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Homework Help: String theory question

  1. Sep 19, 2007 #1
    1. The problem statement, all variables and given/known data
    http://books.google.com/books?id=Xm...5Dw&sig=6cUrZKqmPMoe0QBRTSYNnipNRw4#PPA114,M1

    In the solution to problem 6.3 shown in the attachment, can someone explain to me why [tex] d\vec{X}/dx [/tex] was implicitly set equal to [tex] \partial \vec{X}/\partial{x} [/tex]?



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Sep 20, 2007 #2
    Actually, the trouble begins before that. The unnumbered equation is wrong. They have
    [tex]d\vec{X} = (dx, y' dx) = (1, y') dx[/tex]
    but in the statement of the problem on page 114, Zwiebach has [itex]y' = \partial y / \partial x[/itex]. So the unnumbered equation should read:
    [tex]d\vec{X} = (dx, y' dx + \dot{y} dt)[/tex]
    From this, added to the fact that as you said, he needs [itex] \partial \vec{X}/\partial{x} [/itex] not [itex] d\vec{X}/dx [/itex], I think you can see how to finish up.
     
    Last edited: Sep 20, 2007
  4. Sep 20, 2007 #3
    You are right. Then how do you get dx/ds when the expression in [tex]d\vec{X}[/tex] now has a time differential in it so you cannot use equation 2?
     
  5. Sep 21, 2007 #4
    Don't use equation 2. Don't use equation 3 either. Use the chain rule to find
    [tex]\frac{\partial\vec{X}}{\partial s}[/tex]
     
  6. Sep 21, 2007 #5
    The first equality in equation 3 is the chain rule for that, isn't it? So, I still need dx/ds, don't I?
     
  7. Sep 21, 2007 #6
    It isn't. It doesn't into account the fact that
    [tex]\vec{X}[/tex]
    also depends on time.
     
    Last edited: Sep 21, 2007
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