# Lagrangian Derivation

1. Apr 21, 2014

### BiGyElLoWhAt

Last semester I had intermediate mechanics, and we spent a good amount of the class studying the LaGrangian. One thing that I never got an explaination for was why $L = T-V$, as opposed to $T+V$.

The only reason I can think of is the "give and take" relationship that Kinetic and Potential energy have in an isolated system; but is this correct?

Total energy seems more intuitive to me than the difference ($T-V$). I was just hoping someone could shed some light on this for me.

2. Apr 21, 2014

### Fredrik

Staff Emeritus
Is it not sufficient to know that T+V gives us the wrong equation of motion (F=-ma instead of F=ma)?

\begin{align}
&L=\frac{1}{2}m\dot x^2-V(x)\\
&0=\frac{\partial L}{\partial x} -\frac{d}{dt}\left(\frac{\partial L}{\partial\dot x}\right) =-\frac{dV}{dx} -\frac{d}{dt}(m\dot x)=F-m\ddot x.
\end{align}

3. Apr 21, 2014

### BiGyElLoWhAt

Not really, I was looking for a more in depth explaination.

I guess that works. I was really hoping to grab some warm milk and get ready for story time, because that just isn't something that I would think to try, especially since the Chain Rule relationship holds for any definition of L (or it should mathematically), including defining it as T + V. It just so happens that you get a wrong acceleration value from that.

4. Apr 21, 2014

### George Jones

Staff Emeritus