Flux in spherical coordinates

In summary: Therefore, we can rewrite the integral as:J_{1}= 0.5625* I_{v} \int_{0}^{\frac{\pi}{2}} cos^{2}\theta sin\theta d\thetaNow, to evaluate this integral, we can use the substitution u = cos\theta, which gives us du = -sin\theta d\theta. Substituting this into the integral, we get:J_{1}= -0.5625* I_{v} \int_{1}^{0} u^{2} du = -0.1875* I_{v} = -0.187
  • #1
genloz
76
1

Homework Statement


Calculate the rate at which muons pass through a flat plate of area A.


Homework Equations


[tex]J_{1}=\int_{\theta\leq\frac{\pi}{2}}j(\theta,\phi)cos\theta d\Omega[/tex]
[tex]d\Omega[/tex] = sin[tex]\theta d\phi d\theta[/tex] in spherical coordinates.
[tex]j(\theta,\phi)[/tex] is the angular distribution of muons at sea level.
At sea level [tex]j(\theta,\phi) \approx cos^{2}\theta[/tex]
More specifically [tex]j(\theta = 0,\phi) \equiv I_{v}[/tex] which equals the flux per unit solid angle per unit horizontal area per second about the vertical direction.
[tex]I_{v} = 1.1*10^{2}m^{-2}sec^{-1}sterad^{-1}[/tex] for all penetrating particles at sea level of which 75% are muons.



The Attempt at a Solution


I'm very confused about spherical coordinates... this equation is just a tiny part of my muon lifetime prac but I'm not sure at all how to go about this integral. Any help would be greatly appreciated.

My attempt is as follows:
[tex]J_{1}=0.75* \int_{\theta\leq\frac{\pi}{2}} I_{v} cos\theta sin\theta d\phi d\theta[/tex]
[tex]J_{1}=0.75* 1/2 \int_{\theta\leq\frac{\pi}{2}} I_{v} sin2\theta d\phi d\theta[/tex]

this is where I get stuck... I know that the integral of [tex]sin2\theta[\tex] is [tex]-1/2 cos2\theta[\tex]... but I'm confused about what I'm evaluating between... and what to do with the [tex]\phi[\tex] after that step...

Thanks in advance!
 
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  • #2


To calculate the rate at which muons pass through a flat plate of area A, we can use the equation J_{1}=\int_{\theta\leq\frac{\pi}{2}}j(\theta,\phi)cos\theta d\Omega, where J_{1} represents the flux per unit horizontal area per unit time passing through the flat plate.

To solve this integral, we first need to understand the variables involved. The integral is taken over the solid angle d\Omega, which is defined in spherical coordinates as d\Omega = sin\theta d\phi d\theta. In this case, \theta represents the angle between the direction of the muon and the normal to the flat plate, and \phi represents the azimuthal angle around the vertical direction.

As given in the homework equations, the angular distribution of muons at sea level is j(\theta,\phi)\approx cos^{2}\theta. This means that the flux per unit solid angle per unit horizontal area per second about the vertical direction is I_{v}=1.1*10^{2}m^{-2}sec^{-1}sterad^{-1} for all penetrating particles, of which 75% are muons.

Substituting these values into the integral, we get:

J_{1}=0.75* \int_{\theta\leq\frac{\pi}{2}} I_{v} cos^{2}\theta sin\theta d\phi d\theta

Since we are only considering the muons passing through the flat plate, we can rewrite this as:

J_{1}=0.75* \int_{\theta\leq\frac{\pi}{2}} 0.75I_{v} cos^{2}\theta sin\theta d\phi d\theta

= 0.5625* \int_{\theta\leq\frac{\pi}{2}} I_{v} cos^{2}\theta sin\theta d\phi d\theta

= 0.5625* \int_{0}^{\frac{\pi}{2}} I_{v} cos^{2}\theta sin\theta d\phi d\theta

Since the integral is taken over the solid angle d\Omega, we need to integrate with respect to both \theta and \phi. However, in this case, the integral with respect to \phi is simply the integral
 
  • #3


I would suggest starting by reviewing the concept of flux in spherical coordinates and familiarizing yourself with the equations and units involved. From there, you can approach the problem by breaking it down into smaller steps.

First, you should determine the solid angle dΩ in terms of θ and φ. This can be done using the formula provided in the homework equations section.

Next, you can substitute the given values for j(θ, φ) and I_v into the integral, keeping in mind that j(0, φ) = I_v.

Then, you can evaluate the integral using the limits of integration for θ and φ. In this case, since we are only interested in the flux passing through a flat plate (θ ≤ π/2), the limits of integration for θ will be 0 and π/2, while the limits for φ will depend on the orientation of the plate.

Finally, you can use the calculated value of J_1 to determine the rate at which muons pass through the flat plate by multiplying it by the area of the plate, A.

I hope this helps guide you through the problem. Remember to always double check your units and make sure they are consistent throughout your calculations. Good luck!
 

1. What is flux in spherical coordinates?

Flux in spherical coordinates is a measure of the flow of a vector field through a surface that is defined in terms of spherical coordinates, which are distance from the origin, azimuthal angle, and polar angle.

2. How is flux calculated in spherical coordinates?

Flux in spherical coordinates is calculated using the surface integral of the dot product of the vector field and the surface normal vector over the surface. This can be expressed mathematically as Flux = ∫∫S F · dS = ∫∫S F · n dS, where F is the vector field, dS is the surface element, and n is the unit normal vector to the surface.

3. What is the difference between flux in spherical coordinates and other coordinate systems?

The main difference between flux in spherical coordinates and other coordinate systems, such as Cartesian or cylindrical coordinates, is the form of the surface element. In spherical coordinates, the surface element is expressed in terms of the two angles, while in Cartesian coordinates it is expressed in terms of the two perpendicular sides. This leads to different expressions for calculating flux in each coordinate system.

4. How is flux related to divergence in spherical coordinates?

In spherical coordinates, the divergence of a vector field can be calculated using the divergence theorem, which relates the flux through a closed surface to the volume integral of the divergence of the vector field over the volume enclosed by the surface. This can be expressed mathematically as ∫∫S F · dS = ∭ (∇ · F) dV, where ∇ is the gradient operator and dV is the volume element.

5. What are some real-world applications of flux in spherical coordinates?

Flux in spherical coordinates has many real-world applications, such as in fluid dynamics, electromagnetism, and heat transfer. For example, it can be used to calculate the flow of air or water through a spherical surface, the electric or magnetic field through a spherical surface, or the heat transfer through a spherical surface. It is also used in fields such as astronomy and geology to study the flow of fluids or particles in spherical systems.

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