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Homework Help: Coordinate transformation

  1. Jan 29, 2007 #1
    How can I identify the coordinate transformation that turns

    [tex]ds^2 = \left(1+\frac{\epsilon}{1+c^2t^2}\right)^2c^2dt^2 - \left(\frac{\epsilon}{1+x^2}\right)^2x^2 - \left(\frac{\epsilon}{1+y^2}\right)^2y^2 - \left(\frac{\epsilon}{1+z^2}\right)^2z^2[/tex]

    into the Minkowski metric

    [tex]ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2[/tex]?
     
  2. jcsd
  3. Jan 29, 2007 #2

    George Jones

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    You forgot a bunch of d's in the first equation, and you should use different symbols for the coordinates in the two equations - maybe primes in the second.

    What happens if you identify corresponding terms in the two metrics?
     
  4. Jan 30, 2007 #3
    My equations looks just like that, but I guess they are wrong. (P. Coles, Cosmology)
    What dou you mean by identify?
     
  5. Jan 30, 2007 #4
    Maybe
    [tex]\left(1+\frac{\epsilon}{1+c^2t^2}\right)^2c^2dt^2 = c^2dT^2[/tex]
    [tex]\left(\frac{\epsilon}{1+x^2}\right)^2dx^2 = dX^2[/tex]
    etc?
     
  6. Jan 30, 2007 #5
    Yes that I can see, but that's not a transformation for the whole metric?
     
  7. Jan 30, 2007 #6

    George Jones

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    I'm not sure what you mean by this.

    You're looking for coordinate transformations, i.e., [itex]X = X \left( t, x, y, z)[/itex], etc. Then, e.g.,

    [tex]dX = \frac{\partial X}{\partial{t}} dt + \frac{\partial X}{\partial{x}} dx+ \frac{\partial X}{\partial{y}} dy + \frac{\partial X}{\partial{z}} dz.[/tex]

    A new coordinate does not have to depend explicitly on all of the old coordinates, i.e., some of the terms in the above expansion can be zero.
     
  8. Jan 30, 2007 #7
    So I can just give the answer to the problem as whatta did above?
     
  9. Jan 30, 2007 #8
    no you probably are supposed to solve that in the form of x(X), y(Y)... t(T) or viceversa
     
  10. Jan 30, 2007 #9
    Like

    [tex]t(T)=T+\frac{\epsilon}{c}\arctan{cT}[/tex]
    [tex]x(X)=\epsilon\arctan{X}[/tex]
    [tex]y(Y)=\epsilon\arctan{Y}[/tex]
    [tex]z(Z)=\epsilon\arctan{Z}[/tex]

    so that

    [tex]dt = \left(1+\frac{\epsilon}{1+c^2T^2}\right)dT[/tex]
    [tex]dx = \left(\frac{\epsilon}{1+X^2}\right)dX[/tex]
    [tex]dy = \left(\frac{\epsilon}{1+Y^2}\right)dY[/tex]
    [tex]dz = \left(\frac{\epsilon}{1+Z^2}\right)dZ[/tex]

    and then

    [tex]ds^2=c^2dt^2-dx^2-dy^2-dz^2 = \left(1+\frac{\epsilon}{1+c^2T^2}\right)^2c^2dT^2 - \left(\frac{\epsilon}{1+X^2}\right)^2dX^2 - \left(\frac{\epsilon}{1+Y^2}\right)^2dY^2 - \left(\frac{\epsilon}{1+Z^2}\right)^2dZ^2[/tex].

    Is this correct?
     
  11. Jan 30, 2007 #10

    dextercioby

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    It certainly looks so.
     
  12. Jan 30, 2007 #11
    except that you have lost -1 somewhere
     
  13. Jan 30, 2007 #12
    I don't feel that positive..?
     
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