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

In summary: Lorentz contraction while B... is. So in a frame where E is larger than B, BBB would be smaller and vice versa.Oh I see... This is because EB... is unaffected by Lorentz contraction while BB... is. So in a frame where EB is larger than BB, BBB would be smaller and vice versa.
  • #1
JD_PM
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TL;DR Summary
Understanding how both magnetic and electric fields change in a moving frame
I want to understand how electric and magnetic fields change as measured from an inertial frame ##S## vs. as measured from an inertial frame ##\bar {S}## (which has uniform speed v wrt ##S##). I am working out the following example to do so:

We have a cylindrical symmetric wire of radius R, with constant charge density ##\rho## and current density ##j##. (i.e the current I is uniformly distributed over the wire of circular cross section; assume the flow goes from left to right).

We know that its magnetic field ##\vec B## is ##\vec B = \frac{\mu I}{2\pi s} \hat {z}## (where ##s## is the radius of the Amperian loop) outside the wire while ##\vec B = 0## inside the wire. Let's draw out attention to the external magnetic field.

We know that its electric field ##\vec E## is ##\vec E = E \hat {r}## (where ##\hat {r}## accounts for radial direction).

Now let the wire be in an inertial frame ##\bar {S}##, which has uniform speed v (from left to right) wrt your frame ##S##.

a) Could the electric field become ##\vec E = 0## as measured in frame ##\bar {S}##? Why?
b) Could the magnetic field become ##\vec B = 0## as measured in frame ##\bar {S}##? Why?

I think that time dilation plays no role on changing neither ##\vec E## or ##\vec B## because these are uniform (time-independent) fields. Thus, let's focus on Lorentz contraction (moving objects get shortened).

a) ##E## won't change because Lorentz contraction does not apply. ##E## is perpendicular to the velocity of the frame; dimensions perpendicular to the velocity are not contracted.

b) The magnetic field should be contacted, so it changes. But does that mean that we could make it become zero?

Thanks
 
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  • #2
Yes, ##E## and ##B## are frame-dependent so have different values in different frames. However, ##E^2-B^2## is invariant, meaning that it will always have the same value in all frames; if you find a frame in which ##E## is larger then ##B## will be commensurately smaller so that ##E^2-B^2## comes out the same.

If you think about the implications of ##E^2-B^2## being invariant for a moment, you will see that if ##E^2\gt B^2## there can be a frame in which ##B## is zero but not ##E##, and vice versa.
 
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  • #3
JD_PM said:
Summary: Understanding how both magnetic and electric fields change in a moving frame

I want to understand how electric and magnetic fields change as measured from an inertial frame ##S## vs. as measured from an inertial frame ##\bar {S}## (which has uniform speed v wrt ##S##). I am working out the following example to do so:

The subject of EM and relativity should be covered in most textbooks. You really need a suitable text to learn this from and use PF to ask questions if anything is not clear.
 
  • #4
Nugatory said:
Yes, ##E## and ##B## are frame-dependent so have different values in different frames.

I understand ##B## has a different value in the moving frame (because of Lorentz contraction). But why does E change? Neither Lorentz contraction nor time dilation apply...
 
  • #5
PeroK said:
The subject of EM and relativity should be covered in most textbooks. You really need a suitable text to learn this from and use PF to ask questions if anything is not clear.

I have been reading Introduction to Electrodynamics by Griffiths chapter 12, section relativistic electrodynamics. He deals with line charges (##\lambda##) and plates.
 
  • #6
JD_PM said:
I have been reading Introduction to Electrodynamics by Griffiths chapter 12, section relativistic electrodynamics. He deals with line charges (##\lambda##) and plates.
In a wire, the electric field is generated by the linear density of charges, this is affected by Lorentz contraction.
 
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  • #7
JD_PM said:
I understand ##B## has a different value in the moving frame (because of Lorentz contraction). But why does E change? Neither Lorentz contraction nor time dilation apply...
You really should not be thinking of electromagnetism in terms of Lorentz contraction. You should be considering the E and B fields as parts of the same rank-2 antisymmetric tensor.
 
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  • #9
Gaussian97 said:
In a wire, the electric field is generated by the linear density of charges, this is affected by Lorentz contraction.

