Understanding Variational Calculus: A String Theory Question

[ehrenfest]

Homework Statement

I saw the following equation in my (Zwiebach page 83).

[tex]2 ds \delta (ds) = \delta (ds) ^2 [/tex]

where delta is the variation from variational calculus and ds is the Lorentz invariant spacetime distance.

It seems like they took a derivative from the right to the left but I am really not sure why you can do that because I thought delta was just a very small variation function.

Homework Equations

The Attempt at a Solution

 

[BoTemp]

The delta notation means "change of" which is the same as derivative. If it makes you feel better, replace delta with delta/ dx and then multiply by dx on both sides to remove it. Seems a bit sketchy, but it works as long as the variation is small, which of course is the point of derivatives in the first place.

 

[ehrenfest]

So, in this case it makes the most sense to replace delta by d/ds, right? That does seem sketchy.

 

[Dick]

It's not sketchy. d and delta are two different things. d refers to the differential of s(t) with respect to a parameter as in s(t)->s(t+epsilon). delta refers to the variation of some functional with respect to the variation of s(t) by a arbitrary function. s(t)->s(t)+epsilon*r(t).

 

[Jimmy Snyder]

As Dick wrote, d and [itex]\delta[/itex] are not the same thing. ds as it is said, is the ghost of a departed quantity, while [itex]\delta s[/itex] is a boring old number.
[itex]2dsd(ds) = d(ds)^{2}[/itex] is an exact equation.
[itex]2 ds \delta (ds) = \delta (ds) ^2 [/itex] is approximate. To be exact, it should be
[itex]2ds \delta(ds) + \mathcal{O}((\delta ds)^2) = \delta (ds) ^2[/itex]
In this case, you are expected and indeed required to use only such [itex]\delta[/itex] as to make [itex]\delta^2[/itex] small enough to ignore.

 

[ehrenfest]

The delta notation means "change of" which is the same as derivative. If it makes you feel better, replace delta with delta/ dx and then multiply by dx on both sides to remove it. Seems a bit sketchy, but it works as long as the variation is small, which of course is the point of derivatives in the first place.

Sorry--I encountered this again and I am still confused about it. When you say replace delta by delta/ds (and then multiply both sides by ds), do you mean that I should treat this as the derivative operator with respect to ds? But can you do that if there is more than just one term in there (say [tex] \delta (dx dy dz) ^2 [/tex])? How do you think of it then?