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Help with index notation

  1. Mar 19, 2015 #1
    I'm not sure if this step on my calculation is correct or not.

    [tex]
    A=i\left(M_{\alpha}^{\nu}M_{\beta\nu}-M_{\beta}^{\nu}M_{\alpha\nu}\right)=i\left(M_{\beta}^{\nu}\delta_{\alpha}^{\beta}M_{\beta\nu}-M_{\beta}^{\nu}M_{\alpha\nu}\right)=i\left(M_{\beta}^{\nu}M_{\alpha\nu}-M_{\beta}^{\nu}M_{\alpha\nu}\right)=0[/tex]

    I just change the label in the first generator with the delta and then contract the delta with the other generator. Is it 100% rigorous ?

    Thank you!
     
  2. jcsd
  3. Mar 19, 2015 #2

    Orodruin

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    You will have to modify this, you are not allowed to call a summation index the same thing as a free index and then change which is the free index. Generally, if you have more than two of the samme index, you have done something naughty.
     
  4. Mar 19, 2015 #3
    Thank you for your fast response. The problem is that this thing should be zero ... if it is not zero then there must be an error in the previous calculations. Is A=0 ? And why?
     
  5. Mar 19, 2015 #4

    Orodruin

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    What is M? Does it have some additional properties such as symmetry (as suggested by not caring too much about index ordering with the raised index)? If it is, what stops you from simply raising the nu on one M and lowering it on the other?
     
  6. Mar 19, 2015 #5
    M is total antisymetric. M are the generators of the lorentz group. The index order does matter but I have written it too fast in tex.
     
  7. Mar 19, 2015 #6

    Orodruin

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    I still suspect it is a matter of simply changing which summation index is up and which is down ...
     
  8. Mar 19, 2015 #7
    The expression is [tex]
    A=i\left(M_{\thinspace\thinspace\alpha}^{\nu}M_{\beta\nu}-M_{\thinspace\thinspace\beta}^{\nu}M_{\alpha\nu}\right);\thinspace\thinspace\thinspace\thinspace\thinspace\left(M_{\alpha\beta}\right)_{\thinspace\thinspace\nu}^{\mu}=\delta_{\alpha}^{\mu}g_{\beta\nu}-\delta_{\beta}^{\mu}g_{\alpha\nu}[/tex]

    I don't see how this two terms equal to zero
     
  9. Mar 19, 2015 #8

    Orodruin

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    Are you familiar with how to raise and lower indices?
     
  10. Mar 19, 2015 #9
    Yes with the metric. I've done it to get to this last result.
    Do you mean things like ...

    [tex]
    A^{i}=g^{ij}A_{j};A_{i}=g_{ij}A^{j};g^{ij}g_{ib}=g_{\thinspace\thinspace b}^{j}=\delta_{b}^{j}
    [/tex]

    and this sort of stuff ?
     
  11. Mar 19, 2015 #10

    Orodruin

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    Exactly, so how does ##M^\nu_{\phantom\nu \alpha}## relate to ##M_{\nu\alpha}##?
     
  12. Mar 19, 2015 #11
    I think its

    [tex]
    M_{\nu\alpha}=g_{\nu\gamma}M_{\thinspace\thinspace\alpha}^{\gamma};M_{\thinspace\thinspace\alpha}^{\nu}=g^{\nu\gamma}M_{\gamma\alpha}[/tex]
     
  13. Mar 19, 2015 #12

    Orodruin

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    So what do you get if you apply the latter of these relations to your expression? (And switch the names of the summation indices in one of the terms...)
     
  14. Mar 19, 2015 #13
    I know what are trying to suggest me but I've tried that and I don't see why it should cancel. I feel so stupid right now :P. For example

    [tex]A=i\left(M_{\thinspace\alpha}^{\nu}M_{\beta\nu}-M_{\thinspace\beta}^{\nu}M_{\alpha\nu}\right)=i\left(g^{\nu\xi}M_{\xi\alpha}M_{\beta\nu}-M_{\thinspace\beta}^{\nu}M_{\alpha\nu}\right)=i\left(M_{\xi\alpha}M_{\beta}^{\thinspace\thinspace\xi}-M_{\thinspace\beta}^{\nu}M_{\alpha\nu}\right)[/tex]

    Pffff Sorry for wasting ur time
     
  15. Mar 19, 2015 #14

    Orodruin

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    Do it for both terms! :smile:

    Edit: Or not even that. Rename the summation index in the first term from ##\xi## to ##\nu## and use the antisymmetry of M.
     
  16. Mar 19, 2015 #15
    Yes, this is exactly the part that I do not understand. I thought that this procedure was incorrect...