I have just seen that I was wrong stating that ##E## doesn't change in a moving frame. This is how the wire looks like in the moving frame:

Captura de pantalla (625).png
 
  • #10
JD_PM said:
I understand BBB has a different value in the moving frame (because of Lorentz contraction). But why does E change?
As others have mentioned, length contraction doesn’t apply here. Length contraction refers to the change in the spacelike components of a vector. The electromagnetic field is a rank 2 anti symmetric tensor, so it transforms differently. The E field transforms as the time-space components and the B field transforms as the space-space components.
 
  • #11
Orodruin said:
You really should not be thinking of electromagnetism in terms of Lorentz contraction. You should be considering the E and B fields as parts of the same rank-2 antisymmetric tensor.

Oh I see... This is because E and B are mixed when you go from one inertial frame to the other (I saw this thanks to the transformation rules applying to both ##E## and ##B##).
 
  • #12
JD_PM said:
But why does E change? Neither Lorentz contraction nor time dilation apply...
There aren’t separate magnetic and electric fields. There is a single electromagnetic field, and how we divide it into magnetic and electric components is frame dependent. (This is much easier to see if you use the tensor formulation of Maxwell’s equations).

The properties of the electromagnetic field were discovered first, including the way that both E and B transform between inertial frames. One of the consequences of this transformation rule is that the speed of electromagnetic waves is the same in all inertial frames, and this is what motivated Einstein to develop special relativity - the title of his 1905 paper introducing SR is “On the electrodynamics of moving bodies”. Thus, you may want to think of length contraction, time dilation, and other relativistic effects as things that are required to be consistent with electromagnetism, instead of trying to explain electromagnetism as caused by them.
(Of course you can go either way: if X and Y are necessarily either both true or both false then we can claim with equal justification that X implies Y and that Y implies X).
 
  • #13
JD_PM said:
Summary: Understanding how both magnetic and electric fields change in a moving frame

I want to understand how electric and magnetic fields change as measured from an inertial frame ##S## vs. as measured from an inertial frame ##\bar {S}## (which has uniform speed v wrt ##S##). I am working out the following example to do so:

We have a cylindrical symmetric wire of radius R, with constant charge density ##\rho## and current density ##j##. (i.e the current I is uniformly distributed over the wire of circular cross section; assume the flow goes from left to right).

We know that its magnetic field ##\vec B## is ##\vec B = \frac{\mu I}{2\pi s} \hat {z}## (where ##s## is the radius of the Amperian loop) outside the wire while ##\vec B = 0## inside the wire. Let's draw out attention to the external magnetic field.

We know that its electric field ##\vec E## is ##\vec E = E \hat {r}## (where ##\hat {r}## accounts for radial direction).

Now let the wire be in an inertial frame ##\bar {S}##, which has uniform speed v (from left to right) wrt your frame ##S##.

a) Could the electric field become ##\vec E = 0## as measured in frame ##\bar {S}##? Why?
b) Could the magnetic field become ##\vec B = 0## as measured in frame ##\bar {S}##? Why?

I think that time dilation plays no role on changing neither ##\vec E## or ##\vec B## because these are uniform (time-independent) fields. Thus, let's focus on Lorentz contraction (moving objects get shortened).

a) ##E## won't change because Lorentz contraction does not apply. ##E## is perpendicular to the velocity of the frame; dimensions perpendicular to the velocity are not contracted.

b) The magnetic field should be contacted, so it changes. But does that mean that we could make it become zero?

Thanks
That's an interesting example. Note, however, that your ##\vec{B}## is wrong. It's also a bit unusual that the wire carries both a charge density and a current, but why not ;-).

Let's take the wire in the ##z## direction, and we work in Heaviside-Lorentz units, which are more convenient for relativistic issues than the SI. Let's also keep things as simple as possible and make the wire infinitesimally thin.

We write ##(x^{\mu})=(c t,\vec{x})=(c t,x,y,z)## etc. In frame ##S## we have
$$\rho(\vec{x})=\lambda \delta(x) \delta(y), \quad \vec{j}(\vec{x})=I \delta(x) \delta(y).$$
Here ##\lambda## is the constant charge line-density and ##I## the constant total current through the wire. Using the Jefimenko equations (3.4.6) and (3.4.7) of my SRT manuscript

