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Finding the Lagrangian

  1. Oct 15, 2013 #1
    a particle of mass m is attracted to a center force with the force of magnitude k/r^2. use plan polar coordinates and find the Lagranian equation of motion.

    so i thought for the kinetic energy it would be..

    K=[itex]\frac{1}{2}[/itex]m(r2[itex]\dot{θ}[/itex]2)

    since v2 = r2[itex]\dot{θ}[/itex]2

    but no.. the kinetic energy was actually K=[itex]\frac{1}{2}[/itex]m([itex]\dot{r}[/itex]2+r2[itex]\dot{θ}[/itex]2)

    why???

    the velocity vector of some particle can be described by [itex]\frac{d\vec{r}}{dt}[/itex]. the velocity vector can also be described by rω. Why combine the two? aren't the two velocity terms (in the actual answer) the same? so then wouldn't i end up with 2v? but this is probably wrong. why is this wrong/ why is that the right way?

    thanks all
     
  2. jcsd
  3. Oct 15, 2013 #2
    side question: is the hamiltonian always equal to the total energy when the coordinates are independent of time?
     
  4. Oct 15, 2013 #3

    dextercioby

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    Since x=x(r,phi) this means that in theory both r and phi depend on t, so when differentiating to get v_x you get a sum of 2 terms. Likewise for y. You have in total 3+3 = 6 terms of a sum. You then use that sin^2 phi(t) + cos^2 phi(t) = 1.

    For the second, it's not the coordinates that should be directly independent of time (in that case you'd have a statics problem), but the lagrangian as a whole.
     
  5. Oct 15, 2013 #4
    i can see the reasoning, but if that's the case where S=θr then i don't understand the mathematical reasoning behind
    "[itex]\dot{r}[/itex]2+r2[itex]\dot{θ}[/itex]2" . why isn't it just the chain rule?

    [itex]\frac{dS}{dt}[/itex]= θ[itex]\frac{dr}{dt}[/itex] + r[itex]\frac{dθ}{dt}[/itex]?


    i can see how both of r and theta would be a function of two variables but how how mathematically did they obtain what they did?
     
  6. Oct 15, 2013 #5

    dextercioby

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    v =sqrt (v_x ^2 + v_y ^2) in cartesian coordinates. To switch it to plane polar, first compute v_x, if its 2 variables that you use are r and theta/phi instead of x and y.
     
  7. Oct 15, 2013 #6
    if i were to find the lagrangian why do i just use theta as the coordinate? why don't i use r as well? the answer in the back of the book only writes down the euler-lagrangian formula for the theta but not the r.
     
  8. Oct 15, 2013 #7
     
  9. Oct 15, 2013 #8

    Dick

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    Last edited: Oct 15, 2013
  10. Oct 15, 2013 #9
    thank you very much that actually answered everything :)
     
  11. Oct 15, 2013 #10
    oh one more thing:

     
  12. Oct 15, 2013 #11

    Dick

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    Isn't the Hamiltonian defined as kinetic energy+potential energy? So yes, it's total energy. If it's time dependent that energy doesn't have to be a constant. Been a while since I've done classical mechanics though, so take that with a grain of salt.
     
  13. Oct 15, 2013 #12
    Yes, it is kinetic plus potential energy, but only if the coordinate system is independent of changes in time. Since it's fairly simple to compute both that sum and the more general version, it's fairly easy to double check yourself if you're working on a type of problem that you haven't used the Hamiltonian on before.
     
  14. Oct 15, 2013 #13
    actually no it's defined i believe as [itex]\frac{∂L}{∂\dot{q}}[/itex][itex]\dot{q}[/itex]-L

    it's only equal to the total energy in certain cases.

    oh another question: the hamiltonian of this was..

    [itex]\frac{∂L}{∂\dot{θ}}[/itex][itex]\dot{θ}[/itex]-L (from the back of the book)

    but why was it enough to just use theta as the q (coordinate)? why wasn't it necessary to find the hamiltonian with respect to the radius as the coordinate? wasn't the radius also used as part of the coordinate system?
     
  15. Oct 16, 2013 #14
    Right. As I said, in the case where the coordinate system is independent of changes in time.

    You're also correct about the second part. I'm not sure why the book says that. Since the Hamiltonian is [itex]\sum _i \frac{\partial \mathcal{L}}{\partial \dot{q_i}}\dot{q}-\mathcal{L}[/itex] and your Lagrangian involves first derivatives of each coordinate (you have two [itex]\dot{q}[/itex]'s in the Lagrangian), I see no reason you shouldn't get a term from each.
     
  16. Oct 16, 2013 #15
    if there is no work being done on the system can that also be a criterion for K+U=H? or strictly when q is independent on time?
     
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