Relativistic explanation of electromagnetism

[Karl Coryat]

TL;DR Summary

Why a test charge at rest in the lab frame does not experience a force from a current

I am intrigued by the special-relativity explanation of magnetic force discussed here (linked from the physicsforums FAQ): http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#Length_Contraction

Naively, from this explanation, it seems that a test charge at rest in the lab frame should experience a force from a current-carrying wire, since the electrons' fields are Lorentz-contracted relative to the test charge, but the nuclei fields are not. And, that the test charge should experience no force only if the positive and negative charges in the wire are moving in equal and opposite directions relative to the test charge, i.e., when the test charge is moving along the wire at 1/2 the drift velocity. But that's not what happens. What am I missing?

 

[PeroK]

Summary:: Why a test charge at rest in the lab frame does not experience a force from a current

I am intrigued by the special-relativity explanation of magnetic force discussed here (linked from the physicsforums FAQ): http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#Length_Contraction

Naively, from this explanation, it seems that a test charge at rest in the lab frame should experience a force from a current-carrying wire, since the electrons' fields are Lorentz-contracted relative to the test charge, but the nuclei fields are not. And, that the test charge should experience no force only if the positive and negative charges in the wire are moving in equal and opposite directions relative to the test charge, i.e., when the test charge is moving along the wire at 1/2 the drift velocity. But that's not what happens. What am I missing?

If we assume the test charge is positive:

In the rest frame of the current there is: a) a net postive charge density, which produces an electric field; b) a positive current, which produces a magnetic field; and, c) a moving charge, moving in the same direction as the current.

The electric field produces an outward force on the particle, and the magnetic field produces an inward force on the particle. If you do the maths, these two forces are equal and opposite and the test particle remains at rest.

 

[Karl Coryat]

In the rest frame of the current

Can you please specify what that means — does it mean in the rest frame of the drifting electrons (with the wire moving "backwards"), or in the rest frame of the wire (with the electrons moving "forwards")? Or something else? I'm very weak on electricity.

 

[PeroK]

Can you please specify what that means — does it mean in the rest frame of the drifting electrons (with the wire moving "backwards"), or in the rest frame of the wire (with the electrons moving "forwards")? Or something else? I'm very weak on electricity.

I meant the rest frame of the drifting electrons.

And, of course, bearing in mind that the current is by convention the flow of positive charge, saying the rest frame of the current was wrong. The rest frame of the negative current, perhaps!

 

[Karl Coryat]

Why is there a net positive charge density, producing an electric field? I assume this is independent of the Lorentz effect that produces the counterbalancing attractive magnetic field, and that the same electric field exists in the rest frame of the wire, or any other frame.

 

[PeroK]

Why is there a net positive charge density, producing an electric field? I assume this is independent of the Lorentz effect that produces the counterbalancing attractive magnetic field, and that the same electric field exists in the rest frame of the wire, or any other frame.

The net positive charge comes from length contraction. If we assume the positive charges and negative charges are equally spaced in the rest frame of the wire. Then in the rest frame of the moving charges:

1) They are further apart (in their own rest frame) and this is length contracted in the rest frame of the wire.

2) The positive charges are moving and hence the distance between them is length contracted in the frame of the moving negative charges.

You put these two factors together to get the net positive charge density.

 

[Karl Coryat]

I must be very stupid because I'm still not getting it. I'd like to understand what is producing the net positive charge density in the rest frame of the wire/test charge. (Combining different frames is confusing me.)

In the rest frame of the wire, wouldn't length contraction of the distance between the moving charges in the wire, but not of the stationary charges, result in a net negative charge density?

 

[PeroK]

I must be very stupid because I'm still not getting it. I'd like to understand what is producing the net positive charge density in the rest frame of the wire/test charge. (Combining different frames is confusing me.)

In the rest frame of the wire, wouldn't length contraction of the distance between the moving charges in the wire, but not of the stationary charges, result in a net negative charge density?

The assumption is that in the rest frame of the wire before the current starts there are equal numbers of positive and negative charges, equally spaced. Let's call the spacing between changes $L$.

After the negative charges have started moving (perhaps imagine them hopping from one atom to the next) there are still an equal number of positive and negative charges, equally spaced, but now the negative charges are moving. In the rest frame of the wire the positive and negative charges both have equal spacing $L$.