https://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf
we find
$$\vec{E}=\frac{\lambda}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3} \delta(x') \delta(y') = \frac{\lambda}{4 \pi} \int_{\mathbb{R}} \mathrm{d} z' \frac{1}{[(x^2+y^2)+(z-z')^2]^{3/2}} \begin{pmatrix}x \\ y \\ z-z' \end{pmatrix}.$$
For symmetry reasons ##E_z=0##. For the other components we can set ##z=0## (since the field obviously doesn't depend on ##z## because of translation symmetry in this direction) and then only need the integral
$$\int_{\mathbb{R}} \mathrm{d} z' \frac{1}{(x^2+y^2+z^{\prime 2})^{3/2}}=\frac{2}{x^2+y^2},$$
leading to
$$\vec{E}(\vec{r})=\frac{\lambda}{2 \pi (x^2+y^2)} \begin{pmatrix}x \\ y \\ 0 \end{pmatrix}.$$
For the magnetic field we have
$$\vec{B}(\vec{x})=\frac{I}{c} \vec{e}_3 \times \frac{\vec{E}(\vec{x})}{\lambda} = \frac{I}{2 \pi c (x^2+y^2)} \begin{pmatrix} -y \\ x \\0 \end{pmatrix}.$$
Now let's do the boost in ##z## direction, i.e., let ##\bar{S}## move with velocity ##\vec{\beta}c = \beta c \vec{e}_3## wrt. to the system ##S##.

Since ##x'=x## and ##y'=y## this example is indeed utmost simple :-)). The transformation of the four-current density gives
$$(\bar{j}^{\mu})=\gamma \begin{pmatrix}c \lambda-\beta I \\ 0 \\ 0 \\ I-\beta c \lambda \end{pmatrix} \delta(\bar{x})\delta(\bar{y}).$$
We thus get a field of the very same form but with changed charge-line density and current,
$$\bar{\lambda}=\lambda - \beta I/c, \quad \bar{I}=I-\beta c \lambda.$$
It's clear that you can make ##\vec{\bar{E}}=0## by this boost, if ##\bar{\lambda}=0##, which leads to
$$c \lambda - \beta I=0 \; \Rightarrow \; \beta=\frac{c \lambda}{I}.$$
Now ##\beta## must obey ##|\beta|<1##, i.e., you must have ##|I|>c |\lambda|##.

On the other hand if ##|I|<c |\lambda|## you can find a ##\beta## such that ##\bar{I}=0## and thus ##\vec{\bar{B}}=0##. You only have to make
$$\beta=\frac{I}{c \lambda}.$$

You can also argue without doing the Lorentz transformation with the fields as given wrt. ##\bar{S}##. The Lorentz transformation leave both ##\vec{E} \cdot \vec{B}=0## and
$$\vec{E}^2-\vec{B}^2=\frac{1}{(2 \pi) (x^2+y^2)} (\lambda^2-I^2/c^2)$$
invariant. Thus ##\vec{\bar{E}}=0## implies ##\vec{E}^2-\vec{B}^2<0##, and thus you must have ##\lambda^2<I^2/c^2## or ##|\lambda| c<I## if there should be a frame, where ##\vec{\bar{E}}## vanishes.
 
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  • #14
Under a Lorentz boost in an arbitrary direction ##\hat \phi##:

##\vec E^\prime = \cosh{\phi} \, \vec E + \sinh{\phi} \, (\hat \phi \times \vec B) - 2 \sinh^2 \dfrac{\phi}{2} \, (\hat \phi \cdot \vec E) \hat \phi##

##\vec B^\prime = \cosh{\phi} \, \vec B - \sinh{\phi} \, (\hat \phi \times \vec E) - 2 \sinh^2 \dfrac{\phi}{2} \, (\hat \phi \cdot \vec B) \hat \phi##,

where ##\phi = \tanh^{-1} (v/c)## is the boost parameter (rapidity). Equivalently, though less elegantly IMO:

##\vec E^\prime = \gamma (\vec E + \vec \beta \times \vec B) - \dfrac{\gamma^2 (\vec \beta \cdot \vec E) \vec \beta}{\gamma + 1}##

##\vec B^\prime = \gamma (\vec B - \vec \beta \times \vec E) - \dfrac{\gamma^2 (\vec \beta \cdot \vec B) \vec \beta}{\gamma + 1} ##,

where ##\vec \beta = \vec v / c## and ##\gamma = (1 - \beta^2)^{-1/2}##, and the boost direction is ##\hat \beta## ##[ = \hat v = \hat \phi ]##.
 
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  • #15
Thank you for your replies. I defenitely need to read more before appreciating their beauty :)

Back soon
 
  • #16
JD_PM said:
Thank you for your replies. I defenitely need to read more before appreciating their beauty :)

Back soon

Gah - I totally messed up my original post with an edit, not sure if there's a way to revert. So I'm fixing it up - again.

