# Damped motion in a B field

As a follow-up to the previous post, let’s look at how a charged particle in a uniform magnetic field behaves when its motion is damped. Let’s imagine a gluttonous electron (hereafter a glutton) is travelling through a blob of treacle.

Tiny particles in a viscous medium experience drag which is, in magnitude, proportional to the speed at which they travel. The expression for the drag force on the glutton is

$\bold{F}_d=-m\alpha\bold{v}$

where $m$ is the glutton’s mass, $\bold{v}$ its velocity and $\alpha$ is a positive constant. We are free to express the prefactor of velocity in the form $m\alpha$. The exact form of $\alpha$ is not important in this calculation, but we would expect it to depend on certain physical quantities such as the radius of the particle and the viscosity of the medium.

The equation of motion in vector form then reads

$m\bold{\dot{v}}=-m\alpha\bold{v}+q(\bold{v}\times\bold{B})$

Assuming the magnetic field is uniform and parallel to the $z$ axis, the differential equation in the velocity in the $z$ direction is

$m\dot{v}_z=-m\alpha v_z$

The glutton’s motion in $z$ is completely decoupled from its motion parallel to the $xy$ plane. The solution to this equation is

$v_z (t)=v_z (0)e^{-\alpha t}$

So the velocity of the charge in the direction of the magnetic field decays exponentially with time. This is because only the drag force affects the particle’s motion in the $z$ direction; the Lorentz force acts in a direction perpendicular to the magnetic field, so it has no component in $z$.

As before, we will take the initial velocity of the glutton in the $z$ direction to be 0, so that it is 0 for all time. The motion of the charge will hence be constrained to the $xy$ plane.

The differential equations involving $v_x$ and $v_y$ are coupled:

$m\dot{v}_x=-m\alpha v_x+qB v_y$

$m\dot{v}_y=-m\alpha v_y-qBv_x$

The angular frequency $\omega$ defined by

$\displaystyle \omega=\frac{qB}{m}$

has resurfaced, allowing these differential equations to be expressed in the form

$\dot{v}_x=-\alpha v_x+\omega v_y$

$\dot{v}_y=-\alpha v_y-\omega v_x$

This pair of equations has a noticeable symmetry with respect to $x$ and $y$ – we should expect this, since no particular direction in the $xy$ plane is special. Compare this with the previous problem, where the presence of the gravitational field made the ‘downwards’ direction special, and made the two coupled differential equations asymmetric.

As before, we now go about decoupling these equations.

The first equation can be rearranged to give

$v_y=\frac{1}{\omega}(\dot{v}_x+\alpha v_x)$

We substitute this expression into the second equation. Some manipulation shows that

$\ddot{v}_x+2\alpha\dot{v}_x+(\alpha^2+\omega^2)v_x=0$

Let’s try a trial solution of the form

$v_x=be^{\lambda t}$

We can substitute this ansatz into the differential equation and deduce what value(s) $\lambda$ must take in order for the guess to work. Doing so yields

$b (\lambda^2+2\alpha\lambda+(\alpha^2+\omega^2)) e^{\lambda t} = 0$

Since neither $b$ nor $\exp(\lambda t)$ can be 0, we solve the quadratic equation in $\lambda$:

$\lambda^2+2\alpha\lambda+(\alpha^2+\omega^2)=0$

$(\lambda+\alpha)^2+\omega^2=0$

$\lambda=-\alpha\pm i\omega$

We have successfully found two independent solutions for $v_x$:

$v_x=Ce^{(-\alpha+i\omega)t}+De^{(-\alpha-i\omega)t}$

$v_x=e^{-\alpha t}(Ce^{i\omega t}+De^{-i\omega t})$

which can always be arranged into the equivalent form

$v_x=e^{-\alpha t}(P\cos\omega t+Q\sin\omega t)$

To make our solutions as simple as possible, let’s suppose that at time $t=0$, the particle has a component of velocity only in the $y$ direction. This means $v_x (0)=0$, setting

$P=0$

so

$v_x=D\sin\omega t\ e^{-\alpha t}$

Substituting this expression into the equation we found earlier for $v_y$,

$v_y=D\cos\omega t\ e^{-\alpha t}$

Let the launch velocity of the charge $v_y (0)$ be $v_0$, so

$D=v_0$

This gives us the pair of solutions

$v_x (t)=v_0\sin\omega t\ e^{-\alpha t}$

$v_y (t)=v_0\cos\omega t\ e^{-\alpha t}$

What is the charge doing? We could integrate with respect to time to find an explicit expression for its position, but there is no harm in making a prediction from its velocity. Both velocity components oscillate, but also decay with time – this is to be expected because the Lorentz force cannot do work on the charge, so the charge’s energy is inexorably sapped by the damping force. It is also worth noting that the dependence on time of each velocity component is the same, up to a phase shift of $\frac{\pi}{2}$. When one is 0, the other is non-zero.

