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Proof of fundamental Poisson-brackets

  1. May 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider a cartesian coordinate [tex]q_k[/tex] and conjugate momentum [tex]p_k[/tex]. The Poisson-bracket for two random functions [tex]f=f(q_1,..,q_f,p_1,..,p_f,t)[/tex] and [tex]g=g(q_1,..,q_f,p_1,..,p_f,t)[/tex] is defined as:
    [tex]\{f,g\}=\sum_{i=1}^{f}\biggl(\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i}-\frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i}\biggr)[/tex]

    (a) Prove the validity of the fundamental Poisson-brackets:

    [tex]\{q_k,q_l\}=0[/tex]
    [tex]\{p_k,p_l\}=0 [/tex]
    [tex]\{q_k,p_l\}=\delta_{kl}[/tex]


    2. Relevant equations

    I know from my textbooks and from Wikipedia that the fundamental brackets are calculated with the following relations:

    [tex]\frac{\partial q_k}{\partial p_l}=0[/tex]
    [tex]\frac{\partial p_k}{\partial q_l}=0[/tex]
    [tex]\frac{\partial p_k}{\partial p_l}=\delta_{kl}[/tex]


    3. The attempt at a solution

    [tex]\{q_k,q_l\}=\sum_{k=1}^{q}\biggl(\frac{\partial q_k}{\partial q_k}\frac{\partial q_l}{\partial p_k}-\frac{\partial q_k}{\partial p_k}\frac{\partial q_l}{\partial q_k}\biggr)=\sum_{k=1}^{q}\biggl(\frac{\partial q_l}{\partial p_k}-\frac{\partial q_l}{\partial p_k}\biggr)=0[/tex]

    Problem is I don`t get to even use the above mentioned relations and I get 0 for [tex]\{q_k,p_l\}[/tex], too. What am I doing wrong?
     
    Last edited: May 23, 2010
  2. jcsd
  3. May 22, 2010 #2

    diazona

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    For one thing, you're using an index which is summed over outside of the summation. You can't write {qk,ql} as a sum over k, because k has one fixed value determined by which Poisson bracket you're trying to calculate. Remember that {qk,ql} is just a generic way to express any of {q1,q1}, {q1,q2}, {q2,q2}, etc. and in each of those cases k has a particular numeric value (1,1,2 respectively). So when you write your sum as a sum over k, you wind up with an expression in which some k's are fixed by the Poisson bracket and some k's change for each term in the sum - essentially using the same letter for two totally different variables, and in such a way that you can't even tell them apart. Don't do that.

    The other thing you've got wrong is the partial derivatives. Or some of them, anyway. It's true that
    [tex]\frac{\partial q_k}{\partial p_l} = 0[/tex]
    and
    [tex]\frac{\partial p_k}{\partial q_l} = 0[/tex]
    since momenta and coordinates are taken to be independent, but the other two should be
    [tex]\frac{\partial q_k}{\partial q_l} = \delta_{kl}[/tex]
    [tex]\frac{\partial p_k}{\partial p_l} = \delta_{kl}[/tex]
    After all, it's possible that k and l have the same value in these expresssions, and if you take the derivative of a variable with respect to itself, the answer had better be 1, right?
     
  4. May 22, 2010 #3
    I meant to write [tex]\frac{\partial p_k}{\partial p_l}=\delta_{kl}[/tex], it was a typo.

    So I understand what I shouldn`t do but what should I do after all?
    Use a new index altogether, like m for example or use some combination of k and l?
     
  5. May 22, 2010 #4

    diazona

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    OK (but that's kind of a big typo :wink:)
    Remember that you wrote the formula
    [tex]\{f,g\} = \sum_{i=1}^{N}\biggl(\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i} - \frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i}\biggr)[/tex]
    To compute the Poisson bracket between, say, two coordinates qk and ql, you can use that formula. Plug them in and see what you get.
     
  6. May 23, 2010 #5
    When I plug the conjugate momenta for example, I still get a sumation over only one fixed index and you said I shouldn't do that.

