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Zwiebach page 143

  1. Oct 21, 2007 #1
    1. The problem statement, all variables and given/known data
    On page 143, Zwiebach says that we can define the current tenser to be antisymmetric in mu and nu since it is multiplied by the antisymmetric matrix epsilon^{mu nu}--any symmetric part would drop out of the left hand-side.

    But I thought it already was defined in equation 8.55? What does he mean that the symmetric part would drop out?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 21, 2007 #2
    Equation (8.55) does not completely define j, only the sum. For instance, you could add an amount to [itex]j_{01}[/itex] and add the same amount to [itex]j_{10}[/itex] and the sum [itex]\epsilon j[/itex] would remain the same. This is what he means by "the symmetric part would drop out."
     
  4. Oct 21, 2007 #3
    That makes sense. Then below that he says that "the currents can be read directly from this equation because the factor multiplying epsilon^{mu nu} on the RHS is explicitely antisymmetric"?

    I do not see why the antisymmetry j^alpha_{mu nu} allows you to do that since you are still summing over 2 indices?
     
  5. Oct 22, 2007 #4
    Before you assume that j is antisymmetric, you have some play in the values of j since you can add a symmetric part and equation (8.55) will still hold. However, once you assume that j is antisymmetric, that arbitrariness is gone and you can equate the antisymmetic factors on both sides for each index. For instance
    [tex]j_{01} = \frac{1}{2}(j_{01} - j_{10}) = -\frac{1}{2}(X_{0}{\mathcal P}_{1} - X_{1}{\mathcal P}_{0})[/tex]
     
  6. Oct 22, 2007 #5
    Yes. I understand that is what Zwiebach is claiming; its just that the proof that every antisymmetric matrix is invertible isn't coming to me...
     
  7. Oct 22, 2007 #6
    Zwiebach doesn't claim that every antisymmetric matrix is invertible and it isn't true.
     
  8. Oct 22, 2007 #7
    You're right. He is claiming that if epsilon^{\mu \nu} is antisymmetric and M^{\mu \nu} is antisymmetric, then if [tex] \epsilon^{\mu \nu} M_{\mu \nu} = \epsilon^{\mu \nu} N_{\mu \nu}[/tex], [tex] M_{\mu \nu} = N_{\mu \nu}[/tex], right? How do you prove that?
     
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