The immediate conclusion is that in the rest frame of the moving electrons they must have spacing $\gamma L$. Which is length-contracted to $L$ in the rest frame of the wire.

The next conclusion is the the positive charges, whose spacing is $L$ in their rest frame, have spacing $L/\gamma$ in the rest frame of the moving electrons.

Therefore, the positive charge density is greater in the rest frame of the moving electrons.

 

[Karl Coryat]

Thank you for your infinite patience. This must be what's tripping me up:

now the negative charges are moving. In the rest frame of the wire the positive and negative charges both have equal spacing L.

If (in the rest frame of the wire) they had equal spacing L before the current started, then why (also in the rest frame of the wire) would they still have equal spacing L when the electrons are moving but the positive charges remain at rest? Why (in the rest frame of the wire) isn't the spacing of the moving electrons now L/γ, while the positive charges remain at rest with spacing L?

 

[pervect]

Summary:: Why a test charge at rest in the lab frame does not experience a force from a current

I am intrigued by the special-relativity explanation of magnetic force discussed here (linked from the physicsforums FAQ): http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#Length_Contraction

Naively, from this explanation, it seems that a test charge at rest in the lab frame should experience a force from a current-carrying wire, since the electrons' fields are Lorentz-contracted relative to the test charge, but the nuclei fields are not. And, that the test charge should experience no force only if the positive and negative charges in the wire are moving in equal and opposite directions relative to the test charge, i.e., when the test charge is moving along the wire at 1/2 the drift velocity. But that's not what happens. What am I missing?

It's easiest to analyze a complete loop of wire.

To do the analysis properly, we impose the requirement that the loop of wire is electrically neutral, that the total number of positive charges an negative charges over the whole wire must be equal.

Clearly, we could analyze a charged loop of wire, and we'd expect such a charged loop of wire to have an electric field. But we don't want to analyze a charge loop of wire, we want to analyze a neutral loop of wire.

It's easiest to imagine a square loop of wire. Symmetry says that the + and - charge densities will be independent of angle in the lab frame. So, the densities in the lab frame for + and =- charges must be equal if we are to have overall electrical neutrality.

In the frame where the current loop is moving, the charge density is NOT independent of angle. If you carry out Purcell's analysis, you'll see that some sections of the loop , the sections aligned with the direction of a motion, have a net charge. But the total charge of all sections is still zero, since the wire is neutral.

This can be regarded as a consequence of the relativity of simultaneity, also.

 

[Karl Coryat]

It's easiest to analyze a complete loop of wire.

Oh, that's an interesting way of looking at it. So in the frame in which the loop is moving, say, vertically, one vertical side is net-negative and the other vertical side is net-positive. And the horizontal sides remain neutral. That's a great visual.

If there is no current in the wire, I assume that all sections are electrically neutral in all frames.

 

[pervect]

Yes, with no current density in the wire, given that we've assumed it's neutral, there is no force on a nearby charge - there are no fields of any kind.

 

[PeroK]

If (in the rest frame of the wire) they had equal spacing L before the current started, then why (also in the rest frame of the wire) would they still have equal spacing L when the electrons are moving but the positive charges remain at rest?

That's a different question. And not a bad one. That depends on how the current is sustained. If you imagine that the electrons bunch up in the rest frame of the wire once they are moving, then you have a net negative charge in the rest frame of the wire. The current, in that case, would attract the test positive charge in the rest frame of the wire.

But, that's then a different problem. In that case you would have a charged wire and a current. You could analyse that scenario in the rest frame of the electrons and check that there is still a net attraction of the positive charge.

The assumption for the problem at hand, however, is that we have a neutral wire with a current, hence the electrons retain their spacing in the frame of the wire even once they are moving.

 

[A.T.]

If (in the rest frame of the wire) they had equal spacing L before the current started, then why (also in the rest frame of the wire) would they still have equal spacing L when the electrons are moving but the positive charges remain at rest?

Because the electrons are still repulsing each other, so they still spread out as far as possible. Length contraction doesn't make their E-fields attractive, so they have no reason to reduce their spacing.