The "elegant" way, IMO, involves using tensors. One starts by combiing the E and B fields into the Faraday tensor <wiki link>. I'm not going to even attempt to motivate it, my goal is just to demonstrate the approach.

Some of the steps are a bit abstract and unfamiliar, but - with this approach, the electromagnetic field transform follows _directly_ from the Lorentz transform for time and space, it has the exact same form.

We start off being elegant by choosing units for time and space such that the speed of light, c, is unity, and units of charge so that the permitivity of the vacuum ϵ0 is unity, as is the magnetic suspectibility μ0.

This isn't necessary, really, but it makes things a lot easier to follow and for me to type.

Then we combine the E and B fields into the rank 2 "Faraday Tensor", <<wiki link>>

$$F^{\mu\nu} = \begin{bmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0 \end{bmatrix}$$The way the tensors transform is given by standard transformation laws

$$
F'^{\mu \nu} = \Sigma_{\alpha=0 ... 3 } \, \Sigma_{\beta=0 ... 3} \, \Lambda^{\mu}{}_{\alpha} \, \Lambda^{\nu}{}_{\beta} F^{\alpha \beta}
$$

The double sums are usually implied by the repetition of indices, I'm writing them out explicitly.

Here ##\Lambda## is just the Lorentz transformation matrix. So for a boost in the x direction, we'd have

$$\Lambda = \begin{bmatrix} \gamma & -\beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$Here ##\beta = v/c## and ##\gamma = 1/\sqrt{1-\beta^2}##

Expanding this double sum becomes rather annoying, though the simplicity of the transformation laws as written above can't be beat.

As an example, if we look at how ##E_y## transforms, we would write all the nonzero components as:

$$
F'^{20} = \Lambda^2{}_2 \Lambda^0{}_0 F^{20} + \Lambda^2{}_2 \Lambda^{0}{}_1 F^{21} = \gamma F^{20} - \beta \gamma F^{21}
$$

Which boils down to ##E'_y = \gamma E_y + \beta \gamma E_z##

Hopefully I haven't made any errors, but the principle is sound. The most interesting point is that the electromagnetic field transformations follow _directly_ from the Lorentz transforms , I think, as I mentioned before. If you know how time and space transform, with this approach you also know how the electromagnetic field transforms, once you realize it's a rank 2 tensor.

[add]
It's worth comparing this to the second-most elegant approach, in terms of components.

See <<wiki link>>

Then the transformation properties for the direction perpendicular to the boost for the E-field , which we write as ##E_\bot##, are written

$$E_\bot = \gamma(E_\bot + v \times B)$$

Wiki tells how the parallel and perpendicular components of E and B transform.

With this approach, we have separate transformation laws for the parallel and perpendicular direcitons, and there is no tie-in to the electromagnetic field transformation laws to the transformation law for time and space. But it might be easier to follow.
 
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  • #17
@JD_PM notice how in @pervect’s math the Lorentz transform matrix ##\Lambda## is applied twice. This is why the effect cannot be described simply as length contraction.
 
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  • #18
Another nice way is to use the Riemann-Silberstein vector ##\vec{F}=\vec{E}+\mathrm{i} \vec{B}## (in Heaviside-Lorentz or Gauß units). These complexified fields defines a group automorphism of the proper orthochronous Poincare group with the 3D complex (!) orthogonal group, ##\mathrm{SO}(1,3)^{\uparrow} \rightarrow \mathrm{SO}(3,\mathbb{C})##. The rotations are mapped, of course, to the subgroup ##\mathrm{SO}(3),\mathbb{R})##. Lorentz boosts are rotations with imaginary rotation angles in ##\mathrm{SO}(3,\mathbb{C})##. I have still to add this nice detail in my SRT FAQ.
 
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  • #19
Dale said:
@JD_PM notice how in @pervect’s math the Lorentz transform matrix ##\Lambda## is applied twice. This is why the effect cannot be described simply as length contraction.

I see.

