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I Physical meaning of Lagrangian?

  1. Dec 5, 2015 #1

    Erland

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    Does the Lagrangian L in classical mechanics have any physical meaning?

    In classical mechanics, the Lagrangian is defined as L=T-V, the difference between the kinetic and potential energy of the system. Does this quantity have any meaning apart from that it can be plugged into Euler-Lagrange's equations to find a stationary path?

    Can we say anything about the difference in behavior for systems with high and low values of L, respectively?
     
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  3. Dec 5, 2015 #2

    dextercioby

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    To help you reach conclusions on your own, I may ask you: Is the Lagrangian for a system (let us say free Galilean particle) unique?
     
  4. Dec 5, 2015 #3

    Erland

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    No, not unique. We can, in this case, add an arbitrary function f(t), depending only upon time, explicitly, but not upon position and velocity.
     
  5. Dec 5, 2015 #4

    samalkhaiat

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    No, you can not do that. Two Lagrangians [itex]L_{1}(q, \dot{q})[/itex] and [itex]L_{2}(q, \dot{q})[/itex] are equivalent to each other, if and only if they differ by total time derivative of some arbitrary function of the coordinates only [itex]\frac{d}{dt}F(q)[/itex]. This, actually, is a provable theorem.
     
  6. Dec 5, 2015 #5

    Geofleur

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    Well, it can be the total time derivative of a function of coordinates and time, so ## \frac{d}{dt}F(q,t) ##. For example, if we add such a term to ## \mathcal{L} ##, the action becomes

    ## S = \int_{t_i}^{t_f} \left[ \mathcal{L} + \frac{d}{dt}F(q,t) \right] dt = \int \mathcal{L}dt + F(q_f,t_f)-F(q_i,t_i) ##.

    The last two terms are constants and thus vanish upon variation of ## S ##.
     
  7. Dec 5, 2015 #6

    samalkhaiat

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    Yes, you can stick [itex]t[/itex] in [itex]L_{1}[/itex], [itex]L_{2}[/itex] and in [itex]F[/itex] and the theorem still hold. A function of [itex]q_{a}(t)[/itex], [itex]\dot{q}_{a}(t)[/itex], and [itex]t[/itex] satisfies the Euler-Lagrange’s equations identically (i.e., independent of the [itex]q_{a}(t)[/itex]’s) if, and only if, it is the total time derivative [itex]dF/dt[/itex] of some function [itex]F(q(t) , t)[/itex]:
    [tex]
    \Big\{ G(q , \dot{q} , t) = \frac{d}{dt}F(q , t) \Big\} \ \Leftrightarrow \ \Big\{ \frac{d}{dt} \left( \frac{\partial G}{\partial \dot{q}_{a}} \right) - \frac{\partial G}{\partial q_{a}} \equiv 0 \Big\}
    [/tex]
    Okay, I leave you to prove it.
     
  8. Dec 6, 2015 #7

    Erland

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    Not hard to prove, samalkhaiat. So, the conclusion is that since the Lagrangian is not unique, that it can be altered to such a high degree, there is little hope to find any meaningful physical interpretation of it. It's only useful in calculations.
     
    Last edited: Dec 6, 2015
  9. Dec 6, 2015 #8

    samalkhaiat

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    It will be nice if you can share your “not hard” proof with us. I’m interested to see your proof. Just show us how to prove the “only if” part. That is,
    [tex]
    \Big\{ \frac{d}{dt} \left( \frac{\partial G}{\partial \dot{q}_{a}} \right) \equiv \frac{\partial G}{\partial q_{a}} \Big\} \ \Rightarrow \ \Big\{ G(q,\dot{q},t) = \frac{d}{dt}F(q,t) \Big\} .
    [/tex]
    If you understand the above little theorem, you will have no problem understanding the fundamental importance (or the “meaning”) of the Lagrangian in physics.
     
