Prove a force driven by a cross product between a vector and its velocity gives a spiral movement parallel to the vector

Question

I was given a problem in a Classical Mechanics course that went somehow like the following:

"Consider a particle of mass $m$ moving under the presence of a force $\vec{F} = k\hat{x}\times\vec{v}$, where $\hat{x}$ is an unit vector in the positive direction of the $x$ axis and $k$ any constant. Prove that the movement of a particle under this force is restricted to a circular motion with angular velocity $\vec{\omega}=(k/m)\hat{x}$ or, in a more general case, a spiral movement parallel to the direction of $\hat{x}$."

In an Electrodynamics elementary college course you can see and solve that a magnetic force sort of as:

$$m\ddot{\vec{r}}=\frac{q}{c}\left(\vec{v}\times\vec{B}\right)$$

with a magnetic field, say, $\vec{B}=B_0\hat{z}$ can drive a particle through a spiral movement in the precise direction of that magnetic field you customize, involving a cyclotron frequency and so, if and only if you input further initial conditions to the movement in x, y and z.

My inquiry then is, how can you prove the relation given above for the angular velocity and conclude a spiral movement, from a classical mechanics perspective? I can see there's a link between both procedures, but I cannot try solving the first one without giving a glimpse to the latter.

Answer 1

Write $\vec{v} = \dot{x}\hat{x} + \dot{y}\hat{y} + \dot{z}\hat{z}$, where $\hat{x}, \hat{y}, \hat{z}$ denote unit vectors in the positive direction of the axis $x$, $y$, $z$ respectively. Then,

$$k \hat{x}\times\vec{v} = -k \dot{z}\hat{y} + k\dot{y}\hat{z}.$$

Equating this expression with the analogous for the force, we get a system of three differential equations, which, after dividing by $m$, are as follows:

$$\begin{align*} \ddot{x} &= 0,\\ \ddot{y} &= -\frac{k}{m}\dot{z},\\ \ddot{z} &= \frac{k}{m}\dot{y}. \end{align*}$$

The first equation is readily solved to get the description of a uniform rectilinear movement along the $x$ direction,

$$x = at + b,$$ with $a$, $b$ constants. The remaining pair of equations correspond to the movement of a particle in a central field in the $yz$ plane. To prove that it corresponds to a circular motion, just substitute $v_y = \dot{y}$ and $v_z = \dot{z}$, so that the second and third equations become

$$\begin{align*} \dot{v}_y &= -\frac{k}{m} v_z,\\ \dot{v}_z &= \frac{k}{m} v_y. \end{align*}$$

These equations can be solved to obtain $v_y = A\cos(\frac{k}{m}t + \phi)$, $v_z = A\sin(\frac{k}{m}t + \phi)$. Since $v_y = \dot{y}$, $v_z = \dot{z}$, one last pair of integrals give the final solution for $y$ and $z$:

$$\begin{align*} y &= A\,\frac{m}{k}\,\sin\left(\frac{k}{m} t + \phi\right),\\ z &= -A\,\frac{m}{k}\,\cos\left(\frac{k}{m} t + \phi\right). \end{align*}$$

The equations for $x$, $y$ and $z$ are the parametric equations for a spiral movement, parallel to the direcion of $\hat{x}$.

Answer 2

With an aid of differential geometry, velocity, acceleration and jerk can be written as:

\begin{align*} \mathbf{v} &= \dot{s} \, \mathbf{T} \\ &= v \, \mathbf{T} \\ \mathbf{a} &= \ddot{s} \, \mathbf{T}+ \kappa \, \dot{s}^2\mathbf{N} \\ \mathbf{b} &= (\dddot{s}-\kappa^2 \dot{s}^3) \mathbf{T}+ (3\kappa \dot{s} \ddot{s}+\dot{\kappa} \dot{s}^2) \mathbf{N}+ \kappa \tau \dot{s}^3 \mathbf{B} \\ \end{align*}

Now $$ \mathbf{a}=\boldsymbol{\omega} \times \mathbf{v} \implies \mathbf{a} \perp \mathbf{T} \implies \ddot{s}=0 \implies \dot{s}=\text{constant} \implies v=u$$

and $$ \mathbf{b}= \dot{\mathbf{a}}=\boldsymbol{\omega} \times \mathbf{a} \implies \mathbf{b} \perp \mathbf{a} \implies \mathbf{b} \perp \mathbf{N} \implies \dot{\kappa}\dot{s}^2=0 \implies \kappa=\text{constant}$$

Also, \begin{align*} \int \mathbf{a} \, dt &= \boldsymbol{\omega} \times \int \mathbf{v} \, dt \\ \mathbf{v} &= \mathbf{u}+\boldsymbol{\omega} \times \mathbf{r} \\ \mathbf{r}(0) &= \mathbf{0} \\ \dot{\mathbf{r}}(0) &= \mathbf{u} \\ \boldsymbol{\omega} \cdot \mathbf{v} &= \boldsymbol{\omega} \cdot \mathbf{u} \\ &= \text{constant} \end{align*}

