Electromagnetic field according to relativity

[fog37]

TL;DR Summary

Understanding how the static fields E and B arise in relation to the full EM field

Hello,

I am still trying to fully grasp the general idea of the EM field, which always travels at the speed of light regardless of the reference frame, and is represented by a tensor with 16 components in relativity theory. My understanding is that, depending on the observer's frame of reference and on the value of certain invariants (one of them is $E^2 - B^2$), the observers sees either the E field alone, the B field alone, or both.

There are two speeds: the constant speed of the EM field and the speed of the charges generating the EM field.

In electrostatics, the only field is the E field and the field has no time variation. However, the actual EM field is always time-varying since it is a wave field and oscillates in time and space.

In the scenarios where the E field is time-changing, the B field becomes automatically present. In magnetostatics, we only deal with the B field and
are not concerned with the E field. Where does the actual EM field time variation go in electrostatics and magnetostatics where both fields are static? I guess an observer who is in a reference frame that is at rest relative to the charges will see the EM field as a static E field. Is that correct? However the EM field should still travel at the speed $c$.

When the observer moves at a constant velocity relative to the charges, the only visible field is the B field and it appears statics...

Is my basic understanding correct?

 

[Ibix]

the EM field, which always travels at the speed of light regardless of the reference frame

No - an EM wave propagates at the speed of light. The EM field doesn't have a speed.

is represented by a tensor with 16 components in relativity theory

The tensor formalism is merely a restatement of the more commonly taught vector expressions. It doesn't add anything except a (large) degree of mathematical convenience. Yes, the electromagnetic field tensor has sixteen components, but it's antisymmetric - that is, $F^{\mu\nu}=-F^{\nu\mu}$. That means that the four diagonal components are zero and six of the remaining twelve components are completely determined by the other six - so you only have six free components, which correspond to the six components of the $\vec E$ and $\vec B$ fields.

Your questions about electrostatics and magnetostatics seem to be confused by your belief that the EM field moves. It doesn't - "moving" isn't a concept that makes sense applied to a field. So an electrostatic or magnetostatic situation is simply one in which the field (whether you regard it as an electric field and a magnetic field or as an electromagnetic field) is independent of time in whatever reference frame you are using. Other frames, of course, do see a time-dependent field (the first frame interprets this as the second frame's sensors moving through the field). So they express the field tensor differently, seeing time-varying fields and hence both non-zero electric and magnetic components.

 

[fog37]

Thank you Ibix. But isn't a EM just a propagating EM field traveling at the speed c? Always though that...

There are traveling EM waves, standing EM waves (and hybrid scenarios)...

 

[Ibix]

No. The electromagnetic field does not propagate. EM waves and any other electric, magnetic, or combination field, are excitations of the field - at least, that's a sensible interpretation of classical EM. (Others may provide a quantum field theory view.) An EM wave is a traveling excitation, not a traveling field. I don't think it really makes sense to talk about a field having a velocity.

 

[Nugatory]

Thank you Ibix. But isn't a EM just a propagating EM field traveling at the speed c? Always though that...

No. There’s a field, which may or may not be oscillating. If you want an analogy, you could think about a lake or other body of water: the water is there whether there are waves or not.

However, there’s only so far that analogies can take you (and in this case, it’s not very far). You might try writing down the Faraday tensor for a simple case, the uniform electrical field between the plates of a capacitor that is at rest, then use the Lorentz transformations and tensor transformation rule to convert its components to a frame in which the capacitor is moving at a constant speed... see what that tells you.

 

[fog37]

Thank you to both of you.

I will need to do more reading. So there is the EM field and there is an observer with its reference frame. It is the reference frame speed relative to the EM field that will determine if the observer will see the E field, the B field or both... Is that correct? The static/dynamic E or B field situations are just particular excitations of the EM field?

 

[robphy]

Thank you to both of you.

I will need to do more reading. So there is the EM field and there is an observer with its reference frame. It is the reference frame speed relative to the EM field that will determine if the observer will see the E field, the B field or both... Is that correct? The static/dynamic E or B field situations are just particular excitations of the EM field?

