Direction of E field transforms like direction of a stick

[bcrowell]

A while back ( https://www.physicsforums.com/threa...-transforms-like-direction-of-a-stick.730373/ ), I posted a discussion here of the following seemingly mysterious relationship between two transformation rules:

(1) If a field is purely electric in a particular frame, then its direction in other frames is given by scaling of $\tan \theta$ by $\gamma$.

(2) If a stick has a certain orientation in its own rest frame, then its direction in other frames obeys the same rule.

The methods I knew for proving these facts were completely different, so I couldn't figure out whether there was some simple or deeper reason for the correspondence. I think I finally figured out a simple explanation.

Consider a stick with charges +q and -q fixed at the ends. The stick is nonrotating and moving inertially. In the stick's rest frame K, there is a field line originating from +q and terminating on -q which coincides with the stick.

Now consider frame K' moving in some direction relative to the stick. The field due to each charge points toward or away from its present instantaneous position in K' as well as K (a counterintuitive fact that can be proved in various ways, e.g., see Purcell section 5.6). Therefore each field, at the stick, is parallel to the stick, and we again have a field line in K' that coincides with the stick. Since the transformation of the field is independent of how the field was created, this holds for any field that is purely electric in the original frame.

 

[andrewkirk]

Nice!

 

[vanhees71]

That's a nice derivation, but isn't

The field due to each charge points toward or away from its present instantaneous position in K' as well as K (a counterintuitive fact that can be proved in various ways, e.g., see Purcell section 5.6).

right only for a uniformly moving charge (or a charge density that is static in some inertial frame and then looked at from a different inertial frame which is boosted relative to the rest frame with some constant velocity)?

To derive the "counterintuitive fact" you don't need the overcomplicated treatment in Purcell (I guess, you mean Vol. 2 of the Berkeley Physics Course, which is nice in the sense that it treats the relativistic electromagnetics, but sometimes it's a bit overcomplicating thins; I prefer Landau-Lifshitz Vol. 2 which is usually much more straight forward in its derivations and treats E+M also as a relativistic theory right from the beginning, as it should be).

Let $x'=(t',\vec{x}')$ be the space-time coordinates of the restframe. Then
\begin{equation}
\label{1}
A'(x')=\begin{pmatrix} \frac{q}{4 \pi |\vec{x}'|} \\ 0 \\0 \\0 \end{pmatrix}.
\end{equation}
This we have to write covariantly. As building blocks we have only the four-velocity of the particle $u$ and its four-position vector $x$ to express the four-potential. In the rest frame we have $u'=(1,0,0,0)$ and thus $\vec{x}'^2=(u' \cdot x')^2-x' \cdot x')$ and thus in an arbitrary inertial frame we simply have
\begin{equation}
\label{2}
A(x)=\frac{q}{4 \pi \sqrt{(u \cdot x)^2-x \cdot x}} u.
\end{equation}
The electric field is given by
$$\vec{E}=-\partial_t \vec{A}-\vec{\nabla} A^0=\frac{q \gamma (\vec{x}-\vec{v} t)}{4 \pi \sqrt{(u \cdot x)^2-x \cdot x}^3}.$$
The field at each moment points radially out from the momentaneous position of the point charge.

 

[bcrowell]

That's a nice derivation, but isn't

The field due to each charge points toward or away from its present instantaneous position in K' as well as K (a counterintuitive fact that can be proved in various ways, e.g., see Purcell section 5.6).

right only for a uniformly moving charge (or a charge density that is static in some inertial frame and then looked at from a different inertial frame which is boosted relative to the rest frame with some constant velocity)?

Yes, I stated in #1 that the stick was nonrotating and moving inertially, so each charge is moving inertially.

Thanks for showing Landau's derivation of the instantaneously radial direction of the field -- that's nice, and I hadn't seen it done that way before. I have a simple proof of the same fact in a geometric style, which I'll post here later.

 

[bcrowell]

Here is my attempt at a simple geometrical proof that the electric field of an inertially moving charge points toward its instantaneous position.

Let a charge have inertial world-line W, and let it be at rest in frame K. We make use of three facts:

[1] The field in K is purely electric.

[2] The field in K is static.

