# Can Maxwell's equations be derived from Coulomb's Law and Special Relativity?

As an exercise I sat down and derived the magnetic field produced by moving charges for a few contrived situations. I started out with Coulomb's Law and Special Relativity. For example, I derived the magnetic field produced by a current I in an infinite wire. It's a relativistic effect; in the frame of a test charge, the electron density increases or decreases relative to the proton density in the wire due to relativistic length contraction, depending on the test charge's movement. The net effect is a frame-dependent Coulomb field whose effect on a test charge is exactly equivalent to that of a magnetic field according to the Biot–Savart Law.

My question is: Can Maxwell's equations be derived using only Coulomb's Law and Special Relativity?

If so, and the B-field is in all cases a purely relativistic effect, then Maxwell's equations can be re-written without reference to a B-field. Does this still leave room for magnetic monopoles?

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I have a vague recollection of, when I was in high school, finding a book that did undergrad E&M by assuming SR is correct from the beginning and doing something like this. I don't recall the title, though (or if it was any good), but if you want to see this worked out in detail you might try to look for it? – Mr X Jan 22 '11 at 18:05

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@Jeremy -- The book you're thinking of is probably Electricity and Magnetism by E. Purcell (part of the Berkeley Physics series). A very good book, by the way. – Ted Bunn Jan 22 '11 at 18:32
Yes! I believe it is the one I was thinking of. – Mr X Jan 22 '11 at 18:33
Yes, I used this book in College. Very good book. But it didn't go "all the way" and derive Maxwell's equations (If I recall correctly). – user1247 Jan 22 '11 at 18:58

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You also need the assumption that charge is a scalar and same-charges repel. Then the derivation is contained in Purcell's EM book. – Ron Maimon Jun 15 '12 at 2:06

## 9 Answers

Maxwell's equations do follow from the laws of electricity combined with the principles of special relativity. But this fact does not imply that the magnetic field at a given point is less real than the electric field. Quite on the contrary, relativity implies that these two fields have to be equally real.

When the principles of special relativity are imposed, the electric field $\vec{E}$ has to be incorporated into an object that transforms in a well-defined way under the Lorentz transformations - i.e. when the velocity of the observer is changed. Because there exists no "scalar electric force", and for other technical reasons I don't want to explain, $\vec{E}$ can't be a part of a 4-vector in the spacetime, $V_{\mu}$.

Instead, it must be the components $F_{0i}$ of an antisymmetric tensor with two indices, $$F_{\mu\nu}=-F_{\nu\mu}$$ Such objects, generally known as tensors, know how to behave under the Lorentz transformations - when the space and time are rotated into each other as relativity makes mandatory.

The indices $\mu,\nu$ take values $0,1,2,3$ i.e. $t,x,y,z$. Because of the antisymmetry above, there are 6 inequivalent components of the tensor - the values of $\mu\nu$ can be $$01,02,03;23,31,12.$$ The first three combinations correspond to the three components of the electric field $\vec{E}$ while the last three combinations carry the information about the magnetic field $\vec{B}$.

When I was 10, I also thought that the magnetic field could have been just some artifact of the electric field but it can't be so. Instead, the electric and magnetic fields at each point are completely independent of each other. Nevertheless, the Lorentz symmetry can transform them into each other and both of them are needed for their friend to be able to transform into something in a different inertial system, so that the symmetry under the change of the inertial system isn't lost.

If you only start with the $E_z$ electric field, the component $F_{03}$ is nonzero. However, when you boost the system in the $x$-direction, you mix the time coordinate $0$ with the spatial $x$-coordinate $1$. Consequently, a part of the $F_{03}$ field is transformed into the component $F_{13}$ which is interpreted as the magnetic field $B_y$, up to a sign.

Alternatively, one may describe the electricity by the electric potential $\phi$. However, the energy density from the charge density $\rho=j_0$ has to be a tensor with two time-like indices, $T_{00}$, so $\phi$ itself must carry a time-like index, too. It must be that $\phi=A_0$ for some 4-vector $A$. This whole 4-vector must exist by relativity, including the spatial components $\vec{A}$, and a new field $\vec{B}$ may be calculated as the curl of $\vec{A}$ while $\vec{E}=-\nabla\phi-\partial \vec{A}/\partial t$.

You apparently wanted to prove the absence of the magnetic monopoles by proving the absence of the magnetic field itself. Well, apologies for having interrupted your research plan: it can't work. Magnets are damn real. And if you're interested, the existence of magnetic monopoles is inevitable in any consistent theory of quantum gravity. In particular, two poles of a dumbbell-shaped magnet may collapse into a pair of black holes which will inevitably possess the (opposite) magnetic monopole charges. The lightest possible (Planck mass) black holes with magnetic monopole charges will be "proofs of concept" heavy elementary particles with magnetic charges - however, lighter particles with the same charges may sometimes exist, too.

