Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Elementary question regarding lagrangian

  1. May 2, 2004 #1
    Hi all,

    I'm trying to understand something about the lagrangian.
    My resources for learning a currently limited to Landau's mechanics and anything which is on the internet.

    [tex] L = L(q,\dot{q},t) [/tex]

    Now, here is a simple question: what are these generalized co-ordinated exactly?

    For example, in Landau, does the "component-wise" lagrangian as such:

    [tex] \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_i } -
    \frac{\partial L}{\partial{q_i}} = 0[/tex]

    and several pages later he says:

    [tex] \frac{d}{dt} \frac{\partial L}{\partial \vec{v} } [/tex]

    but in general, the [tex] i^{\mbox{th}} [/tex] component of [tex]\vec{v}[/tex] isn't equal to the time derivative of [tex] q_i [/tex].

    So what is it? [tex] \dot{q_i} \neq \frac{dq_i}{dt} [/tex] ??

    Could someone recommend some books which would better explain this stuff?

  2. jcsd
  3. May 3, 2004 #2


    User Avatar
    Homework Helper

    The book I had for classical mechanics was by Jose and Saletan (I'm not certain of the spelling of the second author). I don't remember the title, but if you search for books by these authors, it should be obvious. I have never heard a bad comment about Goldstien's book (any of his books for that matter).

    A few points concerning the Lagrangian:
    The Lagrangian is a scalar quantity that I like to think of as the "temporal action denisty." It is the amount of action that the system undergoes in an amount of time dt. When you put the parenthesis to the right and some variables in them, then you have demoted the meaning of the Lagrangian to a particular function of those variables, the value of which gives you the "temporal action density." Personally, I prefer to use another letter when I define a function for the Lagrangian, i.e.: L = f(q,q_dot,t). The q's can be whatever dynamical variables you want them to be, as long as you satisfy one stipulation in definining them: they must completely characterize the state of the system (without ambiguity) in themselves and their first time derivatives. This requires the (Lagrangian) function to act not only on the q's, but also on the q_dot's. The q_dot's are indeed the time derivatives of the q's.
  4. May 3, 2004 #3
    I dont think I was clear

    Thanks for the reply.
    I don't think I'm being very clear.
    What I'm trying to say is,

    [tex] \vec{v} = \dot{\vec{r}} [/tex]

    now, lets say, in spherical co-ordiantes,

    \vec{v} = dr/dt\hat{r} + rd\theta/dt\hat{\theta} + r\sin \theta d\phi/dt \hat{\phi}

    what are q and q_dot now???

    Thanks, Amir
    Last edited: May 3, 2004
  5. May 3, 2004 #4


    User Avatar
    Science Advisor

    [itex]q_1=r,q_2=\theta,q_3=\phi[/itex] and the dotted ones are the time derivatives of these e.g. [itex]\dot{q}_2=\dot{\theta}[/itex]
  6. May 3, 2004 #5


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    [tex]q[/tex] and [tex]\dot q[/tex] are coordinates of something the called the "tangent bundle of the configuration space" (sometimes called the "velocity phase space") of the system.

    "Introduction to Mechanics and Symmetry"
    is an advanced text on "geometric mechanics".

    Follow "Mechanics Book (HTML)"

    The text "Applied Differential Geometry" by William Burke
    has as its dedication "To all those who, like me, have wondered how in the hell you can change [tex]\dot q[/tex] without changing [tex]q[/tex]"

    The text "Lagrangian Interaction" by Noel A. Doughty
    has a sentence (p .173) "Thus [tex]\{q_j , \dot q_j \}[/tex] are assumed initially to be mathematically independent variables, despite the fact that knowledge of the physical path [tex]q_j(t)[/tex] traced by the system implies knowldege of the velocities [tex]\dot q_j(t)[/tex].

    Here are some other useful resources:
  7. May 4, 2004 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, [tex]\dot q[/tex] is the time derivative of q but remember that in Lagrangian mechanics they are treated as independent variables.

    (Ooops, I just noted that that is the gist of the Noel Doughty quote that robphy gave!)
  8. May 4, 2004 #7


    User Avatar
    Homework Helper

    One of the immediate reasons for applying the Lagrangian formalism is to avoid vectors. This goes not only for the force vector, but also for the the position vector. The purpose of generalized coordinates is to make the constraints invisible while not having to worry about obscuring the vector quality of the position (which is obscure in itself). To say it again, the q's (and so also the q_dot's) are whatever you want them to be, as long as they completely and uniquely describe the state of the system. For a free particle (in flat space), the Cartesian coordinates are taylor-made for this purpose. Spherical polar coordinates may also be used to describe the state of a free particle. But in the Lagrangian formalism you don't worry about the vector quality (as in the above quote), you only worry about the coordinates, in this case: r, θ, and φ. If these are the values that you want, then:

    q1 = r
    q2 = θ
    q3 = φ

    The subscripts on the qi's are arbitrary, so, for instance you could have just as well defined:

    q1 = θ
    q2 = φ
    q3 = r

    or any other combination. Sticking with the first definition for the generalized coordinates:

    q1_dot = dr/dt
    q2_dot = dθ/dt
    q3_dot = dφ/dt

    It's just that simple. You don't have to worry about the metric even (i.e. multiplying the θ vector by r). So, the Lagrangian is:

    L = f( q1 , q2 , q3 , q1_dot , q2_dot , q3_dot , t ) = f( r , θ , φ , dr/dt , dθ/dt , dφ/dt , t )

    To extend this point, you could have just as well constructed a definition that would lead to:

    q1_dot = dr/dt
    q2_dot = r dθ/dt
    q3_dot = r sinθ dφ/dt

    Of course, this makes the q_dot's more complicated. But, if that's what you want to do, then the formalism allows it. However, you don't lose nor gain any specificity in using this latest definition. You still require 3 dynamical variables (in this case, they are still r, θ, and φ) and their derivatives (for a total of 6), and no more nor less than that are specified in this case.
    Last edited: May 4, 2004
  9. May 4, 2004 #8
    Ok thanks a lot for the replies.
    I get what everyone is saying.

    The trouble I have then is with Landau's use of

    [tex] \frac{d}{dt} \frac{\partial L}{\partial \vec{v} } = 0 [/tex]

    (on page 5) when he clearly defines

    [tex] \vec{v} = \dot{\vec{r}} [/tex]

    Because according to the whole generalized coordinate stuff,
    [tex] \dot{q} \neq \vec{v} [/tex]

    Am I correct on this point?
  10. May 4, 2004 #9


    User Avatar
    Homework Helper

    Are you sure that Landau is using vectors? Does Landau define the derivative wrt a vector? If this is a proper thing to do, then it is above my level. It seems like this is just a short-hand notation.
  11. May 5, 2004 #10


    User Avatar
    Science Advisor

    Yea looks like Landau is using shortand, admittedly confusing.

    We're trying to tell you that it doesn't matter how you define q(dot) so long as you have a corresponding equation of constraint.

    Have you read up on the method of undetermined multipliers yet?
  12. May 5, 2004 #11
    "Have you read up on the method of undetermined multipliers yet?"

    By undetermined multipliers do you mean lagrange multipliers / constrained optimization??

    If so, I know everything they're teaching us here in first year calculus.
    i.e. very little

    I'm thinking I'll learn some more math and come to this stuff later, perhaps with a book not written by a genius. (i.e. Goldstein)

    Thanks alot guys for your help.
  13. May 11, 2004 #12
    My mechanics/EM professor uses the 'derivative with respect to a vector' notation quite a bit--in tensor calculus, it's actually acceptable (and powerful--since the tensor calculus we're using is just a Cartesian space with time tacked on.) I can sort of picture this array of derivatives, but many times I end up writing out one of the components (say, the x-component) of the equation, and use an ellipsis to denote that the rest are similar.

  14. May 11, 2004 #13


    User Avatar
    Homework Helper

    When you say that this is done in tensor calculus, do you mean:

    ∂/∂xμ ?

    This is not the derivative wrt a vector; it is (in the case of spacetime) a component of an object that transforms slightly differently than dxμ. Can you show an example of a derivative wrt a vector that your professor uses?
  15. May 11, 2004 #14


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I wouldn't recommend Jose and Saletan (though it deals very nicely with Tangent bundles, Poincare' maps, etc) as a first text. I suggest you start with Goldstein, or something of that kind.
  16. May 18, 2004 #15

    that's what I meant. He writes it [tex]\delta _\mu[/tex] to indicate that the derivative w.r.t. the contravariant components is a covariant operator.

    I suppose 'derivative w.r.t. a vector' is my terminology, not his--all I meant was that it's a compact way of saying 'take four different derivatives.'

    But I agree with your distinction between taking the derivative with respect to the various components of a vector in one coordinate system and the possibility of taking the derivative w.r.t. the vector itself, which should be independent of the choice of coordinate system.

    These operators remind me of some terms in vector calculus that contain apparently non-sensical derivatives: gradient of a vector, Laplacian of a vector. But if you write out the vector as a column vector and apply the derivative to each term individually, it's unambiguous how to proceed from there.

    Last edited: May 18, 2004
  17. May 18, 2004 #16


    User Avatar
    Homework Helper

    We are consentient.

    I believe the problem lies in a fundamental misinterpretation of these vector operators as composed of the contravariant components of vectors. The operators themselves do not inherently require contraction; it is how they operate that does that. I suppose that's what you meant?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Elementary question regarding lagrangian