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

Schutz, page 115

  1. Oct 22, 2005 #1
    This time I am asking for help on a problem. I think I have an answer for problem 23, part (a), but I would like verificiation. I don't have a solution for part (b).

    Problem 23: Use the identity [itex]T^{\mu \nu}{}_{,\nu} = 0[/itex] to prove the following results for a bounded system (i.e. a system for which [itex]T^{\mu \nu} = 0[/itex] outside a bounded region of space).

    (a)
    [tex]\frac{\partial}{\partial t}\int T^{0 \alpha}d^3x = 0[/tex]

    My solution is:

    [tex]
    \frac{\partial}{\partial t}\int T^{0 \alpha}d^3x
    [/tex]
    [tex]
    = \int T^{\alpha 0}{}_{,0}d^3x
    [/tex]
    [tex]
    = -\int T^{\alpha j}{}_{,j}d^3x[/tex] (by the identity)
    [tex]
    = -\oint \bold{n} \cdot T^{\alpha j}dS[/tex] (Gauss's theorem)
    [tex]
    = 0[/tex](by boundedness, using a sphere that surrounds the support of T)

    I think all this is OK, but when I try to extend this to part (b), the tensor virial theorem, I don't get the right answer:

    (b)
    [tex]\frac{\partial^2}{\partial t^2}\int T^{00} x^i x^j d^3x = 2\int T^{ij} d^3x[/tex]

    The problem for me is that after I convert the first partial in Gauss's theorem, the volume integral turns into a surface integral and I can't do that a second time for the second partial.
     
    Last edited: Oct 22, 2005
  2. jcsd
  3. Oct 22, 2005 #2
    Hi Jimmy

    I'm sorry I can't help you at this time. But I have to say that I really admire you. You are a real go-getter when it comes to learning GR from Shutz and I find that quite impressive. So impressive that I'm going to try to relearn this aspect of GR (Which I learned and quickly forgot :biggrin: ) to try to of more help now and in the future. It won't be anytime soon though. But I am hopeful that I'll be of more help to you in the future or if nobody ever answers this question.

    Keep up the great work Jimmy!

    Pete
     
  4. Oct 22, 2005 #3

    SpaceTiger

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Be careful with your application of gauss' law and remember that:

    [tex]T^{\alpha j}_{~~,j}x^ix^j \neq (T^{\alpha j}x^ix^j)_{,j}[/tex]

    Try expanding the righthand term in the above equation...you ought to get the lefthand term, plus some others. The term on the right will vanish when you convert it to a surface term, but what remains will...well, remain. I haven't worked it out in detail, but it looks like you just have to follow this basic procedure twice (once for each time partial).
     
    Last edited: Oct 22, 2005
  5. Oct 22, 2005 #4
    I appreciate these words of encouragement. Especially as I am thinking of giving up. My objective in reading Schutz was to understand the Einstein equation and I believe that I have accomplished that to the extent that satisfies me. Before I started reading Schutz at the beginning of September, I set myself the goal of finishing the book in a year. I have read through chapter 8 over and over again. Each time I read it, I understand a little more. However, chapter 9 is a bear, and chapter 8 has the answer I was looking for. My goal of finishing in a year no longer seems directly related to my main objective, so I have abandoned it. However, I have not stopped reading and trying to get through chapter 9. With your help and the help of the others who so generously give their time to this project, I may well succeed. I promise myself and you not to ask questions until I have given my all and failed.

    My congratulations to those who have struggled with this book and succeeded in understanding it.
     
  6. Oct 22, 2005 #5
    Jimmy - Do not give up. I also suggest that you don't earmark the time it will take to read it. I bought that book in 1999 and I know that I will never understand the material as good as Schutz! :biggrin: Everyone has hard times with GR - Everyone!. If someone claims otherwise then there is a 99.999% chance that they're lying.

    There have been times where I had to read a paragraph over and over again and still didn't understand it. I recall there was one time I reread a paragraph which seemed to be 100 times. You're working by yourself which is hard. Don't expect to be an expert after you read it. Everyone will have areas where they have an enourmously hard time reading. I had many myeself. But I kept at it and will keep at it. Even Einstein went a littel buggy when learning differential geometry so you're in good company. You can PM me anytime and/or e-mail me. I'll be glad to be of any help I can. Right now you're in areas of the book which I too had problems with and since that was years ago and I haven't used that information constanly since then, then you're probably much better than I with that material.

    Pete
     
    Last edited: Oct 22, 2005
  7. Oct 22, 2005 #6
    Yes, the key is the expansion above and being careful with my application of Gauss' theorem. Here is the full solution:

    [tex]
    \frac{\partial^2}{\partial t^2} \int T^{00} x^i x^j d^3x
    [/tex]
    [tex]
    = \frac{\partial}{\partial t} \int T^{00}{}_{,0} x^i x^j d^3x
    [/tex]
    [tex]
    = -\frac{\partial}{\partial t} \int T^{0k}{}_{,k} x^i x^j d^3x
    [/tex]
    [tex]
    = -\frac{\partial}{\partial t} \int (T^{0k} x^i x^j)_k d^3x
    [/tex]
    [tex]
    + \frac{\partial}{\partial t} \int T^{0k} x^i{}_{,k} x^j d^3x
    [/tex]
    [tex]
    + \frac{\partial}{\partial t} \int T^{0k} x^i x^j{}_{,k} d^3x
    [/tex]

    The first term is 0 by part (a) and the next two terms are symmetric in i and j.

    [tex]
    \frac{\partial}{\partial t} \int T^{0k} x^i x^j{}_{,k} d^3x
    [/tex]
    [tex]
    = \frac{\partial}{\partial t} \int T^{0k} x^i \delta^j{}_{,k} d^3x
    [/tex]
    [tex]
    = \frac{\partial}{\partial t} \int T^{0j} x^i d^3x
    [/tex]
    [tex]
    = \int T^{j0}{}_{,0} x^i d^3x
    [/tex]
    [tex]
    = -\int T^{jk}{}_{,k} x^i d^3x
    [/tex]
    [tex]
    = -\int (T^{jk} x^i)_{,k} d^3x + \int T^{jk}x^i{}_{,k} d^3x
    [/tex]
    [tex]
    = 0 + \int T^{jk}\delta^i{}_{,k} d^3x[/tex] (by part (a))
    [tex]
    = \int T^{ij} d^3x
    [/tex]

    Thanks SpaceTiger.
     
    Last edited: Oct 22, 2005
  8. Oct 22, 2005 #7
    Jimmy

    I sure wish I had the time right now to do Schutz over again. Its a great book. But at the moment I'm reading Classical Charged Particles, by Fritz Rohrlich, (1965). I highly recommend that you pick up this book and read it as soon as possible. There are sections in it on general relativity that are awesome and will make a supurb additional reading for Schutz. By the way, Schutz has a new text book out on gravity. Its pretty good too! There are parts in this text that he ignores in his GR text. In particular he addresses the concept of mass of a body much more and how the mass of the body is effected by stresses on the body. This theory of mass and stress came from Einstein in 1907. It was the beginning of how he'd eventually show that to describe mass completely you need a second rank tensor.

    Pete
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Schutz, page 115
  1. Schutz, page 224 (Replies: 2)

  2. Schutz, page 226 again (Replies: 4)

  3. Schutz, page 246 (Replies: 2)

  4. Schutzs or hurtle? (Replies: 1)

Loading...