Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Time integral of worldsheet current

  1. May 7, 2016 #1

    ShayanJ

    User Avatar
    Gold Member

    I'm reading this paper and have no problem with its calculations. But there is one little thing I'm not sure I understand. At some point, the author tries to find the amount of momentum carried across the spatial direction of the worldsheet in a time ## \Delta t ## by calculating the following integral:

    ## \Delta P_1=\int dt \sqrt{-g} P^r_{\ \ x^1} ##

    where g is the induced metric on the worldsheet and r and t are the coordinates on the worldsheet which are also two of the background spacetime coordinates along with ## x^1 ##.

    My problem is, I'm not sure I understand the reason for the presence of ## \sqrt{-g} ## in the integrand instead of integrating only ## P^r_{\ \ x^1} ##. Could anyone explain?

    Thanks
     
  2. jcsd
  3. May 12, 2016 #2

    MathematicalPhysicist

    User Avatar
    Gold Member

    It's the Jacobian, I guess they change variables in the integration. (I didn't read the paper).
     
  4. May 12, 2016 #3

    ShayanJ

    User Avatar
    Gold Member

    Yeah, Its the Jacobian but there is no change of variable involved. The worldsheet coordinates are r and t which are also two of the coordinates of the background spacetime so no change of variable is needed.
     
  5. May 12, 2016 #4

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It is necessary to make the expression reparametrization invariant. Look up for example the Lagrangian of general relativity, it is basically ##\int \sqrt{-g} R ##.
     
  6. May 12, 2016 #5

    ShayanJ

    User Avatar
    Gold Member

    Thanks, that helps. But just to get it more clearly, I have one question.

    In the case of the EH Lagrangian, We know that ## \int R d^4 x ## is not a scalar because R is a scalar but ##d^4 x ## is a weight ## \pm 1 ## scalar density, so we need to put ##\sqrt{-g} ##(which is a ##\mp 1 ## scalar density) there to make the whole thing a scalar.(I'm not sure about the weights of ## d^4 x ## and ##\sqrt{-g}##).

    But here we're dealing with ## P^r_{\ x^1} ## which is one of the components of a tensor which mixes with other components under a transformation and so is not a scalar. How should we come to the conclusion that putting ## \sqrt{-g} ## besides it, makes the whole integral reparametrization invariant?
     
  7. May 12, 2016 #6

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The "t" that is integrated over is not the physical time of the ambient space, it is just a parameter used to label points along the string. it is under reparametrization of that variable that the integral must be made invariant. With respect to that variable, the P is a scalar.
     
  8. May 13, 2016 #7

    Demystifier

    User Avatar
    Science Advisor

    What is still confusing is that ##g## is the determinant of the 2-dimensional metric (or is it?), while the integral is only 1-dimensional. Shouldn't one have ##\int dt \sqrt{-g_{00}}## instead? When the metric is diagonal, where does the additional factor ## \sqrt{g_{rr}}## comes from? Is there, perhaps, a hidden integration over ##\int dr## involved? (For instance, perhaps ##P^r_1## should really be ##\int dr P^r_1##, or something like that.) In any case, the author seems to be rather sloppy in that equation.
     
    Last edited: May 13, 2016
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Time integral of worldsheet current
  1. Dimensions and time (Replies: 30)

Loading...