In spacetime, there is an antisymmetric tensor field $F_{ab}$ and a [possibly distributional] source tensor field $J_{a}$ (satisfying maxwell's equations) at each event. Suppose there is an observer's worldline. For simplicity, assume an inertial observer.

Given the inertial observer's 4-velocity $u^a$, one can express $F_{ab}$ in terms of two vector fields `$\vec E$ and $\vec B$ that are purely spatial according to that observer, and a scalar field $\rho$ and vector field $\vec j$. These vector fields from $F_{ab}$ evolve in time according to Maxwell's Equations. In general, there are no conditions on the nature of "$\vec E$ being static" or "$\vec E$ being nonzero with $\vec B$ zero", etc... Certain configurations of $J_a$ [which determine $F_{ab}$, up to boundary conditions] and your 4-velocity $u^a$ may lead to such special cases. In general, you get them both, nonzero and dynamic.

A different inertial observer will obtain a different pair from the source $J_a$ and a different set of vector fields from $F_{ab}$, which evolve according to Maxwell's Equations.

 

[PeterDonis]

the reference frame speed relative to the EM field

There is no such thing; this concept doesn't make sense.

if the observer will see the E field, the B field or both

This depends on the observer's 4-velocity and the EM field tensor (as @robphy has pointed out), but there is no way to express this dependence in terms of "the reference frame speed relative to the EM field".

The static/dynamic E or B field situations are just particular excitations of the EM field?

"Static" fields are EM field tensors with particular special properties (which general EM field tensors don't have). Basically, they are fields for which you can find a frame in which observers at rest observe $E$ and $B$ to remain constant in time.

The term "dynamic" is generally used to describe fields that are not static.

 

[pervect]

I'll not that this reply is in the context of SR, not GR.

An observer has both a velocity, and a reference frame, associated with the observer.

The reference frame for different observers with the same velocity are related by simple translation. In cartesian coordinates, if we have a reference frame S with coordinates (x,y,z) and a reference frame S' with coordinates (x', y', z'), and both observers have the same velocity, we can write x' = x+a, y'=y+b, z'=z+c as an example of what we call translation.

At any point in the reference frame of observer S, one can measure the E and B field components of the electromagnetic field. These will be Ex, Ey, Ez, Bx, By, Bz.

If two observers are moving with different velocities, there are well defined formula that can convert time/position coordinates (t,x,y,z) from one frame to another, and formulas that can convert the E and B fields from one frame to another.

The transformations to convert times and distances are the Lorentz transform. The transformations to transform the electromagnetic field are derived from the Lorentz transform by tensor methods, which I'm not sure you are familiar with.

Regardless of whether you are able to derive them directly from the Lorentz transform, they can be summarized by the relationships in the following wiki link:

https://en.wikipedia.org/w/index.ph...netism_and_special_relativity&oldid=950774162

The notation tells how "parallel" and "perpendicular" fields transform - they transform differently. For instance, perpendicular distances are not changed by velocity, while parallel distances are. Similarly, the perpendicular and parallel components of the electromagnetic field transform in different manners.

Most of the OP's posts seem to be seeking some sort of universal reference frame. There is no such thing. However, if physical quantites and/or the compnents of a tensor are known in one reference frame, there are formulas to convert them to another.

 

[fog37]

Thank you pervect.

I think I follow your explanation. In SR, two inertial observers will see two different sets of components (Ex, Ey, Ez) and (Bx, By, Bz). In classical EM theory, it is ok to keep the two vector fields as separate fields that are jointly connected by Maxwell's equations.

As PeterDonis explained,"Static" fields are EM field tensors with particular special properties (which general EM field tensors don't have). Basically, they are fields for which you can find a frame in which observers at rest observe E and B to remain constant in time..."

That was one of my main dilemmas. An electrostatic field E is an example of EM field. No B field around and no time variability.

Electromagnetic waves, I understand, are also an example of a EM field having a particular structure/perturbation propagating through it which travels at the speed $c$. I tended to think that all EM fields had to travel at the speed of light but I guess I was wrong...

 

[Nugatory]

I tended to think that all EM fields had to travel at the speed of light but I guess I was wrong...

A field is something that has a particular value at any gven point in space at any given time. That value may be large at one point and small at another nearby point at one moment, and then a moment later be the other way around... but that doesn't mean that the field is travelling, just that its value at both points is changing over time.