We have

\begin{align*} \mathbf{v} \times \mathbf{a} &= \mathbf{v} \times (\boldsymbol{\omega} \times \mathbf{v}) \\ &= v^2 \boldsymbol{\omega}-(\boldsymbol{\omega} \cdot \mathbf{v})\mathbf{v} \\ \kappa &= \frac{|v^2 \boldsymbol{\omega}-(\boldsymbol{\omega} \cdot \mathbf{v})\mathbf{v}|} {v^3} \\ &= \frac{|\boldsymbol{\omega} \times \mathbf{v}|}{v^2} \\ &= \frac{|\boldsymbol{\omega} \times \mathbf{u}|}{u^2} \\ |\mathbf{a}| &= |\boldsymbol{\omega} \times \mathbf{u}| \\ &= \text{constant} \\ \mathbf{a} \times \mathbf{b} &= a^2 \boldsymbol{\omega}-(\boldsymbol{\omega} \cdot \mathbf{a}) \mathbf{a} \\ &= a^2 \boldsymbol{\omega} \\ \tau &= \frac{\mathbf{v} \cdot a^2\boldsymbol{\omega}} {(\mathbf{v} \times \mathbf{a})^2} \\ &= \frac{\boldsymbol{\omega} \cdot \mathbf{v}}{v^2} \\ &= \frac{\boldsymbol{\omega} \cdot \mathbf{u}}{u^2} \\ &= \text{constant} \end{align*}

Both $\kappa$ and $\tau$ are constants implying the path is helical. If $\mathbf{u} \cdot \boldsymbol{\omega}=0 \implies \tau=0$, then it'll be a circle. While $\mathbf{u} \times \boldsymbol{\omega}=\mathbf{0} \implies \kappa=0$, that'll be a straight line.

Fitting with initial conditions:

$$\fbox{$\quad \mathbf{r}=\mathbf{u}t+ \frac{\mathbf{u} \times \boldsymbol{\omega}}{\omega^2}(\cos \omega t-1)+ \frac{\boldsymbol{\omega} \times (\mathbf{u} \times \boldsymbol{\omega})}{\omega^{3}}(\sin \omega t-\omega t) \quad \\$}$$

Some facts from differential geometry \begin{align*} s &= \int |\mathbf{v}| \, dt \tag{arclength} \\ \dot{s} &= |\mathbf{v}| \tag{speed} \\ &= v \\ \mathbf{T} &= \frac{\mathbf{v}}{v} \tag{tangent vector}\\ \mathbf{B} &= \frac{\mathbf{v} \times \mathbf{a}}{|\mathbf{v} \times \mathbf{a}|} \tag{binormal vector} \\ \mathbf{N} &= \mathbf{B} \times \mathbf{T} \tag{normal vector} \\ \kappa &= \frac{|\mathbf{v} \times \mathbf{a}|}{v^3} \tag{curvature} \\ \tau &= \frac{\mathbf{v} \cdot \mathbf{a} \times \mathbf{b}} {(\mathbf{v} \times \mathbf{a})^2} \tag{torsion} \end{align*}

Answer 3

Take the equations of motion $$ \ddot{\mathbf{r}} = \lambda \;( \hat{k} \times \dot{\mathbf{r}}) $$ and express them in cylindrical coordinates $(r,\theta,z)$

$$\begin{pmatrix} \ddot{r} - r \dot{\theta}^2 \\ r \ddot{\theta} + 2 \dot{r} \dot{\theta} \\ \ddot{z} \end{pmatrix} = \lambda \left( \begin{pmatrix}0\\0\\1 \end{pmatrix} \times \begin{pmatrix} \dot{r} \\ r \dot{\theta} \\ \dot{z} \end{pmatrix} \right) = \lambda \begin{pmatrix} -r \dot\theta \\ \dot{r} \\ 0 \end{pmatrix} $$

which is solved by

$$\begin{align} \ddot{r} & = r \dot{\theta} ( \dot{\theta}-\lambda ) \\ \ddot{\theta} & = \frac{\dot{r} }{r} ( \lambda - 2 \dot{\theta} ) \\ \ddot{z} & =0 \end{align} $$

I see two special cases in the solutions here

• Set $\dot{\theta} = \lambda$ and $\ddot{\theta}=0$ to yield $$ \begin{align} \ddot{r} & = 0 & \dot{r} & = 0 \\ \ddot{z} & = 0 & \dot{z} & = \mbox{const} \end{align} $$

• Set $\dot{\theta} = \frac{\lambda}{2}$ and $\ddot{\theta}=0$ to yield $$ \begin{align} \ddot{r} & = -\frac{\lambda^2}{4} \,r & r & = \mbox{harmonic} \\ \ddot{z} & = 0 & \dot{z} & = \mbox{const} \end{align} $$