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- Thread starter random_soldier
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On page 5 of the above it shows how to remove the angular dependence from angular neutron flux. On the next page the solid angle over which integration takes place is a combination of theta and phi

http://mathworld.wolfram.com/SphericalCoordinates.html

On wolfram above the radius unit vector is shown and it corresponds to what the directional unit vector is shown in the first link.

My question is how/why does omega become a solid angle for the purpose of integration and remove angular dependence of the flux, yet the vector omega corresponds to the radius unit vector which, as far as I recall vectors, does not give an actual direction and thus I don't understand how it gives the direction of the flux and furthermore how it can be integrated out since it is actual the radius and not the variables of integration i.e. phi and theta.

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This is a new approach in the notes, (at least it is new to me), but I think I can see why the author is taking the approach he is. For a simple example, you can have a small cube be the region of interest, and you can have equal and opposite currents of particles incident on the cube from the left as well as from the right. The net current is zero, but the currents will add for how many particles, what he calls flux, are incident on the target per unit time. ## \\ ## I'm not sure how often this will apply in actual scenarios. Most often, I think the neutron beam comes from a single direction. I don't think there are multiple scatterings when this beam encounters a solid target so that the target atom (e.g. in a crystal) normally only sees a beam coming from one direction. ## \\ ## Perhaps he has a reason for this somewhat novel approach. It will be interesting to see as he further develops his concepts, and the various scenarios that he treats. ## \\ ## Edit: The flux ## \Phi ## from a current density ## \vec{J} ## is normally defined as ## \int _A \vec{J} \cdot \hat{n} \, dA ##. ## \\ ## This author is using the definition ## \Phi= -\int_V \vec{J} \cdot \hat{n} \, dA ##, and he's only counting the incident current density, and not considering the current density as it exits the volume.

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@random_soldier Please see the "Edit" of post 6.

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At the same point in space, you can have neutrons moving in direction ##\Omega_1## or ##\Omega_2## (i.e. different angular fluxes, but at the same point in space).

##\Omega## is a unit vector because it is a direction, and does not have a magnitude.

Note that in the Rouben notes, ##r## is not shown as a vector, but it is.

To find reaction rates, or to reduce to the diffusion equation, we want the total flux at a point in space summed over all angles.

This is called the "scalar flux", and it is just the angular flux summed over all directions.

This is the equation on page 5 of the Rouben notes. We are calculating the scalar flux by integrating the angular flux over all directions.

$$ \phi(r,E)=\int_{\Omega} \phi(r,E,\Omega) \, d\Omega$$

Note that the quantity on the left hand side is the scalar flux (no angular dependence) and the quantity in the integral is the angular flux (with an angular dependence).

Now, it becomes a math problem. Since ##\Omega## is a vector, you have to transform it into a coordinate system, either ##(x,y,z)## or ##(r,\theta,\phi)##, or some other system. There are different ways of doing this, and the Wolfram site gives several different coordinate systems in use.

On page 6 of the Rouben notes, he uses the coordinate system with ##\alpha## and ##\theta##.

Integrating over the solid angle is a calculus problem, not a nuclear engineering problem.

Does this answer your question?

- #9

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If you look at my latest edit of post 6, what the author is doing is redefining flux as ## \Phi=-\int_V \vec{J}_{incident} \, \cdot \hat{n} \, dA ## and he is not counting the current density as it exits the volume. This is a completely different definition from the usual definition, and he would do well to state the definition like I just did, instead of thinking you might be able to figure out what he is trying to do.

At the same point in space, you can have neutrons moving in direction ##\Omega_1## or ##\Omega_2## (i.e. different angular fluxes, but at the same point in space).

##\Omega## is a unit vector because it is a direction, and does not have a magnitude.

Note that in the Rouben notes, ##r## is not shown as a vector, but it is.

To find reaction rates, or to reduce to the diffusion equation, we want the total flux at a point in space summed over all angles.