 

[robphy]

In classical EM theory, it is ok to keep the two vector fields as separate fields that are jointly connected by Maxwell's equations.

You don’t have a choice. You must keep them and you can’t forget the source fields $\rho$ and $\vec \jmath$ to satisfy Maxwell’s Equations. (You could package the electric and magnetic fields differently... but it’s just a change of variables.)

 

[fog37]

Thank you robphy.

I think, unless I am wrong, that the two source fields are not independent and are related by the continuity equation. Fundamentally, the current density a spatial point is just the local rate of change of charge density at that point...

 

[robphy]

Yes, not totally independent.
But note that the continuity equation is a algebraic relation among certain derivatives of the 4-current—not among the components themselves. At each event in Spacetime, one has to specify all 4 components of the 4-current.

 

[fog37]

So, all our discussion is about the EM field in the context of special relativity. The inertial observers moves or not relative to the charge density $\rho$ and the current density $j$ so either the B, E, or both fields are presents. What happens in general relativity to the EM field and its tensor?

 

[Nugatory]

What happens in general relativity to the EM field and its tensor?

The beauty of tensor methods is that they work just as well in curved spacetime as in flat spacetime. The tensor has a value at any given point in spacetime, and that's the value we use to calculate the electrical and magnetic fields measured by an observer whose worldline passes through that point.

 

[robphy]

Just do: , —--> ;

 

[Ibix]

Just do: , —--> ;

Translation for anyone not familiar with tensor notation: you replace partial derivatives in the equations with covariant derivatives. Or, more formally, they were always covariant derivatives, but in flat spacetime (and a coordinate basis?) the covariant derivative simplifies to the partial derivative.

 

[vanhees71]

I think, before we discuss the transformation properties of the em. field under Lorentz transformations, we should get the physical concept of an em. field right. It's the very beginning of classical electrodynamics to define the electromagnetic quantities.

As any fundamental physics also electromagnetism starts from observations. The final result of all these observations after some centuries of worked is put into a theory, which are basically Maxwell's equations and the Lorentz-force formula.

So everything starts with the observation that matter has, besides the "mechanical" property of mass, another intrinsic property, the electric charge. In contradistinction to mass, which is always positive, electric charge comes with positive and negative sign and the fundamental property is that for two static charged "pointlike" bodies there's the Coulomb force on body 1, located at $\vec{r}_1$ due to the presence of the body located at $\vec{r}_2$,
$$\vec{F}_{12}=\frac{q_1 q_2}{4 \pi \epsilon_0} \frac{\vec{r}_1-\vec{r}_2}{|\vec{r}_1-\vec{r}_2|^3}.$$
Another important discovery in connection with em. was Faraday's ingenious idea to introduce the field concept into physics. The idea is that there are not "instantaneous actions at a distance" but that another property of electrically charged bodies is that they carry an electric field with them. For our point-like particle 2 this field is given by the Coulomb field,
$$\vec{E}(\vec{r}) = \frac{q_2}{4 \pi \epsilon_0} \frac{\vec{r}-\vec{r}_2}{|\vec{r}-\vec{r}_2|^3}.$$
This is a vector-valued quantity dependent on the point of observation $\vec{r}$, i.e., it is defined as a function of position (and in the general case of non-static situations also time), i.e., a vector field. It can be observed by using a test charge $q_1$ at point $\vec{r}_1$ and measuring the Coulomb force, given above, which now is written as
$$\vec{F}_{12}=q_1 \vec{E}(\vec{r}_1).$$
The important feature is that now this is a local law, i.e., the force is interpreted not as action at a distance but caused due to the electric field at the position $\vec{r}_1$ of the particle under investigation, which field itself is there as another fundamental property of matter carrying a non-zero electric charge.

In a very similar way one can introduce also the magnetic field (caused by either electric-charge currents or ferromagnetism which is due to the presence of spin of the electrons in the matter) by the corresponding force on our test particle $q_1$,
$$\vec{F}=q_1 \vec{v}_1 \times \vec{B}(\vec{r}_1).$$

As it has turned out a little later, electric an magnetic fields are only distinct properties of charged and/or magnetized matter in the static case. The full picture is described by Maxwell's equations as finally formulated in terms of a relativistic (classical) field theory.