  10. Dec 7, 2015 #9

    Erland

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    Ok, not so easy. It is the other way that is quite easy. But please, don't force me to write down the proof. It would take hours getting the TeX-notation right, with all partial derivatves, indices, etc. Just to indicate some important points: using that LHS holds for all possible paths we can use that L doesn't depend upon the accelerations of the coordinates to infer that L is a linear function of the velocities, with coefficient functions depending only on positions and time. Plugging this into the LHS, we obtain in similar way that ##\partial A_i/\partial q_j=\partial A_j/\partial q_i## (##A_i## are the coefficient functions), from which we can find a function ##F(q_1,\dots,q_n,t)## whose partial derivatives are the ##A_i## and its total time derivative is ##L##.

    I suspect that one can also prove this by variational principles.

    Ok, I know it is considered to be of fundamental importance, but this nonuniqueness seems to indicate that its only use is to be plugged into equations, that it is not meaningful to think of it as a measurable quantity...
     
    Last edited: Dec 7, 2015
  11. Dec 8, 2015 #10

    samalkhaiat

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    What do you mean by “[itex]L[/itex]”? The theorem in #6 does not mention Lagrangian or any other dynamical concepts. It is just a mathematical theorem.
    Let us start again. I will restate the theorem and use it to demonstrate the importance of the Lagrangian. And please pay attention to the under-lined words.
    “A function [itex]G(q,\dot{q},t)[/itex] satisfies Euler-Lagrange equation identically, if and only if, it can be written as a total time derivative [itex]dF/dt[/itex] of the arbitrary function [itex]F(q,t)[/itex].”
    The crucial point in here is the statement that the Euler-Lagrange “equation” of [itex]G[/itex] is an identity, i.e., it does not lead to differential equations. This implies that
    [tex]\frac{\partial^{2}G}{\partial \dot{q}_{a}\partial \dot{q}_{b}} = 0 . \ \ \ \ \ \ \ \ \ (1)[/tex] This in turn means that [itex]G[/itex] is at most linear in the [itex]\dot{q}_{a}[/itex]’s
    [tex]G(q,\dot{q},t) = \sum_{a} f_{a}(q,t) \dot{q}_{a} + g(q,t) . \ \ \ \ \ \ (2)[/tex]
    Let’s compare the above with the definition of Lagrangian: “The function [itex]L(q,\dot{q},t)[/itex] is a genuine Lagrangian, if and only if the Euler-Lagrange equation of [itex]L[/itex] leads to differential equations of order no higher than the second in time derivatives.”
    This means that, almost always
    [tex]\frac{\partial^{2}L}{\partial \dot{q}_{a}\partial \dot{q}_{b}} \neq 0 . \ \ \ \ \ \ \ \ \ (3)[/tex]
    This implies that [itex]L[/itex] is, almost always, a quadratic function of the velocities [itex]\dot{q}_{a}[/itex].
    [tex]L(q,\dot{q},t) = \sum_{a,b} g_{ab}(q,t) \dot{q}_{a}\dot{q}_{b} + \sum_{a} f_{a}(q,t)\dot{q}_{a} + g(q,t)[/tex]
    Since [itex]G = dF/dt[/itex] leads to no dynamical equation, the theorem implies the followings:
    (1) A genuine Lagrangian cannot be a total time derivative of some function of the [itex]q_{a}[/itex]’s and time.


    (2) Let [itex]A[/itex] and [itex]B[/itex] be two physical systems described by [itex]L_{A}[/itex] and [itex]L_{B}[/itex] respectively. If for any [itex]F(q,t)[/itex]
    [tex]L_{A} \neq L_{B} + \frac{d}{dt}F(q,t) ,[/tex] then [itex]A[/itex] and [itex]B[/itex] are two distinct physical systems.

    (3) If there exists some function [itex]\Lambda (q,t)[/itex] such that [tex]L_{A} = L_{B} + \frac{d}{dt}\Lambda (q,t) ,[/tex] then [itex]A[/itex] and [itex]B[/itex] are identical physical systems related by a group of symmetry transformations, i.e., [itex]L[/itex] and [itex]L + d\Lambda /dt[/itex] describe one unique physical system whose constants of motion are given by
    [tex]
    C_{\epsilon} = \frac{\partial L}{\partial \dot{q}_{a}} \delta_{\epsilon}q_{a} + \Lambda_{\epsilon}(q,t) .
    [/tex]


    It is, I hope, clear from the above that the correct Lagrangian determines all the measurable dynamical quantities of the physical system. Do physicists wish for a better object than [itex]L[/itex]? No, because there is none.
    We also cannot measure the electromagnetic potential, the wave function in QM, the (phase-space) density of microstates, the entropy, the fields associated with elementary particles etc. Does this cast doubts on the fundamental importance of these quantities?
     