This is called the "scalar flux", and it is just the angular flux summed over all directions.

This is the equation on page 5 of the Rouben notes. We are calculating the scalar flux by integrating the angular flux over all directions.

$$ \phi(r,E)=\int_{\Omega} \phi(r,E,\Omega) \, d\Omega$$

Note that the quantity on the left hand side is the scalar flux (no angular dependence) and the quantity in the integral is the angular flux (with an angular dependence).

Now, it becomes a math problem. Since ##\Omega## is a vector, you have to transform it into a coordinate system, either ##(x,y,z)## or ##(r,\theta,\phi)##, or some other system. There are different ways of doing this, and the Wolfram site gives several different coordinate systems in use.

On page 6 of the Rouben notes, he uses the coordinate system with ##\alpha## and ##\theta##.

Integrating over the solid angle is a calculus problem, not a nuclear engineering problem.

Does this answer your question?

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If you look at my latest edit of post 6, what the author is doing is redefining flux as ## \Phi=-\int_V \vec{J}_{incident} \, \cdot \hat{n} \, dA ## and he is not counting the current density as it exits the volume. This is a completely different definition from the usual definition, and he would do well to state the definition like I just did, instead of thinking you might be able to figure out what he is trying to do.

I'm not sure what you mean by "usual definition". This is common nuclear engineering notation. If your background is in physics or thermal hydraulics, please note that the neutron flux does not have the same meaning as "flux" used in other fields.

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Perhaps someone else is better equipped to answer your question. (Edit: Note: I thought the OP was responding. I didn't realize this is a second responder). This is the best input I can give. The author's approach, IMO, lacks mathematical rigor, but perhaps that is the accepted way of calculating this.I'm not sure what you mean by "usual definition". This is common nuclear engineering notation. If your background is in physics or thermal hydraulics, please note that the neutron flux does not have the same meaning as "flux" used in other fields.

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- #12

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This is a new approach in the notes, (at least it is new to me), but I think I can see why the author is taking the approach he is. For a simple example, you can have a small cube be the region of interest, and you can have equal and opposite currents of particles incident on the cube from the left as well as from the right. The net current is zero, but the currents will add for how many particles, what he calls flux, are incident on the target per unit time. ## \\ ## I'm not sure how often this will apply in actual scenarios. Most often, I think the neutron beam comes from a single direction. I don't think there are multiple scatterings when this beam encounters a solid target so that the target atom (e.g. in a crystal) normally only sees a beam coming from one direction. ## \\ ## Perhaps he has a reason for this somewhat novel approach. It will be interesting to see as he further develops his concepts, and the various scenarios that he treats. ## \\ ## Edit: The flux ## \Phi ## from a current density ## \vec{J} ## is normally defined as ## \int _A \vec{J} \cdot \hat{n} \, dA ##. ## \\ ## This author is using the definition ## \Phi= -\int_V \vec{J} \cdot \hat{n} \, dA ##, and he's only counting the incident current density, and not considering the current density as it exits the volume.

Perhaps someone else is better equipped to answer your question. This is the best input I can give. The author's approach, IMO, lacks mathematical rigor, but perhaps that is the accepted way of calculating this.

It doesn't, but you have to start a lot further back to understand the definition of flux and current that he is using. I think there is confusion because the basic definition of neutron flux is different from a standard "flux" defined in physics. The Rouben notes are what is typically taught in Nuclear Engineering courses and probably what the original question is referring to. (I don't mean any disrespect, I think you just need to go further back in the notes to understand the definitions).

You also made an earlier comment that typical scenarios are beams. This is not correct. In operating reactors, the neutrons will be coming from and scattering in all direction. Your response is what led me to believe that you have a physics background.

Hopefully my response in comment #8 will answer the original question. I would be happy to discuss the other terms, but we need to start earlier in the Rouben notes...

- #13

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At the same point in space, you can have neutrons moving in direction ##\Omega_1## or ##\Omega_2## (i.e. different angular fluxes, but at the same point in space).

