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The following link: http://mathworld.wolfram.com/MoebiusStrip.html shows the Möbius strip parametrized as \begin{eqnarray} x = [ R + s \cos \left ( \frac{1}{2} t \right ) ] \cos t \\ y = [ R + s \cos \left ( \frac{1}{2} t \right ) ] \sin t \\ z = s \sin \left ( \frac12 t \right ) \end{eqnarray} The symbols for $R$ and $s$ and angle $t$ are explained there.

Then they say that from this parametrization we can derive the cubic.

\begin{equation} -R^2 y + x^2 y + y^3 - 2 R x z - 2 x^2 z + y z^2 = 0. \end{equation}

Any ideas about how to do this? I have tried with no success.

Thanks.


This is what I have done so far:

Square the first two equations above and add to find \begin{equation} x^2 + y^2 = \left ( R + s \left ( \cos \frac{t}{2} \right ) \right )^2 \end{equation} Take the square root of this \begin{equation} \sqrt{x^2 + y^2 }= R + s \left ( \cos \frac{t}{2} \right ) \Longrightarrow \sqrt{x^2 + y^2 } - R = s \left ( \cos \frac{t}{2} \right ) \end{equation} Square this and the third equation (for $z$) and add to find \begin{equation} s^2 = \left ( \sqrt{x^2+y^2}- R \right )^2 + z^2 \end{equation}

Now, let us divide the second by the first equation

That is \begin{equation} \frac{y}{x} = \tan t = \frac{2 \tan (t/2)}{1 - \tan^2 (t/2)} \end{equation} multiply numerator and denominator by $\cos^2 (t/2)$ \begin{equation} \frac{y}{x} = \frac{2 \sin(t/2) (\sqrt{1-\sin^2(t/2)} }{\cos^2 (t/2) - \sin^2(t/2)} = \frac{2 \sin(t/2) \sqrt{1 - \sin^2(t/2)}}{1 - 2 \sin^2(t/2)}. \end{equation} Multiply numerator and denominator by $s^2$, then \begin{equation} \frac{y}{x} = \frac{2 z \sqrt{s^2 - z^2}}{s^2 - 2 z^2} \end{equation}

That is \begin{equation} \frac{y}{x} = \frac{2 z (\sqrt{x^2 + y^2} - R)}{ ( \sqrt{x^2 + y^2} - R)^2 - z^2} \end{equation} or \begin{equation} \frac{y}{x} = \frac{2 z (\sqrt{x^2 + y^2} - R)}{ (x^2 + y^2) + R^2 - 2 R \sqrt{x^2+y^2} - z^2} \end{equation}

Or \begin{equation} y (x^2 + y^2) + y R^2 - 2 R y \sqrt{x^2+y^2} - z^2 y = 2 z x \sqrt{x^2 + y^2} -2 R x z \end{equation}

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  • $\begingroup$ What you derived so far ? $\endgroup$ – Cardinal Jul 19 '15 at 16:06
  • $\begingroup$ Do you try to put the transformation in $x^2+y^2+z^2+c$ $\endgroup$ – Cardinal Jul 19 '15 at 16:06
  • $\begingroup$ This is my try: $\endgroup$ – Herman Jaramillo Jul 20 '15 at 20:11
  • $\begingroup$ I find it difficult to represent an object that must have autointersection via an algebraic equation. I don't like having to deal with inequalities and cutting off region of the space. Why don't you derive the equations in a 4 dimensional space? They are way more easier. $\endgroup$ – Lolman Aug 14 '15 at 0:46
  • $\begingroup$ Thanks Lolman: You are right, it will selfintersect if you let $s/2 > R$ otherwise it will not. Do you mean 4D and then reduce it to 3D? Somebody derived the equation above and this puzzles me. I started to study Grobner basis. I know these are not polynomials, but they can be converted to rational expressions (quotients of polynomials) and then there is a topic call $\mu$-basis. I hope to find something there. Otherwise I will not feel that I wasted my time because I would learn something new. Do you think that would help? $\endgroup$ – Herman Jaramillo Aug 14 '15 at 15:27
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Consider the Moebius strip with midline the circle of radius $R$ in the $(x,y)$-plane and having width $2a>0$: $$M:\quad(\phi,s)\mapsto\left\{\eqalign{x&=\bigl(R+s\cos{\textstyle{\phi\over2}}\bigr)\cos\phi \cr y&=\bigl(R+s\cos{\textstyle{\phi\over2}}\bigr)\sin\phi \cr z&=s\sin{\textstyle{\phi\over2}}\ .\cr}\right.\tag{1}$$ The parameter domain is $-\pi<\phi<\pi$, $\ -a<s<a$. For "technical reasons" we have excluded $\phi=\pm\pi$. This guarantees $\cos{\phi\over2}>0$ over the whole parameter domain.

