# Velocity of particle constant under null initial magnetic force

I have got a doubt on the exact application of Newton's first, and third, law.

Let us for example suppose that a particle, electrically charged with charge $q$, is moving at the instant $t=0$ with velocity $\mathbf{v}(0)$ through a uniform magnetic field $\mathbf{B}$ and a uniform electric field $\mathbf{E}$, such that they are orthogonal to each other, as shown in the figure below. The force acting on the particle is $\mathbf{F}(t)=q\mathbf{E}+q\mathbf{v}(t)\times \mathbf{B}$ and, if $\mathbf{v}(0)=\|\mathbf{B}\|^{-2}\mathbf{E}\times \mathbf{B}$, we have $\mathbf{F}(0)=\mathbf{0}$ and therefore the acceleration is $\mathbf{a}(0)=\mathbf{0}$ at $t=0$.

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I intuitively am sure that, if $\mathbf{F}(0)=\mathbf{0}$, in this case nothing modifies the velocity and therefore, for all $t$, $\mathbf{v}(t)\equiv\mathbf{v}(0)$.

Nevetheless, I have got a mathematical problem. Of course, at any time $\tilde{t}$, if $\mathbf{v}(\tilde{t})=\|\mathbf{B}\|^{-2}\mathbf{E}\times \mathbf{B}$, we have $\mathbf{a}(\tilde{t})=\mathbf{0}$, but how can we mathematically be sure that, if $\mathbf{v}(0)=\|\mathbf{B}\|^{-2}\mathbf{E}\times \mathbf{B}$ (and therefore $\mathbf{a}(0)=\mathbf{0}$) the velocity $\mathbf{v}(\tilde{t})$ still is $\mathbf{v}(0)$ if $\tilde{t}>0$ belongs to a neighbourhood of $0$? In fact, if $\mathbf{a}(0)=\mathbf{0}$, the usual continuity assumptions do not guarantee at all that $\mathbf{a}(t)=\mathbf{0}$ in a neighbourhood of $0$ and, while velocity is "istantly constant" at $t=0$ in the sense that $\mathbf{a}(0)=\mathbf{0}$, I am not sure how we can mathematically prove that it does not change. I heartily thank you for any answer.

• You have a guess for the trajectory, right? Just substitute it into the equation of motion and see whether it satisfies it. If it does, there's nothing else to say. Dec 23, 2015 at 12:14
• You have the differential equation of motion in the OP. Solve that equation in the general form, then plug in the initial value. Dec 23, 2015 at 12:14
• @EmilioPisanty The equation(s) of motion determine one and only one solution, which, therefore, is the constant $\mathbf{v}\equiv\mathbf{v}(0)$. Thank you so much!!! Dec 23, 2015 at 13:45
• As an aside, you may notice that $F(v)=0$ has multiple solutions (as F is independent of the y-component/magnetic-field-parallel-component of v). Dec 23, 2015 at 15:09
• I should also suggest changing the title to something that will be easier to find for others with a similar question in the future. I'm not really sure what that title would be, though. Jan 4, 2016 at 13:59

$$\begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix} = \begin{pmatrix} C_1 \cos (\frac{q B t}{m} - \alpha) \\ C_2 \\ C_1 \sin (\frac{q B t}{m} - \alpha) + \frac{E}{B} \end{pmatrix}$$
where $\alpha$, $C_1$, $C_2$ are arbitrary constants. I leave it to the reader to realize what happens at certain initial conditions.