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

Maupertuis' Principle

  1. Jun 9, 2012 #1
    Two Questions:
    1) What do I need to know about it?
    2) During the study of its derivation:I ran across the following:

    They used Lagrangian in its usual form: L=[itex]\frac{1}{2}[/itex][itex]\sum[/itex]a[itex]_{ik}[/itex](q)[itex]\dot{q_{i}}[/itex][itex]\dot{q_{k}}[/itex] - U(q)

    [I had no idea where did this form come from, but if I ignored that and continued my question..]

    Then they mentioned that Momenta are:

    p[itex]_{i}[/itex]=[itex]\partial[/itex]L/[itex]\partial[/itex][itex]\dot{q_{i}}[/itex]=[itex]\sum[/itex]a[itex]_{ik}[/itex](q)[itex]\dot{q_{k}}[/itex],

    So here it comes where did the 1/2 go? Why isn't it present?
     
  2. jcsd
  3. Jun 9, 2012 #2
    Maupertuis principle is that of the least action.
    Was this your first question ?
    The principle says that there is something, called action on a system that is minimized (well, optimized), the action is an integral over the path of something called the Lagrangian.
    You eventually get to prove that under certain conditions, the lagrangian is of the form T-U, with T being the kinetic energy of the system and U the potential energy.
    Anyway, the Lagrangian you provide is of this form, except it is a bit more complicated than it could be since it accounts for some coordinate transformation if I understand correctly.
    Now for the second question, look closely, not only 1/2 disapears, but also q-dot-sub-i.
    Why ?
    because you partial derive with respect to q-dot-sub-i which is going to be 0 for any k!=i (because the velocities are independent of each other) and 1 otherwise
    so you will be left with a sum of squared terms, which will account for, when you take the derivative, the 1/2 half factor you are looking for.

    Cheers...
     
  4. Jun 9, 2012 #3
    Thank you very much, but if my request is annoying, can you please illustrate what you meant by "you will be left with a sum of squared terms" - Just so I can clearly see it (I don't have a good mathematical background)
     
  5. Jun 9, 2012 #4

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

  6. Jun 9, 2012 #5
    Vanhees71, is the answer to my 2nd question true? If yes, illustrate if you may.
     
  7. Jun 9, 2012 #6
    Thank you for this very nice clarification vanhees71,
    But Maupertui's principle *is* the principle of least-action, he coined the term after all :) , at the time solving a strong dilema where Fermat's principle (shortest path for a light ray) was having some philosophical problems (how does light know which path is the shortest to begin with ?)
    It has been generalized since then, yes, but saying that is is different from the least action principle is not too fair.
    Also, M. next deals with a Lagrangian, so we are well over Maupertuis anyway and I guess we are indeed talking about the principle of least action under its historical "father-acknowledging" name
    (but that doesn't take anything from you link which is a nice read, thanks again)
     
  8. Jun 9, 2012 #7
    Hi M. next,
    no I'm sorry if I left you thinking your question was annoying in any way, not at all.
    Now,
    look at your sum
    it is a sum over two indices, i and k of terms (q-dots).
    so you will have two kinds of terms together when you expand the sums: those were i=k, and those were i!=k
    where i=k, this is the same term being multiplied by itself, so it gets squared (you multiply it by itself)
    so when you will take the derivative, 1/2x² -> x, and there went your 1/2 factor.

    Cheers...
     
  9. Jun 9, 2012 #8
    I believe what oli4 has said is quite true and it is somehow mentioned in Landau's book.
    It is a simplified version of Least Action.
    Anyhows, can you please clarify your second answer?
     
  10. Jun 9, 2012 #9
    Ohhh thank you oli4, I get it now, I should note that you handle issues very nicely. Thanks again.
     
  11. Jun 9, 2012 #10

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Ok, historically Maupertuis (or even Leibniz!) where much earlier in their formulation of least-action principles. Nowadays, of course, the least-action principle is, in its most general form, understood as Hamilton's principle of least action in Hamilton's formulation. In classical mechanics, it's equivalent to the Lagrangian version, and that's usually the one which one uses first.

