Do ##\vec E## and ##\vec B## change in a moving frame?

[SiennaTheGr8]

I haven't looked in detail at the problem, so please excuse me if I've mistaken what some of your variables mean, but can't you factor out common terms and reduce the problem to the following?

$1 - \mu^2 v^2$

I've used $j = \rho v$, $V = \pi R^2 l^2$, and $A = \pi R^2$, so of course if any of that isn't what you meant then I'm wrong. Regardless, I'd imagine that some algebra should make the question more manageable.

 

[SiennaTheGr8]

Sorry, should have been $V = \pi R^2 l$, (not $l^2$), so really:

$1 - \mu^2 v^2 l$

(provisionally).

 

[SiennaTheGr8]

Nope, scratch that—yes, it's $V = \pi R^2 l$, but the algebra leads to $1 - \mu^2 v^2$ (without the $l$), as I originally wrote.

I'll stop now

 

[SiennaTheGr8]

(And I think you and I both left out an $\epsilon ^2$, in which case it's $1 - \mu^2 \epsilon^2 v^2$, which looks right and passes dimensional analysis.)

 

[SiennaTheGr8]

Except the $\mu$ and $\epsilon$ shouldn't be squared, so please someone put me out of my misery. The point is the algebra.

 

[JD_PM]

(And I think you and I both left out an $\epsilon ^2$, in which case it's $1 - \mu^2 \epsilon^2 v^2$, which looks right and passes dimensional analysis.)

Let me fix it.

Let's check both $E$ and $B$ (by dimensional analysis):

Dimensional analysis of $E$:

$$\frac{\rho^2 V^2}{4 \pi^2 \epsilon^2 R^2 l^2} = \frac{C^2 N^2}{C^4} = \frac{N^2}{C^2}$$

Which makes perfect sense as it yields Newton squared per Coulomb squared.

Dimensional analysis of $B$:

$$\frac{\mu^2 j^2 A^2}{4 \pi^2 R^2} = \frac{A^2 N^2}{A^4 L^2} = \frac{N^2}{A^2 L^2}$$

Which makes perfect sense as it yields Newton squared per Ampere squared times length squared.

Note that $1 T = \frac{N}{Am}$

@SiennaTheGr8 thanks for pointing dimensional analysis out.

 

[pervect]

I think I mentioned this before - I'm not sure. The answer to the question of whether E or B can become zero in some frame depends on the relativistic invariant $\vec{E} \cdot \vec{B}$. Being invariant, the dot product of E and B is the same in any frame. If you measure the quantity in one frame, while E and B will transform and their components will change, the value of the invariant $\vec{E} \cdot \vec{B}$ will not change. Thus, only if this invariant is zero canone find a frame where $E=0$ - or a frame where $B=0$.

I believe it's very likely that one can go beyond "if" to "if and only if", but I'd need to do some more work to demonstrate that.

There's a short non-tensor proof of the invariance of the invariance of the dot product at <<link>>, though it's likely the link is ephemeral.

Vanheese presented a tensor version earlier in the thread, pointing out that $\vec{E} \cdot \vec{B} = F^{ab} G_{ab}$, where $F^{ab}$ is the Faraday tensor and $G_{ab}$ is it's hodges dual, the Maxwell tensor.

 

[JD_PM]

I think I mentioned this before - I'm not sure. The answer to the question of whether E or B can become zero in some frame depends on the relativistic invariant $\vec{E} \cdot \vec{B}$.

You propose using $\vec{E} \cdot \vec{B}$ Lorentz invariance, which is perfectly fine. My idea was using the other Lorentz invariance though (that's why I am working out the square difference).

After having some thought I got it for this example. Note that the permittivity of free space ($\epsilon$) is of the order of $10^{-12}$ and the permeability of free space ($\mu$) is of the order of $10^{-7}$. That is the key to see that in this case $E^2\gt B^2$, which means that we could find another frame where $B = 0$ but not $E = 0$ (as @Nugatory pointed out in #2).

Now I am thinking why if $E^2\gt B^2$ we could find another frame where $B = 0$ but not $E = 0$.

