I have designed a simple and qualitative thought experiment through which I believe that I have encountered an inconsistency in the relativistic electromagnetism. A point charge $+q$, with respect to the lab observer $A$, is left to fall into a uniform electric field ${E^\prime}_{y^\prime}$ which is surrounded by some walls. It is clear that the charge accelerates along $y^\prime$ without being deflected in $x^\prime$ and $z^\prime$, and thus the charge will never hit the walls which have been erected parallel to the electric field as viewed by the lab observer $A$. [See Figure 1-(a)]

Now consider the scenario from the standpoint of the observer $B$ who moves at $u$ perpendicular to the electric field direction. According to the Lorentz transformation for EM fields, the electric field is increased to $E_y=\gamma_u {E^\prime}_{y^\prime}$ along $y$, and besides, a magnetic field of $B_z=-\gamma_u (u/c^2){E^\prime}_{y^\prime}$ is indeed produced along $-z$. [See Figure 1-(b)]

The weird point is that $B$ measures a resultant velocity $w$ for the charge which complies with the relativistic velocity addition as:

$$w=\sqrt{(\alpha_uv)^2+u^2},$$ where $\alpha_u=1/\gamma_u=\sqrt{1-u^2/c^2}$, and $v$ is the instantaneous velocity of the charge as measured by $A$ when the charge reaches an altitude of $\Delta y^{\prime}$. Observer $B$ asserts that the charge moves at $w$ in a magnetic field $B_z$. Therefore, a Lorentz force of $F_B=qwB_z$ is exerted on the charge, which tends to accelerate the charge along $-x$ as well as $-y$. The horizontal component of this force pushes the charge towards the right wall till they make a contact with each other. (Paradox)

Where did I go wrong?

Before answering this question please remember that:

1- It is assumed that the radiation of the accelerated charge is negligible, or assume that the charge moves at a constant velocity through a fluid as reaching a terminal velocity of $v$.

2- The experiment is carried out away from any gravitational field.

For further discussion, see this link, please.

[Unfortunately, the question (post) at the aforementioned link has been deleted by some moderators! For further study and clarification, see this article.]

Accelerated charge paradox

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    $\begingroup$ ` relativistic electromagnetism`??? $\endgroup$ Commented Aug 26, 2019 at 8:04
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    $\begingroup$ @BЈовић: You're right. Relativity just presented another interpretation for electromagnetism. Therefore, electromagnetism, per se, is compatible with this relativistic interpretation, and thus there is no such a thing as relativistic electromagnetism. By using the phrase relativistic electromagnetism, indeed, I just wanted to emphasize those perfect solutions to the classical electromagnetism equations that are applicable for velocities close to that of light without neglecting any parameter, say the factor $\gamma$, which sometimes happens when velocities are far smaller than light speed $\endgroup$ Commented Aug 26, 2019 at 10:11

3 Answers 3


Everything you've said is perfectly correct. However, there is no contradiction because in special relativity, force is not parallel to acceleration.

To see this formally, note that in special relativity, the definition of force is $\mathbf{F} = d\mathbf{p}/dt$, and the momentum is $\mathbf{p} = \gamma m \mathbf{v}$. Combining these and assuming constant mass, we have $$\mathbf{F} = m \frac{d}{dt}(\gamma \mathbf{v}).$$ This is not necessarily parallel to $\mathbf{a} = d\mathbf{v}/dt$, because $\gamma$ can change in time as well.

You might protest that this makes no sense, and we should instead define force as $\mathbf{F}' = m \mathbf{a}$. Well, you can define the word "force" however you want, but only for the first definition is it true that $$\mathbf{F} = q (\mathbf{E} + \mathbf{v} \times \mathbf{B}).$$ If you use the second definition, this isn't true. You can eventually get the same answer, but only after much more work, which is precisely why the definition $\mathbf{F} = d\mathbf{p}/dt$ is more common.

In your case, the horizontal velocity of the charge should remain constant, while the vertical velocity increases. In order to do this, we need a nonzero horizontal force $$F_x = m v_x \frac{d\gamma}{dt}.$$ With a little more work, you can confirm that this $F_x$ is precisely the horizontal component of the magnetic force you found.

Incidentally, the fact that force isn't parallel to acceleration is one of the reasons the idea of "relativistic mass" is not useful. It was invented to try to recover Newtonian intuition, but it can't get the right acceleration unless you postulate two relativistic masses, one "tranverse" to the direction of motion and one "longitudinal". At that point it's more confusing than it's worth.

  • $\begingroup$ Thanks for your rapid answer. However, as I mentioned in my post, $\gamma$ can be assumed to be constant as the experiment is performed inside a fluid. Moreover, the acceleration vector can eventually show the real direction of the charge motion rather than the force vector. As far as I remember, the ratio of the force components to the corresponding relativistic mass (either transverse or longitudinal) can easily show the final direction of the acceleration as is traditionally explained in right-angle lever paradox. ... $\endgroup$ Commented Aug 25, 2019 at 21:43
  • $\begingroup$ ... As you mentioned, the acceleration direction is not necessarily parallel to that of the force, but I am seriously doubtful that there is no horizontal component of acceleration. $\endgroup$ Commented Aug 25, 2019 at 21:44
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    $\begingroup$ @MohammadJavanshiry If you're working in a fluid, you also have to Lorentz transform the fluid's force, which should now have a horizontal component cancelling the magnetic force. $\endgroup$
    – knzhou
    Commented Aug 25, 2019 at 21:47
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    $\begingroup$ @MohammadJavanshiry I've never heard of the right-angle lever paradox -- unfortunately the statement that $F = ma$ is not correct in relativity, even if $m$ is the relativistic mass. That is, unless that paradox is using a different definition of "force" -- in which case it does not apply because you aren't using that definition here. $\endgroup$
    – knzhou
    Commented Aug 25, 2019 at 21:49
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    $\begingroup$ @MohammadJavanshiry It indeed can and does move at a constant horizontal terminal velocity. That constant velocity is equal to the horizontal velocity of the walls. $\endgroup$
    – knzhou
    Commented Aug 26, 2019 at 22:23