[3] The electric field in any frame must obey Gauss's law on a hypersurface of simultaneity.

Now consider frame K', which is moving relative to K. Because of fact [1] and the structure of the transformation laws for the fields, we have:

[4] The field E' as measured in K' depends in a linear way on the field E.

Construct 3-surfaces of simultaneity S and S' in frames K and K', such that they intersect at a point on W. Take any line L' that lies in S' and intersects W; construct its projection L into S. Take any two points P1' and P2' on L', and construct their projections into S, P1 and P2. Then by [2] the fields are related by E(P1)=E(P1') and E(P2)=E(P2'). By [4], we then have:

[5] E'(P1') is parallel to E'(P2'), and |E'(P1')|/|E'(P2')| = |E(P1')|/|E(P2')|. Therefore E' is a $1/r^2$ field along L'.

By axial symmetry, E' has a vanishing azimuthal component, so the divergence of E' equals $(1/r^2)(\partial (r^2E'_r)/\partial r) + (\ldots)(\partial (\sin \theta E'_\theta)/\partial \theta)$. By [5], the first term in this expression vanishes. Therefore by Gauss's law ([3]), $\sin \theta E'_\theta$ must be independent of $\theta$for fixed r. But $\sin \theta E'_\theta$ equals zero at $\theta=0$, and therefore it must be zero everywhere. QED

 

[Dale]

(1) If a field is purely electric in a particular frame, then its direction in other frames is given by scaling of tanθtanθ\tan \theta by γγ\gamma.

(2) If a stick has a certain orientation in its own rest frame, then its direction in other frames obeys the same rule.

The E field is part of a rank two antisymmetric tensor. Does this hint that the direction of a stick may also be part of a rank two antisymmetric tensor?

 

[bcrowell]

The E field is part of a rank two antisymmetric tensor. Does this hint that the direction of a stick may also be part of a rank two antisymmetric tensor?

Yeah, I've thought about trying to get at it from that angle. Pervect had some thoughts about this, which we talked about a little starting with this post of his: https://www.physicsforums.com/threa...ike-direction-of-a-stick.730373/#post-4615087 . I don't know if he thought he had a definitive argument or if he was just tossing out an idea. It all seemed a little hazy to me. I was never able to flesh it out to my own satisfaction.

There is some nice discussion of geometrical interpretations of various antisymmetric tensors in the books by Schouten (Tensor analysis for physicists) and Burke (Applied differential geometry). A two-dimensional area can be represented as an antisymmetric rank-2 tensor. If you imagine the world-tube of a stick, it looks like a ribbon, and a piece out of it could be represented by such a tensor. I think it could be written as a wedge product of a timelike vector and a spacelike vector. An electromagnetic field that is purely electric in some frame can also be written as such a wedge product, with the timelike vector being the state of motion of an observer who sees it as purely electric. So maybe one could get somewhere in this way, but I'm not fluent enough in this language to fill in the details.

Once one is convinced that the electric field of an inertially moving charge points toward the charge's instantaneous position, the part about transforming like a stick comes out pretty easily by the argument I gave in #1. I would like to figure out how to abstract out the simple geometrical features of the proof of the direction of the field of the moving charge from the other details. Both Landau's argument in #3 and the one in my #5 deal with details that seem inessential, such as the $1/r^2$ nature of the field, or equivalently the fact that it obeys Gauss's law.

For the field of the moving charge, it seems like it ought to be possible just to apply some simple symmetry argument. The naive expectation would be that the field would point toward the retarded position of the charge, i.e., toward the intersection of the charge's world-line with the past light cone of the point of interest. But this means treating the past light cone differently than the future light cone, and that's an asymmetry that's not present in Maxwell's equations.

 

[pervect]

The E field is part of a rank two antisymmetric tensor. Does this hint that the direction of a stick may also be part of a rank two antisymmetric tensor?

This was similar to my thinking. I'm not 100% sure I'm not missing something though. But I'll outline the idea for a more considered evaluation. If one considers all the E-fields from a point charge geometrically, one get the rank-2 dual of the Faraday tensor, the Maxwell tensor, as on pg MTW 109. (There are others, but this seems the most natural to the problem at hand).

