Update 3_drude_model.md - typo fixes

src/3_drude_model.md

@@ -153,20 +153,20 @@ m\mathbf{v}(t + dt) = 0
 $$
 The rest of the electrons $(1 - dt/τ)$ are accelerated by the Lorentz force $F$, so their velocity becomes
 $$
-m\mathbf{v}(t + dt) = m\mathbf{v}(t) + F⋅dt.
+m\mathbf{v}(t + dt) = m\mathbf{v}(t) + \mathbf{F}⋅dt.
 $$
 To find the average velocity, we take a weighted average of these two groups of particles:
 $$
 \begin{aligned}
 m\mathbf{v}(t+dt)  &= [m\mathbf{v}(t) + F dt]\left(1 - \frac{dt}{\tau}\right) + 0⋅\frac{dt}{\tau} \\
-  &= [m\mathbf{v}(t) + F dt] \left(1 - \frac{dt}{\tau}\right) \\
-                  &= m\mathbf{v}(t) + dt [F - m\frac{\mathbf{v(t)}}{\tau}] - \frac{F}{\tau} dt^2
+  &= [m\mathbf{v}(t) + \mathbf{F} dt] \left(1 - \frac{dt}{\tau}\right) \\
+                  &= m\mathbf{v}(t) + dt [\mathbf{F} - m\frac{\mathbf{v(t)}}{\tau}] - \frac{\mathbf{F}}{\tau} dt^2
 \end{aligned}
 $$
 We now neglect the term proportional to $dt²$ (it vanishes when $dt → 0$).
 Finally, we recognize that $\left[\mathbf{v}(t+dt) - \mathbf{v}(t)\right]/dt = d\mathbf{v}(t)/dt$, which results in
 $$
-m\frac{d\mathbf{v}}{dt} = -m\frac{\mathbf{v}}{τ} + F.
+m\frac{d\mathbf{v}}{dt} = -m\frac{\mathbf{v}}{τ} + \mathbf{F}.
 $$
 Observe that the first term on the right-hand side has the same form as a drag force: it always decelerates the electrons.