# Lagrange Multipliers Versus Generalized Coordinates

When forced to explain to someone why one could either set up a general Lagrangian & then incorporate constraints using Lagrange multipliers, as opposed to just setting up a Lagrangian with generalized coordinate built in from the start I found I couldn't do it - I don't actually know why one can use either method other than that it apparently works. Is there some theorem or some substitution that says either method is valid, or is this just stunningly obvious & I'm missing it?

I've re-checked one of my video courses where the guy solves a problem using three different methods but never mentions why they're equivalent, checked both mechanics & calculus of variations books to find an explanation & checked posts on this forum as well as other forums but seem to have missed it, thus I'd really appreciate any commentary & references from you guys on this - thanks for reading!

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## 1 Answer

If you work with a smaller number of coordinates (usually "curved ones" in a sense) and no Lagrangian multipliers, you are simply considering a configuration space that is a submanifold of the full configuration space in the calculation that does include Lagrange multipliers.

Extremizing the action $S_{full}$ with Lagrange multipliers $$\delta S_{full} = 0,\quad S_{full} = S_{orig}+\sum\int \lambda (g(x^i)-c)$$ may be seen to imply $g(x^i)=c$ – that's the derivative of the full action with respect to the Lagrange multiplier(s) $\lambda$. Because $\delta S_{full} = 0$ implies $g(x^i)=c$, among other things, we may assume this relationship while extremizing $S_{full}$ on the subspace of the configuration space that obeys the conditions $g(x^i)=c$. But on this submanifold, $S_{full}=S_{orig}$, so the two extremization conditions are equivalent.

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Wow, that's so much better than what I was expecting. For some reason they seemed like disparate ideas yet now everything's just blatantly & geometrically obvious, thanks man. – bolbteppa Jul 28 '13 at 9:13
That simple could life be, the devil knows why some textbooks make rather a fuzz about certain things ... :-) – Dilaton Jul 28 '13 at 9:25
@bolbteppa but aren't you asking how Lagrange's equations of the first kind are equivalent to the Euler-Lagrange equations? – Physiks lover Jul 28 '13 at 10:48
If you use that terminology then I guess my question was about the direct relationship between the first & second kinds of Lagrange equations. I knew why the equations of the first kind were equivalent to the Euler-Lagrange equations & seen why the second kind were useful but didn't see how working with the second kind was analogous to basically just modifying the domain in which you're working from the get-go, thus relating the two methods - something that should have been obvious since you're modifying variables & the pedant in me should have criticized that a bit more... – bolbteppa Jul 28 '13 at 11:02