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The $delta$ notation in Goldstein's Classical Mechanics on the calculus of variation

Physics Asked on November 25, 2021

In Goldstein’s classical mechanics (page 36) he introduces the basics of the calculus of variation and uses it to effectively the Euler-Lagrange equations. However, there is a step in which the $delta$ notation is defined:

$$delta yequiv left(frac{partial y}{partial alpha}right)text dalpha,$$

in which $alpha$ is the parameter used in the path modification:

$$y(alpha,x)=y(0,x)+alphaeta(x),$$

$x$ is effectively a generalised time parameter. Both of these definitions are fine, however this notation is then introduced into the action integral:

$$frac{text dJ}{text dalpha}=int^{x_2}_{x_1}left( frac{partial f}{partial y} – frac{text d}{text dx}frac{partial f}{partial dot y}right)frac{partial y}{partial alpha}text dx,$$

which becomes:

$$delta J=int^{x_2}_{x_1}left( frac{partial f}{partial y} – frac{text d}{text dx}frac{partial f}{partial dot y}right)delta ytext dx,$$

which seems to imply that:

$$delta ystackrel{?}{equiv}left(frac{partial y}{partial alpha}right)neq left(frac{partial y}{partial alpha}right)text dalpha.$$

I can see that (maybe a little hand-wavingly) this corresponds to a multiplication by $text dalpha/text dalpha$, but I’m not sure if that’s a valid way to think of it. What am I missing here?

2 Answers

$$delta Jequiv dalphafrac{partial J}{partial alpha}=dalphaint_{x_1}^{x_2}(cdots)frac{partial y}{partial x}dx=int_{x_1}^{x_2}(cdots)left(frac{partial y}{partial x}dalpharight) dxequiv int_{x_1}^{x_2}(cdots)delta y dx$$

Answered by Umaxo on November 25, 2021

Using $delta J = frac{text dJ}{text dalpha} mathrm{d}alpha$, we see that:

$$delta J = int^{x_2}_{x_1}left( frac{partial f}{partial y} - frac{text d}{text dx}frac{partial f}{partial dot y}right)left(frac{partial y}{partial alpha} mathrm{d}alpha right) mathrm{d}x $$

Invoking the definition of $delta y$ arrives at the expected result.

Answered by Shrey on November 25, 2021

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