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This is more of a physics-y question, but it's not the physics I care about here -- just the mathematics. In my physics class we learned a particle with constant velocity in a uniform magnetic field perpendicular to it causes it to rotated in a circle of constant radius. The professor told me how to "prove this", in essence solving the following:

$$q \vec{v} \times \vec{B} = m \frac{d\vec{v}}{dt}$$ with $\vec{v}=\langle v_x, v_y, v_z \rangle$ and $\vec{B}=\langle 0,0,B \rangle$ with $B\in\mathbb R_{>0}$, and it worked out great! Got the equation for a circle.

He then told me that if there was an electric field perpendicular to both the moving particle and field, I should get a cycloid from solving this (with $\vec{E}= \langle 0,E,0 \rangle$):

$$q \vec{v} \times \vec{B} + q \vec{E} = m \frac{d\vec{v}}{dt}$$

And, well, I didn't get a cycloid as an answer, but another circle. Here's my attempt at this differential equation, does anyone see anything wrong?


To clarify to those who haven't taken Electromagnetism, $q \in \mathbb R$ is charge, $m,B,E \in \mathbb R_{>0}$, and $v_x,v_y,v_z: \mathbb R \to \mathbb R$.

So let $\vec{v}=\langle v_x, v_y, v_z \rangle$, $\vec{B}=\langle 0,0,B \rangle$, and $\vec{E}=\langle 0,E,0 \rangle$. Then

$\begin{align*} &\ q \vec{v} \times \vec{B} + q \vec{E} = m \frac{d\vec{v}}{dt} \\ &\implies \langle qv_yB, -qv_xB+qE \rangle = \langle m \frac{dv_x}{dt}, m \frac{dv_y}{dt} \rangle \\ &\implies \begin{cases} \frac{dv_x}{dt} = \frac{q}{m} v_y B \\ \frac{dv_y}{dt}=-\frac{q}{m} v_xB+\frac{q}{m}E \end{cases} \\ &\implies \frac{d^2v_x}{dt^2} = \frac{qB}{m} \frac{dv_y}{dt} \\ &\implies \frac{dv_y}{dt} = \frac{m}{qB}\frac{d^2v_x}{dt^2} \\ &\implies \frac{m}{qB} \frac{d^2v_x}{dt^2}=-\frac{qB}{m}v_x+\frac{qE}{m} \\ &\implies \frac{d^2 v_x}{dt^2}=-\frac{q^2B^2}{m^2}v_x+\frac{q^2BE}{m^2} \\ &\implies \frac{d^2 v_x}{dt^2} + \frac{q^2B^2}{m^2}v_x = \frac{q^2BE}{m^2} \end{align*}$

Now the auxiliary equation of the $LHS$ is $$g^2+\frac{q^2B^2}{m^2}=0\implies g=\pm \frac{qB}{m} i$$ so we get $(v_x)_c=c_1 \cos\left(\frac{qB}{m}t\right)+c_2\sin\left(\frac{qB}{m}t\right)$. And

$$RHS=\frac{q^2BE}{m^2} \implies \begin{cases} (v_x)_p = A \\ (v_x)_p' = 0 \\ (v_x)_p''=0\end{cases} \implies (0)+\frac{q^2B^2}{m^2}(A) = \frac{q^2BE}{m^2} \implies A = \frac{E}{B}$$

Thus,

$$v_x = (v_x)_c+(v_x)_p = c_1 \cos\left(\frac{qB}{m}t\right)+c_2\sin\left(\frac{qB}{m}t\right) + \frac{E}{B}.$$

Now back from the top, we knew

$$\frac{dv_x}{dt} = \frac{qB}{m} v_y \implies v_y = \frac{m}{qB} \frac{dv_x}{dt}$$

and

$$\frac{dv_x}{dt} = \frac{c_2qB}{m} \cos \left(\frac{qB}{m}t\right)-\frac{c_1qB}{m} \sin \left(\frac{qB}{m}t\right)$$

so we get

$$v_y = c_2 \cos \left(\frac{qB}{m}t\right)- c_1 \sin \left(\frac{qB}{m}t\right)$$

and in totality

$$\vec{v} = \left[c_1 \cos\left(\frac{qB}{m}t\right)+c_2\sin\left(\frac{qB}{m}t\right) + \frac{E}{B} \right] \hat{i} + \left[ c_2 \cos \left(\frac{qB}{m}t\right)- c_1 \sin \left(\frac{qB}{m}t\right) \right] \hat{j}.$$

This doesn't even look remotely close to an equation of a cycloid. Does someone know where my mistake may be?

Edit: Upon integrating, I arrived at $$\vec{x}= \left[ \frac{c_1m}{qB} \sin \left( \frac{qB}{m}t \right) - \frac{c_2m}{qB} \cos \left( \frac{qB}{m}t \right) + \frac{E}{B}t \right] \hat{i} + \left[ \frac{c_2 m}{qB} \sin \left( \frac{qB}{m}t \right) + \frac{c_1 m}{qB} \cos \left( \frac{qB}{m}t \right) \right] \hat{j}$$ which sadly still doesn't seem to be a cycloid on Wolfram after attempting multiple different values for the constants. Am I missing something?

