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Consider a cylindrical symmetrical infinite wire with radius $R$. The wire has constant charge density $\rho$ and current density $\vec j$ (pointing to the right). This is measured from the stationary frame $S$.

We let the wire move at a uniform speed $v$ to the right; label the moving frame $\bar S$. The observations are made from the stationary frame $S$.

a) Is it possible to find a frame in which the electric field is zero?

b) Is it possible to find a frame in which the magnetic field is zero?

Here I am quite confused.

I know that the set of transformation rules for $E$ and $B$ fields are:

$$\bar{E}_x = E_x$$

$$\bar{E}_y = \gamma(E_y - vB_z)$$

$$\bar{E}_z = \gamma(E_z + vB_y)$$

$$\bar{B}_x = B_x$$

$$\bar{B}_y = \gamma(B_y + \frac{v}{c^2}E_z)$$

$$\bar{B}_z = \gamma(B_z + \frac{v}{c^2}E_y)$$

I have studied the two special cases:

1) If $B = 0$ one gets

$$\bar {B} = -\frac{1}{c^2}(v \times \bar {E})$$

2) If $E = 0$ one gets:

$$\bar {E} = v \times \bar {B}$$

And the two relativistic invariant operations: ($E \cdot B$), $(E^2 - c^2B^2)$

Neither $E$ nor $B$ are zero in $S$ frame. The idea I have is that, as $E$ points radially outwards and B is circumferential then:

$$E \cdot B = 0$$

But I don't see how that information can reveal if it is possible to find frames in which the electric field or/and magnetic field are zero.

For instance I know that if we were to have $E ≠ 0$ and $B = 0$ (not our case of course) in $S$, then it wouldn't be possible to find another frame where $E = 0$ because $E^2 − c^2B^2$ is invariant and it has to be always positive in this scenario.

But here I just know that $E \neq 0$ and $B \neq 0$ in $S$. How can we argue whether we can find another frame in which $E=0$ or/and $B=0$?

PS: Please avoid using Jefimenko equations if possible and explain your argument using just the set of transformation rules and the invariant equations provided above.

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  • $\begingroup$ Hint: there is another Lorentz invariant other than just $E^2-B^2$. $\endgroup$ – jacob1729 Jun 13 at 10:32
  • $\begingroup$ @jacob1729 Do you mean there's a third Lorentz invariant? $\endgroup$ – JD_PM Jun 13 at 11:15
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    $\begingroup$ Oh didn't see that you've mentioned the $E\cdot B$ invariant. These are the only two invariants, so if you can find another field configuration with the same values then it is reachable via a boost. Finding that boost is not hard either. $\endgroup$ – jacob1729 Jun 13 at 11:32
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    $\begingroup$ Don’t know if it has already been mentioned, but if E and B are non-zero in a certain reference frame, you can’t find another frame where BOTH are zero because of the covariance of the electromagnetic tensor $F_{\mu \nu}$ $\endgroup$ – Francesco Arnaudo Jun 13 at 13:52
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In your setup $\vec{E}\cdot \vec{B} = 0$ everywhere. The only two (linearly independent) Lorentz invariants of the EM field are $\vec{E}\cdot \vec{B}$ and $E^2-B^2$. Thus there must be a boost such that you can make $\vec{E}=\vec{0}$ at any given point. In this case it's actually globally true as can be seen by explicitly constructing the boost:

Suppose we are in the case $B^2-E^2>0$ for concreteness. $\vec{E},\vec{B}$ are orthogonal to each other and also to the direction of the wire. Consider a boost along the wire with velocity $\vec{v}$. Since the components of the field along a boost remain unchanged they stay zero. The perpendicular components then transform as:

$$\vec{E}' = \gamma ( \vec{E}+\vec{v}\times\vec{B}) $$ $$\vec{B}' = \gamma ( \vec{B} - \vec{v}\times \vec{E})$$