Then the point is that neither ##E## or ##B## transform as the spatial part of a 4-vector:

$$F'^{\mu} = \Sigma_{\alpha=0 ... 3 } \, \Lambda^{\mu}{}_{\alpha} \, F^{\alpha }$$

But according to the following transformations:

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

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

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

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

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

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

i.e antisymmetric second-rank tensor:

$$F'^{\mu \nu} = \Sigma_{\alpha=0 ... 3 } \, \Sigma_{\beta=0 ... 3} \, \Lambda^{\mu}{}_{\alpha} \, \Lambda^{\nu}{}_{\beta} F^{\alpha \beta}$$
 
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  • #20
JD_PM said:
Then the point is that neither E or B transform as the spatial part of a 4-vector
The 4-potential ##(\phi,\vec A)##, where ##\phi## is the electric potential and ##\vec A## is the magnetic vector potential, is a four vector. But the E and B fields do not, no.
 
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  • #21
JD_PM said:
according to the following transformations:
Looks like you got it!
 
  • #22
JD_PM said:
I see.

Then the point is that neither ##E## or ##B## transform as the spatial part of a 4-vector:

$$F'^{\mu} = \Sigma_{\alpha=0 ... 3 } \, \Lambda^{\mu}{}_{\alpha} \, F^{\alpha }$$

But according to the following transformations:

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

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

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

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

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

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

i.e antisymmetric second-rank tensor:

$$F'^{\mu \nu} = \Sigma_{\alpha=0 ... 3 } \, \Sigma_{\beta=0 ... 3} \, \Lambda^{\mu}{}_{\alpha} \, \Lambda^{\nu}{}_{\beta} F^{\alpha \beta}$$
You can define covariant field-strength vectors,
$$E^\mu=F^{\mu \nu} u_{\nu}, \quad B^{\mu}=^{\dagger}F^{\mu \nu} u_{\nu}.$$
This is the electric and magnetic field vector of an observer moving with velocity ##\vec{v}## relative to the computational frame; ##(u^{\mu})=\gamma(1,\vec{v}/c)##. To get the components in the restframe of this obderver you have to apply the boost matrix to these field components, which also leads to the transformations of ##\vec{E}## and ##\vec{B}##.
 
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  • #23
After getting the six components of the antisymmetric second-rank tensor one sees that there are two really interesting cases:

1) If ##B = 0## one gets:

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

2) If ##E = 0## one gets:

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

But what about if we have, for instance, ##E \neq 0##? Would it be possible to get ##E = 0## in another frame?

The thing is that if we know about ##B## we can play with ##E^2-c^2B^2## (post #1). Imagine that we have ##E \neq 0## and ##B = 0##. Then it wouldn't be possible to find another frame where ##E = 0## because ##E^2-c^2B^2## is invariant and (in this case) has to be always positive.

What to do when I just know about ##E## or ##B##?
 
  • #24
JD_PM said:
But what about if we have, for instance, ##E \neq 0##? Would it be possible to get ##E = 0## in another frame?

It's certainly possible under some circumstances. We can imagine a frame S where E=0 and B>0, then if we boost to frame S', we have |E|>0.

Then a boost from S' back to S will be an example of what you ask for.

I'm not sure if this is always possible, though.

[add]
I believe it's possible if and only if ##\vec{E} \cdot \vec{B} = 0##. Certainly, since ##\vec{E} \cdot \vec{B}## is an invariant, if E=0 in some frame then the invariant dot product must be zero.

I believe we can probably solve for the velocity v required to "cancel out" the E field when E and B are orthogonal, but I haven't actually demonstrated this.
 
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  • #25
There are two invariants, ##F_{\mu \nu} F^{\mu \nu}## and ##^{\dagger}F^{\mu \nu} F_{\mu \nu}## or, in terms of the 3D formulation, ##\vec{E}^2-\vec{B}^2## and ##\vec{E} \cdot \vec{B}##. If you want to have ##\vec{\bar{E}}=0## in some reference frame, you must thus have ##\vec{E} \cdot \vec{B}=0## and ##\vec{E}^2-\vec{B}^2<0##.
 
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  • #26
The following reply is aimed at any reader who is learning about tensors in electrodynamics.

I want to express the tensor invariant ##^{\dagger}F^{\mu \nu} F_{\mu \nu}## in function of both ##\vec E## and ##\vec B##.

The Field tensor is:

$$F^{\mu\nu} = \begin{bmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0 \end{bmatrix}$$

Note that the Field tensor is antisymmetric:

$$F^{\mu \nu} = - F^{\nu \mu}$$

If we upper an index (or lower it; in this case I am going from covariant to contravariant) we have to change the sign if one of the indices is zero.

I am using the convention ##(- + + +)##.