 

[Dale]

Note that the permittivity of free space (ϵ) is of the order of 10^-12 and the permeability of free space (μ) is of the order of 10−7. That is the key to see that in this case $E^2\gt B^2$

Note that in SI units E and B are dimensionally different. So the expression $E^2>B^2$ makes no sense in SI units. Furthermore, in units where E and B share the same dimensions there usually is no $\epsilon_0$ or $\mu_0$

 

[JD_PM]

Note that in SI units E and B are dimensionally different. So this expression makes no sense in SI units.

Absolutely correct my bad. It is $E^2\gt c^2B^2$

 

[Nugatory]

Now I am thinking why if $E^2\gt B^2$ we could find another frame where $B = 0$ but not $E = 0$.

You can. That’s what I meant by the “vice versa” in the other post.

 

[JD_PM]

You can. That’s what I meant by the “vice versa” in the other post.

Oh so you meant that if $E^2\gt c^2B^2$ we could find either a field where $E = 0$ and $B \neq 0$ or a field where $E \neq 0$ and $B = 0$.

How could we tighten the grip then?

 

[PeterDonis]

if $E^2\gt B^2$ there can be a frame in which $B$ is zero but not $E$, and vice versa.

I'm not sure about the "vice versa". Consider the static electric field of a point charge. In the rest frame of the charge $B = 0$ and $E \neq 0$, so $E^2 - B^2 > 0$. But I don't think you can find a frame in which $E = 0$ for this case.

 

[SiennaTheGr8]

Yeah so I had more time just now, and I get:
$$
\begin{aligned}
E^2 - c^2 B^2 &= \frac{\rho^2 V^2}{4 \pi^2 \epsilon^2 R^2 l^2} - \frac{\mu^2 j^2 A^2 c^2}{4 \pi^2 R^2} \\[5pt]
&= \frac{1}{4 \pi^2 R^2} \left( \frac{\rho^2 (\pi R^2 l)^2}{\epsilon^2 l^2} - \frac{\mu^2 (\rho v)^2 (\pi R^2)^2 c^2 }{1} \right) \\[5pt]
&= \frac{\rho^2 R^2}{4 \epsilon^{2}} \left( 1 - \mu^2 \epsilon^2 v^2 c^2 \right) ,
\end{aligned}
$$
which I'm more confident about.

Are $\epsilon$ and $\mu$ supposed to be vacuum permittivity/permeability here? If so, $\epsilon \mu = c^{-2}$, and:
$$
\begin{aligned}
E^2 - c^2 B^2 &= \frac{\rho^2 R^2}{4 \epsilon^{2}} \left( 1 - \frac{v^2}{c^2} \right) \\[5pt]
&= \frac{\rho^2 R^2}{4 \epsilon^{2}} \, \gamma^{-2} .
\end{aligned}
$$
Feels right!

 

[pervect]

You propose using $\vec{E} \cdot \vec{B}$ Lorentz invariance, which is perfectly fine. My idea was using the other Lorentz invariance though (that's why I am working out the square difference).

After having some thought I got it for this example. Note that the permittivity of free space ($\epsilon$) is of the order of $10^{-12}$ and the permeability of free space ($\mu$) is of the order of $10^{-7}$. That is the key to see that in this case $E^2\gt B^2$, which means that we could find another frame where $B = 0$ but not $E = 0$ (as @Nugatory pointed out in #2).

Now I am thinking why if $E^2\gt B^2$ we could find another frame where $B = 0$ but not $E = 0$.

There's no reason we can't do both.

If $\vec{E} \cdot \vec{B}$ is not zero, there is no way to make either E or B zero by any boost to a different frame.

But when $\vec{E} \cdot \vec{B} = 0$, then if E^2 > B^2 >0 we can have B=0 but not E=0, and when E^2 -B^2 < 0, we can have E=0 but not B=0.

Let's consider the case when E^2>B^2 and solve for the required boost that makes B=0.

Let us boost in a direction perpendiculalr to both E and B. Then B transforms like $B_\bot$, (because the boost is perpendicular to B) and we can write in geometric units where c=1

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

Because v is assumed to be perpendicular to both E and B, we can write the equation for the magnitude of the vectors , $E = \sqrt{\vec{E} \cdot \vec{E}}$ and $B = \sqrt{\vec{B} \cdot \vec{B}}$ , up to the sign in v, as:

$$B' = \gamma \left( B - v \,E \right) $$

and we find that when $v = B / E$, B'=0. Because E^2>B^2, |v| < 1 ,i.e. less than the speed of light, and we have just constructed (though I"ve been sloppy about the sign of v) the necessary boost to make B vanish. This boost has a magnitude of |v| = E/B, and is perpendicular to both E and B.