If one takes the wedge product of x^y, one gets a stack of tubes that should represent "sticks" pointing in the z direction, in much the same way as the tubes of the Maxwell tensor represent the electric field.

So, if the anti-symmetric rank 2 tensor can represent both sticks and the E-field, as Dale says, it's not terribly surprising they transform in the same way.

 

[vanhees71]

Yeah, I've thought about trying to get at it from that angle. Pervect had some thoughts about this, which we talked about a little starting with this post of his: https://www.physicsforums.com/threa...ike-direction-of-a-stick.730373/#post-4615087 . I don't know if he thought he had a definitive argument or if he was just tossing out an idea. It all seemed a little hazy to me. I was never able to flesh it out to my own satisfaction.

There is some nice discussion of geometrical interpretations of various antisymmetric tensors in the books by Schouten (Tensor analysis for physicists) and Burke (Applied differential geometry). A two-dimensional area can be represented as an antisymmetric rank-2 tensor. If you imagine the world-tube of a stick, it looks like a ribbon, and a piece out of it could be represented by such a tensor. I think it could be written as a wedge product of a timelike vector and a spacelike vector. An electromagnetic field that is purely electric in some frame can also be written as such a wedge product, with the timelike vector being the state of motion of an observer who sees it as purely electric. So maybe one could get somewhere in this way, but I'm not fluent enough in this language to fill in the details.

Once one is convinced that the electric field of an inertially moving charge points toward the charge's instantaneous position, the part about transforming like a stick comes out pretty easily by the argument I gave in #1. I would like to figure out how to abstract out the simple geometrical features of the proof of the direction of the field of the moving charge from the other details. Both Landau's argument in #3 and the one in my #5 deal with details that seem inessential, such as the $1/r^2$ nature of the field, or equivalently the fact that it obeys Gauss's law.

For the field of the moving charge, it seems like it ought to be possible just to apply some simple symmetry argument. The naive expectation would be that the field would point toward the retarded position of the charge, i.e., toward the intersection of the charge's world-line with the past light cone of the point of interest. But this means treating the past light cone differently than the future light cone, and that's an asymmetry that's not present in Maxwell's equations.

In some sense the "area" interpretation of the electric field is also very physical. In some sense it's related to the invariant charge in a region, enclosed in a 3D spatial volume via Gauss's Law:
$$\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E}(t,\vec{x})=\int_V \mathrm{d}^3 \rho(t,\vec{x})=Q_V.$$
So the electric field is closely related to its flux through a surface.

 

[bcrowell]

If one takes the wedge product of x^y, one gets a stack of tubes that should represent "sticks" pointing in the z direction, in much the same way as the tubes of the Maxwell tensor represent the electric field.

So, if the anti-symmetric rank 2 tensor can represent both sticks and the E-field, as Dale says, it's not terribly surprising they transform in the same way.

I could be wrong, but the following considerations make me think that this probably can't be expanded into a valid proof. The fact that we're trying to prove is a fact that's limited to the special case where the field is purely electric in the original frame. The specialness of this frame is analogous to the specialness of the stick's rest frame. But I don't see any place in this approach where we can squeeze in this necessary assumption that there is some special frame.

MTW's egg crate diagram (figure 4.5, p. 109) depicts a 2-form that represents both the electric and the magnetic field. Suppose we have a field that is not of the special form we're talking about. We could draw an egg crate to represent this field. Maybe we can also say that in a particular frame, the electric field points in a particular direction. Now we transform to another frame, and it could very well happen that in this new frame, the electric field vanishes. The egg crate still exists, but the electric field lines certainly have not transformed like sticks.

In the language of forms, I think the electromagnetic field is a bivector, or the 2-form that is dual to that bivector. A bivector that can be represented as a wedge product of two vectors is called simple. The special property of our field can be expressed by the fact that not only is the field simple, but it can be represented as the wedge product of a timelike vector with a spacelike vector. The timelike vector represents the state of motion of an observer who says that the field is purely electric. I don't think we can make an argument in the language of forms without somehow appealing to the existence of this additional structure.

Let's not lose sight of the fact that the property we're talking about is easy to prove in index-gymnastics language. Certainly we could rewrite any such proof in the language of forms, invoking all the structure I described in the preceding paragraph, but I don't see any reason to think that it would look any simpler.