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String theory question

  1. Sep 14, 2007 #1
    1. The problem statement, all variables and given/known data
    Zwiebach QC 6.4

    Why is

    [tex] \frac{\partial}{\partial{X^{\mu}}} \left( \frac{\partial X}{\partial{\tau}} \cdot \frac{\partial X}{\partial{\sigma}} \right) ^2 = \left( \frac{\partial X}{\partial{\tau}} \cdot \frac{\partial X}{\partial{\sigma}} \right) \cdot \frac{\partial{X^{\mu}}}{\partial{\tau}}[/tex]

    The dot is the relativistic dot product.
    What I am confused about is how you take the derivative with respect to one component X^mu of the spacetime vectors and get a quantity that still has [tex] \left( \frac{\partial X}{\partial{\tau}} \cdot \frac{\partial X}{\partial{\sigma}} \right) [/tex]. Is there an intermediate step someone could show me?


    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Sep 14, 2007
  2. jcsd
  3. Sep 14, 2007 #2

    Hurkyl

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    Staff Emeritus
    Science Advisor
    Gold Member

    Could you explain just what kind of object everything is?

    If I just lively think of X as a function of [itex]\tau[/itex] and [itex]\sigma[/itex] with values in a 4-dimensional vector space with the Minkowski inner product, then the partial derivative w.r.t. the i-th component is

    [tex]
    \frac{\partial}{\partial X_i} \left(
    \frac{\partial X}{\partial \tau}
    \cdot
    \frac{\partial X}{\partial \sigma}
    \right)
    =
    \left( \mathbf{\hat{e}}_i \frac{\partial X_i}{\partial \tau} \right)
    \cdot
    \frac{\partial X}{\partial \sigma}
    +
    \frac{\partial X}{\partial \tau}
    \cdot
    \left( \mathbf{\hat{e}}_i \frac{\partial X_i}{\partial \sigma} \right)
    =
    2\frac{\partial X_i}{\partial \tau} \frac{\partial X_i}{\partial \sigma}
    [/tex]

    ([itex]\mathbf{\hat{e}}_i[/itex] is a standard basis vector) Applying the chain rule would give

    [tex]
    \frac{\partial}{\partial X_i} \left(
    \frac{\partial X}{\partial \tau}
    \cdot
    \frac{\partial X}{\partial \sigma}
    \right)^2 =
    4
    \left(
    \frac{\partial X}{\partial \tau}
    \cdot
    \frac{\partial X}{\partial \sigma}
    \right)
    \frac{\partial X_i}{\partial \tau} \frac{\partial X_i}{\partial \sigma}
    [/tex]

    Not quite what you have, but maybe you can see how your things are different from my things and fill in the gap.
     
    Last edited: Sep 14, 2007
  4. Sep 14, 2007 #3
    Sorry. My post was rather terse.

    [tex] X^{\mu}(\tau,\sigma) [/tex] is the mapping function from parameter space (of the parameters tau and sigma) into spacetime. So, in this context what is the basis vector (it is just some pseudo-Euclidean basis vector probably)? Is there a way to do this without going into basis vectors? Why does [tex]
    \frac{\partial}{\partial X_i} \left(
    \frac{\partial X}{\partial \tau}
    \right)
    =
    \left( \mathbf{\hat{e}}_i \frac{\partial X_i}{\partial \tau} \right)
    [/tex] ?
     
    Last edited: Sep 14, 2007
  5. Sep 15, 2007 #4

    Hurkyl

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    Staff Emeritus
    Science Advisor
    Gold Member

    Because [itex]X = (X_0, X_1, X_2, X_3)[/itex]. If you differentiate it with respect to, say, component 2, you get
    [tex]
    \begin{equation*}
    \begin{split}
    \frac{\partial X}{\partial X_2}
    &= \frac{\partial}{\partial X_2} (X_0, X_1, X_2, X_3)
    \\&= \left(\frac{\partial X_0}{\partial X_2},
    \frac{\partial X_1}{\partial X_2},\frac{\partial X_2}{\partial X_2},
    \frac{\partial X_3}{\partial X_2}
    \right)
    \\&= (0, 0, 1, 0) = \mathbf{\hat{e}}_2
    \end{split}
    \end{equation*}
    [/tex]

    Hrm, I know I did someting wrong last night, but I'm still too sleepy to spot it at the moment...
     
    Last edited: Sep 15, 2007
  6. Sep 16, 2007 #5
     
  7. Sep 16, 2007 #6
    You missed a dot (and/or a prime). The equation is:

    [tex] \frac{\partial}{\partial{\dot{X}^{\mu}}} \left( \frac{\partial X}{\partial{\tau}} \cdot \frac{\partial X}{\partial{\sigma}} \right) ^2 = \left( \frac{\partial X}{\partial{\tau}} \cdot \frac{\partial X}{\partial{\sigma}} \right) \cdot \frac{\partial{X^{\mu}}}{\partial{\tau}}[/tex]
    where [itex]\dot{X}^{\mu}[/itex] is defined in equation (6.40) on page 100.
     
    Last edited: Sep 16, 2007
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