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Your previous answer was only unclear in that it does not address why I am able (in a few contrived examples, admittedly) to solve for the equations of motion without reference to a B-field. All I need to do is show how the E-field transforms under Lorentz boosts, and I can do that without introducing a B-field. Did I not do those examples correctly, or are they lucky exceptions because they are contrived? – user1247 Jan 22 '11 at 18:20

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I read your answer thoroughly, and you are still not answering my questions. When I do my examples, nowhere must I postulate a "new field". I merely start with Coulomb's law and SR, and do the math, and the math shows that a particle experiences forces that can be effectively described by a "new field". This is analogous to the Coriolis force. Does gravity plus a rotating reference frame imply a new "Coriolis field"? Of course not, but it can be effectively described by one. – user1247 Jan 22 '11 at 18:44

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Lubos answer is very good and very precise. I fully subscribe. I am only a bit puzzled by the last paragraph of the answer, where he says that the existence of magnetic monopoles is inevitable in any consistent theory of quantum gravity. The argument given is that two poles of a dumbbell-shaped magnet may collapse into a pair of black holes which will "inevitably possess the (opposite) magnetic monopole charges". If I break a magnet, I do not get two monopoles, of course. I get two magnets. What would prevent the two black holes to do exactly the same, and behave like two magnets, without – Carlo Rovelli Jan 27 '11 at 8:07

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...any monopole around? (that's the rest of Dr. Rovelli's comment, which was cut off by the system) – David Zaslavsky♦ Jan 27 '11 at 8:09

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– Luboš Motl Jan 27 '11 at 10:50
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Lubos Motl's answer is very good, but I think it's worth saying one or two additional things.

You can regard magnetism as simply a byproduct of electricity, in the following sense: if you assume that Coulomb's Law is correct, and that special relativity is correct, and that charge is a Lorentz scalar (so that charge and current density form a 4-vector), then you can derive all of Maxwell's equations. (Actually, you probably also need to assume the theory is linear as well, now that I think about it.) The undergraduate-level textbook by Purcell works this out very explicitly in a nice, pleasing way, and it's also in more advanced textbooks.

Some books gloss over the need to postulate that charge is a scalar. At least one textbook -- I don't remember which -- does emphasize it, and makes a convincing case that it's worth paying attention to. One way to see that it's not a trivial condition to impose is to consider the analogy with gravity -- that is, substitute mass for charge and gravity for electric field, and try to run the same argument. (Assume weak fields so that everything can be treated as linear if you like.) There are "gravitomagnetic" effects, but they're not related to regular gravity in the same way as the magnetic field is related to the electric field -- i.e., the gravitational analogues of Maxwell's equations look different from the regular Maxwell equations). One reason is the sign differences, of course -- like charges repel in one case and attract in the other. But a bigger reason is that the source of gravity is not a scalar: its density doesn't form part of a 4-vector, but rather of a rank-2 tensor.

But on a more philosophical (or perhaps semantic) level, I wouldn't jump from this fact to the conclusion that magnetism is "merely" a byproduct of electricity. At the very least, such language doesn't appear to be useful in understanding the theory or in using it! For instance, understanding how an electromagnetic wave can propagate from a distant galaxy to your eye is much easier and more natural if you look at it from the "usual" point of view.

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Thanks Ted. So, if, as you say, you can derive all of Maxwell's equations as a byproduct of electricity, it seems to follow trivially that one can write Maxwell's equations without reference to a B-field. (Just as I can write, as in the exercise I described, the forces due to a current I without reference to a B-field). This is my question that Lubos seems to refuse to want to address. I understand that this doesn't change the physics, and may not be the most parsimonious way of expressing electromagnetism -- I'm just interested if it can and has been done. – user1247 Jan 23 '11 at 1:39

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Yes, it can be done. With sufficient effort, you can go further and express all of electricity and magnetism without reference to either an E or B field -- just as a very strange and complicated force law between charges, in which the force on each charge depends on the properties of the other charge at the retarded time. Griffiths's textbook writes the force law out explicitly in one of the later chapters. You give up a lot by doing this -- the biggest thing that comes to mind is that I have no idea how you'd even try to talk about energy-momentum conservation in this language. – Ted Bunn Jan 23 '11 at 16:03

by Hans de Vries (*):

The simplest, and the full derivation of Magnetism as a Relativistic side efect of ElectroStatics

He uses only Electrostactic field and the non-simultaneity to obtain the Magnetic Field. He does explain it better than Purcell.

