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Wheeler, Mechanics

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so vanishing δs requires<br />

0 = εh(λ 0 )<br />

⎡<br />

⎣ d<br />

dλ<br />

⎛<br />

⎞⎤<br />

⎝<br />

y 0<br />

′ √ ⎠⎦<br />

1 + (y 0 ′ )2<br />

Since εh (λ 0 ) > 0, we may divide it out, leaving<br />

⎡ ⎛<br />

⎞⎤<br />

⎣ d ⎝<br />

y 0<br />

′ √ ⎠ = 0<br />

dλ<br />

1 + (y 0<br />

⎦λ=λ ′ )2 0<br />

But the point λ 0 was arbitrary, so we can drop the λ = λ 0 condition. The argument holds at every point of<br />

the interval. Therefore, the function y 0 (λ) that makes the length extremal must satisfy<br />

⎛<br />

⎞<br />

d<br />

⎝<br />

y 0<br />

′ √ ⎠ = 0<br />

dλ<br />

1 + (y 0 ′ )2<br />

This is just what we need, if it really gives the condition we want! Integrating once gives<br />

y ′ 0<br />

√1 + (y ′ 0 )2 = c = const.<br />

λ=λ 0<br />

so that<br />

y 0 ′ =<br />

√ c<br />

1 − c ≡ b = const.<br />

y 0 (λ) = a + bλ<br />

Finally, with initial conditions y 0 (0) = y (A) and y 0 (1) = B,<br />

we find<br />

y A = a<br />

y B = a + b<br />

y 0 (λ) = y A + (y B − y A ) λ<br />

This is the unique straight line connection (0, y A ) and (1, y B ).<br />

Find the form of the function x (λ) that makes the variation, δf, of each of the following functionals<br />

vanish:<br />

1. f[x] = ∫ λ<br />

0 ẋ2 dλ where ẋ = dx<br />

dλ .<br />

2. f[x] = ∫ λ (ẋ2 + x 2) dλ<br />

0<br />

3. f[x] = ∫ λ<br />

0 (ẋn − x m ) dλ<br />

4. f[x] = ∫ λ (ẋ2 (λ) − V (x(λ)) ) dλ where V (x) is an arbitrary function of x(λ).<br />

0<br />

5. f[x] = ∫ λ<br />

0 ẋ2 dλ where ẋ = (ẋ, ẏ, ż) and ẋ 2 = ẋ 2 + ẏ 2 + ż 2 . Notice that there are three functions to be<br />

determined here: x(λ), y(λ) and z(λ).<br />

24

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