1. Jul 12, 2013

### Chain

Imagine an box with initial dimension $x_0$ and $y_0$ with a cavity in the centre with area $R^2$. Lets say the box gets stretched in the y-direction but that it's area and the area of the cavity must stay constant.

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

Now let's consider what happens some infinitesimal time $dt$ after the initial setup:

$x=x_0+\dot{x}(t=0)dt=x_0+\dot{x}_0dt \qquad y=y_0+\dot{y}(t=0)dt=y_0+\dot{y}_0dt$

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

Comparing this with the Taylor series for $R^2$:

$R^2(t=dt)=R^2(t=0)+\dot{R^2}(t=0)dt+\frac{\ddot{R^2}(t=0)}{2!}dt^2+ \cdots$

We can see from the coefficients of $dt^2$:

$\ddot{R^2}(t=0)=-2\dot{x_0}^2\frac{y_0}{x_0}$

In fact since the starting time is arbitrary and this formula will hold for all $t$ 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 $x$ and $y$ before evaluating $R^2$ and that this may lead to terms that cancel the $dt^2$ term however doing this would only introduce new terms of order $dt^3$ or higher.

Any help would be much appreciated :)

2. Jul 12, 2013

### Staff: Mentor

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.

3. Jul 12, 2013

### Stephen Tashi

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

Are you claiming $R(t=0) = R(t = dt)$ without setting $dt = 0$?

Is this a simpler version of the paradox:

$f(x) = x^2$
$g(x) = \frac{1}{x^2}$
$H(x) = f(x)g(x) = 1$

$x = x_0 + dx$

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

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

Comparing eq. 1 & 2 and equating coeficients of powers of $dx$ gives:

$H'(x_0) = 0$
$H"(x_0) = - \frac{8}{x_0^2}$ which is not constant.

4. Jul 12, 2013

### Staff: Mentor

@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.

5. Jul 12, 2013

### Chain

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!