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Lagrangian Derivation

  1. Apr 21, 2014 #1

    BiGyElLoWhAt

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    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.
     
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  3. Apr 21, 2014 #2

    Fredrik

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    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}
     
  4. Apr 21, 2014 #3

    BiGyElLoWhAt

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    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.
     
  5. Apr 21, 2014 #4

    George Jones

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