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How prove angular momentum is conserved from the Lagrangian

  1. May 11, 2014 #1
    Hey,

    It's a simple question (hope so). How do you know (analitically) wether angular momentum is conserved based solely on the Lagrangian? Let me elaborate, for example to prove that the linear momentum is conserved you simply look for cyclic coordinates, i.e
    [tex]\frac{\partial L}{\partial q_i}=0[/tex]
    Or if the lagrangian isn't time dependent energy is conserved, i.e
    [tex]\frac{\partial L}{\partial t}=0[/tex]
    Is there a neat way as above to use in order to prove angular momentum is conserved?

    Thanks,

    M.
     
  2. jcsd
  3. May 11, 2014 #2
    Hey loops,

    For the "standard" problems one can usually pick a set of coordinates for which at least one of the generalized momenta can be identified with angular momentum. For instance, potentials which in some coordinates lose their angular dependence (isotropic harmonic oscillators and 2-body problem are two cases). With the same analysis you already described you can show the angular momenta are conserved.
     
  4. May 11, 2014 #3
    If your potential only depends on r, the two angular equations will be first integrals.
     
  5. May 12, 2014 #4
    Thanks guys that was very helpful. Another question I have is: Can the same conclusions be made using symmetry arguments?
     
  6. May 12, 2014 #5

    stevendaryl

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    I think it's actually kind of complicated showing that the angular momentum is conserved using generalized coordinates. (Or at least, it seems complicated to me.) If you switch to spherical coordinates [itex]r, \theta, \phi[/itex], then the Lagrangian doesn't depend on coordinate [itex]\phi[/itex], so the corresponding generalized momentum is constant. This turns out to be [itex]L_z[/itex]. But [itex]L_x[/itex] and [itex]L_y[/itex] aren't the momenta for any coordinate. (There is a momentum associated with the coordinate [itex]\theta[/itex], but it's not conserved, and is not equal to any of the components of the angular momentum).

    I suppose you could prove that [itex]L_z[/itex] is conserved, and then use symmetry to reason that [itex]L_x[/itex] and [itex]L_y[/itex] must be, as well.
     
  7. May 12, 2014 #6

    Orodruin

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    You do not need to worry so much about which coordinates to use rather than how you parametrize the symmetry group. In the case of three dimensional rotational symmetry, the symmetry group is three dimensional (the rotation is uniquely defined by specifying three angles, e.g., the Euler angles).
    Correspondingly, you need three generators of the symmetry and each of these correspond to a conserved quantity (assuming the rotational group is a symmetry of the Lagrangian). Here you can pick the generators of rotations around the x-, y-, and z-axes as the corresponding conserved quantities from Noether's theorem. These will be the x-, y-, and z-components of angular momentum.
     
  8. May 12, 2014 #7

    samalkhaiat

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    Under arbitrary rotation by small angle [itex]\vec{ \theta }[/itex], the position vector changes as follow
    [tex]\delta \vec{ r } = \vec{ \theta } \times \ \vec{ r } . \ \ \ \ \ (1)[/tex]
    This induces the following change in the Lagrangian
    [tex]\delta L = \frac{ \partial L }{ \partial x_{ i } } \delta x_{ i } + \frac{ \partial L }{ \partial \dot{ x }_{ i } } \delta \dot{ x }_{ i } = \frac{ \partial L }{ \partial x_{ i } } \delta x_{ i } + \frac{ \partial L }{ \partial \dot{ x }_{ i } } \frac{ d }{ d t } ( \delta x_{ i } ) .[/tex]
    This can be rewritten as
    [tex]
    \delta L = \left[ \frac{ \partial L }{ \partial x_{ i } } - \frac{ d }{ d t } \left( \frac{ \partial L }{ \partial \dot{ x }_{ i } } \right) \right] \delta x_{ i } + \frac{ d }{ d t } \left( \frac{ \partial L }{ \partial \dot{ x }_{ i } } \delta x_{ i } \right) . \ \ \ (2)
    [/tex]
    Rotational symmetry means that the Lagrangian does not change under arbitrary rotation, i.e. [itex]\delta L = 0[/itex].
    If the [itex]x_{ i }[/itex]’s are solutions to the Euler-Lagrange equations, then eq(2) becomes
    [tex]\frac{ d }{ d t } \left( \frac{ \partial L }{ \partial \dot{ x }_{ i } } \delta x_{ i } \right) = 0 ,[/tex]
    Introducing the momentum, we find
    [tex]\frac{ d }{ d t } ( p_{ i } \delta x_{ i } ) \equiv \frac{ d }{ d t } ( \vec{ p } \cdot \delta \vec{ r } ) = 0 . \ \ \ (3)[/tex]
    Putting eq(1) in eq(3), we find
    [tex]
    \frac{ d }{ d t } \left[ \vec{ p } \cdot ( \vec{ \theta } \times \vec{ r } ) \right] = \frac{ d }{ d t } \left[ ( \vec{ r } \times \vec{ p } ) \cdot \vec{ \theta } \right] \equiv \frac{ d }{ d t } ( \vec{ L } \cdot \vec{ \theta } ) = 0.
    [/tex]
    From this we get
    [tex]\frac{ d \vec{ L }}{ d t } = 0 .[/tex]