    [tex]
    \{p_k,p_l\} = \sum_{i=1}^{N}\biggl(\frac{\partial p_k}{\partial q_i}\frac{\partial p_l}{\partial p_i} - \frac{\partial p_k}{\partial p_i}\frac{\partial p_l}{\partial q_i}\biggr)
    [/tex]

    I feel like the dumbest person on Earth for not getting this. It appears so easy.
     
  7. May 24, 2010 #6
    I finally managed to get this and I did part (b), too, which was to prove:
    [tex]\{L_i ,p_j\}=\epsilon_{ijk} p_k[/tex]
    [tex]\{L_i ,q_j\}=\epsilon_{ijk} q_k[/tex]
    [tex]\{L_i ,L_j\}=\epsilon_{ijk} L_k[/tex]
    [tex]\{L_i , \vec L^2\}=0[/tex]

    I get some difficulties with part (c) though.

    1. The problem statement, all variables and given/known data

    A particle`s motion is described by the Hamilton-function [tex]H = T + V(q_1,q_2,q_3)[/tex]. Calculate the total time derivative of the angular momentum [tex]\frac{d \vec L}{dt}[/tex] using the Poisson-brackets. When is [tex]\vec L[/tex] a conserved quantity?

    2. Relevant equations

    * Total time derivative of some function f:
    [tex]\frac{df}{dt} = \{f,H\} + \frac{\partial f}{\partial t}[/tex]

    [tex]\frac{\partial H}{\partial p_i} = \dot{q_i}[/tex]

    [tex]-\frac{\partial H}{\partial q_i} = \dot{p_i}[/tex]

    3. The attempt at a solution

    [tex]\frac{dL}{dt} = \{L,H\} + \frac{\partial L}{\partial t}[/tex]

    [tex]= \sum_{i=1}^{3}\biggl(\frac{\partial L}{\partial q_i}\frac{\partial H}{\partial p_i} - \frac{\partial L}{\partial p_i}\frac{\partial H}{\partial q_i}\biggr) + \frac{\partial L}{\partial t}[/tex]

    [tex]= \sum_{i=1}^{3}\biggl(\frac{\partial q_i p_i}{\partial q_i}\dot{q_i} - \frac{\partial q_i p_i}{\partial p_i}\dot{p_i}\biggr) + \frac{\partial q_i p_i}{\partial t}[/tex]

    [tex]= p \dot{q} + q \dot{p} + 0[/tex]

    I get this can`t be right, at least because I didn`t use [tex]H = T + V(q_1,q_2,q_3)[/tex] if not anything else.

    About the second question: I realize that L is conserved when [tex]\{\vec L,H\} = 0[/tex] but I cannot do this if I don`t have the first part right.

    Any help would be greatly appreciated as I am really stuck.
     
    Last edited: May 24, 2010
  8. May 24, 2010 #7

    diazona

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    Sorry, I missed your reply. Anyway, as you probably figured out, I didn't say there was anything wrong with what you did there. i would be a "variable index," as opposed to k and l which are "fixed indices" because they appear outside of the summation. That means that for any one given Poisson bracket, k (for example) has only one value, and you can't sum over it because the index being summed over has to be free to take on all possible values.

    For part (c), let me first ask you, what's the definition of [itex]\vec{L}[/itex] in terms of the q's and p's (and/or their derivatives)?

    Also, I believe part (c) wants you to calculate the Poisson bracket [itex]\{\vec{L},H\}[/itex] by using bracket algebra to reduce it to the results you found in part (b).
     
  9. May 24, 2010 #8
    Well the angular momentum is [tex]\vec L = r \times p[/tex] or [tex]\vec L = \epsilon_{ijk} r_j p_k[/tex] and r is q here...

    And by bracket algebra do you mean something like [tex]\{L,H\} = \{L,T+V\} = \{L,T\} + \{L,V\}[/tex]?
     
  10. May 24, 2010 #9

    diazona

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    Okay, so angular momentum is defined by [itex]L_i = \epsilon_{ijk}q_j p_k[/itex]. So why did you plug in [itex]L = q_i p_i[/itex] in your previous post?