    Here the action is defined as a functional of the trajectories in configuration space via

    [tex]S[q]=\int \mathrm{d} t L(q,\dot{q},t).[/tex]

    The trajectories are given by the stationary points of this functional, leading to the Euler-Lagrange equations

    [tex]\delta S=0 \; \Rightarrow \; \frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}}=\frac{\partial L}{\partial q}.[/tex]

    In non-relativistic mechanics in the most simple case of particles with potential forces, one has

    [tex]L=\frac{1}{2} a_{jk}(q) \dot{q}^{j} \dot{q}^{k}-V(q).[/tex]

    Here, Einstein's summation convention is used (i.e., the sum over equal indices is implied). Obviously can also take [itex]a_{jk}=a_{kj}[/itex]. I also use upper indizes for the [itex]q^k[/itex] since [itex]\mathrm{d} q^k[/itex] are contravariant vectors under general coordinate transformations in the sense of general tensor analysis. This is not so important for classical mechanics, but it's a very useful notation to understand the mathematical structure of analytical mechanics better. Then you have

    [tex]\frac{\partial L}{\partial \dot{q}^a}:=p_a=\frac{1}{2} a_{jk} (\dot{q}^j \delta^{k}_{a} + \dot{q}^k \delta^j_k)=\frac{1}{2} (a_{ja} \dot{q}^{j}+a_{ak} \dot{q}^{k})=a_{aj} \dot{q}^j.[/tex]

    In the latter step we have used the symmetry of [itex]a_{jk}[/itex] and have renamed the summation index from [itex]k[/itex] to [itex]j[/itex] in the last term. The [itex]p_a[/itex] is called canonically conjugated momentum of the configuration variable [itex]q^a[/itex].

    I think, in a first course on classical mechanics, you don't need to know about the Maupertuis version of the principle. More important is this Lagrangian formulation and even most important Hamilton's formulation.

    From a philosophical point of view it's interesting that Leibniz and Maupertuis had a teleological interpretation of this principle since nature seems to behave as an intelligent being, minimizing the action in the sense of being "most efficient" in this very specific sense. The same was true for the already mentioned Fermat principle.

    Of course, this doesn't make much sense even in a classical model of the world as is classical mechanics since the very same physics is contained in local equations of motion (namely Newton's differential equations or the Euler-Lagrange equation), and the variational principles can thus be taken as mathematical simplifications of Newton's equations. With simplification I mean that the action principle is a very efficient tool to analyze the mathematical structure of the mechanical system, e.g., with respect to symmetries and conservation laws (Noether's theorem), which notion introduces the most important notion of the importance of (Lie) symmetry groups into physics.

    From a modern point of view, the least-action principle follows as the (semi-)classical approximation of quantum mechanics, most elegantly formulated in terms of the Feynman path integral, but that's another story.
     
  12. Jun 9, 2012 #11
    That was very informative. Thank you, I don't know why we are concentrating on this principle, it is a bit confusing, the way we deal with it.
    Even the easiest problems in this issue: Finding path of a material point of mass m using δSo=0.
    We finally ∫[itex]^{q1}_{q2}[/itex]√2mEdl[itex]^{2}[/itex] = ∫[itex]^{q1}_{q2}[/itex]dl=0
    and I cant figure out why. I just have this very normal problem in not knowing how to apply the formula!! And guess what? We haven't taken tensors yet. Sad, no?
     
  13. Jun 9, 2012 #12
    Hi vanhees71, hi M.next

    vanhees71, I sort of have the feeling that you have some strong feeling about possible confusion between the original statement of Maupertuis' least action principle (maybe first stated by Leibniz according to Koenig, whatever, that is not the point I think)
    Everything you say is correct, but I don't see how it is relevant to the question being asked at the beginning, your own link to wikipedia says that the least action principle is frequently named Maupertuis', or Hamilton's.
    The question being asked involves a Lagrangian, and later general momenta. So we are very clearly talking about Hamilton's Least action principle, now how do you answer "what do I need to know about Maupertuis' principle ?'" without mentioning the least action principle ?
    Anyway, this can be a religious/zealot thing, or you can be a regular victim of people not understanding the difference between those nuances and become very eager to state the distinction at every occasion (or I can miss something deeper that I will thank you for pointing me at, really (no bad humour/sarcasm at all))