As others suggested, it's best to consider the entrie loop. Here is a great figure by @DrGreg:

More explanation here:
https://www.physicsforums.com/threa...-effect-of-electrostatics.577456/post-3768045

See also these threads:
https://www.physicsforums.com/threads/sr-and-magnetism.938812
https://www.physicsforums.com/threa...plain-electromagnetism-with-relativity.932270

 

[jartsa]

An electron

If (in the rest frame of the wire) they had equal spacing L before the current started, then why (also in the rest frame of the wire) would they still have equal spacing L when the electrons are moving but the positive charges remain at rest? Why (in the rest frame of the wire) isn't the spacing of the moving electrons now L/γ, while the positive charges remain at rest with spacing L?

Length contraction does not happen magically at a distance.

Two distant electrons accelerating in an uniform electric field don't care about each other.

OTOH all the particles that form a steel rod do care about each other. Internal stresses in an accelerated rod cause the rod to contract. Lack of internal stresses cause a contracted speeding rod to stay contracted.As for accelerating observer feeling like charges of the universe are pulling his charges more and more to the transverse direction - well we could explain that by the internal magnetic fields in the observer. Right? The observer becomes weak in the transverse direction.

 

[vanhees71]

Before treating that problem, please check out this Insights article, where the correct relativistic treatment of the current conducting straight wire is made. Then you can transform the fields correctly from one frame of reference (moving particle) to the other (particle at rest) properly:

https://www.physicsforums.com/insights/relativistic-treatment-of-the-dc-conducting-straight-wire/

On the other hand it's very easy to see without any such complicated calculations, because in any electromagnetic field, $F_{\mu \nu}$ the four-force on a charge $q$ is
$$K^{\mu}=m \mathrm{d}_{\tau} p^{\mu} = \frac{q} F^{\mu \nu} u_{\nu}$$
with
$$u^{\mu}=\frac{1}{c} \mathrm{d}_{\tau} x^{\mu}=\begin{pmatrix} \gamma \\ \gamma \vec{v} \end{pmatrix} \quad \text{with} \quad \gamma=\frac{1}{\sqrt{1-\vec{v}^2/c^2}}.$$
Thus the four-force is in components
$$K^{\mu} = \begin{pmatrix} \gamma q \vec{E} \cdot \vec{v}/c \\ \gamma (\vec{E}+\vec{v} \times \vec{B}/c) \end{pmatrix}.$$
Thus in a reference frame, where $\vec{v}=0$, there's no contribution to the force of the magnetic field.

 

[Karl Coryat]

Here is a great figure by @DrGreg:

Indeed that is a great diagram!

 

[PeroK]

Indeed that is a great diagram!
Summing up, the source of my confusion seems to be that we must specify that it's the current-carrying wire that's electrically neutral — my assumption was wrong that if it's electrically neutral before the current starts, it'll also be electrically neutral after.

I'm not sure what you mean here. The assumption is that the wire is neutral with or without the current.

We are free to set this problem up any way we want. We could assume that the electrons bunch up once they are moving and then we have a different problem. You might argue about the physical process that would lead to such a configuration, but there is nothing electrodynamically invalid about that. You simply postulate a very long wire with more electrons than protons near the test charge. Perhaps simply that we had a charged wire in the first place.

The important point is that whatever the configuration is in the rest frame of the wire and although the configuration is different in another frame (in terms of E & M fields) it results in the same force on the test charge in all frames.

 

[Karl Coryat]

Duly retracted above. Thank you.

 

[Karl Coryat]

ETA: I've self-answered this below. Perhaps I even did it correctly this time.

As I noted, the diagram by @DrGreg is excellent and clears up my earlier misunderstandings in this thread (and other threads like it). Thank you, @A.T. There's one thing I don't understand, though: Why are only two electrons shown in the lower-right case? Seems like the shape and spacing of the electrons (in the diagram) should be the same as in the upper-left case, since the electrons are at rest in both frames.

However, maintaining such spacing creates a problem between the lower-left and lower-right cases.

What is the explanation for the spreading-out of electrons from the upper-left to lower-right cases, since in both cases the electrons are seen as being at rest? I realize these are physically different situations and you cannot simply transform between them, but I can't get past the idea that the same electrons are seen as being at rest in each case, and yet their spacing is seen as different.

The only difference between the upper-left and lower-right cases seems to be that in the latter, the positive charges are moving to the left. Everything else is the same. Surely this can't be responsible for the electrons spreading out.