Magnetic field is a side effect of movement in the electric field.

(*) Hans de Vries has a very interesting online book (not yet finished) in his site, and he offers another pearl, not related to this post, but I feel compelled to share: The Lorentz contraction is a real effect and not only 'a referential effect' as we are tempted to believe.

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@JxB I can not comment your answer and quoting "To derive the Maxwell equations, you need an additional postulate, and that is provided by the wave equation (for electric potential) in Section 4 of the reference in Helder's answer. Without this additional postulate (that changes in electric potential propagate at the speed of light), you cannot derive the displacement current from Coulomb's law and relativity alone." – Helder Velez Feb 8 '11 at 15:50
@JxB continuing previous comment (trouble with Enter key versus ShiftEnter, sorry)  electric field = light  One can not dissociate electric field from light. "at c speed " here en.wikipedia.org/wiki/Electric_field electr 28 times here: en.wikipedia.org/wiki/Photon_polarization  explore the Radiation2D.exe from here www-xfel.spring8.or.jp  I do not have any doubt that electric field,and gravity,propagates at c speed. – Helder Velez Feb 8 '11 at 16:15
@JxB quoting "arising from time-varying electric field, and not a result of the motion of electric charge." Motion is a relative concept and a 'time-varying' electric field is always resultant of charges in motion. – Helder Velez Feb 8 '11 at 16:35
– Helder Velez Feb 8 '11 at 18:10

With Coulomb's law and special relativity you can derive Ampere's law, which gives you magnetostatics. What's missing for electrodynamics is the displacement current ($\frac{1}{c^2} \frac{\partial E}{\partial t}$), which is a source of magnetic field arising from time-varying electric field, and not a result of the motion of electric charge.

Relativity has only two postulates:

1. The laws of physics are the same in all inertial reference frames
2. All inertial observers measure the same speed for light in vacuum.

Relativity, by itself, does not mandate that electric fields (or electric potential for that matter) must travel at the speed of light. To derive the Maxwell equations, you need an additional postulate, and that is provided by the wave equation (for electric potential) in Section 4 of the reference in Helder's answer. Without this additional postulate (that changes in electric potential propagate at the speed of light), you cannot derive the displacement current from Coulomb's law and relativity alone.

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I know that Purcell and others have used Lorentz symmetry as a pedagogical device to motivate the introduction of magnetic fields, but I do not recall ever having seen an axiomatic derivation of Maxwell's equations. It might be an interesting exercise to see precisely what assumptions beyond Lorentz symmetry and Coulomb's Law are necessary to reconstruct Maxwell's equations.

B fields are not fictitious fields

If you know the electric and magnetic fields in one inertial frame, you can determine the electric and magnetic fields in any other frame via Lorentz transformation. If the magnetic field happens to vanish in a given inertial frame, you could think of magnetic effects in other frames as fictitious. However, it is not always possible to find a frame in which the magnetic fields vanish. The fastest way to see this is to note that E^2 - B^2 c^2 is a Lorentz invariant quantity (see Wikipedia). If we find that B^2 > E^2/c^2 at a given spacetime point in a given inertial frame, it follows that B^2 > 0 at that point in all inertial frames. In fact, you could begin in a frame where the electric field vanishes but the magnetic field does not; the electric fields observed in other frames could then be considered fictitious.

In general, neither the electric field nor the magnetic field can be made to vanish under a Lorentz boost. To see this quickly, note that the dot product of the E field vector with the B field vector at a given spacetime point is a Lorentz invariant quantity (see Wikipedia). If this dot product is nonzero at a given spacetime point in a given inertial frame, the electric and magnetic field vectors will both be nonzero at that spacetime point in all inertial frames.

As Einstein pointed out, you can understand the motion of a charged particle by referring to the electric field in the rest frame of that particle. However, if you have multiple particles with different velocities, you need to keep track of the electric field in the instantaneous rest frame of each particle. Since Lorentz boosts mix the E field with the B field, the only way to keep track of the E field in the rest frame of each of your particles in terms of local quantities in one inertial frame is by reference to the E field and the B field.

Locality

Even if it is possible, it is not clear to me that it would be desirable to use Coulomb's law as an axiom in electromagnetic theory. Maxwell's equations explain the motion of particles by referring to local degrees of freedom, the fields. Coulomb's law, on the other hand, is a form of action-at-a-distance, and is manifestly non-local.

