Understanding Variational Calculus: A String Theory Question

[ehrenfest]

Homework Statement

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

2 ds \delta (ds) = \delta (ds) ^2

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 \delta are not the same thing. ds as it is said, is the ghost of a departed quantity, while \delta s is a boring old number.
2dsd(ds) = d(ds)^{2} is an exact equation.
2 ds \delta (ds) = \delta (ds) ^2 is approximate. To be exact, it should be
2ds \delta(ds) + \mathcal{O}((\delta ds)^2) = \delta (ds) ^2
In this case, you are expected and indeed required to use only such \delta as to make \delta^2 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 \delta (dx dy dz) ^2)? How do you think of it then?