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Wheeler, Mechanics

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We don’t usually need a specific sequence to solve problems using delta functions, because they are only<br />

meaningful under integral signs and they are very easy to integrate using their defining property. In physics<br />

they are extremely useful for making “continuous” distributions out of discrete ones. For example, suppose<br />

we have a point charge Q located at x 0 = (x 0 , y 0 , z 0 ) . Then we can write a corresponding charge density as<br />

ρ (x) = Qδ (x − x 0 ) = Qδ (x − x 0 ) δ (y − y 0 ) δ (z − z 0 )<br />

Perform the following integrals over Dirac delta functions:<br />

1. ∫ f(x)δ (x − x 0 ) dx<br />

2. ∫ f(x)δ (ax) dx (Hint: By integrating over both expressions, show that δ(ax) = 1 aδ(x). Note that to<br />

integrate δ(ax) you need to do a change of variables.)<br />

3. Evaluate ∫ f(x)δ (n) (x) dx where<br />

δ (n) (x) = dn<br />

dx n δ (x)<br />

4. ∫ f(x)δ ( x 2 − x 2 0)<br />

dx (This is tricky!)<br />

Show that<br />

Show that<br />

√ m<br />

δ(x) = lim<br />

m→∞ 2π e− m2 x 2<br />

2<br />

m<br />

δ(x) = lim<br />

m→∞ 2 e−m|x|<br />

Let f (x) be a differentiable function with zeros at x 1 , . . . , x n . Assume that at any of these zeros, the<br />

derivative f ′ (x i ) is nonzero. Show that<br />

δ (f (x)) =<br />

n∑<br />

i=1<br />

1<br />

|f ′ (x i )| δ (x − x i)<br />

The mass density of an infinitesimally thin, spherical shell of radius R may be written as<br />

ρ(r, θ, ϕ) =<br />

M δ(r − R)<br />

4πR2 By integrating ρ over all space, find the total mass of the shell.<br />

Write an expression for the mass density of an infinitesimally thin ring of matter of radius R, lying in<br />

the xy plane.<br />

The definition at last! Using the idea of a sequence of functions and the Dirac delta function, we can<br />

now extract the variation. So far, we have defined the variation as<br />

( )∣ d<br />

∣∣∣α=0<br />

δf [x (λ)] ≡<br />

dα f [x(λ, α)] (11)<br />

where x (λ, α) = x (λ) + αh (λ) , and shown that the derivative on the right reduces to<br />

δf [x (λ)] =<br />

∫ 1<br />

( ∂L (x (λ))<br />

− d ∂L (x (λ))<br />

0 ∂x dλ ∂x (1)<br />

+ . . . + (−1) n d n<br />

)<br />

∂L (x (λ))<br />

dλ n h (λ) dλ (12)<br />

∂x (n)<br />

28

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