Apparent Woodhouse Galilean Maxwell Paradox

Question

I don't understand this Maxwell paradox in Woodhouse's relativity. The first equation implies that there are fields working on the moving charge, but the last equation implies that the magnetic field in question is the one generated by the moving charge, which obviously doesn't work on the charge itself.

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Because I wasn't sure of this example, I've tried to come up with a similar paradox. The set up is the same. Two inertial observers O and O' want to measure the electric field at a point P that has coordinates $\vec{r}$ in O's reference system. Additionally, O knows a priori that there must be some non-magnetic field at the point P because there is some sort of magnetic field generator O knows about. They will use Lorentz's force law to try to do this.

The first test they run goes like this. A particle of charge q follows a trajectory such that it has some velocity $\vec{v}$ relative to O when crossing the point P, while O' also moves with velocity $\vec{v}$ relative to O. $$O: \vec{F}_1=q(\vec{E}(\vec{r})+v \times \vec{B}(\vec{r}))$$ $$O': \vec{F}_2=q \vec{E}'(\vec{r}) $$ Because all inertial observers must measure the same force, $\vec{F}_1=\vec{F}_2$

Now they run another test. A particle of charge q is made to follow a trajcetory such that it has the velocity $\vec{-v}$ relative to O when crossing the point P, while O' also moves with velocity $\vec{-v}$ relative to O. $$O : \vec{F}_1'=q(\vec{E}(\vec{r})-v \times \vec{B}(\vec{r}))$$ $$O': \vec{F}_2'=q \vec{E}'(\vec{r})= \vec{F}_2$$ Because all inertial observers must again measure the same force.

$$\vec{F}_2'=\vec{F}_1'=\vec{F}_2=\vec{F}_1 \Rightarrow q(\vec{E}(\vec{r})+v \times \vec{B}(\vec{r}))=q(\vec{E}(\vec{r})-v \times \vec{B}(\vec{r}))$$ $$v \times \vec{B}(\vec{r})=0 \Rightarrow \vec{B}(\vec{r}) \parallel \vec{v} \vee \vec{B}(\vec{r})=0$$

Because you can repeat this whole experiment with some other velocity $w$ that isn't parallel to v, the only sensible conclusion is that $\vec{B}(\vec{r})=0$ but O knows this to not be the case.

Answer 1

the last equation implies that the magnetic field in question is the one generated by the moving charge

That is incorrect. The charge is at rest for $O'$ and it doesn't generate any magnetic field. What happens is that the field $\mathbf B$ is varying in this reference frame, and it produces a magnetic field due to Faraday's law.

Choose a point $\mathbf r$ in the reference $O$ where we will consider the field $\mathbf B(\mathbf r)$. In reference $O'$, this point is $\mathbf r=\mathbf r'-\mathbf vt$, so we have

$$ \mathbf B'(\mathbf r',t)=\mathbf B(\mathbf r'-\mathbf vt)=\mathbf B(\mathbf r) $$

In reference $O'$ the magnetic field is varying, so we have

$$ \nabla \times \mathbf E' = -\partial_t \mathbf B' = -(\mathbf v\cdot \nabla) \mathbf B = \nabla \times (\mathbf v\times \mathbf B) $$

If we put everything on the same side, we have a field with null rotational. So we can write

$$ \vec E' - \mathbf v\times \mathbf B = -\nabla \Phi' $$

For some scalar potential $\Phi'(\mathbf r')$. We should have $\Phi'=\Phi$, since it only depends on the charge density and it doesn't change for both reference frames. So we have

$$ \mathbf E' = \mathbf E + \mathbf v\times \mathbf B $$