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String motion

  1. Dec 25, 2014 #1
    My question concerns motion of relativistic Strings, as the reader with a great a mount of deductive skills can deduce :).
    The question is: How can I derive the P's. More acuratly speaking why is the numerator how it is.
    I am refering to Zwiebach chapter 6.
    Thanks for any clarifications and sorry about the zoom.

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  2. jcsd
  3. Dec 25, 2014 #2

    Vanadium 50

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    Please type your question in, using LaTex if you need to express a formula. That's better than asking us to try and figure out what is in a tiny blurry picture.
  4. Dec 25, 2014 #3


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    Take the Lagrangian you have, and find its derivative...
    The Nambu Goto lagrangian he has derived is : [itex] L = - \frac{T_0}{c} \sqrt{ - det \gamma } = - \frac{T_0}{c} \sqrt{ ( \dot{X} X' )^2 - \dot{X}^2 X'^2 } [/itex]
    Forgeting the prefactor:
    [itex] \frac{\partial L}{\partial \dot{X}^\rho} \propto \frac{2 ( \dot{X} X' ) X'_\rho - 2 X'^2 \dot{X}_\rho}{2 \sqrt{ ( \dot{X} X' )^2 - \dot{X}^2 X'^2 } }[/itex]
    Similarily for the other...
    If you have some problems with these derivatives of inner products:
    [itex] \frac{\partial }{\partial \dot{X}^\rho} ( \dot{X}^\mu X'_\mu )^2 =\frac{\partial }{\partial \dot{X}^\rho} \dot{X}^\mu X'_\mu \dot{X}^\nu X'_\nu = \delta_{\rho}^{\mu} X'_\mu \dot{X}^\nu X'_\nu + \dot{X}^\mu X'_\mu \delta_{\rho}^{\nu} X'_\nu = X'_\rho \dot{X}^\nu X'_\nu + \dot{X}^\mu X'_\mu X'_\rho =2 (\dot{X}^\nu X'_\nu ) X'_\rho [/itex]

    The only thing you use is the orthonormality relation [itex] \frac{\partial x^a}{\partial x^b}= \delta_b^a[/itex]
    Last edited: Dec 25, 2014
  5. Dec 26, 2014 #4
    Thanks! A like is yours.
  6. Dec 26, 2014 #5
    But why would one have a inner product?
  7. Dec 26, 2014 #6


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    What do you mean by why you have inner products? they are there when you wrote:

    [itex] \gamma = \begin{pmatrix} \dot{X} \cdot \dot{X} & \dot{X} \cdot X' \\ X' \cdot \dot{X} & X' \cdot X' \end{pmatrix}[/itex]
    The [itex]\cdot[/itex] denotes inner product, and in this case the Lorentzian product [itex] x \cdot y = \eta_{\mu \nu} x^\mu y^\nu [/itex]...
    Now why is that the case? because you have already derived the induced metric:
    [itex] \gamma_{ab} = \eta_{\mu \nu} \partial_a X^\mu \partial_b X^\nu \equiv \partial_a X \cdot \partial_b X[/itex]
  8. Dec 26, 2014 #7


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    Or if you didn't ask exactly that, then remember:
    [itex] \frac{d}{dx} f(g(x)) = \frac{d}{dg}f(g) \frac{dg(x)}{dx} [/itex]

    So the [itex]\sqrt{...}[/itex] will give the [itex] \frac{1}{2\sqrt{...}}[/itex]
    and you still have to take the derivatives of [itex](...)[/itex] wrt your initial derivative.
  9. Dec 26, 2014 #8
    Thanks fine from here!
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