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Prove two line elements represent the same space/plane

  1. Nov 12, 2014 #1
    1. The problem statement, all variables and given/known data
    I've been given the line element given below in the relevant equations section and I need to prove that the space it represents is just the same as the 2-D Euclidean plane.

    2. Relevant equations
    [itex] ds^2 = a^2 \frac{d\eta ^2 }{cosh^4(\eta)} + a^2 tanh^2(\eta) d\theta ^2 [/itex]

    Where
    [itex] 0\lt\theta\leq2\pi [/itex]
    [itex] 0\lt\eta\leq\infty [/itex]

    3. The attempt at a solution
    I'm pretty sure that to prove this I need to find the coordinate transform to show that the above line element should be equal to:
    [itex] ds^2 = dx^2 + dy^2[/itex]
    So I believe I'm looking for [itex]x[/itex] and [itex]y[/itex] in terms of [itex]eta[/itex] and [itex]theta[/itex], but I'm not entirely sure how to go about that.

    Any suggestions are much appreciated!
     
  2. jcsd
  3. Nov 13, 2014 #2
    Okay well I can't seem to find a way to edit the original post so I'm going to reply with what I've got so far.
    I've been looking at the trigonomic hyperbolic identities and their differentials.
    If we have
    ## x = a tanh^2(\eta) ## then ## dx = 2tanh(\eta)sech^2(\eta)d\eta##
    ## y = a\theta ## then ## dy = ad\theta ##

    This would then lead to ## ds^2 = 2a^2tanh^2(\eta)sech^4(\eta)d\eta^2 + a^2d\theta^2##
    Where of course ## sech^4(\eta) = \frac{1}{cosh^4(\eta)}##

    So it is sort of getting there, maybe.

    Wondering if the fact that ##tanh^2(\eta)+sech^2(\eta) = 1 ## could be useful.
     
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