##\Omega## is a unit vector because it is a direction, and does not have a magnitude.

Note that in the Rouben notes, ##r## is not shown as a vector, but it is.

To find reaction rates, or to reduce to the diffusion equation, we want the total flux at a point in space summed over all angles.

This is called the "scalar flux", and it is just the angular flux summed over all directions.

This is the equation on page 5 of the Rouben notes. We are calculating the scalar flux by integrating the angular flux over all directions.

$$ \phi(r,E)=\int_{\Omega} \phi(r,E,\Omega) \, d\Omega$$

Note that the quantity on the left hand side is the scalar flux (no angular dependence) and the quantity in the integral is the angular flux (with an angular dependence).

Now, it becomes a math problem. Since ##\Omega## is a vector, you have to transform it into a coordinate system, either ##(x,y,z)## or ##(r,\theta,\phi)##, or some other system. There are different ways of doing this, and the Wolfram site gives several different coordinate systems in use.

On page 6 of the Rouben notes, he uses the coordinate system with ##\alpha## and ##\theta##.

Integrating over the solid angle is a calculus problem, not a nuclear engineering problem.

Does this answer your question?

That's odd. Everywhere I have been seeing r for the flux specifying it's position, it has always been a vector. Even duderstadt and hamilton does that repeatedly. Some clarification, if you don't mind.

Back to the question at hand, I understand how the angular dependence is removed. I am trying to understand why the procedure is carried out as it is as I cannot reconcile it with my previous understanding of vectors, geometry and the definition of the vector omega and the solid angle of integration omega.

Let me try defining first. In a textbook like duderstadt and hamilton or Karl Ott, the unit vector omega which indicates the direction of the flux where the angular flux is a function of this, the vector omega, if we try to convert to cartesian, yields the following x,y,z unit vectors

If we try to derive this mathematically, the terms are derived by taking the dot product of the radius unit vector with the x,y,z unit vectors, implying that the vector omega is actually just the radius unit vector. This is problem 1. As far as I recall the radius unit vector does not give direction, as far as I recall vectors and the actual direction is given by the angle vectors of spherical coordinates.

Then there is the problem that the differential element omega used to integrate over all angles is composed of the angles of spherical coordinates, phi and theta. If the vector omega term is just the radius unit vector, why do we integrate, essentially, over phi and theta to remove the angular dependence? Those would just yield one, since they are not actually present in the flux if my understanding/formulation is correct.

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To integrate ##\Omega## over all directions, you must transform it to a different coordinate system. You proposed ##(x,y,z)##. This is valid, but extremely difficult to work with because the limits on the integrals will be a function of ##(x,y,z)##. An equivalent coordinate system is ##(r,\theta,\phi)##, or any of the other coordinate systems listed on the Wolfram site. They are all equivalent, but using a spherical based coordinate system for a unit vector is much easier than using a Cartesian coordinate system.

I do not understand what you mean by "radius unit vector does not give direction". The whole purpose of vector ##\Omega## is to indicate the direction of neutron travel. It has a magnitude of 1 and shows the direction of neutron travel.

If there was no flux in the integral, the integral over the unit vector would give ##4\pi## (or possibly 1 with some coordinate systems). However, there IS a flux in the integral, so this integral will be something else. Remember, the point of this exercise is to sum the angular flux over all directions in order to obtain the scalar flux.

I hope this helps, there is not anything else I can add.

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I do not understand what you mean by "radius unit vector does not give direction". The whole purpose of vector ##\Omega## is to indicate the direction of neutron travel. It has a magnitude of 1 and shows the direction of neutron travel.

Yeah sorry about that. That is by definition a vector. Though I guess my confusion now becomes what direction is the radius unit vector pointing towards. I thought the whole point of the angles was to show the direction of the overall vector term which makes the direction of the radius unit vector moot or vice versa, IMO.

Thanks for the rest though. I think I understand for the most part.

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