Distance becoming equal to displacement

Question

Consider a charged particle of charge q and mass m being projected from the origin with a velocity u in a region of uniform magnetic field $\mathbf{B} = - B \hat{\mathbf{k}} $ with a resistive force being present given by F = kv. Finally, it comes to rest. Find its dispalcement.

Now, I had a problem with the computations of this. Basically, I approached this in 2 different ways, in one where I found distance, and in the other, I found displacement, yet both answers seemed to match.

METHOD 1:

$$ \mathbf{F} = -v (qB \hat{\mathbf{r}} + k \hat{\boldsymbol{\theta}} )$$ $$\implies m \frac{|\mathbf{dv}|}{dt} = |\mathbf{v}|\sqrt{q^2B^2 + k^2}$$ $$\implies |\mathbf{v}|\frac{|\mathbf{dv}|}{|\mathbf{ds}|} = \frac{|\mathbf{v}|}{m}\sqrt{q^2B^2 + k^2}$$ Simplifying and integrating, we get the required value of distance as the integral of $|\bf{ds}|$ is the distance traversed.

Thus,

$$\Delta S = \frac{um}{\sqrt{q^2B^2 + k^2}}$$

METHOD 2:

Writing the velocity and acceleration in cartesian coordinates, we get,

$$\mathbf{F} = qB(v_x \hat{\mathbf{j}} - v_y \hat{\mathbf{i}}) - k(v_x \hat{\mathbf{i}} + v_y \hat{\mathbf{j}})$$ Thus, $$a_x = \frac{-1}{m}(qBv_y + kv_x)$$ $$a_y = \frac{1}{m}(qBv_x - kv_y)$$

Hence, we get

$$dv_x = \frac{-1}{m}(qB(dy) + k(dx))$$ $$dv_y = \frac{1}{m}(qB(dx) - k(dy))$$

Integrating these 2, we get 2 linear equations for $x$ and $y$, which upon solving gives us the value of the final $x$ and $y$ coordinates.

Thus, finally we get

$$x = \frac{umk}{q^2B^2 + k^2}$$ $$y = \frac{umqB}{q^2B^2 + k^2}$$

Squaring and adding and then taking the square-root, we get, $$\Delta S = \frac{um}{\sqrt{q^2B^2 + k^2}}$$

Now, I feel like the 2nd expression is for displacement and not distance.

Thus, my question is, what am I doing wrong? I don't find any logical flaws in either argument. However, distance clearly will not be equal to displacement. What is the error here?

Answer 1

Method 1 is not correct. You have to be careful with the notation. What does it mean $|\mathbf{dv}|$ and $|\mathbf{ds}|$? I will perform the calculations avoiding this notation. First, we take the norm of $\mathbf{F} = m \frac{d\mathbf v}{dt}$, which yields $$ m\left| \frac{d \mathbf{v}}{dt} \right| = |\mathbf v|\sqrt{q^2B^2 + k^2} $$ Now we use that $\frac{d\mathbf v}{dt} = \frac{d\mathbf v}{ds} \frac{ds}{dt} = \frac{d\mathbf v}{ds} \mathbf v$, where we took the derivative with respect to the length of the trajectory, to obtain $$ |\mathbf v|\left| \frac{d \mathbf{v}}{ds} \right| = \frac{|\mathbf v|}{m}\sqrt{q^2B^2 + k^2}. $$ Assuming $|\mathbf v| \neq 0$ we can simplify this to $$ \left| \frac{d \mathbf{v}}{ds} \right| = \frac{1}{m}\sqrt{q^2B^2 + k^2}. $$ Now, we integrate with respect to $s$ along the entire trajectory, that is, $$ \int_0^{\Delta S}\left| \frac{d \mathbf{v}}{ds} \right| ds = \frac{\Delta S}{m}\sqrt{q^2B^2 + k^2}. $$

At this moment is where I assume the error happened conceptually. Except for some special cases, the derivative of the norm is not equal to the norm of the derivative, i.e., you cannot say that $$ \left| \frac{d \mathbf v}{ds} \right| = \frac{d|\mathbf v|}{ds}. $$ Thus, you cannot conclude that $$ \int_0^{\Delta S}\left| \frac{d \mathbf{v}}{ds} \right| ds = \left|\int_0^{\Delta S} \frac{d \mathbf{|v|}}{ds} \right| = u, $$ from which you got that $\Delta S = \frac{um}{\sqrt{q^2B^2 + k^2}} $.