 

[Karl Coryat]

Okay, so, I've been reading PF posts on this topic for several days now. Everyone seems to have the same confusions as I did. But I think I can now answer my own question above, as well as others:

When the current is on, the electrons do length-contract in the wire frame (bottom-left). Their separation would also length-contract, except that we've specified the boundary condition that the wire is uncharged, and there remain the same number of electrons in the loop as without current, so necessarily the electrons must remain at the same separation and charge density. However, in the co-moving frame (lower right), the electrons and their spacing un-contract — and thus the initial boundary condition results in a lower charge density than exists in the no-current/wire-frame case (upper left). Perhaps one could say that when a current is applied, the electrons acquire a greater proper length between them, which becomes apparent in the co-moving frame. This manifests as a positive charge on that side of the wire.

This also answers why a straight wire uncharged in the lab frame, for example, doesn't spontaneously acquire a charge due to electron length contraction when the current is switched on. With the current, every electron leaving an arbitrary region of the wire is replaced by an electron entering at the other end of the region. They don't contract closer together, the reason for which becomes clear when the wire is a loop.

Finally, this answers the inevitable question of why the test particle doesn't experience zero electrostatic force only when it is moving at half the velocity of the electrons (in which case, in the test particle's frame, the electrons and + nuclei are moving at the same speed in opposite directions). The electrons and the nuclei don't enjoy the same freedom. The electrons can move around; the nuclei can't. This breaks the symmetry. Electrons and their distribution conform to the boundary condition of 0 net charge by acquiring a greater proper distance between them, but the nuclei cannot do this.

Did I get all of that right?

 

[vanhees71]

Again the hint that the usual treatment of the current conducting wire is not correct in the literature. The correct treatment can be found in my insights article and the literature cited therein:

https://www.physicsforums.com/insights/relativistic-treatment-of-the-dc-conducting-straight-wire/

Having the correct relativistic solution you can put a test charge outside the wire and calculate the force (most conveniently in the sense of the relativistically covariant Minkowski four-force vector) on this test particle in any frame of reference you like. You just have to transform the fields of the wire as well as the force, and you see that everything is consistent, because everything is relativistically covariant and thus the Lorentz transformation consistently transforms between the components of the electromagnetic field tensor (Faraday tensor) and the Lorentz force (in covariant Minkowski four-force form) as Minowski-space tensor and vector components.

 

[Karl Coryat]

If I'm understanding this correctly, then the following statement of @PeroK is inaccurate?

The assumption is that the wire is neutral with or without the current.

...And that if a wire is electrically neutral in the wire frame without current, then with current, it will only be neutral in the electrons' rest frame?
And that if the wire is a loop, in this frame the opposite side becomes positively charged when current flows?

 

[vanhees71]

Yes, this statement is wrong, as my Insights article shows in two ways.

 

[PeroK]

Yes, this statement is wrong, as my Insights article shows in two ways.

An assumption is not wrong, per se. It's not physically impossible to accelerate two objects so that they do not get closer together in the original rest frame. The diagram posted in post #14 is not fundamentally physically impossible.

That assumption may not be supported by a valid model of the DC in a wire - although it is an assumption used in the much of the literature - which is what your Insight was aimed at.

Your Insight presents a more realistic model of DC current in a wire, but it takes things beyond a "B" level treatment of EM.

 

[Karl Coryat]

@vanhees71, in one of your comments on your article, you mentioned, "The battery must deliver some net negative charge (i.e., electrons)." How is the charge (in the wire frame) accounted for in the case of a superconducting loop in which a current is induced?

 

[A.T.]

The electrons and the nuclei don't enjoy the same freedom. The electrons can move around; the nuclei can't. This breaks the symmetry. Electrons and their distribution conform to the boundary condition of 0 net charge by acquiring a greater proper distance between them, but the nuclei cannot do this.

Yes, that's the key.

 

[vanhees71]

Well, but all the trouble with this specific question is due to an improper treatment of the relativistic description of this problem. As Einstein said, one should explain things as simple as possible but not simpler!

 

[vanhees71]

@vanhees71, in one of your comments on your article, you mentioned, "The battery must deliver some net negative charge (i.e., electrons)." How is the charge (in the wire frame) accounted for in the case of a superconducting loop in which a current is induced?

That's a very interesting question, I've to think about.

 

[Karl Coryat]

This thread has shown how a neutral current-carrying wire creates an electrostatic force for a test charge moving relative to the wire/current, which we call the magnetic force. Excellent, thank you.