I have encountered an inconsistency in the relativistic electromagnetism

That this scenario poses no paradox becomes very easy to see if you use the usual mathematical notation appropriate for relativistic electromagnetism. In the wall frame the four-force on the charge is given by: $$f_{\mu}=q F_{\mu \nu} U^{\nu}$$ where $f$ is the four force, $q$ is the charge, $F$ is the electromagnetic field tensor, and $U$ is the four velocity. The Lorentz transform from one frame to the other is $x'^{\mu'}=\Lambda_{\mu}^{\mu'}x^{\mu}$ or $x'_{\mu'}=\Lambda_{\mu'}^{\mu}x_{\mu}$ where $x$ is the four-vector or co-vector to be transformed, $\Lambda$ is the Lorentz transform, and the primes represent quantities in the primed frame. So $$f'_{\mu'}=\Lambda_{\mu'}^{\mu}f_{\mu}=\Lambda_{\mu'}^{\mu}\Lambda_{\nu'}^{\nu}\Lambda_{\nu}^{\nu'}f_{\mu}$$ $$=\Lambda_{\mu'}^{\mu}\Lambda_{\nu'}^{\nu}\Lambda_{\nu}^{\nu'}(q F_{\mu\nu}U^{\nu})$$ $$=q(\Lambda_{\mu'}^{\mu}\Lambda_{\nu'}^{\nu}F_{\mu\nu})(\Lambda_{\nu}^{\nu'} U^{\nu})$$ $$=q F'_{\mu' \nu'} U'^{\nu'}$$

So in all frames the transform of the force is exactly what you would expect from transforming the fields and the velocity. It is consistent in all frames, not a paradox. Note that this derivation is not limited to a specific configuration, but applies for all configurations. Whether you include a fluid or not, or how you angle the fields does not matter, it is all guaranteed to be consistent.

Now that we can see that there is no inconsistency/paradox we can look at the root of the problem you encountered. The problem is, as @knzhou mentioned, you have misunderstood the way forces and accelerations transform in relativity. In relativity we have something that looks like Newton's 2nd law $f_{\mu}=m A_{\mu}$ but here $A$ is the four-acceleration, not the three-acceleration. In terms of the three velocity $u$ and the three acceleration $a$ the four-acceleration is given by $(\gamma \dot\gamma c,\gamma^2 a + \gamma \dot\gamma u)$ so $a$ is not generally parallel to the spatial part of $A$.

Also, since the charge is moving at terminal velocity in the unprimed frame then the net force $\Sigma f_{\mu}=0$ so then $\Sigma f’_{\mu’}=\Lambda_{\mu’}^{\mu}\Sigma f_{\mu}=\Lambda_{\mu’}^{\mu} 0=0$


There are a lot of approaches to solve this problem. I appreciate @knzhou answer because it explains everything that I was going to say, however, it doesn't answer the question directly and because of that I will answer this question as well. @Dale 's answer, on the other hand is nice, because it's completely general and one can use it whenever it's needed, but in my humble opinion, it's killing a bird with a nuke or something. So I will introduce two easy ways for this problem in frame work of special relativity.

Note that $v$ is velocity of particle w.r.t moving frame, and $u$ is velocity of moving frame w.r.t rest frame.

The first one is to compute acceleration on x axis directly. We can use this equation: $$a=\frac{1}{\gamma_v m_0}(F-\frac{(v.F)v}{c^2})$$ I will skip its derivation, because it's easy. Check wikipedia for more information. All thing you need to do is going backward from last equation to the first one. Accelration on x axis is: $$a_x=\frac{1}{\gamma_v m_0}(F_x-\frac{(v.F)v_x}{c^2})$$ We will compute $v.F$ soon enough, but before that we should find $F$

$$F=qE_y+qv\times B=qE\hat{j}+qv_yB_z\hat{i}-qv_xB_z\hat{j}$$

So we have: $$v.F=qv_yE + qv_x v_yB_z-qv_x v_yB_z=qv_yE$$ Easy right? We substitute this to equation of acceleration:

$$a_x=\frac{1}{\gamma_v m_0}(qv_yB_z-\frac{qv_yEv_x}{c^2})$$ But we know that $B_z=-\gamma_u \frac{uE_y'}{c^2}$ and $E_y=\gamma_uE_y'$ So: $$a_x=\frac{1}{\gamma_v m_0}(qv_y(-\gamma_u \frac{uE_y'}{c^2})-\frac{qv_y(\gamma_uE_y')v_x}{c^2})$$ Well we do know that $v_x=-u$ because after all, our observer consider himself at rest and relate his velocity to particle and so $$a_x=0$$ No paradox.

The second one however, is even easier! All thing we need to do is using lorentz transformation for acceleration (Actually, this is Dale's answer, but without tensor and four vectors). $$a_x=\frac{a_x'}{\gamma_u^3(1+\frac{v_x' u}{c^2})^3}=0,$$

because $a_x'=0$ obviously.


I replaced lorentz transformation for force with acceleration. ofcourse, $f_x$ is not zero, it's $qv_yB_z$


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