Confused about the principle of least action

1. Dec 15, 2012

To make the confusions both concise and explicit, I will put down some incorrect calculations, and ask for corrections
Take the Lagrangian KE - PE = T - V
Action = S=∫ L dt (with given limits)
Principle of least action: δS= 0: S(t1)-S(t2) =0 if t1-t2 is small (using the (.) as function notation, not multiplication)
I presume δ can be taken as d/dt
Then d(∫ L dt)/dt = L(t1)-L(t2) = T(t1)-T(t2)+V(t2)-V(t1)=0
Next: conservation of energy
KE+PE = k for some k
So d/dt (T+V) = dT/dt + dV/dt = 0
If this is true on all points on the interval (t1,t2), then T(t1)-T(t2)+V(t1)-V(t2) = 0
Putting the two equations together.... but obviously I've already gone way too far into the gross errors, and would be grateful for corrections. Thanks.

2. Dec 15, 2012

DrewD

Here's your error. The $\delta$ denotes change in path, not time. Most decent books on Lagrangian Mechanics or Calculus of variations will do a good job of explaining this. The idea, is that you have your action, $S$. This is a function(al) of the path through space. So if you deform the path a very small amount, that is $\delta S$. The path for which that is 0, like in normal calculus, is a max or min. I'm not comfortable enough to fully describe the process without going back to a book first, but doing this comes up with the Euler-Lagrange equations which are equivalent to Newton's Laws. I'm sure you can find a derivation online if your book isn't clear.

3. Dec 15, 2012

Thank you, DrewD. That helped very much.