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.(adsbygoogle = window.adsbygoogle || []).push({});

[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 :)

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# Paradox with expanding area

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