    Sam
     
  9. May 12, 2014 #8
    I don't mean to derail the thread, but what is the rigorous definition of [tex]\delta[/tex]?
    I know how to use it but never understand what exactly it was.
    Is it simply a more general form of an infinitesimal?
     
  10. May 12, 2014 #9

    samalkhaiat

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    It is the infinitesimal change of some quantity caused by infinitesimal transformation. For example, when you translate (transform) a vector [itex]\vec{ v }[/itex] by an infinitesimal displacement vector [itex]\vec{ \epsilon }[/itex], you write
    [tex]\vec{ v }^{ * } = \vec{ v } + \vec{ \epsilon } ,[/tex]
    so, the change in the vector is
    [tex]\delta \vec{ v } \equiv \vec{ v }^{ * } - \vec{ v } = \vec{ \epsilon } .[/tex]
    In the above example of infinitesimal rotation, if you put in the rotation matrix, [itex]\sin \theta = \theta[/itex] and [itex]\cos \theta = 1[/itex], you find
    [tex]\vec{ r }^{ * } = \vec{ r } + \vec{ \theta } \times \vec{ r } .[/tex]
    Hence
    [tex]\delta \vec{ r } \equiv \vec{ r }^{ * } - \vec{ r } = \vec{ \theta } \times \vec{ r }[/tex]
    Is this ok for you, or you want rigorous mathematical definition?
     
    Last edited: May 12, 2014
  11. May 12, 2014 #10
    My issue is more with the notation than with its purpose.
    Sometimes you see it replacing the partial derivative sign in a functional derivative and other times you don't.
    I guess the delta sign simply serves many purposes ?
     
  12. May 12, 2014 #11

    ChrisVer

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    Of course...
    For that reason we have Noether's theorem which associates symmetries (symmetry transformations) to conserved currents (equivelantly conserved "charges")...
    Symmetry under rotations is equivalent to a conserved "charge"/generator which is the angular momentum (generator of rotations).
     
  13. May 12, 2014 #12

    ChrisVer

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    In which case you see it replacing the partial derivative? in fact how can you define a partial derivative of a functional?
    the functional derivative is more like the variation of the functional if you vary it's function argument...that's why they sometimes put δ...
    However the definition is almost like the one of a normal derivative.

    It's meaning is that of variation...
     
  14. May 13, 2014 #13

    stevendaryl

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    Not that it really makes any difference, but most uses of functional derivatives can be understood in terms of ordinary derivatives of parameters. Instead of

    [itex]q(t) \Rightarrow q(t) + \delta q(t)[/itex]

    you can replace [itex]q(t)[/itex] by a two-argument function [itex]\tilde{q}(t, \alpha)[/itex], where [itex]\alpha[/itex] is a continuous parameter and where [itex]\tilde{q}(t, \alpha)[/itex] satisfies:
    [itex]\tilde{q}(t, 0) = q(t)[/itex]

    Then if the Lagrangian is unchanged (to first order) by changes in [itex]\alpha[/itex], there must be a conserved current:

    [itex]Q = \dfrac{\partial L}{\partial \dot{q}} \dfrac{\partial \tilde{q}}{\partial \alpha}|_{\alpha = 0}[/itex]

    (provided that the order of differentiation makes no difference:
    [itex]\dfrac{\partial}{\partial t}\dfrac{\partial}{\partial \alpha} \tilde{q} = \dfrac{\partial}{\partial \alpha}\dfrac{\partial}{\partial t} \tilde{q}[/itex]
     
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