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Lagrangian and its derivatives

  1. Dec 7, 2013 #1
    Question 1

    When I take the derivatives of the Lagrangian, specifically of the form:

    [itex]\frac{\partial L}{ \partial q}[/itex]

    I often find myself saying this:

    [itex]\frac{\partial \dot{q}}{ \partial q}=0[/itex]

    But why is it true? And is it always true?
     
    Last edited: Dec 7, 2013
  2. jcsd
  3. Dec 7, 2013 #2
    [STRIKE]Question 2[/STRIKE] Answer Below

    When solving a double pendulum problem I built a Lagrangian of the form:

    [itex]L(\theta_1, \theta_2, \dot{\theta_1}, \dot{\theta_2})[/itex]

    And found that my Euler Lagrangian equations for each coordinate where coupled to each other, as expected.

    But I was a little confused about the direction of motion...

    Does

    [itex]\frac{\partial L}{\partial q_\alpha} = \frac{d}{d t} \frac{\partial L}{\partial \dot{q_\alpha}}[/itex]

    Describe a motion only in the

    [itex]\hat{q_\alpha}[/itex]

    Direction? Even if the Euler Lagrangian is coupled?

    Answer

    According to Taylor's text, Classical Mechanics, yes. The [itex]\hat{q_\alpha}[/itex] direction is the only one an Euler Lagrangian equation specifies. But, a true understanding of the motion in that direction comes from:

    [itex]- \frac{\partial U}{\partial q_\alpha} = \frac{d}{d t} \frac{\partial L}{\partial \dot{q_\alpha}} - \frac{\partial T}{\partial q_\alpha}[/itex]

    Which is used to build:

    [itex]\vec{F}=-\nabla U[/itex]
     
    Last edited: Dec 7, 2013
  4. Dec 7, 2013 #3
    Many a student has been confused by this. The short answer is that the Euler-Lagrange equation is derived from a variational principle using the technique of variations, and ##q## and ##\dot q## are merely labels for the arguments of the Lagrangian.

    Consider that you have a function ##F(x, y)##. At some ##(x_0, y_0)## its value is ##z_0##. Now you want to find out its value close to that point. Using the Taylor expansion, you get ##F(x_0 + \delta x, y_0 + \delta y) = z_0 + {\partial F \over \partial x} \delta x + {\partial F \over \partial y} \delta y ##. Now if you had ##x = f(t) ## and ##y = \dot f(t) ##, you would still use the expression above, where ##x## and ##y## are assumed independent. This is exactly what happens when you derive the Euler-Lagrange equation.
     
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