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Zwiebach page 113

  1. Sep 16, 2007 #1
    1. The problem statement, all variables and given/known data
    http://books.google.com/books?id=Xm...42EpgKaxsi5Dw&sig=6cUrZKqmPMoe0QBRTSYNnipNRw4
    In problem 6.1, I am trying to show "why the following relations hold"

    for the first one, that would just be the definition of ds^2 if you had (cdt)^2 in there but I cannot figure out what happened to it

    for the the second one, I am confused about how that can possible be write because the expression for v_perp on 110 implies that it is a vector with the same number of components as X^mu but the y has two fewer components than X. What gives?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 18, 2007 #2
    Is this even the same ds that is used in the equation ds^2 = cdt^2 +dx^2 + dy^2 + ...?

    Or is this just an infinitesimal length of the string? Are we holding time constant in this problem?
     
    Last edited: Sep 18, 2007
  4. Sep 18, 2007 #3

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    This is a non-relativistic problem, so by ds they probably mean the ordinary distance measure in Euclidean space. For the second part, it seems they are breaking up all the (n dimensional) vectors into the form [itex](x_1, \vec x_T)[/itex] (where [itex]\vec x_T[/itex] is n-1 dimensional). So by [itex]\vec v_\perp[/itex], they probably mean the last n-1 components of the n dimensional perpendicular velocity (the first component being zero to first approximation).
     
  5. Sep 18, 2007 #4
    If this is so, then isn't the first part true by definition?

    Yes. Maybe this is just a really minor issue but y is a d-1 dimensional vector (where d is the number of spatial dimensions) and v_T is a dimensional vector and Zwiebach sets the two equal.
     
  6. Sep 19, 2007 #5
    Does my question make sense to people?
     
  7. Sep 19, 2007 #6

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    Sorry, didn't see your reply.

    Pretty much. I think the reason they're asking you about it is because you'll need the arclength in constructing the lagrangian.

    It is a little confusing. It's not clear if he wants [itex]\vec v_\perp[/itex] to be a d-dimensional vector (as seems to be the case when he writes above [itex]|\vec v_\perp|<<|c|[/itex], since here its the entire perpendicular velocity vector that's relevant), or a d-1 dimensional projection of this vector, as would be necessary for the equation you're asked to show. He probably means to ask you to show that it is only these d-1 coordinates that are non-zero, but there's not an easy way to express this with his notation.
     
    Last edited: Sep 19, 2007
  8. Sep 22, 2007 #7
    A related thing that is confusing me is Figure 6.9. X-vector is defined on page 106 as containing only the spatial string coordinates, but in this figure its time-partial clearly has a time-component.
     
    Last edited: Sep 22, 2007
  9. Sep 23, 2007 #8
    Does this make sense to anyone else?
     
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