But I seek a more complete understanding.

If we replace the test charge with a chunk of ferromagnetic material, we find that it experiences a magnetic force even if both it and the wire are uncharged, and even if it is stationary relative to the wire.

How is this explained in terms of electrostatic forces and the Lorentz transform? The PF threads I've reviewed, and the linked resources (such as http://physics.weber.edu/schroeder/mrr/MRRtalk.html), only discuss test charges; and all explanations I've found of ferromagnetic materials appeal directly to the magnetic force, with none of that relativistic goodness. Perhaps Purcell covers it, but I don't have that book. Does it have to do with eddy currents?

 

[Orodruin]

a neutral current-carrying wire creates an electrostatic force for a test charge moving relative to the wire/current, which we call the magnetic force.

We call it electromagnetic force. This is because, as you have noted, how the electromagnetic field splits into an electric and a magnetic field is frame dependent.

If we replace the test charge with a chunk of ferromagnetic material, we find that it experiences a magnetic force even if both it and the wire are uncharged, and even if it is stationary relative to the wire.

The force is created due to how the material reacts to the external magnetic field by creating induced currents.

 

[Karl Coryat]

The force is created due to how the material reacts to the external magnetic field by creating induced currents.

Ordinarily we associate induction with a changing magnetic field. Here, we have a static magnetic field. Is it that the field is always changing for the moving electrons in the ferromagnetic material? And, that the resulting currents set up a calculable electric-charge buildup on the surface of the material, and an opposite charge on the wire, which causes the wire to attract the material?

I realize this is more complicated than the test-charge case, so if anyone knows of a good online resource, I'd love to save you the time of having to explain the details.

 

[vanhees71]

This thread has shown how a neutral current-carrying wire creates an electrostatic force for a test charge moving relative to the wire/current, which we call the magnetic force. Excellent, thank you.

But I seek a more complete understanding.

If we replace the test charge with a chunk of ferromagnetic material, we find that it experiences a magnetic force even if both it and the wire are uncharged, and even if it is stationary relative to the wire.

How is this explained in terms of electrostatic forces and the Lorentz transform? The PF threads I've reviewed, and the linked resources (such as http://physics.weber.edu/schroeder/mrr/MRRtalk.html), only discuss test charges; and all explanations I've found of ferromagnetic materials appeal directly to the magnetic force, with none of that relativistic goodness. Perhaps Purcell covers it, but I don't have that book. Does it have to do with eddy currents?

Ferromagnetism is explained by the fact that elementary particles (in this case the electron) do not only carry electric charge but also a magnetic dipole moment, related to their spins. A permanent magnet is a material, where a macroscopic number of spins is oriented in one direction, because (at the given temperature) it is energetically more favorable for the associated magnetic moments being directed in one direction than being in random orientation as is the case in usual materials. To understand this completely from first principles you need quantum many-body theory (in this case the non-relativistic version is sufficient).

I also do not think that one can derive in a logical way from electrostatics the complete electrodynamics just using the relativistic spacetime structure. A more convincing argument is the analysis of relativistic quantum field theory in view of the symmetry group of Minkowski space, which is Poincare symmetry (i.e., symmetry under translations in space and time, rotations of space, and Lorentz boosts; to be more precise the here relevant symmetry group is the proper orthochronous symmetry group since it's known that the weak interaction breaks the discrete symmetries of space reflections and time reversal; only the "grand reflection" CPT is to the best of our knowledge a symmetry in accordance with the predictions from local relativistic QFT). Then you find out that causal theories can be built via representations with local fields of a given mass with $m^2 \geq 0$ and spin $s \in \{0,1/2,1,\ldots \}$.

The massless case $m=0$ is special, and for $s=1$ you necessarily get a gauge theory, if you don't want to have continuous intrinsic polarization-degrees of freedom. Since such a thing has never been observed, that's a plausible additional assumption, but as soon as you have the necessity of a gauge theory electromagnetism follows (together with using only the minimal number of necessary field-derivatives in the Lagrangian, i.e., keeping the corresponding QFT Dyson-renormalizable) quite inevitably. Also the generalization to non-Abelian gauge groups (a la Yang and Mills) is pretty obvious. These considerations, together with a lot of empirical input from the last decades of experimental HEP physics, lead to the Standard Model of elementary particle physics, which is the most robust theory of matter ever.

