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Calculating energy from the Lagrangian

  1. Nov 18, 2013 #1
    1. The problem statement, all variables and given/known data

    While doing a problem I have found the Lagrangian to be [itex]L=\frac{1}{2}m \dot{r}^2 \left( 1 + 4a^2r^2 \right) + \frac{1}{2}mr^2 \dot{\theta}^2 -mgar^2[/itex]. I have also shown that the angular momentum l is constant and is equal to [itex]l=mr^2 \dot{\theta}[/itex]. I want to calculate the energy given this. I am getting two different answers based on how I do the problem and I'm not sure I understand why.

    2. Relevant equations

    [tex]E=\sum_{\alpha} \dot{q}^{\alpha} \frac{\partial L}{\partial \dot{q}^{\alpha}} - L[/tex]


    3. The attempt at a solution

    If I substitute the angular momentum into the Lagrangian first my Lagrangian becomes

    [tex]L=\frac{1}{2}m \dot{r}^2 \left( 1 + 4a^2r^2 \right) + \frac{l^2}{2mr^2}-mgar^2.[/tex]

    To get the energy I only have to sum the r variable in the energy equation, which gives me an energy equal to

    [tex]E=\frac{1}{2}m \dot{r}^2 \left( 1 + 4a^2r^2 \right) - \frac{l^2}{2mr^2} + mgar^2.[/tex]

    However, if I take the first Lagrangian, plug it into the energy equation this time summing for both r and theta, and then substitute in the angular momentum I get

    [tex]E=\frac{1}{2}m \dot{r}^2 \left( 1 + 4a^2r^2 \right) + \frac{l^2}{2mr^2} + mgar^2.[/tex]

    Basically the sign of my second term is different. Intuitively, the second one seems correct but I don't quite understand what is wrong with the first approach. If I look at the solutions manual for the problem the angular momentum is plugged in directly to the Lagrangian, but the solution still comes out with an energy where all terms are positive. Can someone help me out with what's going on here?

    Thanks
     
  2. jcsd
  3. Nov 19, 2013 #2
  4. Nov 19, 2013 #3

    ehild

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    Homework Helper
    Gold Member

    Your first method is not legal. You cannot eliminate a variable before taking the partial derivative.

    See the example: f(x,y)= x+y. Determine the partial derivatives in case when y=x.

    With the correct method [itex]\partial f/ \partial x =1[/itex], [itex]\partial f/ \partial y =1[/itex]

    With your method, [itex]f(x,y)=2x[/itex], [itex]\partial f/ \partial x =2[/itex]. Is it correct? Now you calculated the directional derivative along the line y=x. It is not the partial derivative.

    ehild
     
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