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Cusps in the evolution of closed strings

  1. Nov 14, 2013 #1
    1. The problem statement, all variables and given/known data
    This is problem 7.7 in Zwiebach's book, 2ed ed.
    In (b) he want us to show that near the cusp, ##y\sim x^{2/3}.##
    In (d), Check that the period of the motion of the closed string is ##\sigma_1/4c##. How many cusps are formed during a period?

    2. Relevant equations
    (b) ##\overrightarrow X(t_0,\sigma) = \overrightarrow X_0+\frac{1}{2}(\sigma-\sigma_0)^2\overrightarrow T+\frac{1}{3!}(\sigma-\sigma_0)^3\overrightarrow R
    \\=\frac{1}{2}(\sigma-\sigma_0)^2 T\hat y+\frac{1}{3!}(\sigma-\sigma_0)^3 R(\cos\theta\hat x+\sin\theta \hat y)##, where ##|\overrightarrow T|=T \mbox{ and } |\overrightarrow R|=R##.
    (d)##\overrightarrow X(t,\sigma)
    =\frac{1}{2}[\overrightarrow F(u)+\overrightarrow G(v)]
    =\frac{\sigma_1}{4\pi}(\sin \frac{2\pi u}{\sigma_1}+\frac{1}{2}\sin\frac{4\pi v}{\sigma_1},-\cos\frac{2\pi u}{\sigma_1},-\frac{1}{2}\cos\frac{4\pi u}{\sigma_1})
    =\overrightarrow X(t+T,\sigma)##

    3. The attempt at a solution
    (b) ##y=\frac{1}{2}(\sigma-\sigma_0)^2[T+\frac{1}{3}(\sigma-\sigma_0)R\sin\theta]##
    ##x=\frac{1}{3!}(\sigma-\sigma_0)^3R\cos\theta##
    But, I don't see very clearly why ##y\sim x^{2/3}##.
    (d) I observe that ##\overrightarrow F(u) \mbox{ and } \overrightarrow G(v)## has periods ##\sigma_1/c \mbox{ and }\sigma_1/2c##, respectively. But I don't know why the period of ##\overrightarrow X## is smaller.

    Any advice would be very appreciated.
     
    Last edited: Nov 14, 2013
  2. jcsd
  3. Nov 14, 2013 #2

    George Jones

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    These aren't quite correct. Try finding them again. Don't factor out any powers of ##\sigma-\sigma_0##.
     
  4. Nov 14, 2013 #3
    I corrected it.
    Or, I still have something lost?
    Thanks a lot!
     
  5. Nov 14, 2013 #4

    George Jones

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    It looks like you're still missing something in the expression for y.
     
  6. Nov 14, 2013 #5
    Oh, sorry for the mistake!
    Now, can I conclude that since ##y\sim (\sigma-\sigma_0)^2 \mbox{ and } x\sim (\sigma-\sigma_0)^3## so ##y\sim x^{2/3}##?
    I think it's yes, because the factors, T, and x-component of R are just numbers, they don't affect the main feature of the line.
    Thank you very much!
     
  7. Nov 14, 2013 #6

    George Jones

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    Roughly, yes.

    Maybe better is to solve the ##x## equation for ##(\sigma-\sigma_0)##, and substitute this into the ##y## equation, so as to obtain ##y## as a function of ##x##. The expression for ##y## should contain two terms that involve ##x##. Argue that, near a cusp, one of these terms dominates.
     
  8. Nov 14, 2013 #7
    OK, then I know how to get the answer.
    But now I have a question in (d). I don't know how to derive the period of the function of the sum of 2 periodic functions. I thought it was the smaller one between the two, but it seems like something was missing.
     
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