    In any case, and this is becoming offtopic, you say:

    "it's interesting that Leibniz and Maupertuis had a teleological interpretation of this principle since nature seems to behave as an intelligent being, minimizing the action in the sense of being "most efficient" in this very specific sense. The same was true for the already mentioned Fermat principle"

    Well, that's interesting, I don't know about Leibniz side, (well neither do I in fact for the other side ;)) but I was under the impression that Maupertuis' own victory was precisely getting rid of it, that is, this problem was coming from Fermat's expression, and Maupertuis' was an answer that got rid of this overlooking intelligence. I have read a nice book about 'least princple action history' but it was a while ago and I can mix things, but it was my feeling/memory so thanks for fixing it :)

    M. next:
    I'm sorry I didn't understand your last question, but as far as not seeing tensors yet, no this is not really sad, you don't need them for what you are doing now (of course, it wouldn't hurt either) so you don't have to feel bad or 'under-prepared'
     
  14. Jun 9, 2012 #13
    I just meant what would you answer if you were asked to find the trajectory of a mass m using Maupertuis' Principle?
     
  15. Jun 10, 2012 #14

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I think there's a bit missing. If there were only squared terms left then you'd end with:

    [itex]\partial[/itex]L/[itex]\partial[/itex][itex]\dot{q_{i}}[/itex]=a[itex]_{ii}[/itex](q)[itex]\dot{q_{i}}[/itex]

    The terms [itex]\dot{q_{i}}[/itex][itex]\dot{q_{k}}[/itex] and [itex]\dot{q_{k}}[/itex][itex]\dot{q_{i}}[/itex], k≠i, don't disappear. a[itex]_{ik}[/itex] being symmetric, these combine in pairs and differentiate to give [itex]2a_{ik}(q)\dot{q_{k}}[/itex]
     
  16. Jun 10, 2012 #15
    Hi haruspex, of course !
    Indeed you're right I wasn't careful enough, thanks for correcting me :)

    Cheers
     
  17. Jun 10, 2012 #16
    What now?
     
  18. Jun 10, 2012 #17
    I told you before some terms were disappearing, but that was wrong, I just confused myself, and had that been the case you wouldn't have the same result anyway
    Instead, the diagnoal terms are indeed squared terms whose derivative will make the 1/2 go away, but for the other terms, they just pair together because the matrix is symemtric
    a00 q0q0 + a11q1q1 + a01q0q1+a10q1q0
    which gives a00q0²+2a01q0q1 because a10=a01
    take the derivative and you are back yo your expression.
    with my wrong explanation, you would just have had the aii(q) in the sum

    Cheers...
     
  19. Jun 10, 2012 #18
    Thanks for explaining this once again. Endless thread :p!
     
  20. Jun 10, 2012 #19

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Again, the Maupertuis principle is a different variational principle than Hamilton's, and I have no strong feeling at all. Also the history is not so important. I'm after the important fact that these principles are different variational principles for the same mechanical problems. On an exact level of course all are equivalent.

    The Maupertuis principle is a variational principle for the special case of conserved systems (i.e., such Hamiltonian systems that have a Lagrangian and then also Hamiltonian that is not explicitly dependent on time) at fixed energy. The variation is under the constraint of fixed endpoints in configuration space but not at fixed times. It leads to equations of the trajectories alone, not to dynamics.

    That's a very elegant principle to , and I've nothing against it. It can be very useful to calculate the trajectories of systems, particularly to get non-trivial approximate solutions from a variational principle. The idea here is to use an ansatz function for the trajectories with some parameters in it and then make the action stationary with respect to variations of these parameters, i.e., you simplify a variational principle (variations of functions) to a usual extremal problem (variations of parameters).

    This is also a powerful tool to get non-perturbative approximations in to quantum mechanical problems within the path-integral formulation of quantum mechanics.
     
  21. Jun 10, 2012 #20
    :) good thing haruspex came by too :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Maupertuis' Principle
  1. Hamiltons principle (Replies: 1)

  2. Principle of Gyroscope (Replies: 3)

  3. Principles of a Hookah (Replies: 9)

  4. D'alembert's Principle (Replies: 4)

  5. Principle of moments (Replies: 2)

Loading...