I Neutron transport equation

[eneacasucci]

I have a question about the neutron transport equation, my question is more about mathematics, from the book Duderstadt Hamilton

I tried to make the calculations, it should be quite simple but still I don't understand where the 2\pi terms went...
the integral over Omega in spherical coordinates has a part that is shown with the polar angle theta and then there is another integral from 0 to 2\pi for the azimuthal angle. This last integral should have as an integrand 1, because nothing depends on it and so it should result in a 2\pi that here i don't see.

(another question: when we say that the flux phi is dependent on theta, is it because it is dependent on Omega theta that is the cos\theta right?)

 

[Astronuc]

I have a question about the neutron transport equation, my question is more about mathematics, from the book Duderstadt Hamilton
View attachment 356272
View attachment 356273
I tried to make the calculations, it should be quite simple but still I don't understand where the 2\pi terms went...
the integral over Omega in spherical coordinates has a part that is shown with the polar angle theta and then there is another integral from 0 to 2\pi for the azimuthal angle. This last integral should have as an integrand 1, because nothing depends on it and so it should result in a 2\pi that here i don't see.

(another question: when we say that the flux phi is dependent on theta, is it because it is dependent on Omega theta that is the cos\theta right?)

I have D&H from 1976, which shows the same equations. The second equation is (4-46). Also, see Figure 4-9, which should be a vertical plane and two arrows, one being the normal to the plane parallel to the reference axis (1 dimension) and the second the direction of travel of the particle. The figure is entitled Coordinates characterizing plane symmetry.

In one dimension, one is concerned about the particles traveling with respect to that one direction. Particles can travel in any direction, particularly after being scattered by a collision with a nucleus, or proton. One is only concerned about the projection to the reference axis.

Note that when θ is 0, the direction of travel is parallel to the reference axis, and when θ is π, the direction is in the opposite direction (equivalent to a backscatter). One is not concerned with the direction transverse to the reference axis, and one could visual a 'cone' in which the particle travels. In that cone, the poloidal angle (φ) can be 0 to 2π (think of the base of the cone, but that doesn't matter since one is only concerned with the direction of travel along the axis of the cone, and the projection onto the cone's axis is not dependent of φ. Note that the 'one dimensional' problem is actually quasi 2-dimensional, or 1.5 dimensional as some call it.

μ of course is cos θ. For calculations, it is more efficient to solve the equation in terms of μ rather than calculating cos θ, which is why one solves the equation in terms of μ. Regardless of one or three dimensions, one is calculating with respect to a reference coordinate system, and individual particles can scatter in any direction, but depending on the energy, direction of travel and the nuclei interacting with the particle, but it is the collective behavior of a population of particles of interest (e.g., neutrons) that one is attempting to describe.

 

[eneacasucci]

I have D&H from 1976, which shows the same equations. The second equation is (4-46). Also, see Figure 4-9, which should be a vertical plane and two arrows, one being the normal to the plane parallel to the reference axis (1 dimension) and the second the direction of travel of the particle. The figure is entitled Coordinates characterizing plane symmetry.

In one dimension, one is concerned about the particles traveling with respect to that one direction. Particles can travel in any direction, particularly after being scattered by a collision with a nucleus, or proton. One is only concerned about the projection to the reference axis.

Note that when θ is 0, the direction of travel is parallel to the reference axis, and when θ is π, the direction is in the opposite direction (equivalent to a backscatter). One is not concerned with the direction transverse to the reference axis, and one could visual a 'cone' in which the particle travels. In that cone, the poloidal angle (φ) can be 0 to 2π (think of the base of the cone, but that doesn't matter since one is only concerned with the direction of travel along the axis of the cone, and the projection onto the cone's axis is not dependent of φ. Note that the 'one dimensional' problem is actually quasi 2-dimensional, or 1.5 dimensional as some call it.

μ of course is cos θ. For calculations, it is more efficient to solve the equation in terms of μ rather than calculating cos θ, which is why one solves the equation in terms of μ. Regardless of one or three dimensions, one is calculating with respect to a reference coordinate system, and individual particles can scatter in any direction, but depending on the energy, direction of travel and the nuclei interacting with the particle, but it is the collective behavior of a population of particles of interest (e.g., neutrons) that one is attempting to describe.

Thank you so much! Could I ask you also about this mathematical part "the integral over Omega in spherical coordinates has a part that is shown with the polar angle theta and then there is another integral from 0 to 2\pi for the azimuthal angle. This last integral should have as an integrand 1, because nothing depends on it and so it should result in a 2\pi factor that here i don't see." ? I'm stuck in the mathematical derivation from the first to the second

shouldn't it be like this:
$$\int_{4\pi} d\Omega' \to 2\pi \int_{-1}^1 d\mu'$$

 

[Astronuc]

$\int_{4\pi} d\Omega' \to 2\pi \int_{-1}^1 d\mu' = 4\pi$, which is the solid angle that subtends a point source in the center of a sphere, or is the area of a sphere of radius r = 1 (units not specified). The entire solid angle surrounding a point (center of a sphere) is 4π steradians.

