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Invariant quantities of a lagrangian?

  1. Sep 21, 2014 #1
    Given a basic Lagrangian, how would I determine invariant quantities? My hunch says it would be quantities that do not depend on position or time? Saying that, perhaps using the Lagrange equation to solve for equations of motion and along the way whatever terms disappear would be my invariant quantities? This seems like a harsh oversimplification, but I can not tell? Any help understanding the invariant quantities would be appreciated. I don't want to solve the actual problem here so to speak, but rather understand the theory.

    I did solve a problem a few weeks back showing that electromagnetic force was invariant under a gauge transformation, but this seems to be a different concept?
  2. jcsd
  3. Sep 21, 2014 #2
    You need to define "invariant", otherwise your question is meaningless. If that means "constant with respect to time", you might find it useful to consider that when ##x## is constant with respect to time, then $$ {dx \over dt} = 0 .$$ Would that remind you something in Lagrange equations?
  4. Sep 21, 2014 #3
    Yes, that is what I was thinking, conserved quantities. So my first term regarding some time derivative of a partial derivative of a generalized coordinate might give me zero. The wording I was missing was that the Lagrangian is invariant under symmetry.
  5. Sep 21, 2014 #4
    I am not sure what you actually mean now. Have you figured this out?
  6. Sep 21, 2014 #5
    No, I haven't. It asks: a Lagrange with kinetic energy in cartesian coordinates KE = [m/2]*(dx^2, dy^2, dz^2), and a potential in cartesian V = V(x^2 + y^2 , z) is invariant under some symmetry, find the corresponding invariant quantity. What I see is that the potential (V) is a function of some cylindrical shape. But I do not know what the "invariant quantities" are?
  7. Sep 21, 2014 #6
    Let's go back to the Euler-Lagrange equation: $$ {d \over dt} {\partial L \over \partial \dot q} - {\partial L \over \partial q} = 0. $$ Under what condition does it easily result in something that is conserved, i.e., whose time-derivative is zero?
  8. Sep 21, 2014 #7
    I think you should look into Landau's mechanics for inspiration. There, Landau shows that since the Lagrangian of a system of free particles is time-independent, energy is conserved. GIven translational invariance, we know that the total momentum is conserved. So all these symmetries lead to conservation laws.

    In your problem, by "invariant quantities" they mean things like momentum and energy, which are conserved as a result of symmetries in the Lagrangian. They want you to find, given that the Lagrangian is invariant in some way, what quantity is conserved.
  9. Sep 21, 2014 #8
    Yes, this is where my confusion takes place! The word 'quantities' is a little vague to me. We did see that under a gauge transformation that the Lagrangian differs by a time derivative, and that the Electromagnetic force was invariant under these transformations. We saw that the equations of motion in the field was unchanged given the transformation.

    But, here, we are not given any gauge transformation, but rather some seemingly general case. So my first instinct was to use the Lagrange equation. An important question is, my potential is a function of some cylinder, so what does my actual V term look like?

    Regarding Landau's book, I hear nothing but good things about this, and I don't have it! :/ but I am using Goldstein, and coming across vague suggestions as well as horrible typos. Anybody who has Landau in PDF is welcome to send it my way :)
  10. Sep 21, 2014 #9
    Here's the Mechanics PDF, it's great :D
    And maybe you want to look into Emmy Noether's theorem (not in full detail, just to get an idea of it)- that's where all this wonderful stuff comes from, after all.
  11. Sep 21, 2014 #10
    Landau & Lifschitz is good stuff, yet I would suggest you should try to answer the question in #6.
  12. Sep 21, 2014 #11