Classical electrodynamics thus indeed follows quite convincingly from the mathematical structure of Minkowski space, i.e., the special-relativistic spacetime model, but not from electrostatics alone though electrostatics gives a good hint at the fact that the electromagnetic field should be most simply be describable by a (Lorentz-)vector field. That it is massless is an empirical fact and cannot be derived from more fundamental (symmetry) assumptions.

 

[Hans de Vries]

Summary:: Why a test charge at rest in the lab frame does not experience a force from a current

I am intrigued by the special-relativity explanation of magnetic force discussed here (linked from the physicsforums FAQ): http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#Length_Contraction

Naively, from this explanation, it seems that a test charge at rest in the lab frame should experience a force from a current-carrying wire, since the electrons' fields are Lorentz-contracted relative to the test charge, but the nuclei fields are not. And, that the test charge should experience no force only if the positive and negative charges in the wire are moving in equal and opposite directions relative to the test charge, i.e., when the test charge is moving along the wire at 1/2 the drift velocity. But that's not what happens. What am I missing?

This question is a simpler version of the always recurring question:

"How to explain Magnetism as a relativistic side effect of the Electric Field"

----------------------------------------------------

The answer to the OP's question:

- A test-charge at rest is only subject to an Electric Lorentz force.

- In the rest-frame the Lorentz force is calculated by integrating over all relativistic transformed Electric fields.

- Charge is Lorentz invariant. A wire with an equal number of negative and positive charges has a net charge of zero

- The Electric field of a moving charge changes under Lorentz transform.

- But the integral over all Lorentz transformed electric fields of all electrons in an infinitely long straight wire does not change regardless of the velocity of the electrons as long as the electron density in the wire stays the same. (The relativistic calculation is http://www.physics-quest.org/Magnetism_from_ElectroStatics_and_SR.pdf in section 2)

- Electrons move in principle independently from each other through the wire. The electron density is therefore not necessarily subject to Lorentz contraction. The field of each individual electron is transformed but the electron density is in principle arbitrary. The wire is neutral if the electron density is the same as the positive charge (proton) density.

This why the test charge at rest in the lab frame does not experience a force from a neutral current.

Also note this logical fallacy: The drift-speed of electrons is spread over a wide range of different velocities and the speed of each individual electron changes all the time. A Lorentz contraction of the electron density based on some average electron velocity makes no sense.

----------------------------------------------------

Next:

Explain the Magnetic Lorentz force on a test-charge moving in parallel with a neutral current carrying wire. The by far simplest way to explain this (using pure electric fields) is non-simultaneity and going to the rest-frame of the test-charge- In its rest-frame the test-charge is only subject to Electric Lorentz forces.

- Due to non-simultaneity one end of the wire lays in the future and the other end in the past

- Therefore a net current has streamed into (or out from) the wire when viewed from the rest-frame of the test-charge.

- The wire is thus not electrically neutral anymore in the rest-frame of the test-charge

- The integral over all electric fields gives us the non-zero Lorentz force. (See http://www.physics-quest.org/Magnetism_from_ElectroStatics_and_SR.pdf in section 1)

----------------------------------------------------

we can also calculate the Lorentz force on a test-charge moving perpendicular to a neutral current carrying wire by integrating over all electric fields as seen in the test-charge's rest-frame. (Again http://www.physics-quest.org/Magnetism_from_ElectroStatics_and_SR.pdf in section 3)

----------------------------------------------------

A detailed derivation of the Lorentz transform of the Electromagnetic Potentials and Fields can be found here in my book:
http://www.physics-quest.org/Book_Chapter_EM_LorentzContr.pdf
following the original work of Liénard and Wiechert in 1898-1900.

 

[vanhees71]

As I said, you only can derive the Lorentz force from the force in a static electric field if you know the transformation rules for the electromagnetic field components under Lorentz transformations, which you cannot derive from electrostatics alone. I think it's best to start from the full set of Maxwell's equations to understand electrodynamics and then deriving the Lorentz-transformation properties of the fields from the Lorentz-transformation properties of the four-current density, which follows from the assumption that electric charge is a Lorentz scalar (as it must be for any intrinsic quantity of matter, which are nowadays always defined as scalars and operationally in the (local) rest frame of the matter) and the transformation properties of space-time four-vector components.