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(^118) THE DIRAC &FUNCTION
Y
dx Figure 5.19 The Differential Length ds Along the Surface x = y in the xy-Plane
where the last step follows because d x = dy for this problem. Thus Equation 5. becomes
(5.65)
If we use the expression in 5.62 for pm. the volume integral on the RHS of Equa- tion 5.60 is
= J d x / d z u o , (5.66)
where the last step results from performing the integration over y. Comparing Equa- tions 5.66 and 5.65, we see that they are off by a factor of fi! Where does this discrepancy come from? The problem was our assumption in Equation 5.62 to use a 8-function in the form 8 ( y - x). We could have very well chosen [? ] 8 ( y - x ) , where [?I is some function that we need to determine. This still makes all the mass lie in the y = x plane. The correct choice of [?I is the one that makes both sides of Equation 5.60 equal. For example, when we make our “guess” for the distribution function for this example, we write
Pm(x, y , z > = “?luo6(x - y ) - (5.67)
Then, when we evaluate the RHS of Equation 5.60, we get
In this case, we already showed that the value of [?I is simply the constant &.In more complicated problems, [?] can be a function of the coordinates.This can happen
SINGULAR DENSITY FUNCTIONS 119
if either the mass per unit area of the sheet is not constant, or if the sheet is not flat. You will see some examples of this in the next section and in the problems at the end of this chapter.
5.5.3 Line Distributions
As a final example of singular density functions, consider the mass per unit volume of a one-dimensional wire of uniform mass per unit length A,. The wire is bent to follow the parabola y = Cx2 in the z = 0 plane, as shown in Figure 5.20. The factor C is a constant, which has units of l/length. We will follow the same procedure in constructing the mass density for this wire as we did for the previous example. In this case, the volume integral of the mass density must collapse to a line integral along the wire
L d. r p , ( r ) = ds A,(s). (5.69)
h
In this equation, s is a variable which indicates parametrically where we are on the wire, and A,(s) is the mass per unit length of the wire at the position s. Because all the mass must lie on the wire, we write the mass density function as
Here we have used two &functions. The 6(z) term ensures all the mass lies in the z = 0
plane, while S ( y - Cx2) makes the mass lie along the parabola. As before, we include an unknown factor of [?I, which we will have to determine using Equation 5.69. In terms of Cartesian coordinates, the general expression for the differential arc length ds is
Y
Z Figure 5.20 Parabolic Line Distribution
