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Lagrangian of an isolated particle, independant from Newtonian Mechanics?

  1. May 1, 2010 #1

    fluidistic

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    I don't have the books in front of me so I only use my memory. According to my professor and if I remember well, Landau and Lifgarbagez, the Lagrangian of an isolated particle can in principle depend on [tex]\vec q[/tex], [tex]\vec \dot q[/tex] and t. Therefore one can write [tex]L(\vec q, \vec \dot q , t)[/tex]. With some non mathematical arguments, one can reach [tex]L(\vec \dot q)[/tex] because it doesn't matter where the particle is, the Lagrangian should be unchanged and it doesn't matter if we study the particle today or in 10⁹ years, the Lagrangian which describes the motion of the particle should be unchanged. Ok fine until here.
    Now they go on to say that since the Lagrangian is a real number, L must depends on [tex]\vec \dot q ^2[/tex]. This argument doesn't convince me. I don't think the implication is true. There are a lot of functions that takes [tex]\vec \dot q[/tex] and transform it into a real number, for instance the one that assignates [tex]\vec \dot q[/tex] to [tex]|\vec \dot q|[/tex] or [tex]\vec \dot q ^4[/tex]. I don't see why one function should be more natural than the others. To me it looks like we choose the one that allow us to later find out that [tex]L=T-V[/tex], thus to coincide with notions we already knew before.

    Then they do some algebra and find out that [tex]L=k \vec \dot q ^2[/tex] where k is a constant. Then they choose [tex]k=\frac{m}{2}[/tex] to make the Lagrangian coincide with the kinetic energy of a particle in Newtonian's mechanics. But if we didn't know anything from Newtonian's mechanics, we couldn't have chose such an "k" nor such a function for the Lagrangian, right? I mean, all in all Lagrangian mechanics as we use it daily is not totally independent of Newtonian's mechanics, am I right?
    I don't remember where I had read that the Lagrangian mechanics is totally independent of Newtonian's one... What do you think?
     
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  3. May 2, 2010 #2

    physicsworks

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    I stuck on this also when I first read L&L. Then I checked up what would be if [tex]L[/tex] were the function of, say, [tex]\mathbf{v}^4[/tex]. Well, if, according to L&L, we consider two inertial reference frames K and K' moving with an infinitestimal velocity [tex]\mathbf{\varepsilon}[/tex] relative to each onther and, hence, [tex]\mathbf{v' = v + \varepsilon}[/tex], then, since the equations of motion must be the same in both frames (Galileo's principle), the Lagrangian [tex]L(\mathbf{v}^4)[/tex] must be converted to a function [tex]L'[/tex] which differs from [tex]L[/tex] only by the total derivative of a function of time and coordinates, right?
    We have
    [tex]L' = L(\mathbf{v'}^4) \approx L(\mathbf{v}^4 + 4 \mathbf{v}^3 \mathbf{\varepsilon}) \approx L(\mathbf{v}^4) + \frac{\partial L}{\partial v^4} 4\mathbf{v}^3 \mathbf{\varepsilon}[/tex]
    The second term in the above equation is the total time derivative only if it is a linear function of [tex]\mathbf{v}[/tex]. Hence
    [tex] \frac{\partial L}{\partial v^4} \sim \frac{1}{v^2}[/tex]
    And from this [tex]L = L(v^2)[/tex].
    It is easy to show that for all functions [tex]L(\mathbf{v}^{n})[/tex] where [tex]n=2k, k \in \mathcal{N}[/tex] this dependence [tex]L= L(v^2)[/tex] holds. For [tex]L(|\mathbf{v}|)[/tex] I'm sure the calculations are not very differ from above.

    P.S. My notations may differ from yours because I read the original book in Russian.
     
  4. May 2, 2010 #3

    fluidistic

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    Thank you so much, I get the idea it. That's really convincing.
     
  5. May 9, 2010 #4
    So what are absolutely all assumptions?
    I guess one is that Langrangian formalism works at all, but why does it imply already? It's not easy to picture what minimizing action should mean.

    And where do counter examples fail this? For example what if I imagine a physical world where momentum is conserved, but forces are 1/r? What about the prediction of energy conservation?

    For 3D this would fail energy conservation, so what does the Lagrangian say about this?
    But I guess for 2D with 1/r forces you still would have energy conservation?
    So where does the above proof speak about dimensionality?
     
  6. May 10, 2010 #5

    Jano L.

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    Hi Gerenuk,
    in fact the energy conservation holds for every possible function of particle distance
    [tex]
    \boldsymbol{F}(r).
    [/tex]
    What is important, is that force is determined only by the position, not by velocities or time.
     
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