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Paradox with expanding area

  1. Jul 12, 2013 #1
    Imagine an box with initial dimension [itex] x_0 [/itex] and [itex] y_0 [/itex] with a cavity in the centre with area [itex] R^2 [/itex]. Lets say the box gets stretched in the y-direction but that it's area and the area of the cavity must stay constant.

    [itex] A=x_0y_0-R^2=x(t)y(t)-R^2 [/itex] Differentiating with respect to time gives [itex] 0=\dot{x}y+x\dot{y} \Rightarrow \dot{y}=-\dot{x}\frac{y}{x} [/itex]

    Now let's consider what happens some infinitesimal time [itex] dt [/itex] after the initial setup:

    [itex] x=x_0+\dot{x}(t=0)dt=x_0+\dot{x}_0dt \qquad y=y_0+\dot{y}(t=0)dt=y_0+\dot{y}_0dt[/itex]

    [itex] R^2(t=dt)=xy-A=(x_0+\dot{x}_0dt)(y_0+\dot{y}_0dt)-A=(x_0y_0-A)+(\dot{x_0}y_0+x_0\dot{y_0})+\dot{x_0}\dot{y_0}dt^2 = R^2(t=0)+0-\dot{x_0}^2\frac{y_0}{x_0} [/itex]

    Comparing this with the Taylor series for [itex] R^2 [/itex]:

    [itex] R^2(t=dt)=R^2(t=0)+\dot{R^2}(t=0)dt+\frac{\ddot{R^2}(t=0)}{2!}dt^2+ \cdots [/itex]

    We can see from the coefficients of [itex] dt^2 [/itex]:

    [itex] \ddot{R^2}(t=0)=-2\dot{x_0}^2\frac{y_0}{x_0} [/itex]

    In fact since the starting time is arbitrary and this formula will hold for all [itex] t [/itex] and this clearly contradicts the starting assumption that the area of the cavity stays constant! You could argue that maybe I should include higher order terms in the expansion of [itex] x[/itex] and [itex] y [/itex] before evaluating [itex] R^2 [/itex] and that this may lead to terms that cancel the [itex] dt^2 [/itex] term however doing this would only introduce new terms of order [itex] dt^3 [/itex] or higher.

    Any help would be much appreciated :)
     
  2. jcsd
  3. Jul 12, 2013 #2

    mfb

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    You cannot compare second derivatives in R with linear approximations for the area, that does not give a meaningful result. Include the second order in R^2, and the result should be fine.
     
  4. Jul 12, 2013 #3

    Stephen Tashi

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    [itex] (\dot{x_0}y_0+x_0\dot{y_0}) [/itex] should be [itex] (\dot{x_0}y_0+x_0\dot{y_0}) dt [/itex] (not that it changes the fact the term is zero).

    Are you claiming [itex] R(t=0) = R(t = dt) [/itex] without setting [itex] dt = 0 [/itex]?


    Is this a simpler version of the paradox:

    [itex] f(x) = x^2 [/itex]
    [itex] g(x) = \frac{1}{x^2} [/itex]
    [itex] H(x) = f(x)g(x) = 1 [/itex]

    [itex] x = x_0 + dx [/itex]

    [itex] H(x) \approx (f(x_0) + f'(x_0)dx)\ (g(x_0) + g'(x_0) dx) [/itex]
    [itex] = (x_0^2 + 2x_0dx)(\frac{1}{x_0^2} + \frac{-2}{x_0^3}dx ) [/itex]
    [itex] = 1 - \frac{2}{x_0} dx + \frac{2}{x_0}dx - \frac{4}{x_0^2} dx^2 [/itex]
    [itex] = 1 - \frac{4}{x_0^2} dx^2 [/itex] [eq. 1]

    [itex] H(x) \approx H(x_0) + H'(x_0) dx + \frac{1}{2!} H"(x_0) dx^2 + ...[/itex]
    [itex] = 1 + H'(x_0) dx + \frac{1}{2} H"(x_0) dx^2 + ...[/itex] [eq. 2]

    Comparing eq. 1 & 2 and equating coeficients of powers of [itex] dx [/itex] gives:

    [itex] H'(x_0) = 0 [/itex]
    [itex] H"(x_0) = - \frac{8}{x_0^2} [/itex] which is not constant.
     
  5. Jul 12, 2013 #4

    mfb

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    @Stephen Tashi: I think that should be H(dx).

    We can resolve the issue if we expand f and g up to second order: we get additional ##f(x_0)g''(x_0) dx^2 + f''(x_0)g(x_0) dx^2 = \frac{6}{x^2}dx + \frac{2}{x^2}dx = \frac{8}{x^2}dx##, which cancels the ##\frac{8}{x^2}## calculated in post 3. In a similar way, the missing second derivatives will cancel the term in post 1.
     
  6. Jul 12, 2013 #5
    Ah okay so my mistake was to simply expand x and y only to first order then, for some reason I thought expanding to higher order would only produce terms of higher order but don't worry I'm convinced otherwise now.

    Thank you for the replies!
     
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