The final properties according to this theory are that indeed the electric and magnetic field components together build one entity, the electromagnetic field and the decomposition into electric and magnetic field components is dependent on the reference frame of the observer, but the field as a whole is a frame-independent concept, described by an antisymmetric 2nd-rank tensor field.

Another consequence of the relativistic nature of the em. field is that indeed it turns out that temporal changes due to changes of the charge-current-magnetization distributions, which are the sources of the field, can not propagate faster than the speed of light, which leads to electromagnetic waves. As should have become clear, an electromagnetic field itself doesn't have a speed, but they are a fundamental entity due to the presence of the sources, i.e., they are themselves part of the properties of matter, and temporal changes of the field show wave properties, for which you can define various speeds as for any wave like phase and group velocities etc.

It's indeed a bit like water waves as mentioned above: The water is always there, no matter whether it's in wave-like motion or not. The most important distinction for the em. field is that it can be 0, i.e., there's no conservation law for the electromagnetic field (only for electric charge), i.e., the electromagnetic field can get also 0, when no (relevant) sources are present.

 

[fog37]

Thank you! I am enjoying this thread.
As far as tensors go, I understand they are a generalization of the basic concept of tensor. Tensors have components which are numbers that change value, according to specific transformations. For example Lorenz transformation change the components value (same Cartesian coord. system) for different observers moving at different constant velocities.

The tensor components can change with these scenarios:

1. Change of the reference frame (a difference observer observes the phenomenon) keeping the same coordinate system.
2. Change of the coordinate system while keeping the same reference frame
3. Change both the reference frame and its associated coord. system

Is that correct? Also, do the transformation we discussed in this thread refer to CASE 1?

My definition of reference frame: an origin O where a hypothetical observers sits together with a set of other spatial points which are at rest relative to the origin O. The frame is also equipped with a way to measure time (clocks located at every spatial position) and a way to measure position of each spatial point part of the reference frame (a coordinate system, three axes, orthogonal or not).

Thanks

 

[PeterDonis]

these scenarios

Your 1. and 2. are impossible. "Reference frame" as the term is being used in this thread is a synonym for "coordinate system", so it's impossible to change one but not change the other.

 

[PeterDonis]

My definition of reference frame

Is the same as "coordinate system". Coordinate systems include coordinates for time as well as space. What you describe is just a physical realization of how coordinates in a reference frame (more precisely a global inertial reference frame in flat spacetime) are measured.

 

[fog37]

Your 1. and 2. are impossible. "Reference frame" as the term is being used in this thread is a synonym for "coordinate system", so it's impossible to change one but not change the other.

Thank you. As far as case 1, I envisioned two observers moving relative to each other and both using Cartesian coordinates to specify the spatial locations. For case 2, I envision one single observer considering a specific tensor and changing from Cartesian to, say, spherical coordinates. The components of the same tensor will change.

 

[PeterDonis]

As far as case 1, I envisioned two observers moving relative to each other and both using Cartesian coordinates to specify the spatial locations.

Then they are using two different coordinate systems, so changing from one observer to the other changes both the coordinate system and the reference frame. The fact that they are both using Cartesian coordinate systems does not mean they are using the same Cartesian coordinate system; they're not.

For case 2, I envision one single observer considering a specific tensor and changing from Cartesian to, say, spherical coordinates.

Then the definition of "spatial points" changes, because you change the spatial coordinates attached to them, so you are changing both the coordinate system and the reference frame.

The difference between this case and case 1 is that you only have one observer, not two observers in relative motion; but the proper way to say that is to simply say "one observer". There's no need to drag in anything about "reference frames" at all.

 

[fog37]

Then they are using two different coordinate systems, so changing from one observer to the other changes both the coordinate system and the reference frame. The fact that they are both using Cartesian coordinate systems does not mean they are using the same Cartesian coordinate system; they're not.
Then the definition of "spatial points" changes, because you change the spatial coordinates attached to them, so you are changing both the coordinate system and the reference frame.

The difference between this case and case 1 is that you only have one observer, not two observers in relative motion; but the proper way to say that is to simply say "one observer". There's no need to drag in anything about "reference frames" at all.

