Variational principle and energy conservation in the action

[nomadreid]

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.

 

[DrewD]

I presume δ can be taken as d/dt

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.

 

[nomadreid]

Thank you, DrewD. That helped very much.