We’ll now go ahead and integrate these equations to give the dependence of the particle’s position on time. This is of more mathematical interest than physical, but adding another algebraic trick to your arsenal never does any harm!

Consider the general integral

$\displaystyle I=\int e^{i\omega t}e^{-\alpha t}\ dt$

We combine the exponents:

$\displaystyle I=\int e^{(i\omega-\alpha)t}\ dt$

$\displaystyle I=C_0+\frac{e^{(i\omega-\alpha)t}}{i\omega-\alpha}$

where $C_0$ is a complex constant of integration. We want to find the real and imaginary parts of this expression. To do this, we multiply the fraction by the complex conjugate of its denominator, yielding

$\displaystyle I=C_0-\frac{e^{-\alpha t} e^{i\omega t}(\alpha+i\omega)}{\alpha^2+\omega^2}$

which when written out in trigonometric form is

$\displaystyle I=C_0-\frac{e^{-\alpha t}(\alpha\cos\omega t-\omega\sin\omega t+i\omega\cos\omega t+i\alpha\sin\omega t)}{\alpha^2+\omega^2}$

We then separate this into real and imaginary parts:

$\displaystyle \mathrm{Re}[I]=A_0-\frac{\alpha\cos\omega t-\omega\sin\omega t}{\alpha^2+\omega^2}e^{-\alpha t}$

$\displaystyle \mathrm{Im}[I]=B_0-\frac{\omega\cos\omega t+\alpha\sin\omega t}{\alpha^2+\omega^2}e^{-\alpha t}$

where both $A_0$ and $B_0$ are real arbitrary constants of integration.

Now comes the punchline. Consider again the original expression for the integral:

$\displaystyle I=\int e^{i\omega t}e^{-\alpha t}\ dt$

$\displaystyle I=\int (\cos\omega t+i\sin\omega t)e^{-\alpha t}\ dt$

$\displaystyle \mathrm{Re}[I]=\int \cos\omega t e^{-\alpha t}\ dt$

$\displaystyle \mathrm{Im}[I]=\int \sin\omega t e^{-\alpha t}\ dt$

Hence

$\displaystyle \int \cos\omega t e^{-\alpha t}\ dt=A_0-\frac{\alpha\cos\omega t-\omega\sin\omega t}{\alpha^2+\omega^2}e^{-\alpha t}$

$\displaystyle \int \sin\omega t e^{-\alpha t}\ dt=B_0-\frac{\omega\cos\omega t+\alpha\sin\omega t}{\alpha^2+\omega^2}e^{-\alpha t}$

We now know how to integrate the expressions for $v_x$ and $v_y$. It can be assumed without loss of generality that the particle starts at the origin (because the magnetic field is homogeneous). Choosing the constants of integration appropriately, we get

$\displaystyle \bold{r}(t)=\frac{v_0}{\alpha^2+\omega^2} \begin{pmatrix} \omega-(\omega\cos\omega t+\alpha\sin\omega t)e^{-\alpha t} \\ \alpha - (\alpha\cos\omega t-\omega\sin\omega t)e^{-\alpha t} \\ 0 \end{pmatrix}$

These equations are quite opaque unless you already know how this story ends, so here’s an animation showing how the trajectory of the charge changes as the damping term is slowly reduced.

The horizontal axis is $x$ and the vertical $y$. Remember the magnetic field is pointing out of the plane of the screen, and we have taken $q>0$.

For very heavy damping, the glutton moves only a short distance before all of its kinetic energy is dissipated. As the damping is switched off, the charge’s trajectory distends into a spiral. Were we to continue the animation, the path would become circular in the limit that the damping term vanishes.

This result makes good physical sense, and its shape should be familiar to anyone who has seen images of the motion of charged particles in a bubble or cloud chamber. You might like to think about the effects of changing $m$ and $q$, and see if the maths agrees with your intuition.

Since we have gone to the trouble of deriving $\bold{r}(t)$, it is worth exploring a little further its mathematical properties.