Thank you, PeterDonis. In regards to your reply to case 2, for example, a rotation of the Cartesian coord. system about the same origin $O$ seems to not represent a change in reference frame but re-labels the same physical points. I agree that the unrotated and rotated coord. system, although both Cartesian, are different but appear to pertain to the same reference frame (same observer). Unless, as you mention, a change of coord. system automatically changes the reference frame as well since they are intimately connected...

 

[Ibix]

a rotation of the Cartesian coord. system about the same origin $O$ seems to not represent a change in reference frame but re-labels the same physical points.

I would regard this as a change of reference frame. Think of Einstein's physical implementation of an inertial reference frame as a grid of rods equipped with clocks. Two grids in different orientations are not the same, so these are different reference frames. They share a notion of coordinate time, and potentially one spatial axis, but that's all.

It's worth noting that this is just arguing over names, not anything of any physical significance.

 

[PeterDonis]

the same reference frame (same observer)

Stop using "reference frame" when you mean "observer". They're not the same thing.

 

[PeterDonis]

It's worth noting that this is just arguing over names, not anything of any physical significance.

While, strictly speaking, this is true--changing the names of things doesn't change the physics--it is also true that physicists have a common agreed-upon vocabulary, and if one wants to be understood when talking about physics, one needs to learn that vocabulary and use it correctly. In that vocabulary, "reference frame" and "observer" mean different things, and conflating the two will only cause confusion.

 

[PeterDonis]

Stop using "reference frame" when you mean "observer". They're not the same thing.

To illustrate the difference with respect to tensors specifically, consider the following:

In a reference frame (which, as I noted before, as the term is being used in this thread is synonymous with "coordinate system"), tensors have components. Those components transform (as has been discussed in this thread) in particular ways when the coordinate system is changed.

Observers do not observe (or measure) tensor components. They observe/measure scalar invariants which are constructed by contracting tensors with vectors that describe the state of motion and the spatial orientation of the observer. These observed/measured values are, as just noted, invariants, and do not change when you change the coordinate system. To change them, you have to change the observer, and that is a different, separate thing from changing the coordinate system.

For example, the stress-energy tensor $T_{ab}$ describes the energy, momentum, and other properties of a continuous distribution of matter. This tensor has components in a given coordinate system.

An observer with a particular 4-velocity $u^a$ at a given event in spacetime measures the energy density of a continuous distribution of matter whose stress energy tensor at that event is $T_{ab}$ to be $\rho = T_{ab} u^a u^b$. This is a scalar invariant.

Consider how the above works in two different coordinate systems. First, consider coordinates in which the observer is at rest at the given event. Then the observer's 4-velocity in these coordinates is $u^a = (1, 0, 0, 0)$, and if we assume that the stress-energy tensor is that of "dust" (matter with zero pressure), then we will have $T_{ab} = \text{diag}(\rho, 0, 0, 0)$. So the energy density measured will be $T_{ab} u^a u^b = \rho \times 1 \times 1 = \rho$.

Now change coordinates by a boost in the $x$ direction, so that the observer's 4-velocity is now $u^a = (\gamma, \gamma v, 0, 0) = \gamma (1, v, 0, 0)$. The stress-energy tensor components in this new coordinate system will change; if you work out how the SET transforms, you will see that the components will become

$$
T_{ab} = \gamma^2 \begin{bmatrix}
\rho & - \rho v & 0 & 0 \\
- \rho v & \rho v^2 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}
$$

But the energy density the observer observes will still be the same:

$$
T_{ab} u^a u^b = {\gamma^2} \gamma^2 \left( \rho \times 1 \times 1 - 2 \rho v \times 1 \times v + \rho v^2 \times v \times v \right) = \rho \gamma^4 \left( 1 - 2 v^2 + v^4 \right) = \rho
$$

 

[Ibix]

it is also true that physicists have a common agreed-upon vocabulary

I was actually thinking of the difference between coordinate system and reference frame when I wrote that, specifically in the context of a spatially rotated Cartesian system. Perhaps context did not make that as clear as I thought. They are absolutely different coordinate systems. Are they different reference frames? I'd think the answer to that one is a little vaguer, because often I only care about the relative velocity of my reference frames and some (or all) of the details of the spatial coordinates fall under "don't care".

