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Wheeler, Mechanics

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Moreover, taking a regular derivative with respect to α then setting α = 0 gives the same expression as the<br />

first term of a Taylor series in small h(x). This happens because the only dependence f has on α is through<br />

x :<br />

d<br />

d<br />

f [x(λ, α)] =<br />

dα dα<br />

=<br />

=<br />

∫<br />

(<br />

)<br />

L x (λ, α) , x (1) (λ, α) , . . . dλ<br />

∫ ( ∂L ∂x<br />

∂x ∂α +<br />

∫ ( ∂L<br />

∂x h +<br />

∂L ∂x (1)<br />

∂x (1) ∂α + . . . + ∂L ∂x (n) )<br />

dλ (8)<br />

∂x (n) ∂α<br />

∂L<br />

∂x (1) h(1) + . . . +<br />

∂L )<br />

∂x (n) h(n) dλ (9)<br />

so that when we set α = 0, all dependence on x (λ) + αh (λ) reduces to dependence on x (λ) only.<br />

therefore define the variation of f [x] as<br />

We<br />

δf [x (λ)]<br />

≡<br />

=<br />

=<br />

( )∣ d<br />

∣∣∣α=0<br />

dα f [x(λ, α)]<br />

∫ ( ∂L (x (λ, α))<br />

h + . . . +<br />

∂x<br />

∫ ( ∂L (x (λ))<br />

h +<br />

∂x<br />

)∣<br />

∂L (x (λ, α)) ∣∣∣α=0<br />

h (n) dλ<br />

∂x (n)<br />

)<br />

∂L (x (λ))<br />

h (1) ∂L (x (λ))<br />

+ . . . + h n dλ<br />

∂x (1)<br />

∂x (n)<br />

Notice that all dependence of L on h has dropped out, so the expression is linear in h (λ) and its derivatives.<br />

Now continue as before. Integrate the h (1) and higher derivative terms by parts. We assume that h (λ)<br />

and its first n − 1 derivatives all vanish at the endpoints. The k th term becomes<br />

∫ 1<br />

∣ ∫ (<br />

∂L ∣∣∣x(λ,α)=x(λ) 1<br />

∣ )<br />

h (k) (λ) dλ = (−1) k dλ dk ∂L ∣∣∣x=x(λ)<br />

∂x (k) dλ k h (λ)<br />

∂x (k)<br />

so we have<br />

0<br />

0<br />

δf [x (λ)] =<br />

=<br />

( )∣ d<br />

∣∣∣α=0<br />

dα f [x(λ, α)]<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 (10)<br />

∂x (n)<br />

where now, h (λ) is fully arbitrary.<br />

Fill in the details of the preceeding derivation, paying particular attention to the integrations by parts.<br />

The functional derivative We finish the procedure by generalizing the way we extract the term in<br />

parentheses from the integrand of eq.(10), without demanding that the functional derivative vanish. Recall<br />

that for the straight line, we argued that we can choose a sequence of functions h(x) that vanish everywhere<br />

except in a small region around a point x 0 . It is this procedure that we formalize. The required rigor is<br />

provided by the Dirac delta function.<br />

The Dirac delta function To do this correctly, we need yet another class of new objects: distributions.<br />

Distributions are objects whose integrals are wel-defined though their values may not be. They may be<br />

identified with infinite sequences of functions. Therefore, distributions include functions, but include other<br />

objects as well.<br />

26

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