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

Exercise 2.5 in Misner, Thorne and Wheeler

  1. Jul 4, 2013 #1

    TerryW

    User Avatar
    Gold Member

    If the squared length of a 4-velocity is -1, how can you have a component v0 =0?

    I've played with equation (2.35) and produced a result of v = γ(-u2, -u1, -u2, -u3), which doesn't have a squared length of -1.

    Can anyone help out?



    TerryW
     
  2. jcsd
  3. Jul 4, 2013 #2

    Bill_K

    User Avatar
    Science Advisor

    In Exercise 2.5, u is the particle's 4-velocity with squared length -1, whereas v is a newly defined spacelike vector, apparently the projection of u onto the observer's space section.
     
  4. Jul 4, 2013 #3

    TerryW

    User Avatar
    Gold Member

    Thanks

    Hi Bill_K,


    So my result is probably OK. A few words along those lines in the text hinting at the special nature of v might have helped, it does come out of left field a bit.

    Thanks for your prompt reply. You obviously don't subscribe to the view I saw in one post that if you are working through MTW, you should need any help!


    Regards

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




Similar Discussions: Exercise 2.5 in Misner, Thorne and Wheeler
Loading...