A steradian is defined as the solid angle which, having its vertex at the center of the sphere, cuts off a spherical surface area equal to the square of the radius of the sphere.

https://en.wikipedia.org/wiki/Steradian
https://en.wikipedia.org/wiki/Solid_angle


Integrating over a solid angle is only significant if there is an angular dependency on the angular neutron density.

If one has a point source S at the center of a sphere, then one has a mean angular flux of S/(4π), or a mean (scalar) flux of S/(4πr²) at distance r (from a point source).

 

[eneacasucci]

$\int_{4\pi} d\Omega' \to 2\pi \int_{-1}^1 d\mu' = 4\pi$, which is the solid angle that subtends a point source in the center of a sphere, or is the area of a sphere of radius r = 1 (units not specified). The entire solid angle surrounding a point (center of a sphere) is 4π steradians.

A steradian is defined as the solid angle which, having its vertex at the center of the sphere, cuts off a spherical surface area equal to the square of the radius of the sphere.

https://en.wikipedia.org/wiki/Steradian
https://en.wikipedia.org/wiki/Solid_angle


Integrating over a solid angle is only significant if there is an angular dependency on the angular neutron density.

If one has a point source S at the center of a sphere, then one has a mean angular flux of S/(4π), or a mean (scalar) flux of S/(4πr²) at distance r (from a point source).

$2\pi \int_{-1}^1 d\mu' = 4\pi$ we can't write this, in our case because we have a function $\varphi$ that depends on $\mu$ (or $\theta$) so the only integral that we can directly solve is the one related to the azimuthal angle (that gives the $2\pi$), because none of the functions depends on it.
What I think is that this $\int_{4\pi} d\Omega' \to 2\pi \int_{-1}^1 d\mu'$ must hold as a transformation in which, as in our case, we don't have any dependency of the functions of the azimuthal angle (since we are in 1D), therefore should appear the factor $2\pi$


As here:

$\hat\Omega$ is still defined as that (with all the components) but in the case of the angular neutron flux only depending on x the other terms (with the derivative) becomes 0, but this doesn't mean that the $\hat\Omega$ is not defined as having all the 3 components; the same should apply to the integration of the scattering term, in which the integral over also the azimuthal angle should persist, giving a $2\pi$ as mentioned before.

Not that i think I can do better maths than the book ahah but I love understanding maths derivations and here I can't figure it out why it isn't like that:

 

[vibrationalmodes]

I have a question about the neutron transport equation, my question is more about mathematics, from the book Duderstadt Hamilton
View attachment 356272
View attachment 356273
I tried to make the calculations, it should be quite simple but still I don't understand where the 2\pi terms went...
the integral over Omega in spherical coordinates has a part that is shown with the polar angle theta and then there is another integral from 0 to 2\pi for the azimuthal angle. This last integral should have as an integrand 1, because nothing depends on it and so it should result in a 2\pi that here i don't see.

(another question: when we say that the flux phi is dependent on theta, is it because it is dependent on Omega theta that is the cos\theta right?)

I think you could reason that they must have absorbed the 2$\pi$ into the cross section. If you look closely, the top expression has the cross section for particles coming in from direction $\Omega$' and scatters into direction $\Omega$ (and you integrate over all incoming angles to get the inscattering from all angles, into the angle you are considering/evaluating flux for). Then in slab geometry, the problem reduces and you typically assume the system is fully isotropic across the entire phi domain (like the flux is constant for a given theta and x, across phi's from 0 to 2$\pi$....basically it doesn't matter if you are talking about whether or not the particles scattered along the -x or the x axis for example, the flux will be the same for those two directions since $x=cos(phi)sin(theta)$...but the flux only varies with theta so for theta=pi/2, on xy plane, the flux will be the same regardless of what phi you are looking at), hence the cross section can be defined to absorb the phi dependency. Basically what I'm saying is, they could be doing: the cross section for slab geo is $\Sigma_{s}(E' \rightarrow E, \theta' \rightarrow \theta)=2\pi\Sigma_{s}(E' \rightarrow E, \Omega' \rightarrow \Omega)$

 

[eneacasucci]

I think you could reason that they must have absorbed the 2$\pi$ into the cross section. If you look closely, the top expression has the cross section for particles coming in from direction $\Omega$' and scatters into direction $\Omega$ (and you integrate over all incoming angles to get the inscattering from all angles, into the angle you are considering/evaluating flux for). Then in slab geometry, the problem reduces and you typically assume the system is fully isotropic across the entire phi domain (like the flux is constant for a given theta and x, across phi's from 0 to 2$\pi$....basically it doesn't matter if you are talking about whether or not the particles scattered along the -x or the x axis for example, the flux will be the same for those two directions since $x=cos(phi)sin(theta)$...but the flux only varies with theta so for theta=pi/2, on xy plane, the flux will be the same regardless of what phi you are looking at), hence the cross section can be defined to absorb the phi dependency. Basically what I'm saying is, they could be doing: the cross section for slab geo is $\Sigma_{s}(E' \rightarrow E, \theta' \rightarrow \theta)=2\pi\Sigma_{s}(E' \rightarrow E, \Omega' \rightarrow \Omega)$

thank you so much, that is exactly what i was asking

 

[vibrationalmodes]

thank you so much, that is exactly what i was asking

Happy to help!