    Anyway, I don't think that matters so much, because:
    Yes, and don't forget about
    [tex]\{A,BC\} = \{A,B\}C + B\{A,C\}[/tex]
    I think there are some others like that, formulas that let you decompose a Poisson bracket into other Poisson brackets without having to actually plug anything into the definition (the formula with the summation and the derivatives).
     
  11. May 25, 2010 #10
    I believe I have it now. It appears that they really wanted me to use the results from part (b).

    [tex]\frac{d \vec L}{dt} = \{ \vec L,H\} + \frac{\partial \vec L}{\partial t} =[/tex]

    [tex]= \{q_i p_i , H\} + 0 =[/tex]

    [tex]= \{q_i,H\}p_i + q_i\{p_i,H\} =[/tex]

    Using H = T + V

    [tex]= \biggl\{q_i, \frac{p_i^2}{2m} + mgq_i\biggr\}p_i + q_i\biggl\{p_i,\frac{p_i^2}{2m} + mgq_i\biggr\} =[/tex]

    [tex]= \frac{p_i}{2}\{q_i,p_i^2\} + mgq_i\{p_i,q_i\} =\frac{p_i^2}{2} + mgq_i [/tex]

    Which then means that [tex]\vec L[/tex] is conserved when [tex]\frac{p_i^2}{2} + mgq_i = 0[/tex] or when [tex]p_i = \sqrt{-2mgq_i}[/tex]

    I`m farely sure this has to be correct but I`d still appreciate an opinion.
    By the way, thanks for all the help!
     
  12. May 25, 2010 #11

    diazona

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    No, that's not correct. You've got a couple of problems in that derivation:

    - Again, you substituted in [itex]\vec{L} = q_i p_i[/itex]. Why did you do that? I know you know it's not correct.
    - Also, you've used [itex]T = p_i^2/2m[/itex] and [itex]V = mgq_i[/itex]. Where did those come from? Is there additional information given that you haven't included in the problem statement above?
     
  13. May 25, 2010 #12
    I see that I was too quick to announce victory...
    Well, I got "inspired" by an example problem with a harmonic oscillator from my textbook. It was:
    [tex]\dot{x} = \{x,H\}= \biggl\{x, \frac{p^2}{2m}+\frac{D}{2}x^2\biggr\} = \frac{1}{2m}\{x,p^2\}=\frac{p}{m}[/tex]

    [tex]\dot{p} = \{p,H\}= \biggl\{p, \frac{p^2}{2m}+\frac{D}{2}x^2\biggr\} = \frac{D}{2}\{p,x^2\}=-Dx[/tex]

    I figured on my own to try using potential energy [tex]mgq_i[/tex] like in [tex]V=mgh[/tex]. There is really no additional information given.
    Oh, and I didn't add the [tex]\epsilon_{ijk}[/tex] infront of qp because I didn't think it would make any difference in this case, does it?
     
    Last edited: May 25, 2010
  14. May 25, 2010 #13

    diazona

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    It absolutely does make a difference. Do you know what the definition of [itex]\epsilon_{ijk}[/itex] is? Because I get the sense there are some things you don't know about subscripted variables, and you really should get that sorted out before you work on problems like this one. Here's a question to check yourself: using the definition you know, [itex]L_i = \epsilon_{ijk}p_j q_k[/itex], what is the formula for [itex]L_1[/itex] in terms of [itex]p_1,p_2,p_3[/itex] and [itex]q_1,q_2,q_3[/itex]?

    The advantage of the Hamiltonian approach is that you can use different expressions for kinetic and potential energy, and the procedure still works. In other words, even if [itex]T \neq p^2/2m[/itex] or [itex]V \neq mgh[/itex], Hamiltonian mechanics is still valid, and for strange definitions of T and V it can get you answers that you would be hard pressed to find any other way.

    Now, it's true that T is usually equal to [itex]p^2/2m[/itex] in realistic problems, but V can be pretty much anything. The way this question is phrased, it sounds like you're supposed to compute the result for arbitrary T and V.
     
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