    Last edited: Dec 8, 2015
  12. Dec 8, 2015 #11

    Erland

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    Ok, so I wrote L instead G, and my proof was along the same lines as yours...

    Thanks for your explanations. Yes, the Lagrangian(s) is very useful, it contains all information we can extract. But still, it seems to lack physical intuition, just as the wave function in QM, but at least the electrostatic potential has the interpretation that a positively charged particle tends to move from high to low potential...

    So I agree, it is very useful, but there is the pedagogical difficulty that it is hard to grasp...
     
  13. Dec 8, 2015 #12

    ShayanJ

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    I don't want to hijack this thread, so if Erland is satisfied with the answers, I want to ask this question.
    Could you explain about those situations which made you say "almost always"?(You could point to a book or paper in case that's preferable)
    Thanks
     
  14. Dec 8, 2015 #13

    samalkhaiat

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    Well I don't think you need book or a paper, you should know it. If the coordinates are c-numbers taking values in Grassmann algebra, i.e., anticommuting numbers, the Lagrangian (which is bosonic) must be linear in the "velocities" [itex]\dot{q}_{a}[/itex], just like the classical Dirac Lagrangian.
     
  15. Dec 8, 2015 #14
    I was puzzled for long before finally understanding what you mean by "not lead to differential equations": [itex]G[/itex] always satisfies the Euler-Lagrange equations, which is not to be solved for [itex]q(t)[/itex] while the Lagrangian [itex]L[/itex] only satisfies the Euler-Lagrange equations when the actual path of motion in the configuration space (q, t) between two fixed time points is chosen, and this actual path [itex]q(t)[/itex] is the solution of the differential equations given by the Euler-Lagrange equations of [itex]L[/itex].

    If [itex]\frac{\partial^2 L}{\partial\dot{q}_a\partial\dot{q}_b}=0[/itex], then
    [itex]L(q,\dot{q},t) = \sum_{a} f_{a}(q,t)\dot{q}_{a} + g(q,t)[/itex], and then the canonical momentum [itex]p_a=\partial L/\partial\dot{q_a}=f_{a}(q,t)[/itex] isn't a function of the generalized velocity [itex]\dot{q}[/itex]. Isn't this weird? (Consider the mechanical momentum is defined to be [itex]p_x=m\dot{x}[/itex], where [itex]x[/itex] is the Cartesian coordinate.)
     
