I've heard that special relativity makes the concept of magnetic fields irrelevant, replacing them with relativistic effects between charges moving in different velocity frames. Is this true? If so, how does this work?

  • 11
    $\begingroup$ Perhaps the simplest thing is to note that the two relativistic invariants of the electromagnetic field are $\vec{E}\cdot\vec{B}$ and $E^2 - B^2$ (natural units yada yada; in tensor notation they are $\epsilon_{\mu\nu\rho\lambda}F^{\mu\nu}F^{\rho\lambda}$ and $F_{\mu\nu}F^{\mu\nu}$ resp.). So in any configuration such that the latter quantity is negative then there is no frame in which the magnetic field disappears. $\endgroup$
    – Michael
    Commented Sep 24, 2013 at 5:33

2 Answers 2


Although the relationship between special relativity and magnetic fields is often stated as making magnetic fields irrelevant, this is not quite the correct way to say it.

What actually disappears is the need for magnetic attractions and repulsions. That's because with the proper choice of motion frames a magnetic force can always be explained as a type of electrostatic attraction or repulsion made possible by relativistic effects.

The part that too often is overlooked or misunderstood is that these changes in the interpretation of forces does not eliminate the magnetic fields themselves. One simple way to explain why this must be true is that if it was not, a compass would give different readings depending on which frame you observed it from. So to maintain self-consistency across frames, magnetic fields must remain in place, even when they no longer play a role in the main attractive or repulsive forces between bodies.

One of the best available descriptions of how special relativity transforms the role of magnetic fields can be found in the Feynman Lectures on Physics. In Volume II, Chapter 13, Section 13-6, The relativity of magnetic and electric fields, Feynman describes a nicely simplified example of a wire that has internal electrons moving at velocity v through the wire, and an external electron that also moves at v nearby and parallel to the wire.

Feynman points out that in classical electrodynamics, the electrons moving within the wire and the external electron both generate magnetic fields that cause them to attract. Thus from the view of human observers watching the wire, the forces that attract the external electron towards the wire are entirely magnetic.

However, since the external and internal electrons move in the same direction at the same velocity v, special relativity says that an observer could "ride along" and see both the external and internal electrons as being at rest. Since charges must be in motion to generate magnetic fields, there can in this case be no magnetic fields associated with the external electron or the internal electrons. But to keep reality self-consistent, the electron must nonetheless still be attracted towards the wire and move towards it! How is this possible?

This is where special relativity plays a neat parlor trick on us.

The first part of the trick is to realize that there is one other player in all of this: The wire, which is now moving backwards at a velocity of -v relative to the motionless frame of the electrons.

The second part of the trick is to realize that the wire is positively charged, since it is missing all of those electrons that now look like they are sitting still. That means that the moving wire creates an electric current composed of positive charges moving in the -v direction.

The third and niftiest part of the trick is where special relativity kicks in.

Recall than in special relativity, when objects move uniformly they undergo a contraction in length along the direction of motion called the Lorentz contraction. I should emphasize that Lorentz contraction is not some kind of abstract or imaginary effect. It is just as real as the compression you get by squeezing something in a vice grip, even if it is gentler on the object itself.

Now think about that for a moment: If the object is also charged at some average number of positive charges per centimeter, what happens if you squash the charged object so that it occupies less space along its long length?

Well, just what you think: The positive charges along its length will also be compressed, resulting in a higher density of positive charges per centimeter of wire.

The electrons are not moving from their own perspective, however, so their density within the wire will not be compressed. When it comes to cancelling out charge, this is a problem! The electrons within the wire can no longer fully cancel out the higher density of positive charges of the relativistically compressed wire, leaving the wire with a net positive charge.

The final step in the parlor trick is that since the external electron has a negative charge, it is now attracted electrostatically to the wire and its net positive charge. So even though the magnetic fields generated by the electrons have disappeared, a new attraction has appeared to take its place!

Now you can go through all of the details of the math and figure out the magnitude of this new electrostatic attraction. However, this is one of those cases where you can take a conceptual shortcut by realizing that since reality must remain self-consistent no matter what frame you view if from, the magnitude of this new electrostatic attraction must equal the magnetic attraction as seen earlier from the frame of a motionless wire. If you do get different answers, you need to look over your work.

But what about the other point I made earlier, the one about the magnetic field not disappearing? Didn't the original magnetic field disappear as soon as one takes the frame view of the electrons?

Well, sure. But don't forget: Even though the electrons are no longer moving, the positively charged wire is moving and will generate its own magnetic field. Furthermore, since the wire contains the same number of positive charges as electrons in the current, all moving in the opposite (-v) direction, the resulting magnetic field will look very much like the field originally generated by the electrons.

So, just as the method of attraction switches from pure magnetic to pure electrostatic as one moves from the wire frame to the moving electron frame, the cause of the magnetic field also switches from pure electron generated to pure positive-wire generated. Between these two extremes are other frames in which both attraction and the source of the magnetic field become linear mixes of the two extreme cases.

