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.

  • $\begingroup$ 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

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.


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 $E=0$.

  • $\begingroup$ 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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