    Last edited: Dec 8, 2015
  16. Dec 8, 2015 #15

    samalkhaiat

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    For [itex]G[/itex], the Euler-Lagrange equation is not a equation. So there is no equation to solve, it simply gives you [itex]X = X[/itex]. This is why I used the sign ([itex]\equiv[/itex]) instead of the equal sign to write the E-L equation for [itex]G[/itex]. On the other hand, for a genuine Lagrangian, the E-L equation produces a set of dynamical differential equations [itex]\Phi_{a}(\ddot{q},\dot{q},q) = 0[/itex] which, in principle, can be solved given an appropriate set of initial conditions [itex]q_{a}(t) = f_{a}(q(0), \dot{q}(0);t)[/itex].
    Nothing is weird about it. The Dirac Lagrangian and all gauge constraint systems have singular Hessian matrix. The treatment of constraint system is a very rich and annoying (i.e., hard) subject.
    Let me just few words about this. In detailed form, the E-L equations are
    [tex]\frac{\partial^{2}L}{\partial \dot{q}_{a} \partial \dot{q}_{b}} \ddot{q}_{b} = \frac{\partial L}{\partial q_{a}} - \frac{\partial^{2}L}{\partial \dot{q}_{a} \partial q_{b}} \dot{q}_{b} .[/tex]
    From this, we see that the accelerations [itex]\ddot{q}_{b}[/itex] at a given time are uniquely determined by the [itex]\dot{q}_{a}[/itex]’s and the [itex]q_{a}[/itex]’s at that time if, and only if, the Hessian matrix [tex]M_{ab} = \frac{\partial^{2}L}{\partial \dot{q}_{a} \partial \dot{q}_{b}} , \ \ a = 1, 2 , …, n \ \ \ (1)[/tex] can be inverted, i.e., [itex]\det |M| \neq 0[/itex]. If, on the other hand, the Hessian is singular, i.e., [itex]\det |M| = 0[/itex], the accelerations will not be uniquely determined and the solution to the equations of motion could contain arbitrary functions of time. For the canonical momenta [tex]p^{a} = \frac{\partial L}{\partial \dot{q}_{a}} ,[/tex] singular Hessian is just the condition for non-solvability of the velocities as functions of the coordinates and momenta. This means that the [itex]p^{a}[/itex]’s are not all independent, rather, there are some relations
    [tex]\Psi_{i}(q,p) = 0 , \ \ i = 1,2, …, r \ \ \ \ \ \ (2)[/tex] that follow from the defining relations (1). That is to say that replacing the [itex]p[/itex]’s in (2) by their definitions (1) in terms of the [itex]q[/itex]’s and [itex]\dot{q}[/itex]’s, equation (2) becomes an identity. The relations (2) are called primary constraints to indicate that the equation of motion are not used to obtain these conditions and that they implies no restrictions on the local coordinates [itex](q_{a},\dot{q}_{b})[/itex] in the configuration space. However, they do define a submanifold smoothly embedded in the [itex](q_{a},p^{b})[/itex] phase-space. This submanifold is called the primary constraint surface.
     
  17. Dec 8, 2015 #16

    Erland

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    You don't need to know anything of Grassmann algebras etc. to prove this

    Ok, so assume that ##\frac{d}{dt} \left( \frac{\partial G}{\partial \dot{q}_{a}} \right) \equiv \frac{\partial G}{\partial q_{a}}##.
    This can be expanded and rewritten as

    [tex]\sum_b\frac{\partial^2 G}{\partial\dot q_b\partial\dot q_a}\ddot q_b\equiv \frac{\partial G}{\partial q_a}-\sum_b\frac{\partial^2 G}{\partial q_b\partial\dot q_a}\dot q_b-\frac{\partial^2 G}{\partial{t}\partial q_a}.[/tex]
    This should hold for all ##q_b##:s, ##\dot q_b##:s, ##\ddot q_b##:s, and ##t##. Fix the ##q_b##:s, ##\dot q_b##:s, and ##t##. If we put ##\ddot q_b=0## for all ##b##, the LHS vanishes. Thus the RHS vanishes too. But since the RHS does not depend on the ##\ddot q_b##:s, this means that the LHS vanishes for all choices of the ##\ddot q_b##:s, for example if one of them is ##1## and the all the other ##0##, which means that ##\frac{\partial^2 G}{\partial\dot q_b\partial\dot q_a}=0##. This holds for all ##a## and ##b##. From this one can infer, as indicated in previous posts, that
    [tex] G(q,\dot{q},t) = \frac{d}{dt}F(q,t). [/tex]
     
    Last edited: Dec 8, 2015
  18. Dec 9, 2015 #17

    Erland

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    Ooops, should be
    [tex]\sum_b\frac{\partial^2 G}{\partial\dot {q_b}\partial\dot {q_a} }\ddot {q_b}\equiv \frac{\partial G}{\partial q_a}-\sum_b\frac{\partial^2 G}{\partial q_b\partial\dot {q_a}}\dot {q_b}-\frac{\partial^2 G}{\partial{t}\partial\dot {q_a}}.[/tex]
    Continuing, writing
    [tex]G(q_1,\dots,q_n,\dot{q_1},\dots,\dot {q_n},t) = \sum_{b} f_{b}(q_1,\dots,q_n,t) \dot{q_b} + g(q_1,\dots,q_n,t),[/tex]we plug this into the RHS of the former equation, which is identically ##0##, and using that the ##\dot{q_b}##:s are independent, we obtain ##\partial f_b/\partial q_a=\partial f_a/\partial q_b## and ##\partial g/\partial q_a=\partial f_a/\partial t## for all ##a## and ##b##.
    This means that there is a function ##F(q_1,\dots,q_n,t)## such that ##\partial F/\partial q_a=f_a## for all ##a##, and ##\partial F/\partial t=g##.

