# Solid angle confusion

1. Dec 21, 2014

### albega

In deriving the pressure of a gas, my book states that
'if all molecules are equally likely to be travelling in any direction, the fraction whose trajectories lie in an elemental solid angle dΩ is dΩ/4π'.

This initally made sense to me, but then thinking about it, I wrote dΩ=sinθdθdφ and this means that the fraction whose trajectories lie in an elemental solid angle dΩ is sinθdθdφ/4π. This is confusing me, because the fraction whose trajectories lie in an elemental solid angle varies with θ, which I find a little contradictive given what I have quoted in the first paragraph.

One thing I have noted is that if you do the φ integral, you still have the θ dependence, and this makes sense because as θ varies the remaining annular region gets bigger.

This is just something I have never thought about before, because I then noted that the surface area of a unit sphere is
∫dA=∫dΩ=∫sinθdθdφ
and thus the area element at the top and bottom of the sphere do not contribute to the area as θ=0,π.

So, before I confuse myself even further, would anybody be able to explain why the solid angle element is varying with θ, and how this makes sense when the molecules are supposed to be equally likely to be travelling in any direction and the fraction in some elemental solid angle is dΩ/4π (which varies). Thankyou :)

2. Dec 21, 2014

### Staff: Mentor

It does not. The way your coordinates change within a small solid angle changes, and the sine term accounts for this. It is purely an issue of the coordinate system.

Yes, you have more solid angle per θ at θ=pi/2 than you have at θ=pi/4.

3. Dec 21, 2014

### albega

But the fraction whose trajectories lie in an elemental solid angle dΩ is dΩ/4π, so at θ=0 the fraction is zero and at θ=π/2 the fraction is non-zero. I don't understand how that is because of the coordinates?

4. Dec 21, 2014

### Staff: Mentor

The amount for a single angle is always zero because a single angle does not span any solid angle.
The fraction between some θ and $\theta+\delta \theta$ depends on θ, that should not be surprising and the sine in the solid angle element reflects that.

5. Dec 22, 2014

### albega

Why should it not be surprising?

I suppose I'm mostly confused about
'if all molecules are equally likely to be travelling in any direction, the fraction whose trajectories lie in an elemental solid angle dΩ is dΩ/4π'
now.

What does it actually mean, to say have a gas in a volume, choose some axis and origin with a spherical coordinate system and angle θ, and find the amount whose trajectories lie in a the range θ,θ+dθ to that axis? What confuses me is, the gas particles are moving all over the place, it's not as though they all start out at the origin and move in their required directions. I guess it makes more sense in my head if I imagine we collect all the particles at the origin at t=0 and let them disperse.

Last edited: Dec 22, 2014
6. Dec 22, 2014

### Staff: Mentor

There is no reason to assume it to be the same, and you already found the "rings" of different size for different small regions of θ.

The total solid angle is 4 pi. This is just a normalization.

"If a value is equally likely to be any real value between 3 and 5, the fraction of values within a range dx is dx/2".
Don't show that to mathematicians because it is an improper way to use a differential, but it should show the idea.

7. Dec 22, 2014

### Stephen Tashi

Instead of thinking about the volume as a volume of gas, you could think about the result applied to a surface that was the result of plotting data. You pick a volume of gas. You measure the velocity of each molecule in the gas and you plot the the result as a unit vector drawn from the origin of an (x,y,z) cartesian coordinate axes. Your data points fall uniformly on the surface of a unit sphere since you are using unit vectors. A given solid angle with vertex at the center of the sphere tells you about the fraction of the total data points that are insde the angle. To get the number of data points inside a given angle, you have to multiply that fraction times number of molecules that were in the volume of gas you measured.

8. Dec 29, 2014

### albega

Thanks - this was the sort of thing I wanted... I managed to satisfy myself before seeing this anyway by imagining freezing all the particles, bringing them to my origin and then letting them go, which is similar to what you said.