# Force on a moving charge in 2 frames

The following simple problem seems to lead to a contradiction when analyzed in different frames.

Consider an infinitely long stationary wire with positive line charge density $$\lambda_0$$ and negative charge density $$-\lambda_0$$. In its rest frame, the wire is electrically neutral.

Now consider driving a current so that the negative charges move to the right with uniform speed $$v$$ while the positive charges stay fixed. At a distance $$y$$ from the wire, a positive charge $$q$$ is moving to the right with the same speed $$v$$ parallel to the wire.

In the lab frame, the wire now is negatively charged, because the motion of the negative charges leads to a Lorentz contraction, making net charge density $$(1-\gamma)\lambda_0$$. In addition, the moving charges also create a current, with $$I=-\gamma \lambda_0 v$$, which creates a magnetic field. Therefore, by the right hand rule, the magnetic force on the positive charge points away from the wire and has magnitude $$F_B=qvB= \frac{\mu_0 q\gamma\lambda_0 v^2}{2\pi y}$$. In addition, the contribution from the electric field points in the oppositive direction, and has magnitude $$F_E=qE=\frac{q(\gamma-1)\lambda_0}{2\pi\epsilon_0 y}$$. The total force, analyzed in the lab frame, is thus $$F=\frac{\mu_0 q\gamma\lambda_0 v^2}{2\pi y}-\frac{q(\gamma-1)\lambda_0}{2\pi\epsilon_0 y}=\frac{\lambda_0 q}{2\pi\epsilon_0 y}(\epsilon_0\mu_0v^2 \gamma -\gamma+1) =\frac{\lambda_0 q}{2\pi\epsilon_0 y}(1-\frac{1}{\gamma})$$, using $$\epsilon_0\mu_0=1/c^2$$ and $$1/\gamma=\sqrt{1-v^2/c^2}$$. Since $$1-\frac{1}{\gamma} >0$$, the charge will be repelled from the wire.

However, in the frame of the moving charge, there is no magnetic force since the charge itself is stationary in its own frame. As for the electric field, in this frame the negative charges in the wire are stationary (because they move at the same speed $$v$$ in the lab frame), whereas the positive charges moves at speed $$-v$$. This means that viewed in this frame, the wire has a net positive charge $$\lambda=(\gamma-1)\lambda_0$$, which results in a force $$F'=\frac{(\gamma-1)\lambda_0 q}{2\pi\epsilon_0 y}$$ on the charge.

Clearly, the expression for $$F$$ does not agree with $$F'$$ ($$F'=\gamma F$$), which leads to different physical pictures. What is wrong here? (I'm guessing that F should also transform under a change of frame?)

Let's begin by noting the transformation equations for force. We define 3-force $$\bf f$$ by its relation to 3-momentum: $${\bf f} = \frac{d \bf p}{d t}$$ where $$t$$ is the time in some inertial frame, and $$\bf p$$ the momentum of a body on which $$\bf f$$ is the net force acting. Consider such a force $$\bf f$$ acting on a body whose velocity is $$\bf u$$ relative to frame S. Then the force as observed in a frame S' having velocity $$\bf v$$ relative to S is $${\bf f}'_\parallel = \frac{{\bf f}_\parallel - ({\bf v}/c^2) dE/dt} {1 - {\bf u} \cdot {\bf v}/c^2}, \;\;\; {\bf f}'_\perp = \frac{{\bf f}_\perp} {\gamma_v (1 - {\bf u} \cdot {\bf v}/c^2)}$$ where $$E$$ is the energy (in S) of the body acted upon and the equations refer to the components parallel and perpendicular to $$\bf v$$.

If the rest mass of the body acted upon is not changing, then $$dE/dt = {\bf f} \cdot {\bf u}$$ so then one has

$${\bf f}'_\parallel = \frac{{\bf f}_\parallel - {\bf v}({\bf f} \cdot {\bf u})/c^2} {1 - {\bf u} \cdot {\bf v}/c^2}$$

There are two easy-to-remember cases:

1. If $$\bf u$$, $$\bf v$$ and $$\bf f$$ are all aligned then $${\bf f}'_\perp = {\bf f}_\perp = 0$$ and $${\bf f}'_\parallel = {\bf f}_\parallel$$ so $${\bf f}' = {\bf f}$$.

2. If $$\bf v$$ is perpendicular to $$\bf u$$, or if $${\bf u}=0$$, then $${\bf f}'_\perp = {\bf f}_\perp / \gamma_v$$. Thus a transverse force gets weaker when observed in a frame which is moving relative to the body on which the force is acting. Equally, the transverse force is larger (by a factor $$\gamma$$) in the rest frame of the body on which the force acts (where by 'transverse' we mean 'orthogonal to the direction of relative motion of the frames').

In your example it is your primed frame which is the rest frame of the body on which the force you want to calculate is acting, so this is the frame where the force is larger. So your result is consistent with the general formula, so on this evidence your result is correct. The answer to your final question, 'what is wrong here?' is: nothing is wrong (except perhaps our intuition when we are not familiar with this aspect of relativity).