    For this ##F##
    [tex]G(q_1,\dots,q_n,\dot{q_1},\dots,\dot {q_n},t) =\frac d{dt}F(q_1,\dots,q_n,t)[/tex]
    as desired.

    But I have a question. This ##F## might exist only locally, if the domain (phase space) does not have the "right" type of connectedness, right? If e.g. the domain is 2-dimensional, then the domain must be simply connected (without "holes"), otherwise it might not be possible to "glue together" the locally defined ##F##:s, right?
     
  19. Dec 9, 2015 #18

    samalkhaiat

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    No, you don't, but who said you do need Grassmann numbers to prove the theorem? The question from Shyan to me was about the reason for using the phrase "almost always" in post #10.
     
  20. Dec 10, 2015 #19
    Though the Lagrangian seems not to correspond to an informative physical observable, the action integrand [itex]Ldt=p_adq^a-Hdt[/itex] represents the energy-momentum 1-form in Minkowski spacetime, where [itex]p_a's[/itex] are the canonical momenta and [itex]H[/itex] is the Hamiltonian, if the energy-momentum 4-vector is given by [itex]P=H\partial/\partial t+p_a\partial/\partial q^a[/itex], which applies to a nonrelativistic system as well as a relativistic system. This may give a little justification for that the equations of motion are given by [itex]\delta\int^{t_2}_{t_1} Ldt=0[/itex].

    Why is this called a group of symmetry transformations? This is just a shift of the boundary term of the action integral. In gravitational field theories I learnt one can utilize the adjustment of the boundary term of the action integral or the Hamiltonian integral to change the boundary condition, such as from the Dirichlet to Neumann boundary conditions, because different boundary terms correspond to different boundary conditions, which is in some sense like the Legendre transformation. But [itex]\Lambda(q, t)[/itex] here seems not to have this function because [itex]\Lambda[/itex] is not a function of [itex]\dot{q}[/itex]. In gravitational field theories the added boundary term can be a function of [itex]\partial_\mu{Q}[/itex] as well as [itex]{Q}[/itex]. Is it possible to extend [itex]\Lambda(q, t)[/itex] to [itex]\Lambda(q, \dot{q}, t)[/itex] here? Since Hamilton's principle has restricted the generalized coordinates at the time endpoints [itex]t_1[/itex] and [itex]t_2[/itex] not to be varied, the variation of the added boundary term should vanish even [itex]\Lambda[/itex] is a function of [itex]\dot{q}[/itex] as well: [itex]\delta\big(\Lambda(q(t_2), \dot{q}(t_2) ,t_2)-\Lambda(q(t_1), \dot{q}(t_1), t_1)\big)=0[/itex].

    Then in field mechanics, do we have
    [tex]
    \Big\{ G(Q , \partial_\mu{Q} , x^\mu) = \frac{d}{dt}F(Q , x^\mu) \Big\} \ \Leftrightarrow \ \Big\{ \frac{d}{dx^\mu} \left( \frac{\partial G}{\partial(\partial_\mu{Q}_a)} \right) - \frac{\partial G}{\partial Q_{a}} \equiv 0 \Big\}
    [/tex]?
     
    Last edited: Dec 10, 2015
  21. Dec 10, 2015 #20

    samalkhaiat

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    The reason is in the word "some".
    Which means that not all [itex]F(q,t)[/itex], but rather some function [itex]\Lambda(q.t)[/itex], i.e., you obtain the Lagrangian [itex]L_{B}[/itex] plus [itex]d\Lambda_{\epsilon}/dt[/itex] when you transform [itex]L_{A}[/itex] under some transformation [itex]T(\epsilon)[/itex] of the coordinates and time.
    That is completely different story, diffeomorphism is a "gauge" group which is the symmetry group in the second Noether theorem.
    If by "here" you mean general relativity and diffeomorphism invariance, then my answer is Yes. You can split the H-E action of GR this way.


    Yes.
     
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