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My question comes from reading about distances on Wikipedia, and seeing the phrasing Variational formulation of distance. I am not very comfortable with vector notation, so I've rephrased the math slightly (hope it is correct).

Basically, the Euclidean distance between two points on $\mathbb{R}$ $A$ and $B$ can be written in variational form where the distance is the minimum value of an integral:

$$D=\int_0^T\sqrt{\left( \frac{\partial|h^*-h_t|}{\partial t}\right) ^2}dt$$

where $|h^*-h_t|$ is the simple (Euclidian) distance between two points at time $t$. This distance is the minimal value of the integral when $h^*=h_t$.

Questions:

  1. Is my reformulation done correctly?

  2. Why, intuitively, is the above formulation a partial derivative with respect to time?

  3. How would this look if generalised to $\mathbb{R}^n$, still keeping my naïve and space consuming notation?

  4. Is there a simple way to prove that the integral is at its minimum when $h^*=h_t$?

  5. How does this, intuitively, relate to the principle of least action?

This is a crosspost from Math.SE, but I think that the question is closer to physics.

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tmo
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1 Answers1

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  1. No, when adding $h^*(t)$, you are modifying the integrand which is not anymore equal to speed $\times$ time increment, which it should be. Besides, you added an unnecessary $t$ subscript to $h$.
  2. It is wrongly written as partial derivative, there is only one argument in $h$ so you can write as well $\frac{d}{dt}$.
  3. The wikipedia article is already written in $\mathbb{R}^n$ (see arrow on $r$), you wrongly interpreted it as being in $\mathbb{R}$.
  4. the integral (the correct one from the wikipedia article) is at its minimum when $h=h^*$ by the very definition of $h^*$.
  5. It relate to the principle of least action because the varying integral is minimized.
  • Thanks for the answers! Is there a way to formulate the correct integral not using tensor notation? E.g. referring to two points in dimension i. I am interested in both $\mathbb{R}$ and $\mathbb{R}^n$. – tmo Dec 02 '16 at 08:46
  • I see no simpler formulation than the one in the wikipedia article you linked to. The points $\vec{r}(t)$ describe a trajectory in $\mathbb{R}^n$, what's the problem with this notation? You can remove the arrow if you want, but the object remains the same, a vector in $\mathbb{R}^n$. –  Dec 02 '16 at 11:29