    M is antisymmetric so
    [tex]
    M_{ij}=-M_{ji}
    [/tex]
    I guess that
    [tex]
    M_{\thinspace\thinspace j}^{i}=-M_{\thinspace\thinspace i}^{j}
    [/tex]
    [tex]A=i\left(M_{\xi\alpha}M_{\beta}^{\thinspace\thinspace\xi}-M_{\thinspace\beta}^{\nu}M_{\alpha\nu}\right)=i\left(M_{\nu\alpha}M_{\beta}^{\thinspace\thinspace\nu}-M_{\thinspace\thinspace\beta}^{\nu}M_{\alpha\nu}\right)=i\left(-M_{\alpha\nu}M_{\beta}^{\thinspace\thinspace\nu}-M_{\thinspace\thinspace\beta}^{\nu}M_{\alpha\nu}\right)[/tex]

    And this is the farthest I could arrive before posting this thread and I don't get why this is zero. M generators does not even commute so ...

    One thing that I'm not 100% sure arei the indexs on the generators. They are not matrix indexs so it does not matter the order, right ?

    [tex]M_{\thinspace\thinspace j}^{i}=M_{ j}^{\thinspace\thinspace i}[/tex]
     
  17. Mar 19, 2015 #16

    samalkhaiat

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    Assuming you meant to write [tex]A_{\alpha \beta} = i \left( M^{\nu}{}_{\alpha} \ M_{\beta \nu} - M^{\nu}{}_{\beta} \ M_{\alpha \nu} \right) ,[/tex] then by lowering the index [itex]\nu[/itex] [tex]A_{\alpha \beta} = i \left( \eta^{\nu \rho} M_{\rho \alpha} \ M_{\beta \nu} - \eta^{\nu \rho} M_{\rho \beta} \ M_{\alpha \nu} \right) .[/tex] In the second term, change the dummy indices [itex]\nu \leftrightarrow \rho[/itex] and find [tex]A_{\alpha \beta} = - i \eta^{\nu \rho} [ M_{\alpha \rho} , M_{\beta \nu} ] .[/tex] Now, using the Lorentz algebra [tex]i [M_{\alpha \rho} , M_{\beta \nu} ] = \eta_{\alpha \beta} M_{\rho \nu} - \eta_{\rho \beta} M_{\alpha \nu} + \eta_{\alpha \nu} M_{\beta \rho} - \eta_{\rho \nu} M_{\beta \alpha} ,[/tex] you find [tex]A_{\alpha \beta} = ( 2 - n ) M_{\alpha \beta} ,[/tex] where [itex]n[/itex] is the dimension of space-time.
    Clearly, [itex]A_{\alpha \beta} = 0[/itex], only for [itex]n = 2[/itex], i.e., in 2-dimensional space-time.
    However, if by [itex]A[/itex] you meant the trace [itex]A = \eta^{\alpha \beta} A_{\alpha \beta}[/itex], then yes [itex]A = 0[/itex] identically.

    Sam
     
  18. Mar 20, 2015 #17

    Mentz114

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    if ##M## is antisymmetric then ( I believe) ##M_{\thinspace\thinspace\alpha}^{\nu}M_{\beta\nu}## is symmetric in ##\alpha,\ \beta##
     
  19. Mar 20, 2015 #18
    What I am trying to prove is that the commutator of the operator W^2 constructed from the Pauli-Lubanski pseudo vector is a Casimir operator for the poincare group.

    The whole commutator [W^2,M_ij] splits up into two parts, my textbook says that the first part should be zero and I've revised multiple times my calculations and I always get what I have written in my first post...

    @Mentz114 if you are correct then of course is 0 but I can not prove it.
     
  20. Mar 20, 2015 #19

    Orodruin

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    It is really only a matter of doing what I proposed you to do. I suggest first switching the indices of the M with both indices down, then you switch which index is up and which is down, then you switch the indices of the other. You can never have ##M^\mu_{\phantom\mu \nu} = -M^{\phantom \mu \nu}_\mu##, the different sides have different indices contra/covariant.
     
  21. Mar 20, 2015 #20
    If I play with the indexs, lowering them etc. I just walk in circles. For example I get

    [tex]i\left(M_{\thinspace\alpha}^{\nu}M_{\beta\nu}-M_{\thinspace\beta}^{\nu}M_{\alpha\nu}\right)=i\left(g^{\nu\xi}M_{\xi\alpha}M_{\beta\nu}-g^{\nu\xi}M_{\xi\beta}M_{\alpha\nu}\right)=i\left(M_{\xi\alpha}M_{\beta}^{\thinspace\thinspace\xi}-M_{\xi\beta}M_{\alpha}^{\thinspace\thinspace\xi}\right)[/tex]
     
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