Integration ## f(\theta, \phi) = \frac{sin \theta}{4\pi}##

[happyparticle]

Homework Statement

integral

Relevant Equations

$f(\theta, \phi) = \frac{sin \theta}{4\pi}$

Hi,

I have this formula $f(\theta, \phi) = \frac{sin \theta}{4\pi}$

I have this statement that say if I integrate this formula above on a sphere then p = 1.

what does integrate on a sphere means? I know $\int_0^{2\pi}$ is used for the circle.

 

[haruspex]

How are θ and φ defined, what variable or variables are you integrating over, and what is p?
It sounds like this is integrating over the surface of a sphere using spherical coordinates.

 

[happyparticle]

I can't really answer your questions. That's all I know.
I have to use spherical coordinates.

I'm not sure how to work with this formula to get p = 1.

 

[haruspex]

I'm not sure how to work with this formula to get p = 1.

If p is not mentioned anywhere else, of course you cannot know how to get p=1.
Neither can you get x=1.3, nor h = a banana.

 

[Delta2]

You have got to tell us at least what p is otherwise we cannot help you.

Homework Statement:: integral
Relevant Equations:: $f(\theta, \phi) = \frac{sin \theta}{4\pi}$

what does integrate on a sphere means?

To integrate a function $f(r,\theta,\phi)$ over a sphere of radius $R$ means that we have to calculate the integral $$\int_0^{\pi}\int_0^{2\pi}f(R,\theta,\phi)R^2\sin\theta d\phi d\theta$$

 

[happyparticle]

I think P means the radius.

 

[Delta2]

I think P means the radius.

The radius of the sphere upon which's surface we integrate must be given as initial data, it cannot be something that we ll calculate.

 

[haruspex]

I think P means the radius.

Please provide the question exactly as given to you, even if it is not in English.

 

[happyparticle]

Alright, it is not in english.

La simulation de dispersion est un processus en deux étapes. On commence par un test probabilistedu même genre que décrit précédemment pour l’absorption, afin de décider si le neutron sera dévié ou non. S’il est dévié, nous devons calculer un angle de déviation. Dans le cas de l’interaction d’un neutron avec un noyau atomique non-excité, tous les angles sont équiprobables. En trois dimensions spatiales, ettravaillant en coordonnées sphériques avec la direction horizontale comme axe de symétrie (disons), la probabilité de dévier le neutron vers les angles polaire et azimutal (θ,φ) est donnée par :p(θ,φ) =sinθ4π.
Je vous laisse également le soin de vérifier qu’intégrer cette distribution sur la sphère conduit bien à p= 1.

 

[PhDeezNutz]

I assume by integrating over a sphere you literally mean integrating over the surface of a sphere and not its interior. If that is the case I'm getting that the radius of the sphere must be $\frac{2}{\sqrt{\pi}}$ in order for the integral to be 1. Is there anything in your book or notes that indicate that this would be the case?

It looks like a familiar normalization constant.

 

[PhDeezNutz]

@EpselonZero Please disregard my last post. I think I now understand your question. Maybe the more reputable posters in this thread can confirm (or refute).

If we have an incoming neutron it gets scattered as an outgoing spherical wave which has the form

$\Psi = f \left( \theta, \phi \right) \frac{e^{ikr}}{r}$

As you probably know from your Quantum Mechanics classes, to find the probability density you take the quantity

$\left| \Psi \right|^2 = \Psi \times \Psi^* = \frac{1}{16 \pi^2} \frac{\sin^2 \theta }{r^2}$ and integrate over the domain (in this case a spherical surface)

The only problem is that when I integrate this over the surface of a sphere I'm getting $\frac{1}{16}$. I feel like there is a normalization constant that is off in the original problem statement.

 

[haruspex]

Google translation:
—
Dispersion simulation is a two-step process. We start with a probability test of the same kind as described above for absorption, in order to decide whether the neutron will be deflected or not. If it is deflected, we need to calculate a deflection angle. In the case of the interaction of a neutron with an unexcited atomic nucleus, all angles are equally likely. In three spatial dimensions, and working in spherical coordinates with the horizontal direction as the axis of symmetry (say), the probability of deflecting the neutron towards the polar and azimuthal angles (θ, φ) is given by: $p (θ, φ) = \frac{\sin(θ)}{4π }$.
I also leave it to you to verify that integrating this distribution on the sphere does indeed lead to p = 1.
—
So it appears the task is to show that
$$\int_0^{\pi}\int_0^{2\pi}\frac{\sin(θ)}{4π }\sin\theta d\phi d\theta=1$$

 

[Charles Link]

It looks like we are working with an extra $\sin{\theta}$ factor. With only one $\sin{\theta}$ the integral is normalized to unity. With a second power, it does not give unity, but rather $\pi/4$. Perhaps the author intends $p(\theta,\phi)$ to be weighted simply by $d \theta$ and $d \phi$.

 

[haruspex]

Rereading the translation, this statement is meaningless:

The probability of deflecting the neutron towards the polar and azimuthal angles (θ, φ) is given by: $p (θ, φ) = \frac{\sin(θ)}{4π }$.

The probability of any such precisely specified deflection is zero. I believe what it intends to say is that is the probability of deflecting into a solid angle $d\phi\times d\theta$.
If so, the integral is
$$\int_0^{\pi}\int_0^{2\pi}\frac{\sin(θ)}{4π } d\phi d\theta=1$$,
as @Charles Link suggests.

 

[Charles Link]

In reading the translation more carefully, the author says that "all angles are equally likely", i.e. the $p$ will be a constant when taking $p \, d \Omega$, where $d \Omega=\sin{\theta} \, d \theta \, d \phi$. It appears that is what the author is doing, but he didn't provide extensive description. This is a spherically symmetric scattering, which is the simplest case possible.