

I believe, that as you pointed out, all technical vacuum surfaces are Lambertian reflectors. Any other case would invalidate basic vacuum theory: analytic conductance formulas couldn't be derived from an infinite sum of "random walk" rebounces.

As for gas sources, including effusion from a volume in equilibrium, they are also Lambertian except in some special cases, as I'll explain below. If you would take any geometry with uniformly outgassing walls, for example a box, and deviate from Lambertian desorption, pressure distribution wouldn't be uniform but you would have local peaks in corners / in the center, depending if you deviate towards isotropic or beamed distributions.

For example, this cube has uniform isotropic desorption on its walls:

Image

Molflow includes the Uniform desorption option to approximate spherical gas sources (like a spark spraying molecules at an RF discharge) with a small, 2-sided facet. This is closest to a point source.

It also includes beamed distributions, in Cos^N form, to allow to "collimate" desorption direction at the entrance of tubes, where due to the beaming effect the prevailing desorption direction is the axis of the tube.

While we sometimes use the Uniform desorption option, as above, we never use the Uniform reflection option: in fact, I included it on Roberto's request who wanted to prove that a to be reviewed paper miscalulcated because it used Uniform reflection.

The Lambertian desorption is established by Knudsen's experiments in the 1800s, whether the microscopical explanation is the micro-roughness or something else, I have to think about.

Could you explain a bit your sentence "However, when I assume particles leave a surface with a 1-sided MWBD (Maxwell-Boltzmann velocity Distribution) I obtain an isotropic source."? I didn't fully get it.

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Particles leaving a surface don't have velocity component toward the surface. I call a MWBD missing the particles moving away in one of the velocity components a 1-sided MWBD. Here are my notes in derriving the equations for implimenting a random 1-sided MWBD in code. In this case I take 'z' to be the direction normal to surface and only integrate over Vz velocities for particles moving away. The distribution is normailzed to one over this distribution. The functions for the angles should lead to an isotropic distribution of particle trajectories. For some reason the image loader is rotating my image.

1-sided MWBD

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Marton Thanks for your reply,

Technically, I didn't check that it was isotropic, but I think it will, and I'll double check this. I started with a 1-sided MWBD and got isotropic. However, I also know that effusion through a hole has a 1-sided MWBD too. I guess they both have 1-sided MWBD, but in the case of the hole you pickup the extra Cos factor from the volume of particles behind that hole that could "possible" reach the hole at the angle. This includes particles going away from the hole, but those get removed when you only multipy this by the 1-sided MWBD. In the case of a particles coming from a "flat" surface rather than from a volume behind the hole you wouldn't have this extra Cos factor that makes it cos/lambertian. Something is still not making sense to me about this difference. Like I said previously, maybe a true "flat" surface is really isotropic, but we don't have true flat surfaces. Here is a decent derivation of effusion through a hole.

FYI, here is the general procedure I use for generating random particles with a certain distribution function. It includes examples of exponetial decay, 1D & 3D MWBD. Obviously, molflow must do the same. Hopefully your got your normalizations done correctly.

Random Note 1

Random Note 2

Random Note 3

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Yes, Molflow also uses the numerical inversion method to convert a RND number to a distribution. I use exactly the same formula for exponential decay ("pro tip": 1-RND is the same as RND :). For the particle speeds Molflow doesn't generate according to the M-B distribution, but according to the speed distribution of particles crossing a plane in equilibrium. The difference is small, but this new distribution is shifted towards faster velocities.

All of this is written in this - unfortunately rejected - article:

molflow.pdf