Can Area Remain Constant While Dimensions Expand? A Paradox in Geometry

[Chain]

Imagine an box with initial dimension x_0 and y_0 with a cavity in the centre with area R^2. Let's 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 :)

 

[mfb]

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.

 

[Stephen Tashi]

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 <br /> <br />

<br /> <br /> (\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).<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> = R^2(t=0)+0-\dot{x_0}^2\frac{y_0}{x_0} </div> </div> </blockquote>
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&#039;(x_0)dx)\ (g(x_0) + g&#039;(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&#039;(x_0) dx + \frac{1}{2!} H&quot;(x_0) dx^2 + ...
= 1 + H&#039;(x_0) dx + \frac{1}{2} H&quot;(x_0) dx^2 + ... [eq. 2]

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

H&#039;(x_0) = 0
H&quot;(x_0) = - \frac{8}{x_0^2} which is not constant.

 

[mfb]

@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.

 

[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!