##^{\dagger}F^{\mu \nu} F_{\mu \nu} = F^{0 0} F^{0 0} - F^{0 1} F^{0 1} - F^{0 2} F^{0 2} - F^{0 3} F^{0 3} - F^{1 0} F^{1 0} - F^{2 0} F^{2 0} - F^{3 0} F^{3 0} + F^{1 1} F^{1 1} + F^{2 2} F^{2 2} + F^{3 3} F^{3 3} + F^{1 2} F^{1 2} + F^{1 3} F^{1 3} + F^{2 1} F^{2 1} + F^{2 3} F^{2 3} + F^{3 1} F^{3 1} + F^{3 2} F^{3 2}##

Let's explain one case: ##-F^{0 1} F^{0 1}##

##F^{0 1} = \frac{E_x}{c}## by itself. But now I have to multiply it by the second ##F^{0 1}##. I have brought the indices up, so that means changing sign. Thus the second ##F^{0 1} = -\frac{E_x}{c}##, getting:

$$F^{0 1} F^{0 1} = -(\frac{E_x}{c})^2$$

The rest follows:

$$^{\dagger}F^{\mu \nu} F_{\mu \nu} = -\frac{2\vec E^2}{c^2} + 2 \vec B^2 = 2(\vec B^2 - \frac{\vec E^2}{c^2})$$

Please let me know if more details are required.
 
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  • #27
JD_PM said:
The following reply is aimed at any reader who is learning about tensors in electrodynamics.

...

more details are required.

It looks good, though I'm not sure what you mean by ##^{\dagger}{F}##. I'm pretty sure that vanhees71 used the notation to describe the hodges dual of the Faraday tensor, the so-called Maxwell tensor.

Informally, the process of taking the hodges dual can be described as swapping the roles of E and B. This description is oversimplified, though, due to sign issues. Formally, the duality operation and hence the dual tensor is defined using the Levi-Civiti tensor.

Wiki talks about the Maxwell tensor in <<link>>., though they use the symbol G to denote it.

On the topic of invariants, Google finds the following course notes <<link>>. They're rather nice, but I've noticed such things tend to disappear, so at some future date it may not around. The notes do nicely explain why ##\vec{E} \cdot \vec{B}## is invariant though.

For an interesting but somewhat irrelevant point, Maxwell's equations in flat space-time are basically

$$\partial_a F^{ab} = J^b \quad \partial_a G^{ab}=0$$

where I've used geometric units, and I've also used G to denote the Maxwell tensor, following wiki. Vanhees71 used ##^{\dagger}F## to denote this tensor, and the other web reference I mentioned used ##\mathcal{F}## to denote it.
 
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  • #28
Let me recap after some reading.

I am interested in knowing whether ##E## or ##B## can become zero in a moving reference frame. (we know that that could be the case, but let me discuss it to see if I understand why).

The original problem was about a cylindrical wire with radius ##R##, ##\rho## and ##\vec j## (to the right).

Firstly let's focus on ##S## frame (non moving frame):

1) What's ##E## in ##S##?

Well, I just drew a Gaussian cylinder of radius ##s## and length ##l## and using Gauss' law one gets:

$$\oint \vec E \cdot d \vec a = \frac{\rho V}{\epsilon}$$

$$E l \oint_0^{2\pi R} ds = \frac{\rho V}{\epsilon}$$

$$E = \frac{\rho V}{2\pi \epsilon R l} \hat {r}$$

2) What's ##B## in ##S##?

Now it's time to use Ampere's Law

$$\oint \vec B \cdot d \vec l = \mu I_{enc}$$

Using an Amperian loop so that we enclose the current density:

$$B \oint_0^{2\pi R} dl = \mu j A$$

$$B = \frac{\mu j A}{2 \pi R} \hat {\phi}$$

Now I want to qualitatively know which is greater in ##S##; ##E## or ##B##? This is key when using the invariant ##E^2-B^2##. Is it possible with the info we have at this point?

Besides, there's something that bothers me. The current density is ##\vec j = \rho v##, but frame S is at rest, meaning that we'd get ##\vec j = 0## and thus ##B = 0##, something that we know is not the case; the magnetic field of a cylindrical current-carrying wire in a rest frame ##S## is non-zero. What am I missing here?
 
