Wikipedia:Reference desk/Archives/Mathematics/2011 June 6

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June 6

Fermat's Principle

Can Fermat's principle be used as a definition of the straight line? 128.232.240.223 (talk) 22:58, 6 June 2011 (UTC)

Plato's definition of a line was basically that which blocks the view between two points, or in a more modern interpretation, the path followed by a ray of light. (See Heath's commentary on Euclid Book I Definition 4.) Since Euclid however this kind of definition is frowned upon because it appeals to physical theory rather than mathematical truth. This is especially true in more recent times, where space itself is considered curved. The modern viewpoint is that a line is an undefined concept whose behavior is determined by the axioms of geometry.--RDBury (talk) 00:22, 7 June 2011 (UTC)

Fermat's principle is just a physics-y way of putting what is more commonly phrased "the shortest distance between two points is a straight line". That holds in Euclidean geometry, due to the triangle inequality. (In some respects, the triangle inequality is the math-y way of putting what is more commonly phrased "the shortest distance between two points is a straight line".) For non-Euclidean geometry, at least for metric spaces, the triangle inequality is a convenient property to have (although not essential). For those metric spaces with the triangle inequality, it's perfectly feasible to use it to define a line-like object, although that might not match the conventional definition of a straight line that others use for that space. -- 174.31.219.218 (talk) 15:57, 7 June 2011 (UTC)

What is the definition of the straight line that does not require discussion of linear equations then? Your article is somewhat unclear. 128.232.240.223 (talk) 19:46, 7 June 2011 (UTC)

A straight line in Euclidean space is an example of a geodesic. — Fly by Night (talk) 20:01, 7 June 2011 (UTC)

I appreciate that yours is a correct answer but I would ask, is it a helpful one? If pressed for the definition of a real number, one might define it as an example of a tensor. I don't think we have learnt anything from this definition though. Is there a definition, which mentions linear equations neither explicitly nor implicitly (unless you regard any two, apparently, distinct definitions of a mathematical object as implicitly mentioning the other by virtue of the fact that they are equivalent definitions), of a more basic nature that is still unambiguous? Or are we in rather murky murky waters here, since the straight line underpins Euclidean geometry, and, as such, is challenging to rigorously define? 128.232.240.223 (talk) 21:25, 7 June 2011 (UTC)

If you're willing to accept the "distance between two points" as a primitive concept, then the definition of a line as a geodesic is a natural one. This is, by definition, the shortest continuous curve between two points. I don't see what your issue is. Sławomir Biały (talk) 22:16, 7 June 2011 (UTC)

"A straight line is the shortest distance between two points" is the colloquial way of putting it. "128.232.240.223", what do you find so unhelpful about that? Michael Hardy (talk) 01:25, 9 June 2011 (UTC)

Surely we are relying on concepts here that themselves depend upon the straight line. I would argue that 'shortest' implies a minimisation, which implies calculus. Elementary calculus necessitates an understanding of the straight line, ie when you let consider the gradient of the straight line passing through two points on a curve and then let one point tend towards the other. Similarly, surely 'distance' can only properly be discussed once we have an understanding of the straight line. I like the OP's point about scalars and tensors. Generally, we proceed from the special case to the general case, eg we start with scalars, move onto vectors, onto matrices, onto tensors, and so it strikes me unhelpful to define a simpler concept in terms of a more complicated one. Or is the question itself just unhelpful; can a concept as fundamental as the straight line ever have an entirely rigorous definition in terms of simpler concepts, since no such concepts exist? meromorphic [talk to me] 16:50, 12 June 2011 (UTC)

As the geodesic article says: "In the presence of an affine connection, geodesics are defined to be curves whose tangent vectors remain parallel if they are transported along it." Don't worry too much about the technical language if it's too much. In Euclidean space (parallel) transport is just translation. So a geodesic in Euclidean space is a smooth curve whose tangent vectors are all parallel (in the ordinary sense of the word). Physically, a tangent vector represents the velocity of an object. So a geodesic in Euclidean space, i.e. a straight line, is given by the path of a particle whose velocity is parallel at each and every moment. — Fly by Night (talk) 19:19, 9 June 2011 (UTC)

You can use an infinitesimal argument to show that this curve is a straight line. We want the velocity vectors to be parallel at each point. Consider a parametrisation: $\scriptstyle \gamma (t):=(x(t),y(t)).$ Consider two nearby velocity vectors, say $\scriptstyle {\dot {\gamma }}(t)$ and $\scriptstyle {\dot {\gamma }}(t+\delta t).$ Notice that

${\dot {\gamma }}(t+\delta t)={\dot {\gamma }}(t)+{\ddot {\gamma }}(t)\,\delta t+O(\delta t^{2})\,.$

If you want $\scriptstyle {\dot {\gamma }}(t)$ and $\scriptstyle {\dot {\gamma }}(t+\delta t)$ to be parallel you need

$\det \left[{\begin{array}{cc}{\dot {x}}(t)&{\dot {y}}(t)\\{\dot {x}}(t)+{\ddot {x}}(t)\,\delta t+O(\delta t^{2})&{\dot {y}}(t)+{\ddot {y}}(t)\,\delta t+O(\delta t^{2})\end{array}}\right]=0\,.$

Expanding we see that we need $\scriptstyle ({\dot {x}}(t){\ddot {y}}(t)\,-\,{\ddot {x}}(t){\dot {y}}(t))\,\delta t\,+\,O(\delta t^{2})\,=\,0.$ For the limiting process we divide by $\delta t$ and then let $\delta t$ tend to zero. This gives $\scriptstyle {\dot {x}}(t){\ddot {y}}(t)\,-\,{\ddot {x}}(t){\dot {y}}(t)=0,$ i.e. the velocity and acceleration are parallel for each t, i.e. the plane curve curvature of γ is zero for all t, i.e. we have a straight line. — Fly by Night (talk) 19:45, 9 June 2011 (UTC)