Perhaps I am being a bit casual. But I've certainly written of two observers at rest with respect to each other having the same rest frame without considering details of their spatial coordinates and basis vectors.

 

[PeterDonis]

I was actually thinking of the difference between coordinate system and reference frame

What difference? The only definition of "reference frame" that I can see being used in this thread is the one that is synonymous with "coordinate system". If you have a different definition in mind, then please state it explicitly (and, ideally, give a reference for it).

often I only care about the relative velocity of my reference frames and some (or all) of the details of the spatial coordinates fall under "don't care".

"Don't care" is very different from "don't exist".

 

[Ibix]

What difference? The only definition of "reference frame" that I can see being used in this thread is the one that is synonymous with "coordinate system".

MTW refer me to local Lorentz frames on p. 217, which defines a frame as a coordinate system with orthonormal coordinates with coordinate basis such that the metric is $\eta_{\alpha\beta}$ and $\Gamma^\alpha_{\beta\gamma}=0$, plus some other restrictions that I think are only germane to curved spacetime. That is more restrictive than "coordinate system", more restrictive than "frame" as used by fog37 in post #23 (and I think implicitly accepted by you in your response) which allows spherical polars for the spatial coordinates, and less restrictive than the definition of "frame" implied by pervect in #9 where he relates frames by translations and does not mention rotations.

"Don't care" is very different from "don't exist".

Fair enough - but even in this thread there's a degree of slop in how people are using "frame". So learning to accept a degree of variation in definitions (i.e., not getting too hung up on the precise definition of a term when it may not be generally used) and recognising when that does and does not have consequences is an important skill. You may recall resolving my confusion over Kruskal diagrams by pointing out that different sources label regions III and IV differently...

 

[PeterDonis]

MTW refer me to local Lorentz frames

Yes, this is a particular kind of coordinate system/frame. It is not a general definition of "coordinate system" or "reference frame".

even in this thread there's a degree of slop in how people are using "frame"

Which is why I asked you to state explicitly what definition you are using.

not getting too hung up on the precise definition of a term when it may not be generally used

I don't think the term "reference frame" can be said to be not generally used.

I agree that sometimes differences in definitions don't matter and aren't worth getting hung up on. In this thread, however, I think confusion over the meanings of terms is contributing to the OP's confusion over the physics involved. So I think in this case it's worth being more precise and explicit about the definitions of terms, and also about the physical implications. I tried to do some of that in post #29.

 

[vanhees71]

I was actually thinking of the difference between coordinate system and reference frame when I wrote that, specifically in the context of a spatially rotated Cartesian system. Perhaps context did not make that as clear as I thought. They are absolutely different coordinate systems. Are they different reference frames? I'd think the answer to that one is a little vaguer, because often I only care about the relative velocity of my reference frames and some (or all) of the details of the spatial coordinates fall under "don't care".

Perhaps I am being a bit casual. But I've certainly written of two observers at rest with respect to each other having the same rest frame without considering details of their spatial coordinates and basis vectors.

I'd also say that inertial reference frames which just differ by a static spatial rotation and/or a static space-time translation are to be taken as different reference frames, because the components of four-vectors and four-tensors change, but that's also kind of semantics.

Further an observer defines local (not necessarily inertial) reference frames, because you can always define a set of tetrades along his time-like worldline with the temporal basis vector given by $u^{\mu}$ (his four-velocity with $u_{\mu} u^{\mu}=1$) and three Minkowski-orthonormal space-like vectors. You can even make these "non-rotating" by Fermi-walker transporting an arbitrary initial tetrade.

 

[PeterDonis]

an observer defines local (not necessarily inertial) reference frames, because you can always define a set of tetrades along his time-like worldline with the temporal basis vector given by $u^{\mu}$ (his four-velocity with $u_{\mu} u^{\mu}=1$) and three Minkowski-orthonormal space-like vectors.

Note that this is a different definition of "reference frame" (the more technical term is "frame field") from the one that has been used up to now in this thread (and which you also used in the first paragraph of your post). This definition has nothing to do with coordinates (it can be defined without even making a coordinate choice at all).