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 $t=0$.

$\hskip2in$ enter image description here

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.

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    $\begingroup$ 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. $\endgroup$ Dec 23, 2015 at 12:14
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    $\begingroup$ You have the differential equation of motion in the OP. Solve that equation in the general form, then plug in the initial value. $\endgroup$
    – LLlAMnYP
    Dec 23, 2015 at 12:14
  • $\begingroup$ @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!!! $\endgroup$ Dec 23, 2015 at 13:45
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    $\begingroup$ 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). $\endgroup$
    – LLlAMnYP
    Dec 23, 2015 at 15:09
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    $\begingroup$ 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. $\endgroup$ Jan 4, 2016 at 13:59

1 Answer 1


The general solution of motion in crossed EM fields (for the equation in the OP, that is) is:

$$ \begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix} = \begin{pmatrix} C_1 \cos (\frac{q B t}{m} - \alpha) \\ C_2 \\ C_1 \sin (\frac{q B t}{m} - \alpha) + \frac{E}{B} \end{pmatrix}$$

where $\alpha$, $C_1$, $C_2$ are arbitrary constants. I leave it to the reader to realize what happens at certain initial conditions.

PS Of course, I assume E along x and B along y here.

  • $\begingroup$ Useful (and provable even by me) formula. Thank you very much! $\endgroup$ Dec 23, 2015 at 13:49

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