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    $\begingroup$ Your forget to integrate $\vec{v}$ over $t$ to get $\vec{x}$. $\endgroup$ – achille hui Sep 30 '17 at 4:01
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    $\begingroup$ $$g^2+\frac{q^2B^2}{m^2}=0\implies g=\pm \frac{qB}{m} i$$ $\endgroup$ – Nosrati Sep 30 '17 at 4:04
  • $\begingroup$ @MyGlasses Thank you so much for pointing that out! It was surprising how forgetting to simply get rid of some squares seemingly made the rest of work look so messy. $\endgroup$ – Andrew Tawfeek Sep 30 '17 at 4:50
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    $\begingroup$ @achillehui I felt silly reading your comment! Can't believe I was wondering "How in the world does this represent moving like a cycloid?" while staring at it's velocity and not position, haha! $\endgroup$ – Andrew Tawfeek Sep 30 '17 at 4:51
  • $\begingroup$ @AndrewTawfeek Did you use anywhere that velocity is constant? $\endgroup$ – Nosrati Sep 30 '17 at 5:33
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$$q \pmb{v} \times \pmb{B} + q \pmb{E} = m \frac{\mathrm d\pmb{v}}{\mathrm dt}\tag 1$$ If $\pmb{B}=B\hat z$ an $\pmb E=E\hat y$, we have $$\left\{ \begin{align} \dot v_x&=\omega v_y\\ \dot v_y&=-\omega v_x+\gamma \end{align}\right.\tag 2 $$ with $\omega=\frac{qB}{m}$ (cyclotron frequency) and $\gamma=\frac{qE}{m}$. The system (2) can be written as

$$\left\{ \begin{align} \ddot v_x+\omega^2 v_x&=\omega\gamma\\ \ddot v_y+\omega^2 v_y&=0 \end{align}\right.\tag 3 $$ and the solution might be written as $$\left\{ \begin{align} v_x(t)&=a\cos(\omega t+\phi)+\tfrac{E}{B}\\ v_y(t)&=b\cos(\omega t+\theta) \end{align}\right.\tag 4 $$ and finally $$\left\{ \begin{align} x(t)&=\frac{a}{\omega}\sin(\omega t+\phi)+\tfrac{E}{B}t\\ y(t)&=\frac{b}{\omega}\sin(\omega t+\theta) \end{align}\right.\tag 5 $$ Assuming for simplicity that initially $(t = 0)$ the particle is at rest at the origin of a Cartesian coordinate system, we have $$\left\{ \begin{align} v_x(0)=0&=a\cos(\phi)+\tfrac{E}{B}\\ x(0)=0&=\frac{a}{\omega}\sin(\phi) \end{align}\right. $$ $$ \phi=k\pi, \, k\in\Bbb Z\quad\text{and}\quad a=-\tfrac{E}{B} $$ that is $$ x(t)=\frac{E}{B}\left[t-\frac{1}{\omega}\sin(\omega t)\right]\tag 6 $$ Analogously we find $$ y(t)=\frac{E}{B}\frac{1}{\omega}\left[1-\cos(\omega t)\right]\tag 7 $$ Putting $\omega t=\psi$ and $\frac{E}{B\omega}=\rho$ we have $$\left\{ \begin{align} x&=\rho\left[\psi-\sin(\psi)\right]\\ y&=\rho\left[1-\cos(\psi)\right] \end{align}\right.\tag 8 $$ which gives us the parametric equations of a cycloid.

Observe that from (6) and (7) we have $$ (x-\rho \omega t)^2+(y-\rho)^2=\rho^2\tag 9 $$ that is a circle of radius $\rho$ whose center $C:=(\rho \omega t,\rho)$ travels in the $x$ direction at a constant speed $$v_c=\omega \rho=\frac{E}{B}$$ The particle moves as though it were a spot on the reem of a wheel, rolling down the $x$ axis at speed $v_c$. And the curve generated this way is a cycloid.

Notice that the overall motion is not in the direction of $\pmb E$, as one might suppose, but perpendicular to it.

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You just missed one more step. You need to find the $x$ and $y$ coordinates. Just integrate $\vec{v}$ with respect to time. The cycloid contains the sine and cosine terms, plus in the $\hat{i}$ direction you have a term proportional to time: $$\frac{E}{B}t$$

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  • $\begingroup$ Thank you, I felt silly knowing now I was staring at velocity and not position this entire time! I'll try integrating! Also, could you clarify how I have $\frac{E}{B}t$ and not $\frac{E}{B}$ when $(v_x)_p=A$ (a constant) satisfies the particular solution? $\endgroup$ – Andrew Tawfeek Sep 30 '17 at 4:52
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    $\begingroup$ The velocity has the $E/B$ factor, which is a constant. The position is then linear in $t$ $\endgroup$ – Andrei Sep 30 '17 at 13:55
  • $\begingroup$ Ohh, sorry for the misunderstanding -- thought you meant that for the velocity not the position! $\endgroup$ – Andrew Tawfeek Sep 30 '17 at 14:43
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As others have already discussed your solution's math, I'm just going to answer the "cycloid" part. It is possible your professor was speaking in a not very strict (mathematically) sense.

The solution is not a cycloid in general. It might be for particular values of $\bf E$, $\bf B$, and the initial velocity of the particle. In fact, it is often shown in physics textbooks that when the electric field is small compared to the magnetic (as in the electrical component of the force is much smaller in magnitude to the magnetic) the particle will perform an almost-circular motion in which the centre of the circle slowly drifts in the direction of ${\bf E}\times \bf B$ (this drift velocity is basically the $\frac{E}{B}$ part of your $v_x$).

EDIT: (I had the wrong initial velocity at first) You will get a cycloid with $c_1 = -E/B$ and $c_2 = 0$. This means the initial velocity has to be $v_{0x} = v_{0y} = 0$. Conversely, if the particle starts at rest, the trajectory will be a cycloid, since the equations are deterministic. Other values for initial velocity may also work (essentially equivalent to choosing another moment in time as "initial"). For example, ${\bf v}_0 = \langle{\frac{2E}{B}, 0}\rangle$.

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  • $\begingroup$ This makes it all clear, thank you very much! $\endgroup$ – Andrew Tawfeek Sep 30 '17 at 14:27

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