First, consider the final resting place of the glutton, given by

$\displaystyle \bold{r}(\infty)=\frac{v_0}{\alpha^2+\omega^2} \begin{pmatrix} \omega \\ \alpha \\ 0 \end{pmatrix}$

This gives us the $x$ and $y$ coordinates of the position at which the charge has stopped moving as a function of the parameter $\alpha$. Let’s take $v_0=1$ without loss of generality. Then

$\displaystyle x=\frac{\omega}{\alpha^2+\omega^2}$

$\displaystyle y=\frac{\alpha}{\alpha^2+\omega^2}$

The curve that this parametrises $(x(\alpha),y(\alpha))$ can be found in Cartesian form by eliminating $\alpha$. Doing so yields

$\displaystyle \bigg(x-\frac{1}{2\omega}\bigg)^2+y^2=\bigg(\frac{1}{2\omega}\bigg)^2$

This is a semicircle with centre $(\frac{1}{2\omega},0)$ and radius $\frac{1}{2\omega}$.  To clarify what is meant by this curve, look at the following animation.

The outermost circle is the path that would be taken by the glutton were its motion undamped. The inner circle is the curve given above, on which every final position of the charge is found. You’ll notice that the inner circle has exactly half the diameter of the outer circle.

This is quite an interesting result but is still not particularly informative. We’ll now go about characterising the curve itself.

It will be easier to translate our coordinate axes such that the origin coincides with the centre of the spiral, that is

$\bold{r}(t)\to\bold{r}(t)-\bold{r}(\infty)$

This removes the constant terms:

$\displaystyle \bold{r}(t)=-\frac{v_0}{\alpha^2+\omega^2} \begin{pmatrix} \omega\cos\omega t+\alpha\sin\omega t \\\alpha\cos\omega t-\omega\sin\omega t \\ 0 \end{pmatrix}e^{-\alpha t}$

This can be written in the form

$\displaystyle \bold{r}(t)=-\frac{v_0}{\alpha^2+\omega^2} \begin{pmatrix} \sqrt{\alpha^2+\omega^2}\cos\big(\omega t-\arctan(\frac{\alpha}{\omega})\big) \\\sqrt{\alpha^2+\omega^2}\cos\big(\omega t-\arctan(-\frac{\omega}{\alpha})\big) \\ 0 \end{pmatrix}e^{-\alpha t}$

These results can be proven algebraically from the compound angle formulae, or from a geometric argument. This form shows the two coordinates vary in almost exactly the same way, but differ by a phase. We should expect this, since we found a similar result for the corresponding velocities.

Further manipulation can be done using the fact that

$\tan\phi=-\cot(\frac{\pi}{2}-\phi)$

Again, you might want to verify this for yourself. This means that if

$\tan\phi=-\frac{\omega}{\alpha}$

then

$\tan(\phi-\frac{\pi}{2})=\frac{\alpha}{\omega}$

On substitution, we find that

$\displaystyle \bold{r}(t)=\frac{v_0}{\sqrt{\alpha^2+\omega^2}} \begin{pmatrix} -\cos(\omega t-\psi)\\\sin(\omega t- \psi) \\ 0 \end{pmatrix}e^{-\alpha t}$

where $\psi=\arctan\big(\frac{\alpha}{\omega}\big)$

We’re almost there. We now define the angle the glutton’s position vector subtends with the $x$ axis through

$\displaystyle \tan\theta = \frac{y}{x}$

By inspection we see this means

$\theta(t)=\psi-\omega t$

Note that the distance of the charge from the origin $|\bold{r}|$ is

$\displaystyle |\bold{r}|=r(t)=\frac{v_0}{\sqrt{\alpha^2+\omega^2}}e^{-\alpha t}$

But from the expression above, we can replace the variable $t$ with the angle $\theta(t)$:

$\displaystyle r(\theta)=\frac{v_0}{\sqrt{\alpha^2+\omega^2}}e^{-\frac{\alpha}{\omega}(\phi-\theta)}$

or, in its most fundamental form

$r(\theta)=ae^{b\theta}$

This is the definition of a logarithmic spiral.

On that note, I’ll draw this long post to a close. The Lorentz force is a subtle one; it has a strange way of perturbing a charge’s path while never actually doing any work itself! In my next post I will probably talk more about the magnetic field, and if possible, I will try to bring in some concepts from electromagnetism.