Feynman briefly mentions the magnetic field generated by the moving positive wire, but focuses his discussion mostly on the disappearance of the electron-generated magnetic fields. That's a bit unfortunate, since it can leave a casual reader with the incorrect impression that the magnetic field as a whole disappears. It does not, since that would violate self-consistency by making a compass (e.g., the magnetic dipole of that external electron) behave differently depending on the frame from which you observe it. The preservation of the magnetic field as the set of particles generating it changes from frame to frame is in many ways just as remarkable as the change in the nature of the attractive or repulsive forces between objects, and is worth noting more conspicuously.

Finally, all of these examples show that the electromagnetic field really is a single field, one whose overt manifestations can change dramatically depending on the frame from which they are viewed. The effects of such fields, however, are not up for grabs. Those must remain invariant even as the apparent mechanisms change and morph from one form (or one set of particles) to another.

  • 4
    $\begingroup$ Great explanation! I like how you point out the incompleteness in Feynman's discussion $\endgroup$ Commented Sep 25, 2012 at 15:11
  • $\begingroup$ I have realized that though there will be a magnetic field, it will not affect the still charge. I mean that you say that there is no frame of reference where there is no magnetic field. Based on that you conclude that it is still relevant, despite it is a relativistic manifestation of electric field. But, when we say that magnetic field is irrelevant we do not mean that it is absent totally in some frame of reference. So, argument showing that it does not does not make the magnetic field relevant. $\endgroup$
    – Val
    Commented May 1, 2013 at 19:31
  • $\begingroup$ In your answer to your own question about Special Theory and magnetism, there was an explanation that you can't quite get rid of magnetic fields. Even though I didn't get it, it seems like you can call that up as a ubiquitous palliative ("cop-out" in the vernacular) for the following dilemma... (which is a darn shame because if you could explain it without the cop-out, it would be beautiful.) Here it goes.... If we put a grounded shield around the current carrying conductor we know from experience that the force would still be felt by our moving test charge. $\endgroup$
    – user31977
    Commented Nov 4, 2013 at 0:39
  • 2
    $\begingroup$ Shreyansh, I appreciate your question, but must confess that I don't think I can agree on the initial premise. Two electric charges moving in the same direction alongside each other in the same inertial frame, if in the vacuum and far from any other matter, do not experience any diminution of the electrostatic force. The more complicated forces occur only when they are moving relative to a background of charged matter, which relativity then compresses to create the effect of a higher density of whatever that background charge is. So: in a vacuum there is no such diminution of charge. $\endgroup$ Commented Aug 15, 2017 at 11:37
  • 1
    $\begingroup$ I totally get your answer, but it completely blew my mind. What you said is very intuitive if one observes the charges in the frame in which they are at rest. But if I observe them from a frame in which the they are moving, wont Ampere's circuital law say that there is a magnetic field due to a changing electric field, and hence there will be a magnetic (according to the non-relativistic theory) force on the moving charges? Or do we always need a surrounding (such as a wire) to apply Ampere's circuital law? What am I missing here? Thanks again. $\endgroup$ Commented Aug 18, 2017 at 17:16

Special relativity makes the existence of magnetic fields an inevitable consequence of the existence of electric fields. In the inertial system B moving relatively to the inertial system A, purely electric fields from A will look like a combination of electric and magnetic fields in B. According to relativity, both frames are equally fit to describe the phenomena and obey the same laws.

So special relativity removes the independence of the concepts (independence of assumptions about the existence) of electricity and magnetism. If one of the two fields exists, the other field exists, too. They may be unified into an antisymmetric tensor, $F_{\mu\nu}$.

However, what special relativity doesn't do is question the independence of values of the electric fields and magnetic fields. At each point of spacetime, there are 3 independent components of the electric field $\vec E$ and three independent components of the magnetic field $\vec B$: six independent components in total. That's true for relativistic electrodynamics much like the "pre-relativistic electrodynamics" because it is really the same theory!

Magnets are different objects than electrically charged objects. It was true before relativity and it's true with relativity, too.

It may be useful to notice that the situation of the electric and magnetic fields (and phenomena) is pretty much symmetrical. Special relativity doesn't really urge us to consider magnetic fields to be "less fundamental". Quite on the contrary, its Lorentz symmetry means that the electric and magnetic fields (and phenomena) are equally fundamental. That doesn't mean that we can't consider various formalisms and approximations that view magnetic fields – or all electromagnetic fields – as derived concepts, e.g. mere consequences of the motion of charged objects in spacetime. But such formalisms are not forced upon us by relativity.

  • 1
    $\begingroup$ There are electric charges but no magnetic. So, electricity generates magnetic fields. Ok? $\endgroup$
    – Val
    Commented May 1, 2013 at 10:05
  • $\begingroup$ The added section on symmetry is appropriate and rounds out the discussion nicely - thanks. $\endgroup$ Commented Sep 30, 2013 at 1:45

Not the answer you're looking for? Browse other questions tagged or ask your own question.