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Lagrangian anyone know their Lagrangian mechanics?

  1. Apr 27, 2008 #1
    1. The problem statement, all variables and given/known data

    A particle of mass m moves in a potential of the form

    U(r) = (1/2)kr^2

    k = const greater than zero

    1) Determine the possible orbits r = r(theta) and show that they are closed

    2) Solve the equations of motion (although it is sufficient to derive the time dependence
    of the radial variable, r = r(t), in accordance with the general theory of the motion in
    central potentials, there is a simpler way to address the problem !!!).

    3) Analyze whether this potential respects Kepler’s first law: The orbits are ellipes with
    the force centre at one focus.

    4) Demonstrate that this potential does not respect Kepler’s third law.

    2. Relevant equations

    The lagrange equations

    (d/dt) (dL/dr) - (dL/dr_dot) = zero (sorry for the crappy formatting, you know what mean)

    The Newton equation

    m*acceleration = -(d/dr) U (r)

    'Reduction to quadrature'

    t-t0 = the integral of 1/ sqrt (E - U(r)) dr (again... sorry!)
    This last one is helpful for finding r in terms of t but often yields tricky integrals

    3. The attempt at a solution

    I think this is a doozy. WHat coordinates do I switch too? And what is the Lagrangian for a spherical harmonic oscillator????

    I tried modelling my answer on a spherical pendulum problem (no oscillations)... There's a detailed worked pendulum problem in our course notes. Here the coordinate system was changed to

    x = r sin (theta) cos (phi)

    y = r cos (theta) cos (phi)

    z = -rcos (theta)

    why would you bother, when you can just work with r? Do you have to switch coordinate systems for the Lagrange equations to be valid?

    Anyways I located the equation for a sperical pendulum Lagrangian in our text. It's:

    L = (1/2)m (x˙^2 + y˙^2 + z˙^2) − mgz

    (Note the dots... That rate in change of x squared plus rate of change in y squared plus...)

    In the new coordinate system this is

    ml2(˙θ2 + ˙φ2 sin2 θ) + mgl cos θ

    For my spherical oscillator can I just replace the potential energy (mgz, the final term) with the given potential, (1/2)kr^2?

    Also, is angular momentum conserved? That would mean I can set the change in in phi to a constant (Noether's theorem, for Lagrange aficionados). SOmehow I doubt I can... They do it in the pendulum example but... is angular momentum conserved in the spherical oscillator case? I'm finding it hard to picture.

    I proceeded with my Lagrangian and assumed delta phi was constant and found

    r = (+/-) 4nth root of (1/sin^2 theta) * (1 / (Veff + k theta/2m))

    I have no idea if this is even close... that initial Lagrangian could be way off.

    Also not sure how to ascertain whether this path is 'closed'. sin^2 theta moves between 0 and 1 and theta can go between a and pi so... I think the radius changes periodically, as an oscillator should.

    On the other hand, that theta in the denominator on the right hand side is troubling... it suggests the radius goes to infinity when theta (the angle of the radial vector to z) is zero.

    Help! Help, help, help. ANy advice about Lagrange or links to good resources would be enormously appreciated. Sorry for the scrappy summary- it's 2am here in Australia and I want to get this posted before all the AMericans start waking up :)
  2. jcsd
  3. Apr 27, 2008 #2
    Well, you're not too far, but not too close. You can't just re-use your solution for the spherical pendulum here, as the sphecial pendulum as a constraint (namely, r=l). Furthermore, is doesn't have the same potential. I suggest your start the problem from scratch so you have a better understanding of Lagrangian mechanics, which is a very powerful and beautiful way to solve problem. For further reading, I suggest you read Taylor's Classical Mechanics who gives some good examples (3-4) on how to use lagrangian dynamics. If you want more after that, there's Goldstein who's pretty good and is a lot more rigorous than Taylor (but it's on a higher level).

    The best way to start is from scratch. First, you know that the Lagrangian, L, is written like
    [itex] $ L=T-U$ [/itex] with T the kinetic energy and U the potential energy. Now, you're looking for a good choice of coordinates... the spherical one seems OK to me, as your potential is initially written in spherical coordinates.

    Now, you need to write the kinetic energy in spherical coordinates. What is it? Well, you got the first step right ( writing x, y and z in their respective spherical representation). As [itex] T =\frac{1}{2}m( (\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}+(\frac{dz}{dt})^{2}) [/itex], you need to differentiate your three components with respect to time. Don't worry, the kinetic term will come out nicely with some trigonometric identities (i.e. when you'll have squared your terms and put them together).

    From Wikipedia, your lagrangian will look like
    [​IMG] Now, just replace V(r) by your potential, and you'll be on the right track to solve the problem.

    Hope it helped.