It is claimed that $M$ is part of a certain cubic surface $S\subset{\mathbb R}^3$. This surface results from revoking the condition $-a<s<a$ in $(1)$, so that now the parameter $s$ runs from $-\infty$ to $\infty$. Consider a fixed value $\phi\in\ ]{-\pi},\pi[\ $. Then $$g_\phi:\quad s\mapsto\left(\bigl(R+s\cos{\textstyle{\phi\over2}}\bigr)\cos\phi, \ \bigl(R+s\cos{\textstyle{\phi\over2}}\bigr)\sin\phi , \ s\sin{\textstyle{\phi\over2}}\right)\tag{2}$$ describes a straight line lying on $S$. We now replace the parameter $s$ in $(2)$ by the new parameter $u:=R+s\cos{\phi\over2}$. In this way $g_\phi$ appears in the form $$g_\phi:\quad u\mapsto\bigl(u\cos\phi, u\sin\phi, \tan{\textstyle{\phi\over2}}(u-R)\bigr)\qquad(-\infty<u<\infty)\ .\tag{3}$$ When we let $\phi$ vary as well in $(3)$ we obtain another parametrization of our surface $S$. In order to get rid of the trigonometric functions we introduce the new parameter $t:=\tan{\phi\over2}$, which runs from $-\infty$ to $\infty$. In this way we obtain the following rational representation of $S$: $$S:\quad(t,u)\mapsto\left\{\eqalign{x&=u\ {1-t^2\over 1+t^2} \cr y&=u\ {2t\over1+t^2}\cr z&=t\ (u-R)\cr}\right.\qquad\qquad\bigl((t,u)\in{\mathbb R}^2\bigr)\ .$$ One has $2(z+Rt)=2ut=y(1+t^2)$ and $$2tx=y(1-t^2)\ .\tag{4}$$ Adding these two equations leads to $$t={y-z\over R+x}\ .$$ We insert this value of $t$ into $(4)$ and obtain after clearing denominators $$2x(R+x)(y-z)=y\bigl((R+x)^2-(y-z)^2\bigr)\ .$$ This equation is valid for all points $(x,y,z)\in S$, and as well on the line $g_{\pm \pi}$ omitted from consideration. It expands to $$-R^2 y + x^2 y + y^3 - 2 R x z - 2 x^2 z - 2 y^2 z + y z^2=0\ ,$$ as given in the quoted Wikipedia link (in your question you have forgotten one term).

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  • $\begingroup$ Excellent! your answer is correct. I was in the right track when I said that I should convert to a rational parametrization and use $mu$-basis, but I was behind on the theory of $mu$-basis. Thanks. It is weird that you use the notation "a" and then switch to "R". $\endgroup$ – Herman Jaramillo Aug 16 '15 at 18:40
  • $\begingroup$ I see your "a" is not "R" is the width of the Mobius tape..... $\endgroup$ – Herman Jaramillo Aug 16 '15 at 18:42
  • $\begingroup$ Two more things: It was not a "Wikipedia" link it is a "wolfram" link. Yes, I forgot one term. I realized about that a couple of days after I posted the question.. I did not want to correct because I wanted to use this as a way to test those responding to my question. $\endgroup$ – Herman Jaramillo Aug 16 '15 at 18:46
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I'd just substitute the given coordinates into the equation and show that it's satisfied. The terms can be grouped according to the powers of $R$ and $s$ they contain, with the two exponents always summing to $3$, and the equation has to be satisfied for all four groups separately; that makes the calculation more manageable. For example, for $R^3s^0$ I get

$$-\sin t+\cos^2t\sin t+\sin^3t=0\;,$$

which is indeed satisfied.

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  • $\begingroup$ I believe you. Instead of a verification I would like to see a derivation. $\endgroup$ – Herman Jaramillo Jul 21 '15 at 14:17
  • $\begingroup$ @HermanJaramillo: I'm afraid I don't understand that distinction. Could you elaborate? $\endgroup$ – joriki Jul 21 '15 at 15:49
  • $\begingroup$ Verify implies to plug the answer into the equation and get something like 0=0. Derive means that you do not even know the answer and gets to it by using valid identities. Here is an example: Solve $x^2 + x - 1=0$. Yes, the solutions are $x=\frac12 ( \pm \sqrt{5}-1 )$. One thing is to plug this into the equation and get 0=0 and one is to derive this solution by factorization or even using the quadratic formula. How did the first person solved this problem? he did not know the Cartesian equation before hand. $\endgroup$ – Herman Jaramillo Jul 21 '15 at 15:53

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