Proof: Force always perpendicular and motion in a plane implies that the trajectory is a circle

Question

I am looking for a proof for a physics problem. Consider a particle which is subject to a force $\vec{F}(t)$ with $|\vec{F}(t)| = \text{const}$ which is always perpendicular to the velocity $\vec{v}(t)$. Further assume that the motion takes places in a plane.

To put it in a mathematical problem:

Let $x\colon \mathbb{R} \to \mathbb{R}^2$ (2 because of the "plane" condition) be smooth.

Suppose $<x''(t),x'(t)>= 0$ for all $t$ and $|x''(t)| = \text{const}$. Then $x(\mathbb{R})$ is a circle. $<\cdot,\cdot>$ denotes the standard scalar product on $\mathbb{R}^2$ and $'$ the derivative.

How to prove this?

Does the theorem remains correct if one drops the assumption $|x''(t)| = \text{const}$?

Note this is not a homework problem, I just want to know how to prove this physical result mathematically. In physics books the statement above is claimed sometimes, but in every instance I have seen they proved just the logically converse statement.

Answer 1

Yes, you do need $|x''|$ to be constant.

Since the acceleration is normal to the velocity, the speed $|x'(t)|$ is constant, say $|x'(t)| = s$. In the plane of motion there are two directions normal to the velocity vector, and if $x''$ is continuous and never $0$ one of these two possible orientations must be chosen for the whole trajectory. Thus we have the differential equation $$x'' = n \times x'$$ where $n$ is a normal vector to the plane (and, in order to have $|x''|$ constant, $n$ is constant). Now motion with a certain constant speed $s$ (which I'll leave to you to calculate) in a circle of radius $r$ will satisfy this differential equation, and we can choose the radius and initial position to match any initial condition $x(0)$, $x'(0)$. By the Existence and Uniqueness Theorem for differential equations, these circular motions give you every possible solution.

If you were allowed to switch suddenly from one $n$ to another, you could have a motion that starts off on one circle and then suddenly transfers to another circle of the same radius tangent to it. But at the instant of transfer, the acceleration would not exist (the one-sided limits of $\dfrac{x'(t) - x'(t_0)}{t - t_0}$ as $t \to t_0+$ and as $t \to t_0-$ would be different).

Answer 2

One easy way to see this is to identify the plane $\Bbb{R^2}$ with $\Bbb{C}$, so that $x(t)$, $x'(t)$ and $x''(t)$ are now complex numbers. The derivative of $t\mapsto |x'(t)|^2$ is $2{\rm Re}(x''(t)\overline{x'(t)})=2\langle x'(t),x''(t)\rangle=0$, so the velocity is constant: $\forall\, t\in\Bbb{R}, |x'(t)|=v>0$ (there nothing to prove if $v=0$). Now, look at the next function $$g:\Bbb{R}\to\Bbb{C}:g(t)=\dfrac{x''(t)}{x'(t)}$$ Since ${\rm Re}(g(t))=\frac{1}{v^2}{\rm Re}(x''(t)\overline{x'(t)})=0$ and $|g(t)|=\frac{a}{v}$, (where by assumption $\forall\, t\in\Bbb{R}, |x''(t)|=a$,) we conclude that the continuous function $g$ takes its values in $\{i\frac{a}{v},-i\frac{a}{v}\}$ and the continuity of $g$ implies that, in this case, it must be constant. That is either $$\forall\, t\in\Bbb{R}, g(t)=i\frac{a}{v}\quad\hbox{or}\quad\forall\, t\in\Bbb{R}, g(t)=-i\frac{a}{v}.$$ So, $x''=\epsilon i\frac{a}{v} x'$ where $\epsilon\in\{+1,-1\}$. This can easily be integrated to get $x'(t)=x'(0)\exp(\epsilon i\frac{a}{v}t)$ and $x(t)=x(0)-\epsilon i\frac{v}{a}x'(0)\exp(\epsilon i\frac{a}{v}t)$. This is clearly the parametric representation of a circle.$\qquad\square$

Answer 3

A charged positive particle enters a region of uniform magnetic field $B \vec{k}$. Initially $\vec v=v_0\vec i$.

At general time $t$, $\vec v=v_1\vec i+v_2\vec j$ and $\vec a=a_1\vec i+a_2\vec j$.

$$\vec F=q.\vec v\times \vec B=qB(-v_1\vec j+v_2\vec i)$$

Also by conservation of energy, $|\vec v|$ remains constant.

$$v_1^2+v_2^2=v_0^2$$

$$a_1=\frac{qBv_2}{m}$$ $$v_1\frac{dv_1}{dx}=\frac{-qB\sqrt{v_0^2-v_1^2}}{m}$$

$$\int_{v_0}^{v_1}\frac{v_1dv_1}{\sqrt{v_0^2-v_1^2}}=\frac{-qBx}{m}$$ $$\sqrt{v_0^2-v_1^2}=\frac{qBx}{m}$$ $$\int\frac{dx}{\sqrt{v_0^2-\frac{q^2B^2x^2}{m^2}}}=t$$

$$x=\frac{mv_0}{qB}\sin{\frac{qBt}{m}}$$

Similarly you will obtain $$y=-\frac{mv_0}{qB}(1-\cos{\frac{qBt}{m}})$$

Which are clearly parametric coordinates of a circle.

Answer 4

Imagine that the motion is in a complex plane, described by position $f \in \mathbb{C}$ and time $t$.

Suppose your motion starts at $f(0) = 1$. Assume your velocity is perpendicular to position and speed is a constant factor of position (ignoring unit conversions for simplicity):

$$f'(t) = i\,f(t)$$

(Remember that when you multiply by $i$ in a complex plane, it rotates a vector by 90 degrees).

So what function is described by $f(0) = 1$ and $f'(t) = i\,f(t)$?

$$f(t) = e^{it} = \cos(t) + i\sin(t)$$

which is the motion of a circle. On the other hand, if you only assume velocity proportional to position, but make no assumption about speed by introducing arbitrary function $g(t)$:

$$f'(t) = i\,g(t)\,f(t)$$

solving:

$$f(t) = e^{i\int\,g(t)dt} = \cos\left(\int\,g(t)dt\right) + i\,\sin\left(\int\,g(t)dt\right)$$

which you can use to make many kinds of shapes.