If E^2-B^2=0, then E=B and the construction fails, giving a velocity v equal to the speed of light, which is not allowed. So the construction only applies when E^2 > B^2, not when E^2 = B^2. I believe it's correct to say that when E^2 = B^2, we can't make either vanish even if $\vec{E} \cdot \vec{B} = 0$.

More precisely, if E=0 and B=0, in one frame, then they are zero in all frames, and if they are equal and nonzero, I don't believe we can find a boost to make them vanish, because if we could make them vanish in one frame, they'd have to vanish in all frames, and we know they don't vanish in one frame.

We can use similar arguments to demonstrate how to make B vanish when the necessary conditions are met. So the conditions on the two invariants are both necessary and sufficient, when both invariants are satisfied we can construct the velocity needed to make E or B vanish.

 

[JD_PM]

But when $\vec{E} \cdot \vec{B} = 0$, then if E^2 > B^2 >0 we can have B=0 but not E=0, and when E^2 -B^2 < 0, we can have E=0 but not B=0.

In this case we have $\vec{E} \cdot \vec{B} = 0$, because $E$ goes radially outwards and $B$ goes circumferential. Having a look at the above equations for them, we can guess that $E^2 > B^2 >0$ . In such scenario I understood, as you it seems, that when we are in such conditions, then we could find another frame where $B=0$ but not $E=0$. (see #38).

But it seems I misunderstood Nugatory's words and it could also be that when $\vec{E} \cdot \vec{B} = 0$ and $E^2 > B^2 >0$, we could find another frame where $E=0$ but not $B=0$...

I am still wondering why when we have $E^2 > B^2 >0$ we are able to find another frame where $B=0$ but not $E=0$... how can we prove it mathematically?

 

[JD_PM]

I am still wondering why when we have $E^2 > B^2 >0$ we are able to find another frame where $B=0$ but not $E=0$... how can we prove it mathematically?

OK let me answer myself (I think I got it).

The transformation for $E$:

$$E' = \gamma (E + vB)$$

So $E'=0$ is not possible because we'd require that $v=\frac{-E}{B}$... and that's greater than the speed of light when $E^2 > B^2 >0$ (this is so funny! :) )

@Nugatory did I convince you that when we have $E^2 > B^2 >0$ we are able to find another frame where $B=0$ but not $E=0$ and not vice versa?

 

[Ibix]

I'm in hurry-up-and-wait mode today so I did a general transformation to kill time. Like @pervect I may have been sloppy about minus signs. Keeping track of details on a small phone screen with no paper to hand is tricky.

For a general electromagnetic field boosted in the $x$ direction, the transformed fields in terms of the components of the original frame are:
$$E'=\pmatrix{E_x\cr
\gamma(B_z v+E_y)\cr
\gamma(E_z-B_y v)\cr }$$
$$B'=\pmatrix{B_x\cr
\gamma(B_y-E_z v)\cr
\gamma(E_y v+B_z)\cr }$$
So we cannot ever make either field zero if it has a non-zero component parallel to the boost because that component doesn't change. Otherwise, we can zero the magnetic field if $vE_z=B_y$ and $vE_y=-B_z$ and zero the electric field if $E_y=-vB_z$ and $E_z=vB_y$. You can't satisfy both of those pairs of constraints unless $v=\pm 1$, the speed of light, in which case the Lorentz transforms I used in the derivation are invalid - so that answer is meaningless, and the fields can't be zeroed simultaneously unless they were zero to start with. Which is nice to know because we'd have a problem if the invariants said it was impossible and this method said it was...

 

[Orodruin]

Oh so you meant that if $E^2\gt c^2B^2$ we could find either a field where $E = 0$ and $B \neq 0$ or a field where $E \neq 0$ and $B = 0$.

How could we tighten the grip then?

This is not what he meant. He was saying that if $E^2 > c^2 B^2$ and $\vec E \cdot \vec B = 0$, then you can find a frame where $B = 0$ and $E \neq 0$ and, conversely, that if $E^2 < c^2 B^2$ and $\vec E \cdot\vec B = 0$, then you can find a frame where $E = 0$ and $B \neq 0$.

 

[JD_PM]

Oh now everything makes sense.