enter image description here

The other day i was watching an episode of Tom and Jerry in which a similar situation was present toms head comes out of his own mouth. My head hurts when i think how is that even possible so i googled couldn't find any good answer, all the answers were vague without any proof. So i'm asking here.

Consider the torus as shown in the image eating itself. Now...

a)Assuming the torus to be finite length (When it is not eating itself). A mathematical proof for how far it can go.

Note: Intuitively i think it will devour itself until the hole in the torus reduces to a zero value.

b) What if we consider the entire surface of the torus at its hole can pass through without any physical boundaries. Then how far would it devour itself ? will it disappear into a point ? or will it eat itself forever without any end.

c) Now what will the characteristics be if both ends of this torus would increase its length infinity.

please try not to give any answers based on logical reasoning without a mathematical proof.

  • $\begingroup$ Unfortunately, the question is also vague. I think I can make most things rigorous except what you mean by "how far it can go" in part (a). $\endgroup$ – Eric Stucky Nov 28 '14 at 6:01

I know that the OP probably isn't on this site anymore but I'm going to answer anyway.

This is actually a fairly simple geometry question, no differential topology needed.*

The figure shown can be described as 'an ordinary torus whose minor radius is variable'. As per definition of 'torus', the torus will intersect itself when the minor radius exceeds the major radius - all that is left is to define the minor radius (the major radius $R$ may be arbitrary).

Consider the plane of the major radius. WLOG, let $(0,R)$ be the 'starting point' of the torus in polar coordinates and $f(\theta)=r:\theta\in[0,\infty)$ be the minor radius, with $f(0)=0$. If there exists points $\theta_1$ and $\theta_2$, where $\theta_2\geq\theta_1+2\pi n:n\in\mathbb{Z}$, then the torus will intersect itself. Hence, the minor radius must increase with each 'pass' (I will leave the proof of this to others).

In the image shown, the minor radius increases as a linear function of $\theta$ - in general, this need not be the case; however, assuming otherwise makes the answer arbitrary, since the minor radius can then asymptotically approach the major radius. Hence, let $f(\theta)=m\theta$ for some real number $m$.

The maximum value of $\theta$ (i.e. 'how far it can go') is then the solution for $\theta$ to the equation $R=f(\theta)$.

$$R=f(\theta)\implies R=m\theta\implies\frac{R}{m}=\theta$$

This should answer (a). As for (b), the answer is in the question - permitting self intersection removes the restriction on the minor radius, the result being that the minor radius continues to grow indefinitely and the surface intersects itself.

(c) is slightly ambiguous, since the context would suggest that 'characteristic' is meant to be vernacular, but it could also refer to Euler characteristic. In the case of the latter, it is $0$ (since the figure hommotopy equivalent to a cylinder).

In the vernacular sense, expanding the domain of $\theta$ to $(-\infty,\infty)$ does not result in anything spectacular, it just leads to more intersections. Here is a section through the plane of the major radius:

enter image description here

I think the confusion arises from trying to force the question into a particular formal context without first having a clear idea of what information you're after.

It's better to focus on the substance of the question rather than the format (you can always go back and make the question more rigorous later); take the time to find the right question and the answer will come naturally. If done right, it should take less time to answer the question than it should to ask it.

*At least, I assume this is a geometry question. Considered from a purely topological standpoint the figure shown is just a plane (or a cylinder, if I interpreted (c) correctly).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.