    P.S. Generally,
    [itex] z=rcos(\theta) [/itex],
    your teacher just turned the coordinate system to make the problem of a pendulum more insightful. I suggest your start with that more general definition of the three coordinates in spherical coordinates.
    Last edited: Apr 27, 2008
  4. Apr 27, 2008 #3
    Ooh thank you Erythro, that's a great summary not just for this but all my Lagrange problems. I was a bit overwhelmed by pages of subtly different examples and needed just what you gave: a template for how to address a problem generally.

    The only thing I don't understand now is how to get a time derivative for x,y and z when there is no time dependence! That would make all the derivatives zero and then my kintetic energy would be zero (which is wrong, of course ).

    I guess I have to derive a time dependence from somewhere else? Not the 'coordinate' equation.

    Thanks again! :)

  5. Apr 27, 2008 #4
    You very much HAVE time dependance, but it is not explicit. Partial derivative of x with respect to time would be 0, right, but not total derivative as implied in velocity. If some variable were to have no dependance on time, it would be written in the problem (e.g. the case of the pendulum where r=l which implies that every time derivative of r are equal to 0 : there's no speed nor acceleration in r because it's constant).

    As an example, [itex] x=rsin(\theta)cos(\phi) [/itex].
    Here, for your problem, [itex] r, \theta, \phi [/itex] are variables who MAY have time dependance (if they wouldn't, well, the problem would be pretty dull as this would imply that all variable are constants in time : you'd have a particle who doesn't move at all). Remember all the derivation rules from your calculus course. So, if I do the derivative with respect to time, I need to consider the three variables to be variable of time. Hence,

    [itex] \frac{dx}{dt}=\frac{dr}{dt}sin(\theta)cos(\phi) + rcos(\theta)\frac{d\theta}{dt}cos(\phi)-rsin(\theta)sin(\phi)\frac{d\phi}{dt}[/itex]
    . Do the same with y and z, squared them and when you'll put them in your kinetic energy term, you should have a pretty form of your equation, as I posted above.

    The good part is, you only have to do this once (finding the kinetic energy in spherical coordinates), because after that, you'll know that this equation is right in spherical coordinates, and this equation is very general, meaning you can place your constraints in the kinetic energy term directly, without doing it over and over again.

    Finally, you must understand the difference between partial derivatives and total derivatives. This is something that is very easy to mix up. With partial derivatives, you have to derive the variable explicitly. Meaning the partial derivative with respect to time of x would be 0, as there's no explicit time dependance in the expression (i.e. no "t"). Total derivative means you have to derive it all with respect to time, even if it is not explicit.

    As a brief example, imagine a simple equation such as g=x, with x=x(t). The partial derivative with respect to time would be 0, as the time variable doesn't appear anywhere in the equation. Yet, if you do the total derivative, then you have [itex]\frac{dg}{dt}=\frac{dx}{dt}[/itex].

    Hope this help :).

    To be honest, lagrangian dynamics is, to some extent, almost boring, because you don't have to think that much about the problem. You just have to do each steps in order and you'll have your equations of motion quite easily.

    1- You find a good coordinates system
    2- You (usually) know the general kinetic energy term in that system, because it is written in your textbook or you did it once. You can always do it all again in the worst case scenario, which is not that bad.
    3- You place your constraints in the kinetic energy term if your system has any constraint.
    4- You write your potential energy in these coordinates
    5- You write your lagrangian.
    6- You apply the Euler-Lagrange/Lagrange equations.
    7- You have your motion equations.
    Last edited: Apr 27, 2008
  6. Apr 27, 2008 #5
    Ah I've been suffering from the age old partial vs. full derivative confusion :) Thanks again Erythro!
  7. Jun 8, 2008 #6
    Can someone please explain to me what Euler- LaGrangian equations are and how they are solved or interpreted in both a quantum field and a classical field? What is the idea or point behind these equations and what math do I need to know to solve them?

    Any help would be appreciated...thank you. Also some background on the "action principle" would be helpful as well. I am taking a class in relativity at Stanford U. and am confused about what my teacher is saying.
  8. Jun 8, 2008 #7


    User Avatar
    Staff Emeritus
    Science Advisor

    My first thought was "Ask your teacher!"

    But then I had this horrible image: Another student is asking your teacher a question and you run up, shove the other student out of the way and insist that your teacher ignore the other student and answer your question!

    Because that's just what you did here. It is extremely rude to interrupt another person's thread in order to ask your own question. You are welcome to start a thread of your own to ask your question- and it is a good idea to make the question as specific as possible. No one here is going to give you a one-semester course on "Lagrange mechanics" here!
  9. Aug 1, 2008 #8
    or is s/he trying to say that s/he is at stanford?
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