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Schwarzschild extension coordinate transformation algebra

  1. Mar 18, 2015 #1
    So I have the metric as ##ds^{2}=-(1-\frac{2m}{r})dt^{2}+(1-\frac{2m}{r})^{-1}dr^{2}+r^{2}d\Omega^{2}##*

    I have transformed to coordinate system ##u,r,\phi, \theta ##, where ##u=t-r*##(2),
    where ##r*=r+2m In(\frac{r}{2m}-1)##
    and to the coordinate system ##v,r,\phi, \theta ##,
    where ##v=t+r*##,(1)

    From (1) and (2) I see that ##dt=dv-\frac{dr}{(1-\frac{2m}{r})}## and ##dt=du+\frac{dr}{(1-\frac{2m}{r})}##
    (On a side note, what is the proper name of these types of derivative expressions?)

    Substituting these into * in turn it is easy enough to get the metrics:


    (which I believe are correct?).

    Question:

    I am now want to get the metric using both \(v\) and \(u\) in favour of \(r\) and \(t\).
    To do this I make use of:
    ##\frac{1}{2}(v-u)=r+2M In(\frac{r}{2M}-1) ##
    therefore ##\frac{1}{2}(dv-du)(1-\frac{2m}{r})=dr##

    and I sub this into either (1) or (2),
    say (1) , I then get:
    ##ds^{2}= - (1-\frac{2M}{r}) dudv+r^{2}d\Omega^{2}##

    And the first term is a minus sign out.
    (in accord to source sean m carroll lecture notes on general relativity eq.7.73.)

    I have no idea why I am a sign out,

    Thanks,your assistance is greatly appreciated !
     
  2. jcsd
  3. Mar 19, 2015 #2
  4. Mar 19, 2015 #3

    PeterDonis

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    I don't see any metrics here. Did you leave them out by mistake?
     
  5. Mar 19, 2015 #4
    Apologies!
    metrics are:
    ##ds^{2}=-(1-\frac{2M}{r})dv^{2}+2dvdr + r^{2}(d\theta^{2}+sin^{2}\theta d\phi^{2} ##
    ##ds^{2}=-(1-\frac{2M}{r})du^{2}-2dudr + r^{2}(d\theta^{2}+sin^{2}\theta d\phi^{2} ##
     
  6. Mar 19, 2015 #5

    PeterDonis

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    Ok, those look correct.

    I don't think you are. I think you are just using an opposite sign convention from Carroll for the definition of ##r## in terms of ##u## and ##v##. If you look at Carroll's equation 7.74, it has ##u - v## where you have ##v - u##. That's why he has an opposite sign in 7.73 from yours.

    (Carroll also has a factor of 1/2 in front of 7.73, whereas you do not. That's because he also scales ##u## and ##v## differently than you are. None of these differences affect the physics; they're just different conventions for the math.)
     
  7. Mar 20, 2015 #6
    I used a different notation than Carrol ## u bar=v##.
    I cancelld the ##1/2## using ##dudv=dvdu##
     
  8. Mar 20, 2015 #7

    PeterDonis

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    Hm, yes, I see. So much for that theory. :oops:

    I'm not sure what's going on. I get the minus sign in the ##du dv## term the same way you do. Physically, the minus sign makes sense: a line element with ##du## and ##dv## both the same sign should lie inside one of the light cones (future for ##du## and ##dv## both positive, past for ##du## and ##dv## both negative), and so should be timelike and have a negative squared length. Unless Carroll is using a different sign convention somewhere else that I haven't spotted, the only other thing I can think of is that his equation 7.73 has a typo.
     
  9. Apr 20, 2015 #8
    Looking at the next metric, equation 7.77, it has a negative sign. (And I belive this to be corect as it agrees with http://www.damtp.cam.ac.uk/user/hsr1000/black_holes_lectures_2014.pdf , page 33, eq 2.35). Looking at how u' and v' are defined as functions of u,v, I am getting a sign change when going from the metric in u,v to u'v'. (whilst the cambridge notes, equation 2.32, defines u' with a neg sign compared to Carroll, either definition gives arise to a neg sign) and so it appears that there should be a sign change and so equation 7.73 in carroll should have a positive.

    Hopefully with this information someone may find it easier to spot why I am getting a negative - been llooking for a long time.

    Thanks

    (Cambridge ntes and Carroll have defined their u and v, u' and v' the other way around).
     
    Last edited: Apr 20, 2015
  10. Apr 22, 2015 #9
    Sorry just to add both sources are using the same metric signature - (-,+,+,+)).
     
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