NE406
Radiation Protection and Shielding

Lesson #4  Directional Properties of RadiationReading Assignment: Section 2.3Distributions vs. FunctionsOur 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 Zenolike 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 socalled 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 FlowPaying particular attention to the socalled 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 deltatime 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:
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 cosinebased 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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