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Once upon a time I was travelling by train and noticed an intresting optic effect I started to think about in terms of math. Let's consider two examples of curves: 1)The curve defined by the differential equation in polar coordinates $r'(t)^2 + r(t)^2 = t^2$ with $r(0)=0$, $r"(0)>0$

It is easy to see that the junction point of the curve and the ray uniformly rotated in the origin coordinates moves uniformly accelerated.

2) The curve defined by differential equation

$y'=\sqrt{A^2x^2-1}$, where $A$ - const >0; the junction point of the curve and the ray uniformly moved along abscissa axis moves unifromly accelarated -

enter image description here

3) let me explain how I see the 'uniformly accelerated surface": the curve of intersection of the two surfaces - the 'uniformly accelerated surface" and a predefined surface which is moved uniformly moves uniformly accelerated. For eg. let's extend eg 2 above: a plane moves along abscissa axis and crosses the uniformly accelerated surface according to the lines of the example above.

One may to create similar examples where we get the unifromly accelerated curves. My question what is the simplest way to describe all kind of curves ( or even surfaces)?

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I don't know what you might mean by 'uniformly accelerated surface', but I think that, by 'uniformly accelerated curve', you mean a curve in the plane parametrized in such a way that its velocity at time $t$ is a(n inhomogeneous) linear function of $t$, i.e., in Cartesian coordinates, it would be a curve $\bigl(x(t),y(t)\bigr)$ that satisfies an equation of the form $$ x'(t)^2 + y'(t)^2 = (at+b)^2, $$ where $a$ and $b$ are constants (not both zero).

This is a classical problem in differential geometry, and it can be solved by the method of Monge, which reduces it to the integration of an Engel system in dimension $4$. Explicitly, the method goes like this: The curves you seek, when 'graphed' in $\mathbb{R}^3$ in the form $\bigl(x(t),y(t),t\bigr)$ are null curves for the homogeneous Monge equation $$ \mathrm{d}x^2+\mathrm{d}y^2-(at+b)^2\mathrm{d}t^2 = 0. $$ Introducing the coordinate $z = \tfrac12a\,t^2 + bt$, one has $\mathrm{d}z = (at{+}b)\,\mathrm{d}t$, so this can be written in more standard form as $$ \mathrm{d}x^2+\mathrm{d}y^2-\mathrm{d}z^2 = 0. $$ Introduce a parameter $u$ and write the equation on $\mathbb{R}^4$ as the pair of Pfaffian equations $$ \mathrm{d}x - \cos u\ \mathrm{d}z = \mathrm{d}y - \sin u\ \mathrm{d}z = 0. $$ Now, this is a system of Engel type on $\mathbb{R}^4$ and so can be integrated by a change of variables: Setting $$ p = x\ \cos u + y\ \sin u - z,\quad q = -x\ \sin u + y\ \cos u, \quad r = -x\ \cos u - y\ \sin u $$ the Pfaffian system in these coordinates takes the standard Engel form $$ \mathrm{d}p - q\ \mathrm{d}u = \mathrm{d}q - r\ \mathrm{d}u = 0. $$ Thus, on any integral curve in $\mathbb{R}^4$ of this system on which $\mathrm{d}u$ is nonvanishing, there must exist a function $f$ such that $$ \begin{aligned} x\ \cos u + y\ \sin u - z = p\ &= f(u)\\ -x\ \sin u + y\ \cos u = q\ &= f'(u)\\ -x\ \cos u - y\ \sin u = r\ &= f''(u) \end{aligned} $$ Solving these equations for $x$, $y$ and $z = \tfrac12a\,t^2 + bt$ now gives a parametric formula for 'uniformly accelerated curves' in terms of one arbitrary function $f$ of the parametrizing function $u$.
$$ \begin{aligned} x\ &= -\sin u\ f'(u)-\cos u\ f''(u)\\ y\ &= \phantom{-}\cos u\ f'(u) - \sin u\ f''(u)\\ \tfrac12a\,t^2 + bt = z\ &= -f(u) - f''(u) \end{aligned} $$

The curves for which this formula does not work are the ones for which $u$ is constant, but these are just the straight lines in the plane, parametrized so that the velocity changes linearly in time, and these are easily written down.

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  • $\begingroup$ Dear Robert, Thank you, let me explain how I see the 'uniformly accelerated surface": the crossing line of the 'uniformly accelerated surface" and a predefined surface moves uniformly accelerated. For eg. let's extend eg 2 of my questions: a plane moves along abscissa axis and crosses the uniformly accelerated surface according to the lines of the example above. I hope my explanation is clear ( I could draw a picture). $\endgroup$ – Mikhail Gaichenkov Aug 2 '14 at 17:13
  • $\begingroup$ @MikhailGaichenkov: I don't understand what you mean by 'crossing line'. Do you mean the curve of intersection of the two surfaces? If so, I don't see how you are getting a parametrization of the intersection curve in order to speak of it being 'uniformly accelerated'. Finally, do you agree, in the case of curves, that you don't need to speak of moving lines or rays, only the velocity of a parametrized curve? I assumed that this was so in my answer, but you didn't confirm this. If the parametrization of the curve is all that matters to you, could you confirm this? $\endgroup$ – Robert Bryant Aug 3 '14 at 8:46
  • $\begingroup$ I added more comments in eg 2 to see if I understand you correctly. Could you look at notes in eg. 2 please. If I confirm, can you show how we get the curve of eg. 2 from your answer? ( I would postpone "surface" question untill confirmation with curves). $\endgroup$ – Mikhail Gaichenkov Aug 3 '14 at 10:24
  • $\begingroup$ @MikhailGaichenkov: It seems that, in your Example 2, you are claiming that the curve $$\bigl(x(t),y(t)\bigr) = (\cosh(t){-}1, \ \tfrac14\sinh(2t){-}\tfrac12t)$$ satisfies $x'(t)^2+y'(t)^2 = t^2$ (since you were solving with $A=1$ and $B=0$), but this is obviously not true since $x'(t) = \sinh(t) > t$ when $t>0$, so we must have $x'(t)^2+y'(t)^2 > t^2$. Can you check your calculations? I'm afraid that you are using the variable $t$ in two different senses, and this might have confused you. $\endgroup$ – Robert Bryant Aug 3 '14 at 12:42
  • $\begingroup$ I added apicture to show initial 'physical' effect - the picture explains why x=t. I double checked my calculations, but still cannot find an error. $\endgroup$ – Mikhail Gaichenkov Aug 3 '14 at 14:16

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