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Zwiebach page 210

  1. Oct 25, 2007 #1
    1. The problem statement, all variables and given/known data
    equation 12.28 is only true if [tex]\dot{X'^j} [/tex] is equal to 0, right? Why is that true?


    2. Relevant equations



    3. The attempt at a solution
     
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  3. Oct 25, 2007 #2

    nrqed

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    No. It follows from 12.27 which does not require anything like that
     
  4. Oct 25, 2007 #3
    but isn't the only way that

    [tex] \frac{d}{d\sigma} (X^I \dot{X^J}-\dot{X}^J X^I) = X^I' \dot{X^J}-\dot{X}^J X^I'[/tex]

    if \dot{X^J}' equal 0

    ?
     
  5. Oct 25, 2007 #4

    nrqed

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    No. I am not sure why you conclude this. Notice that for I not equal to J, the expression in parenthesis is zero even before taking a derivative.
     
  6. Oct 25, 2007 #5
    I think I see now. It should really be

    [tex] \frac{d}{d\sigma} (X^I (\tau, \sigma) \dot{X^J}(\tau, \sigma')-\dot{X}^J(\tau, \sigma') X^I(\tau, \sigma)) [/tex]

    and the sigma derivative of a function of sigma prime is 0.
     
  7. Oct 25, 2007 #6

    nrqed

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    I am sorry, maybe I am too slow tonight (after 8 hours of marking) but I don't quite see what you mean. The expression in parenthesis is zero whenever I is not equal to J. Those X's are operators. And they commute when I is not equal to J. when I = J, they don't commute but give a delta function of sigma - sigma'.
     
  8. Oct 25, 2007 #7
    Note that [itex]\sigma[/itex] and [itex]\sigma'[/itex] are independent variables in this equation.
     
  9. Oct 25, 2007 #8
    I think maybe you are missing the dot on top of the x^J. That represents a tau derivative.
     
  10. Oct 25, 2007 #9

    nrqed

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    You are right, I forgot to say that it's the conmmutator of X with dot X. But my point is the same: X^I and (dot X)^J commute whenever I is not equal to J. So the derivative is zero trivially because the expression in parenthesis is zero before even taking thee derivative and not because there is no sigma dependence (well, there is sigma dependence trivially because it's zero!)
     
  11. Oct 25, 2007 #10
    [tex]X^I (\tau, \sigma) \dot{X^J}(\tau, \sigma')-\dot{X}^J(\tau, \sigma') X^I(\tau, \sigma) = 2 \pi \alpha' \eta^{IJ} \delta(\sigma - \sigma ') [/tex]
    is equation 12.27 expanded

    [tex] \frac{d}{d\sigma} (X'^I (\tau, \sigma) \dot{X^J}(\tau, \sigma')-\dot{X}^J(\tau, \sigma') X'^I(\tau, \sigma)) = 2 \pi \alpha' \eta^{IJ} \frac{d}{d \sigma}\delta(\sigma - \sigma ') [/tex]
    is equation 12.28 expanded

    12.28 follows from 12.27 only if [tex] \frac{d}{d\sigma}(\dot{X}^J(\tau, \sigma')) = 0[/tex] which is true only because the sigma argument has a prime but the sigma in the derivative operator has no prime.
     
    Last edited: Oct 25, 2007
  12. Oct 25, 2007 #11

    nrqed

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    There si no d/dsigma on the left side
    Ok, I see what your question was. yes, of course, [itex] \frac{d}{d \sigma} f(\sigma') = 0 [/itex]. Sorry, I did not understand your question because this was implicit for me.
     
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