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Lagrangian/ Noether's Theorem

  1. Oct 20, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider the following Lagrangian of a particle moving in a D-dimensional space and interacting with a central potential field

    L = 1/2mv2 - k/r

    Use Noether's theorem to find conserved charges corresponding to the rotational
    symmetry of the Lagrangian.

    How many independent charges are there?


    Hint: infinitesimal rotations are parametrized by a skew-symmetric matrix Eij, that is
    xi --> x'i = xi + Eijxj, Eij + Eji = 0

    2. Relevant equations



    3. The attempt at a solution

    My lecturer gave solutions and I'm trying to follow them but I'm getting lost at one point;

    xi --> xi + Eijxj => δxi = Eijxj

    Then he says 1/2JijEij = (δL/δvi)δxi

    I've no idea where this is coming from. I'm assuming Jij is the conserved charges? Where is the factor of 1/2 coming from??

    He then says piEijxj = 1/2(pixj - pjxi)Eij
    Again, not sure where this comes from...

    Advice? :smile:
     
    Last edited: Oct 20, 2011
  2. jcsd
  3. Oct 20, 2011 #2

    I like Serena

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    I thought you were studying math - not (advanced) physics? :confused:


    I can't answer your questions, but I can give you the meanings of a couple of the symbols and the theories behind them.


    pi is the generalized momentum:
    [tex]p_i={\partial \mathcal L \over \partial \dot x_i}[/tex]
    The generalized momentum is conserved for every generalized coordinate of which the Lagrangian is not dependent. (Why?)

    xi represents a path that the particle may follow, that is, a full set of coordinate-functions defined over a certain time interval.
    x'i is an alternate path that lies "close" to xi.
    The difference δxi is the "variation".
    The variation consists of an offset at each point in each possible direction.
    In particular Eijxj is an implied summation. (Did you already know those?)

    The Principle of least action states that the action-integral takes on a minimum.
    The "action" of a path xi is defined as S(xi) = ∫L(xi, vi, t) dt
    If L is not dependent on time, as in your case of a particle in an orbit, we have the "abbreviated action" which is:
    S0(xi) = ∫pi dxi = pi δxi (again with an implied summation).

    According to the wiki article Jij are indeed the conserved charges, but I'm not familiar with them.
    I'm not sure of the factor 1/2, but I think it derives from the fact that the kinetic energy T is:
    T = (1/2)mv2 = (1/2)pv
    Note that p=mv is the momentum of a linear coordinate.


    Does this help?
    Or did you already know all this?
     
  4. Oct 20, 2011 #3

    George Jones

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    This part is fairly easy. Start with "identity"
    [tex]p_i x_j E_{ij} = \frac{1}{2} p_i x_j E_{ij} + \frac{1}{2}p_i x_j E_{ij} ,[/tex]
    use antisymmetry to flip the indices in the second [itex]E_{ij}[/itex], and then relabel dummy indices in the second term.
     
  5. Oct 21, 2011 #4
    the half is simply a factor that can be absorbed into the infinitesimal parameter so that you can cleanly extract the angular momentum as the conserved current associated with rotational symmetry i.e. it's there because it is
     
  6. Oct 21, 2011 #5

    dextercioby

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    In other words, under a full summation E_ij picks up only the antisymmetrization of the product x_i p_j.

    The 1/2 is a weight factor coming from

    x_i p_j = 1/2 (x_i p_j + x_j p_i) + 1/2 (x_i p_j - x_j p_i)

    between the brackets you have either a symmetric expression (first), or an antisymmetric one (2nd).
     
  7. Oct 23, 2011 #6
    Thanks! I understand it now! :smile:

    One other question,

    For a D-dimensional harmonic oscillator with L = 1/2mijvivj-1/2kijxixj

    my e.o.m. work out as mijaj + kijxj = 0
    where a = d(v)/dt

    but my lecturer gets (mij + mji)aj + (kij + kji)xj = 0

    :confused:
     
  8. Oct 23, 2011 #7

    vela

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    Did you use the product rule when you differentiated vivj wrt vk? The same sort of thing happens when you differentiate the potential term wrt xk.
     
  9. Oct 23, 2011 #8

    I like Serena

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    It's those implicit summations that are creating a problem here.

    You need to separate your matrix mij into 4 regions:
    1. Column i with the exception of row i.
    2. Row i with the exception of column i.
    3. Row i and column i (aka mii).
    4. The rest.

    Take the derivative to vi of each region and sum them up afterward.

    Same thing for kij.
     
  10. Oct 23, 2011 #9
    What do you mean?
    Since I'm differentiating partially all other variables are constant apart from vi.

    :confused:
     
  11. Oct 23, 2011 #10

    I like Serena

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    [tex]{\partial \over \partial v_k}(v_i v_j) = {\partial v_i \over \partial v_k} v_j + v_i {\partial v_j \over \partial v_k} = \delta_{ik} v_j + v_i \delta_{jk}[/tex]
     
  12. Oct 23, 2011 #11

    vela

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    Take a 2x2 matrix for instance. The kinetic term will be equal to
    [tex]\frac{1}{2}(m_{11} x_1^2+m_{12} x_1 x_2 +m_{21}x_2x_1+m_{22}x_2^2)[/tex]
    If you differentiate wrt x1, you'll get three contributions, but your expression only picks up the first two terms, sort of. Do you see why you're missing the third term?
     
  13. Oct 23, 2011 #12
    Okay I understand this now. :smile:

    I also understand this. However, when I differentiate w.r.t xi I get
    2m11x1 + m12x2 + m21x2 + m12x1 + m21x1 + 2m22x2

    I don't see how this will give me the required expression when I differentiate with respect to t.
     
  14. Oct 23, 2011 #13

    I like Serena

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    Hmm, if you double it and differentiate to x1, you should get:
    2m11x1 + m12x2 + m21x2

    Which is:
    m1jxj + mj1xj
    when summed over j.

    The kinetic term should actually have v instead of x.
    Taking the derivative to time gives you a.


    EDIT: In particular, you should not sum over i in this case! (Why?)
    (Yes, those implicit summations are nasty! :wink:)
     
    Last edited: Oct 23, 2011
  15. Oct 23, 2011 #14
    Okay I understand it now! :smile:

    As for this part, I shouldn't sum over i because we differentiated with respect to xi so xi will vanish from each of the terms (except the first one, but that can be written in terms of xj)
     
  16. Oct 23, 2011 #15

    I like Serena

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    Not quite.

    You're supposed to calculate [itex]{d \over dt}{\partial \mathcal L \over \partial v_i} - {\partial \mathcal L \over \partial x_i} = 0[/itex].
    This equation holds for each individual i, so you're not supposed to sum over i.

    The problem here is that you also use i when you write [itex]{\mathcal L}=m_{ij} v_i v_j - k_{ij} x_i x_j[/itex].
    But this is a different i over which you are supposed to sum.

    This is why vela introduced the index k (effectively used as the first i), to distinguish it from the second i.
     
  17. Oct 23, 2011 #16
    Alright, thank you. :smile:

    Also, to answer your earlier question, my course is called "Mathematics", by there are quite a few physics modules (especially next year) which I can study if I want to. I plan on splitting my degree up with pure mathematics and theoretical physics. :smile:
     
  18. Oct 23, 2011 #17

    I like Serena

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    Sounds a bit like what I did. :wink:
    (Although I never learned Noether's Theorem.)
     
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