Since all three things are perpendicular the cross products reduce to a product of magnitudes:

$$E' = \gamma (E + vB) $$ $$B' = \gamma (B - vE) $$

So we see that to make $E'=0$ we require $v=-E/B$ which is possible in the case $B^2-E^2>0$. Conversely we cannot make $B'=0$ since that would require a boost by $v=B/E$ which is greater than the speed of light. Note that this analysis so far has all been discussing the fields at a point, using the fact $\vec{E}\cdot\vec{B}=0$ at that point. But in fact for the setup given, the fields are orthogonal everywhere and since both fall off as $\frac{1}{r}$ they have the same ratio everywhere too. Thus one boost can set the smaller field equal to zero everywhere.

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According to the moving frame, i.e. the wire itself, the equations for $\bar{E}$ and $\bar{B}$ outside the wire (for simplicity) are as follows (with cylindrical coordinate systems with $+\hat{z}$ being towards right):

$$\bar{E} = \frac{\rho R^2}{2\epsilon_0r} \hat{r},\ \ \bar{B} = \frac{\mu_0jR^2}{2r} \hat{\phi}.$$

Using your results,

1) If $B=0$ $$\bar{B} = \frac{\mu_0jR^2}{2r} \hat{\phi} = -\frac{1}{c^2}(v \times \bar{E}) = -\frac{v}{c^2}\frac{\rho R^2}{2\epsilon_0r}\hat\phi$$ $$\epsilon_0\mu_0c^2j=-v\rho\ \therefore\ v=-\frac{j}{\rho}$$

2) If $E=0$ $$\bar{E} = \frac{\rho R^2}{2\epsilon_0r}\hat{r} = v\times\bar{B} = -v\cdot\frac{\mu_0jR^2}{2r}\hat{r}$$ $$v=-\frac{c^2\rho}{j}.$$

The case for which $B=0$ is trivial, it is the case where we introduce a new current density $j'=v\rho$ such that it diminishes the original current density and its effects. But I don't have any intuition for the $E=0$ case, so I am not really sure if that one is correct.

Also note that I haven't made use of invariants, so any feedback and new (and cleaner) answers are more than welcome.

Edit: Alright, looking into this system further, I noticed that the system has to be one of the three possibilities:

1) $\frac{j}{\rho}<c$: There exists a $v<c$ s.t. $B=0$ but no $v<c$ with $E=0$.

2) $\frac{j}{\rho}>c$: There exists a $v<c$ s.t. $E=0$ but no $v<c$ with $B=0$.

3) $\frac{j}{\rho}=c$: There doesn't exist a $v<c$ s.t. $B=0$ or $E=0$.

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  • $\begingroup$ Note that $\frac{j}{\rho}>c$ cannot be correct (it surpasses the speed of light). $\endgroup$ – JD_PM Jun 13 at 11:20
  • $\begingroup$ @JD_PM We could make a system with additional negative $\rho$ as well as a positive $\rho$ so that $\rho$ is even zero (as in normal conductors). This was just a mathematical analysis $\endgroup$ – acarturk Jun 13 at 11:40
  • $\begingroup$ ...by moving the positive density frame only, and leaving the negative one static. $\endgroup$ – acarturk Jun 13 at 11:42
  • $\begingroup$ I see but I am afraid I do not see physical sense on it... $\endgroup$ – JD_PM Jun 13 at 11:52
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    $\begingroup$ @JD_PM The problem is, if $j$ was only produced by moving $\rho$ around (which is the reason you put the limit $\frac{j}{\rho}<c$) the solution is indeed trivial, just un-move the charge :) But, since current can flow through neutral wires ($\rho=0$), I think it is safe to assume a scenario with $j$ having produced by relative movement of two charge densities with a total of $\rho$. It's like a metal where the nuclei framework is not moving, but the free electrons can. Note that the overall charge density is zero, but $j$ is not, then $\frac{j}{\rho}\to\infty>c$ $\endgroup$ – acarturk Jun 13 at 12:02

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