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

#### JD_PM

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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#### Nugatory

Mentor
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.

#### PeroK

Homework Helper
Gold Member
2018 Award
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.

#### JD_PM

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...

#### JD_PM

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.

#### Gaussian97

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.

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
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.

Gold Member
See Sect. 3.3 of

#### JD_PM

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:

#### Dale

Mentor
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.

#### JD_PM

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 $B$).

#### Nugatory

Mentor
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).

#### vanhees71

Gold Member
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

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
$$\int_{\mathbb{R}} \mathrm{d} z' \frac{1}{(x^2+y^2+z^{\prime 2})^{3/2}}=\frac{2}{x^2+y^2},$$
$$\vec{E}(\vec{r})=\frac{\lambda}{2 \pi (x^2+y^2)} \begin{pmatrix}x \\ y \\ 0 \end{pmatrix}.$$
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.

#### SiennaTheGr8

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 $\hat \beta$ $[ = \hat v = \hat \phi ]$.

#### JD_PM

Thank you for your replies. I defenitely need to read more before appreciating their beauty :)

Back soon

#### pervect

Staff Emeritus
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 vaccum 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.

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

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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#### Dale

Mentor
@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.

#### vanhees71

Gold Member
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.

#### JD_PM

@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:

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

#### Ibix

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.

#### Dale

Mentor
according to the following transformations:
Looks like you got it!

#### vanhees71

Gold Member
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 $\vec{B}$.

#### JD_PM

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 $B$?

#### pervect

Staff Emeritus
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.

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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#### vanhees71

Gold Member
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 $\vec{E}^2-\vec{B}^2<0$.

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

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