NE406 Radiation Protection and Shielding

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Lesson #4 - Directional Properties of Radiation

Reading Assignment: Section 2.3

Distributions vs. Functions

Our interest in flux and current often takes us beyond the simple dependence of position that we have talked about thus far, i.e.,

to an interest in the energy or direction dependence as well, e.g., or . 

Unfortunately, these considerations raise Zeno-like paradoxes.  For example, if we are interested in the flux "at" an energy of 1 MeV, we have to acknowledge that NO particles will have EXACTLY 1.0000000.... MeV of energy.

We get around this by going to a energy distribution which is a density in energy and requires the specification of an energy range in order to return to the units that we are used to (particles/cm2/sec):

with

  = the particles/cm2/sec in a differential energy range dE about the energy E

or

= the particles/cm2/sec in the energy range E1 to E2.

Similarly, if we are interested in the angular distribution, we have the same sort of distribution in direction, e.g.,

or

Notice that the so-called scalar flux, , can be obtained by integrating ANY of these over their ENTIRE range, e.g.,

As mentioned in the book, one must be very careful to keep up with the units (as has been the case throughout your engineering education!).  For example, the angular flux in units of particles/cm2/sec/MeV will be a number 1,000,000 times bigger than the flux at the same physical energy in units of particles/cm2/sec/eV.  Be careful.

Angular Properties of Flow

Paying particular attention to the so-called angular flux, or , we have already seen how we can integrate it over all directions to get the scalar flux (and I assume you could multiply it by a delta-time to get fluence), but we need to relate it to the other variables: current and flow.

Figure 2.4 on page 21 shows the relationship between:

• The direction of particle travel, ;

• The area of the "flux sphere", , perpendicular to ; and

• The area on the "current surface" (where the plane is defined by the normal vector, ) that the beam passes through.

As is shown in this figure, the area on the surface is "stretched" to a larger size than the area because of the obliqueness of the direction.  So, if the beam's current contribution ends up being smaller -- since it is on a "per area" basis -- because it is spread over this larger area. 

Using the notation from the previous section of these notes, if we have an angular flux, , that is in all directions, then we can break it into differential "beams" of strength particles/cm2/sec, each of which will contribute to the current an amount:

We can then integrate these differential amounts that depend on to get:

The result if we start with a cosine-based angular distribution,, is simpler since :

Breaking this down into the + and - contributions reduces to breaking this integral into two parts -- one for positive and one for negative:

which gives us the familiar definition:

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