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  • #29
JD_PM said:
Is it possible with the info we have at this point?
You've just written down E and B. The difference of squares should be straightforward.
JD_PM said:
Besides, there's something that bothers me. The current density is \vec j = \rho v, but frame S is at rest, meaning that we'd get \vec j = 0 and thus B = 0, something that we know is not the case;
In a wire, what are the protons doing in the frame where the electrons are at rest?
 
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  • #30
Ibix said:
You've just written down E and B. The difference of squares should be straightforward.

We have:

$$\frac{\rho^2 V^2}{4 \pi^2 \epsilon^2 R^2 l^2} - \frac{\mu^2 j^2 A^2}{4 \pi^2 R^2}$$

I am not able to qualitatively guess if the difference of squares would be positive or negative.

Ibix said:
In a wire, what are the protons doing in the frame where the electrons are at rest?

They move. So what you are suggesting is that in one frame we see the electrons moving and the protons at rest while in the other is the other way around. Thus ##j \neq 0## in both frames
 
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  • #31
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.
 
  • #32
Sorry, should have been ##V = \pi R^2 l##, (not ##l^2##), so really:

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

(provisionally).
 
  • #33
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 :smile:
 
  • #34
(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.)
 
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  • #35
Except the ##\mu## and ##\epsilon## shouldn't be squared, so please someone put me out of my misery. The point is the algebra.
 
<h2>1. What is the relationship between ##\vec E## and ##\vec B## in a moving frame?</h2><p>In a moving frame, the electric and magnetic fields are related by the Lorentz transformation. This means that they are not independent of each other and will change in relation to each other as the frame moves.</p><h2>2. How do the components of ##\vec E## and ##\vec B## change in a moving frame?</h2><p>The components of ##\vec E## and ##\vec B## will change in a moving frame due to the Lorentz transformation. The electric field will experience a transformation in both magnitude and direction, while the magnetic field will only experience a transformation in direction.</p><h2>3. What is the significance of the Lorentz transformation in relation to ##\vec E## and ##\vec B## in a moving frame?</h2><p>The Lorentz transformation is significant because it allows us to understand how the electric and magnetic fields are related in a moving frame. It also helps us to understand how these fields behave in different reference frames and how they are affected by the motion of the observer.</p><h2>4. How does the speed of the moving frame affect the changes in ##\vec E## and ##\vec B##?</h2><p>The speed of the moving frame will affect the changes in ##\vec E## and ##\vec B##. As the speed of the frame increases, the magnitude of the electric field will decrease and the magnetic field will increase in magnitude. This is because the Lorentz transformation is dependent on the relative velocity between the frame and the observer.</p><h2>5. Can the changes in ##\vec E## and ##\vec B## be observed in everyday life?</h2><p>Yes, the changes in ##\vec E## and ##\vec B## can be observed in everyday life. For example, the Doppler effect is a real-life phenomenon that demonstrates the changes in these fields as a result of relative motion between the source and the observer. Other examples include the behavior of charged particles in a magnetic field and the functioning of devices such as MRI machines.</p>

1. What is the relationship between ##\vec E## and ##\vec B## in a moving frame?

In a moving frame, the electric and magnetic fields are related by the Lorentz transformation. This means that they are not independent of each other and will change in relation to each other as the frame moves.

2. How do the components of ##\vec E## and ##\vec B## change in a moving frame?

The components of ##\vec E## and ##\vec B## will change in a moving frame due to the Lorentz transformation. The electric field will experience a transformation in both magnitude and direction, while the magnetic field will only experience a transformation in direction.

3. What is the significance of the Lorentz transformation in relation to ##\vec E## and ##\vec B## in a moving frame?

The Lorentz transformation is significant because it allows us to understand how the electric and magnetic fields are related in a moving frame. It also helps us to understand how these fields behave in different reference frames and how they are affected by the motion of the observer.

4. How does the speed of the moving frame affect the changes in ##\vec E## and ##\vec B##?

The speed of the moving frame will affect the changes in ##\vec E## and ##\vec B##. As the speed of the frame increases, the magnitude of the electric field will decrease and the magnetic field will increase in magnitude. This is because the Lorentz transformation is dependent on the relative velocity between the frame and the observer.

5. Can the changes in ##\vec E## and ##\vec B## be observed in everyday life?

Yes, the changes in ##\vec E## and ##\vec B## can be observed in everyday life. For example, the Doppler effect is a real-life phenomenon that demonstrates the changes in these fields as a result of relative motion between the source and the observer. Other examples include the behavior of charged particles in a magnetic field and the functioning of devices such as MRI machines.

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