It is certainly possible to rewrite both the E and B fields in terms of integrals over charge density and current density (I can't post another link, so google "Jefimenko's equations"), and then to use these expressions to interpret electromagnetic forces as a form of retarded action-at-a-distance. However, to obtain these expressions requires assumptions about the boundary conditions on the E and B fields. We can always obtain another valid solution of Maxwell's equations by simply changing the boundary conditions on the fields, which demonstrates that the fields have independent existence, and are not mere book-keeping variables to simplify a more fundamental non-local interaction.

Monopoles

As usually written, Maxwell's equations do not contain terms corresponding to magnetic charge, but it would be consistent to add such terms. In fact, Dirac showed that the quantization of electric charge could be due to the existence of magnetic monopoles (I can't post another link, so google "magnetic monopole dirac quantization condition"). Maxwell's equations do not tell us whether magnetic monopoles exist or could exist, but the quantization of electric charge could be evidence that magnetic monopoles exist somewhere in the universe.

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In my answer I linked a paper by Hans de Vries where he did as user1247 is saying, and you can check his validity. In motionmoutain ch 18 - Motion in GR, we find also that GravitoElectric is a fundamental field and the GravitoMagnetic is a relativistic effect by the same reason. The motion induces it. To me is a force and not a 'field' (there is no Coriolis field, but force). How the particle can do this? "multiple particles with different velocities, you need to keep track of the electric field in the instantaneous rest frame of each particle" and 'in advance'? – Helder Velez Mar 8 '11 at 3:39

Not a direct answer to your question but still a surprising derivation of Maxwells equations:

Feynman's proof of the Maxwell equations (FJ Dyson - Phys. Rev. A, 1989) shows, that it is possible to derive Maxwells equations from Newtons second law of motion and the uncertainty principle.

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No, you can't. For several reasons. First, if you have E, to get the B field, you need additional assumptions about the structure of the theory, ie in more detail the field strength tensor, see above reply by Lubos. But in addition to this, even if you had the solution for a point charge, to get Maxwell's equations you need to know more than just having one solution. For example that they're linear, second order, and what the symmetry group is. And if you've added that, you can derive the Maxwell equations from these assumptions anyway without even starting with the Coulomb field.

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I have to agree that some extra assumptions are needed. – Philip Gibbs Jan 23 '11 at 14:45

You cannot. B is not just a relativistic side-effect of E. Jackson, Electrodynamics, Section 12.2 has a nice discussion, in which he refutes the "proofs" given in some undergraduate texts.

"The confusion arises chiefly because the Lorentz transformation properties of the force are such that a magnetic-like force term appears when the force in one inertial frame is expressed in terms of the force in another frame. It is tempting to give this extra force term an independent existence and so identify the magnetic field as a separate entity. But such a step is unwarranted without additional assumptions."

Jackson goes on to exhibit an explicit counterexample, based on a Lorentz scalar potential. This field looks like electrostatics (or even Newtonian gravitation!) in the non-relativistic limit. It also has "an apparent magnetic-like force. But there is no independent entity B." So in this "theory" B is indeed only a relativistic effect, but this theory does not apply to Nature.

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Yes, you can make it, but you also need to use a superposition principle.

1. You determine that Couloms's law, $$\mathbf F = \frac{qQ\mathbf r}{|\mathbf r |^{3}},$$ is a boundary case of the relativistic force, which acts on the charge q by the field of a Q-charge.
2. Using Lorentz transformation for the force and for the radius-vector, $$\mathbf F = \mathbf F' + \gamma \mathbf u \frac{(\mathbf F' \cdot \mathbf v')}{c^{2}} + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf F')}{c^{2}},$$ $$\mathbf r' = \mathbf r + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf r)}{c^{2}} - \gamma \mathbf u t = \mathbf r + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf r)}{c^{2}} (t = 0),$$ where u is the speed of inertial system, v is the charge speed, you can assume, that relative to the other inertial system with relative speed u the force looks as $$\mathbf F = q\mathbf E + \frac{q}{c}[\mathbf v \times \mathbf B],$$ where $$\mathbf E = \frac{\gamma Q \mathbf r}{(r^{2} + \frac{\gamma^{2}}{c^{2}}(\mathbf r \cdot \mathbf u)^{2})^{\frac{3}{2}}}, \quad \mathbf B = \frac{1}{c}[\mathbf u \times \mathbf E].$$ Of course, magnetic field is a relativistiс kinematic effect, but a procedure described above are the relativistic kinematiс transformation of Coulomb's law. So some people made a mistake by giving negative answer.
3. After that, using primary theoremes of vector analisys and regularization procedure, you can "take" rot and div of the E and B expressions above. After that you can earn Maxwell's equations. You must use superposition principle, when you move from a field of one charge to multi-charge continuously distribution.
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