PDF of uniform distribution on surface of a sphere

[sanctus]

Hi all,

 

I am looking for as the title says for the probability density function of the uniform distribution on the surface of the sphere.

 

More in detail I have symmetry in [math] \phi[/math] ([math] \phi[/math] ranges from 0 to 2pi and [math] \theta[/math] from 0 to pi) and would like to know what

[math] f(\theta)=\int f(\theta,\phi)[/math] is.

 

My guess is that it is wrong to say that it is just [math]\frac{1}{\pi}[/math] but I am not sure.

 

Maybe to explain why I need this, I want to show how far from the uniform distribution the integral (second moment) I have is:

[math]\int_0^\pi \theta^2 \vert\hat{\Psi}(\theta)\vert[/math]

where [math]\vert\Psi\vert[/math] should be a density since defined such that:

[math]\int_0^\pi \vert\hat{\Psi}(\theta)\vert=1[/math].

 

Also, what I am not sure is if the marginalized density can be integrated just wrt to [math]d\theta[/math] or not...

Hope my question is understandable.

[Jay-qu]

Wouldnt the probability density by something like total probability divided by the area

[math]\frac{1}{4\pi r^2}[/math]

[sanctus]

But isn't that for a surface element? I.e. for having a point in the area [math] [\theta, \theta+d\theta]\times [\varphi,\varphi+d\varphi][/math]

? But in my case I fix [math]\varphi=0[/math]

 

But overnight I convinced myself that it is [math]\frac{1}{\pi}[/math], but still not 100% sure

[sanctus]

Now I am sure, it is 1/pi...because I just want an uniform distribution over the interval [0,pi] and hence do not need to care if I am on the surface of a sphere or not (since the increase in radians is directly proportional of the increase in angle).

This would not be the case if I want an uniform distribution over the surface of the sphere because then to calculate a probability the solid angle (with its sine) enters with the result that if you make just an uniform distribution on the two angles you get a bigger concentration at the poles...

[Qfwfq]

Isn't it a matter of using the Jacobian determinant?

[sanctus]

Yes, I read that somewhere, but I do not think that for my case with varphi fixed it is important only when you want an unifrom distribution over the whole surface of the sphere...

[Qfwfq]

  sanctus said:

...with \varphi fixed it is important only when you want an unifrom distribution over the whole surface of the sphere...

Indeed, cuz the Jacobian for unit r is due only to the [imath]d\varphi[/imath] but that, of course, means that you're not really talking about the spherical surface after all, only about a circumferemce! :hyper:

 

Edit: Actually, the matter is subtle. It depends on whether you must restrict the spherical distribution to a meridian, or actually need the distribution on the circumference itself. Are you sure of what you really need to do? The fact that you are dividing by [imath]\pi[/imath] suggests you should also have the cosine of [imath]\theta[/imath].

[sanctus]

Yep, talking about a circumference, or actually about half a circumference--> a meridian.

 

Then nothing changes and 1/pi is right...me thinks.