Physics Asked on August 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?
$$delta Jequiv dalphafrac{partial J}{partial alpha}=dalphaint_{x_1}^{x_2}(cdots)frac{partial y}{partial x}dx=int_{x_1}^{x_2}(cdots)left(frac{partial y}{partial x}dalpharight) dxequiv int_{x_1}^{x_2}(cdots)delta y dx$$
Correct answer by Umaxo on August 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 August 25, 2021
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