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I E cross B field, with E>B

  1. Feb 26, 2016 #1
    I have been unable to find a satisfactory explanation of this problem, elsewhere.

    Consider an uniform electric field, E, along the y axis. Consider also a uniform magnetic field, B, along the z axis. If we release a particle (charge=q, mass=m) at rest on the origin at time t=0, what will be its dynamics in the case that E>B (in cgs units)?

    The manner in which my classmates and I solved the problem was to consider a reference frame boosted along the x axis by speed v=cB/E. Looking at how this transforms the fields, we see that the magnetic field goes to zero, and so in the moving frame, the particle accelerates along the y axis due to the electric field, only. Transforming back to the LAB frame, we see that the particle motion remains solely in the y direction.

    This seems to be in violation of the Lorentz force equation, which dictates that a charged particle moving in the y direction in the presence of a magnetic field in the z direction should experience a non-zero force in the x direction (note: this is the scenario in the lab frame, which is where the contradiction seems to exist, not in the boosted frame, which seems self-consistent).

    Does anyone understand the resolution of this problem? I am hesitant to conclude that special relativity is wrong.
  2. jcsd
  3. Feb 26, 2016 #2


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    This is not correct. If relativistic effects are important, the acceleration will generally not be parallel to the force.
  4. Mar 3, 2016 #3
    My point here is that in the boosted frame, there is only an electric field. I am only claiming that the boosted motion should follow the boosted electric field (which it indeed must). I understand your concern that transforming back to the lab frame may not, generally, keep acceleration in this same direction. However, this is the confusing aspect of the problem. When you work out the Lorentz transformation of the acceleration in the boosted frame back to the lab frame, it remains purely in the y-direction. This does not have to happen in all cases, but doing the math for this problem, we see that it does.

    Just to reiterate, within an inertial reference frame, force and acceleration must be parallel. Transforming between a boosted and lab frame may not keep boosted and lab accelerations parallel, but in this special case it does (don't take my word for it, see for yourself, the calculation is trivial).

    I suspect that the issue with my analysis does lie in transforming back to the lab frame from the boosted frame. However, I am not going to feel satisfied until I understand the explicit, mathematical nature of this error, and the true dynamics of such a system (since motion purely along the y-axis clearly cannot be the true dynamics).
  5. Mar 3, 2016 #4


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    Not true except in certain special cases. Assuming m is constant: $$\vec{F}=\frac{d}{dt}\vec{p}= \frac{d}{dt}(\gamma m\vec{v})= m\frac{d\gamma}{dt}\vec{v}+m\gamma\vec{a}$$Unless ##\vec{v}## and ##\vec{a}## are parallel, or ##|\vec{v}|=0##, (edit: or ##d\gamma/dt=0##) force and acceleration are not parallel.
    Last edited: Mar 3, 2016
  6. Mar 4, 2016 #5


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    No, this is just wrong.

    In the beginning of the motion the force and acceleration in the boosted frame are parallel because acceleration is perpendicular to the motion. However, as soon as you acquire a velocity component in the electric field direction acceleration will no longer be parallel to the force, resulting in a non-zero acceleration in the x-direction.
  7. Mar 4, 2016 #6


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    Perhaps one should start from scratch to clarify things. In the original frame (let's call it "lab frame" for definiteness) you have a homogeneous electromagnetic field with ##\vec{E}=E \vec{e}_y## and ##\vec{B}=B \vec{e}_z##. Further you have also given that ##E>B## (fortunately you seem to work in sensible units, i.e., Gaussian or Heaviside-Lorentz, which doesn't make a difference here anyway).

    Now you have to remember that the electromagnetic field can be characterized by two Lorentz scalars, which are given in the here more convenient (3+1) notation by ##\vec{E}^2-\vec{B}^2## and ##\vec{E} \cdot \vec{B}##. Since ##\vec{E}^2-\vec{B}^2>0## and ##\vec{E} \cdot \vec{B}=0## in your case indeed you can aim at Lorentz boosting to a reference frame, where ##\vec{B}'=0##, and this is indeed achieved by a boost in ##x## direction. The general transformation of the electromagnetic field in the (3+1) notation is
    $$\vec{E}'=\gamma [\vec{E}+\vec{\beta} \times \vec{B}]-\frac{\gamma^2}{1+\gamma} \vec{\beta}(\vec{\beta} \cdot \vec{E})\\
    \vec{B}'=\gamma[\vec{B}-\vec{\beta} \times \vec{E}]-\frac{\gamma^2}{1+\gamma} \vec{\beta}(\vec{\beta} \cdot \vec{B}).$$
    Now since we want to make ##\vec{B}'=0## indeed we should have ##\vec{\beta} \perp \vec{B}##. Since ##\vec{B}=B \vec{e}_z## and ##\vec{E}=E \vec{e}_y## we should thus indeed have ##\vec{\beta}=\beta \vec{e}_x##. Then
    $$\vec{B}'=\gamma (B-\beta E) \vec{e}_x \stackrel{!}{=}0 \; \Rightarrow \beta=\frac{B}{E},$$
    which makes sense since then ##|\beta|<1##.

    Now you can solve the equation of motion in the frame ##\Sigma'##, where you have only an electric field, leading to "hyperbolic motion". You find the explicit solution in my relativity FAQ (p. 30)


    Then you boost back to get the solution in the lab frame. The boost matrix for the space-time coordinates is
    $$\Lambda^{-1} = \begin{pmatrix} \gamma & +\gamma \vec{\beta}^{\mathrm{T}} \\
    +\gamma \vec{\beta} & (\gamma-1) \vec{\beta} \vec{\beta}^{\mathrm{T}}+\mathbb{1}_3 \end{pmatrix}.$$
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