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    If you know the symmetry transformations, (the transformations which leave the Lagrangian invariant), you can easily deduce the corresponding constants of motion. The term “invariant quantities” is not accurate.
    Consider your Lagrangian
    [tex]L = \frac{ 1 }{ 2 } \left( ( \dot{ x }_{ 1 } )^{ 2 } + ( \dot{ x }_{ 2 } )^{ 2 } + ( \dot{ x }_{ 3 } )^{ 2 } \right) - V ( x_{ 1 }^{ 2 } + x_{ 2 }^{ 2 } , x_{ 3 } ) .[/tex]
    By inspection, we see that both KE and PE are invariant under rotations in the [itex]x_{ 1 } x_{ 2 }[/itex]-plane. Therefore, the corresponding constant of motion is given by 3rd component of the angular momentum [itex]L_{ 3 } = x_{ 1 } p_{ 2 } - x_{ 2 } p_{ 1 }[/itex].
    To see this, we need the explicit form of the transformations, i.e.,
    [tex]\bar{ x }_{ 1 } = x_{ 1 } \cos ( \theta ) - x_{ 2 } \sin ( \theta ) ,[/tex]
    [tex]\bar{ x }_{ 2 } = x_{ 2 } \cos ( \theta ) + x_{ 1 } \sin ( \theta ) ,[/tex]
    [tex]\bar{ x }_{ 3 } = x_{ 3 } .[/tex]
    To find the constant of motion, it is sufficient to consider the infinitesimal version of the above transformations. We obtain this by setting [itex]\cos ( \theta ) = 1[/itex] and [itex]\sin ( \theta ) = \theta[/itex],
    [tex]\bar{ x }_{ 1 } = x_{ 1 } - x_{ 2 } \theta , \ \ \Rightarrow \ \ \delta x_{ 1 } = - \theta x_{ 2 } , \ \ (1a)[/tex]
    [tex]\bar{ x }_{ 2 } = x_{ 2 } + x_{ 1 } \theta , \ \ \Rightarrow \ \ \delta x_{ 2 } = \theta x_{ 1 } , \ \ (1b)[/tex]
    [tex]\delta x_{ 3 } = 0 . \ \ \ \ \ (1c)[/tex]
    From these infinitesimal transformations, we find the following important relation
    [tex]\frac{ d }{ d t } ( \delta x_{ i } ) = \delta ( \frac{ d }{ d t } x_{ i } ) \equiv \delta \dot{ x }_{ i } . \ \ \ \ (2)[/tex]
    The fact that the transformations leave the Lagrangian invariant (i.e., symmetry transformations) mean that
    [tex]0 = \delta L = \sum_{ i = 1 }^{ 3 } \frac{ \partial L }{ \partial x_{ i } } \delta x_{ i } + \sum_{ i = 1 }^{ 3 } \frac{ \partial L }{ \partial \dot{ x }_{ i } } \delta \dot{ x }_{ i } . \ \ \ (3)[/tex]
    Now, if we use the Euler-Lagrange equation of motion in the first tern, and equation (2) in the second term, we find
    [tex]\frac{ d }{ d t } \left( \sum \frac{ \partial L }{ \partial \dot{ x }_{ i } } \delta x_{ i } \right) = \frac{ d }{ d t } ( \sum p_{ i } \delta x_{ i } ) = 0 .[/tex]
    Or, using the transformations (1a)-(1c),
    [tex]\frac{ d }{ d t } ( x_{ 1 } p_{ 2 } - x_{ 2 } p_{ 1 } ) \equiv \frac{ d L_{ 3 } }{ d t } = 0 .[/tex]
    Thus rotational invariance means that [itex]L_{ 3 } = x_{ 1 } p_{ 2 } - x_{ 2 } p_{ 1 }[/itex] is time-independent on the physical trajectories (constant of motion corresponding to the invariance under rotations in the xy-plane).

  13. Sep 22, 2014 #12
    It seemed to me that KleZMeR wanted to understand the theoretical foundation for this "therefore" step.
  14. Sep 22, 2014 #13


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    Did you read ONLY this paragraph? The detailed proof of "therefore step" was given just after that paragraph.
  15. Sep 22, 2014 #14
    Thank you for the book guitarphysics!! Ok, so I'm beginning to understand this I hope. I am curious about the explicit transformations shown by Sam, are those a general form of some rotational symmetry I would use after seeing by inspection that my potential is dependent on a rotation in the x, y plane? I am just expressing rotational symmetry?

    If this is the case, I am guessing that although I was not given an explicit function of V, I should still be able to detect the motion of my system?
    Is this called a cyclic coordinate? I understand I can explicitly put my Lagrangian into the Lagrange equations and find what is not conserved, but I was not given an explicit function.

    From equation 3 to the next step, why was the partial derivative in respect to position dropped?
  16. Sep 22, 2014 #15
    Oh wait, I see that [tex]P_j = dL/d \dot q_j[/tex] , being the canonical momentum.

    Ok, so why isn't my tex working? Firs timer here, trying to get with the program...
  17. Sep 22, 2014 #16
    oh, it's working, happy!!
  18. Sep 22, 2014 #17
    Also, what is the difference of finding these quantities in cartesian and cylindrical coordinates? Perhaps my [tex]R x L [/tex] could be written in cylindrical? I'm not sure.
    Last edited: Sep 22, 2014
  19. Sep 23, 2014 #18
    A problem with that detailed proof is that it focuses on the details of that particular Lagrangian, and leaves open the general question. Trees vs forest, so to speak.

    KleZMeR, a cyclic coordinate is one that is not present in the Lagrangian (but its time-derivative may), so $$ {\partial L\over \partial q} = 0, $$ and the E. L. equation then gives you a conserved quantity.

    Practically, change coordinates in your problem so that one them is cyclic, and you have a conserved quantity.

    (You can go without changing coordinates like samalkhaiat demonstrated, but that is far less obvious.)
  20. Sep 23, 2014 #19
    I see, so there are symmetry shifts in a translation, rotation, or time, as well as many kinetic and potential functions which can be conserved under symmetry.

    I am wondering about the differences in cartesian and cylindrical coordinates? Would I just then convert my [tex](x, y, z)[/tex] accordingly to [tex](r, theta, z)[/tex].
  21. Sep 23, 2014 #20
    Try that and see what happens with the Lagrangian. Pay attention to what I said in #18.
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