It always puzzles me whenever I work on Lagrangian equations. It is easy to see that $L=T-V$ yields the correct equations of motion, but the question is, how do you get to that formula? Is it trial and error(making guesses and seeing if they actually give the correct equations), or does it have a deeper, more logical reasoning and thought behind it? Same is for all Lagrangians, from Hilbert action to the special relativistic one-how do you "derive" the lagrangian to get the correct equations. I hope you understand what I mean.
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It comes from d'Alembert's principle of virtual works: http://en.wikipedia.org/wiki/D%27Alembert%27s_principle – Phoenix87 Jan 01 '15 at 15:36
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Possible duplicate by OP: http://physics.stackexchange.com/q/117017/2451 Possible duplicates: http://physics.stackexchange.com/q/5648/2451 , http://physics.stackexchange.com/q/78138/2451 and links therein. – Qmechanic Jan 01 '15 at 15:47
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@Phoenix87, I still don't see it.... – GRrocks Jan 01 '15 at 15:57
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See the last equation: $ \frac{d}{dt} \frac{\partial T}{\partial \dot{q}j} -\frac{\partial T}{\partial q_j} = Q_j, \quad j=1,\ldots,m.$. Assume the generalised forces come from a potential through $\mathbf Q = -\nabla{\mathbf q}U$ and put $L = T - U$. Then replace $T - U$ by $L$ in that equation. – Phoenix87 Jan 01 '15 at 16:01
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@Phoenix87 yeah but thats what I am asking-WHY to use L=T-V....Hope I am not being too stupid ....haha – GRrocks Jan 01 '15 at 16:04
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because $Q_j = -\frac{\partial U}{\partial q_j}$ and $U$ is assumed to depend on the $q_j$s alone. So $\partial U/\partial\dot q_j=0$ for any $j$. Put all of this in the above equation and see what you get. – Phoenix87 Jan 01 '15 at 16:19