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I need some help deriving the equation

  1. Oct 5, 2011 #1
    1. The problem statement, all variables and given/known data
    This is from Exploring Black Holes, Query 1, page G-3. It involves a one dimensional creature on a two dimensional circle.

    Constant dl but different dx. Collect terms in equation [5]. Show that the result can be written dl[itex]^{2}[/itex] = [itex]\frac{dx^{2}}{1 - x^{2}/r^{2}}[/itex]
    Equation [5], involving dl, is for finding the length of a measuring rod of length dl along the circle, laid by the one dimensional creature.


    2. Relevant equations
    Equations in text for the chapter so far:
    [1] r[itex]^{2}[/itex] = x[itex]^{2}[/itex] + y[itex]^{2}[/itex]
    [2] y[itex]^{2}[/itex] = r[itex]^{2}[/itex] - x[itex]^{2}[/itex]
    [3] 2ydy = -2xdy r = constant
    [4] dy = -[itex]\frac{xdx}{y}[/itex] = -[itex]\frac{xdx}{(r^{2} - x^{2})^{1/2}}[/itex] r = constant
    [5] dl[itex]^{2}[/itex] = dx[itex]^{2}[/itex] + dy[itex]^{2}[/itex] = dx[itex]^{2}[/itex] + [itex]\frac{x^{2}dx^{2}}{(r^{2} - x^{2}}[/itex] r = constant

    3. The attempt at a solution
    I've realized fairly easily that the bottom part of the equation I'm trying to get has an r[itex]^{2}[/itex] factored from the bottom. I haven't quite figured out where it went, though. The closest I've gotten is:

    dy = -[itex]\frac{xdx}{(r^{2} - x^{2})^{1/2}}[/itex]
    Solving for dx, I get dx = -[itex]\frac{dy(r^{2} - x^{2})^{1/2}}{x}[/itex]
    I then input this into equation [5], getting dl[itex]^{2}[/itex] = [itex]\frac{dy^{2}(r^{2} - x^{2})^{2}}{x^{2}}[/itex] + [itex]\frac{x^{2}dx^{2}}{(r^{2} - x^{2})}[/itex] = [itex]\frac{dy^{2}(r^{2} - x^{2})^{2} + x^{4}dx^{2}}{x^{2}(r^{2} - x^{2})}[/itex]
    Then I moved the denominator over, dl[itex]^{2}[/itex]x[itex]^{2}[/itex](r[itex]^{2}[/itex] - x[itex]^{2}[/itex]) = dy[itex]^{2}[/itex](r[itex]^{2}[/itex] - x[itex]^{2}[/itex])[itex]^{2}[/itex] + x[itex]^{4}[/itex]dx[itex]^{2}[/itex]
    Substituting the dy gives dl[itex]^{2}[/itex]x[itex]^{2}[/itex](r[itex]^{2}[/itex] - x[itex]^{2}[/itex]) = x[itex]^{2}[/itex]dx[itex]^{2}[/itex] + x[itex]^{4}[/itex]dx[itex]^{2}[/itex]
    Dividing the things by the dl gives dl[itex]^{2}[/itex] = [itex]\frac{dx^{2}(1 + x^{2})}{(r^{2} - x^{2})}[/itex] = [itex]\frac{dx^{2}(1 + x^{2})}{r^{2}(1 - x^{2}/r^{2})}[/itex]
    Clearly, the only stuck thing is the [itex]\frac{(1 + x^{1})}{r^{2}}[/itex]. Does this mean anything, like it equals 1 or something, and I've just forgotten an identity? I've also tried a couple of other methods, but I would say this is pretty much what I get to. It may not make perfect sense, but I was running out of ideas, and this was the last thing I tried.
     
  2. jcsd
  3. Oct 5, 2011 #2
    Just look at [5] above and see where you can make the whole thing a single fraction.
     
  4. Oct 5, 2011 #3
    Aha! I got it now!

    I did:
    dl[itex]^{2}[/itex] = [itex]\frac{dx^{2}(r^{2} - x^{2}) + x^{2}dx^{2}}{(r^{2} - x^{2})}[/itex]

    [tex]dl^{2} = \frac{dx^{2}(r^{2} - x^{2} + x^{2})}{(r^{2} - x^{2})}[/tex]

    Canceling out the x[itex]^{2}[/itex]s and dividing r[itex]^{2}[/itex] from top and bottom gets the answer.

    I knew there was an obvious answer, I just couldn't see it. I guess I had a brain fart. Thanks for the help.
     
    Last edited by a moderator: Oct 5, 2011
  5. Oct 5, 2011 #4

    Mark44

    Staff: Mentor

    MrNerd - a tip
    Instead of using a slew of itex /itex pairs of tags, put one itex tag at the beginning of your equation, and a /itex tag at the end of the equation.

    For the 2nd equation above, I used a tex and /tex pair of tags. This makes fractions a little larger and easier to see.
     
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