# Is the presence of a magnetic field frame-dependent?

I do not have a strong background in physics, so please refrain from using complex mathematics in any answers :)

I was on this site, and I read:

When an electrical charge is moving or an electric current passes through a wire, a circular magnetic field is created.

I am trying to grasp the notion of magnetic fields under different reference frames.

Suppose I have a point charge next to a compass needle. It is experimentally verifiable that the compass needle is not affected by the point charge - i.e., the needle does not move. It can be concluded that the point charge does not create any magnetic field.

But since the point charge is on the Earth, and the Earth is moving, shouldn't the point charge produce a magnetic field? Why doesn't the compass needle move in response to this magnetic field? Is there a magnetic field in the room?

You are indeed correct about the frame-dependence of magnetic fields. The reason the point charge doesn't affect the compass is because the compass and the charge are both moving at the same speed, both being on the Earth, and therefore, the compass sees the charge as stationary. This means no magnetic field is produced.

As a side note, you hit upon an important realization: in order for electrodynamics to be consistent, you must adopt the same set of assumptions as in special relativity! In other words, special relativity is a necessary consequence of electrodynamics. Some books even derive the phenomenon of time dilation by considering the magnetic field experienced by a point charge moving parallel to a line charge.

• Actually, there are two ways for electrodynamics to be consistent: either Special Relativity, or a privileged reference frame ("luminiferous aether"). – Mark Jun 28 '17 at 20:38

This is a fantastic question, and you'd probably become famous for trying to solve it if it weren't for the fact that Einstein asked the same question over 100 years ago. The first sentence in his 1905 paper on Special Relativity says (translated into English, of course):

"It is known that Maxwell’s electrodynamics—as usually understood at the present time—when applied to moving bodies, leads to asymmetries which do not appear to be inherent in the phenomena."

He was referring to something along the lines of this: if a charged particle makes a magnetic field, then if I move at the same speed and in the same direction of this charge, I shouldn't see a magnetic field (he was actually considering moving a magnet through a metal loop, but it's essentially the same idea). This realization inspired him to essentially devise special relativity! HE found, in particular, that magnetism is actually an entirely relativistic effect. That is, magnetism only exists because of special relativity. If you walk past a stationary charge, you should see a magnetic field in your frame of reference.

This realization absolutely revolutionized the field of physics and indeed the world (Einstein would be considered a genius even if he never did another thing in his life -- but he did so much more). If you're interested in physics, you should pursue it. You ask the right kind of questions.

I hope this helps!

• It's also an eye-opener as to the strength of the electromagnetic field versus gravity. The electro-magnetic field that can suspend tons of iron against the gravitational attraction of the entire Earth disappears in the relativistic correction when one moves into the rest frame of the moving electrons creating it. And those electrons are moving nowhere near the velocity of light - typically at mere centimeters per second! Everyday experience is blind to the strength of electrostatic interaction, because negative and positive charges are everywhere almost perfectly balanced. – nigel222 Jun 28 '17 at 8:10
• Could you please clarify: With zero relative motion Maxwell predicts a magnetic field where SR predicts none? – cumfy Jun 10 '20 at 12:59

In the context of special relativity, the electric and magnetic fields are not distinct vector fields related through Maxwell's equations but are, rather, part of a more general higher rank (than a vector) electromagnetic tensor

The components of this tensor 'mix' when transforming between inertial reference frames so that both the electric and magnetic components are frame dependent.

Is the presence of a magnetic field frame-dependent?

In general, it is not possible to 'transform away' the magnetic field. Consider the simple case of two point charges in relative uniform motion. Since there is no inertial reference frame in which both charges are at rest, there is no inertial frame in which there is only an electric field.

When examining quantities in special relativity, it is helpful to find some combination of them that is frame invariant. In the case of momentum/energy, that quantity is the rest mass $\left(mc^2 = \sqrt{E^2 - (\mathbf{p}c)^2}\right)$. In the case of coordinates, that quantity is the proper time/interval $\left(s^2 = \mathbf{x}^2 - (ct)^2 \right)$. The importance of such invariant quantities is that all inertial observers will agree on the value they have.

There are two analogous quantities for electromagnetism. The first is $\mathbf{E}^2 - (c\mathbf{B})^2$, and the second is $\mathbf{E}\cdot\mathbf{B}$. The fact that there are two invariant quantities produces some interesting consequences. First, it is only possible to make the magnetic field vanish at a point if $\mathbf{E}\cdot\mathbf{B} = 0$ and $\mathbf{E}^2 - (c\mathbf{B})^2 < 0$. If both of those hold, then (up to a minus sign) an observer moving with velocity $$\mathbf{v} = \pm \frac{\mathbf{E}\times\mathbf{B}}{E^2},$$ will see no magnetic field. Similarly, $\mathbf{E}\cdot\mathbf{B} = 0$ and $\mathbf{E}^2 - (c\mathbf{B})^2 > 0$ the observer moving with velocity $$\mathbf{v} = \pm c^2 \frac{\mathbf{E}\times\mathbf{B}}{B^2}$$ will see no electric field. If $\mathbf{E}\cdot \mathbf{B} \neq0$ or if $\mathbf{E}^2 - (c\mathbf{B})^2 = 0$ then there is no way to make either field vanish at the point in question for an observer moving with $v < c$.

Note that it is only possible to cancel the field at a single point and time, with the exceptions being the constant field cases, when it is possible at all.

For an interesting application of this, see this question about crossed $\mathbf{E}$ and $\mathbf{B}$ fields.

• it might be good to emphasize your second last line is the reason why your answer is not telling against Alfred's answer, which, naturally, is also correct; maybe the "apparent" contradiction is why yours isn't more upvoted. – Selene Routley Jul 11 '17 at 3:03
• @SeleneRoutley There is no contradiction - I just provided more technical detail. – Sean E. Lake Oct 7 '20 at 8:53

You have to make a distinction between the generation and the detection of a magnetic field. If you generate a magnetic field, but can't detect it, for all intents and purpose, it is the same as not generating it.

In the example you give, even though there is motion of both the charged particle and the compass, there is no motion between them, therefore, no magnetic field between them, is detected.

Since the magnetic effect of a charged particle is an effect that depends on the relative motion between the particle and the "medium," the presence of a magnetic field (and its detection), is frame dependent!

This is related to the Trouton-Noble paradox. As you have stated, an observer in the frame where the charge is moving would note that the electric charge is in motion so that a magnetic field is expected. This means that the observer would expect there be a force acting on the compass.

How can this be reconciled with the fact that an observer in the reference frame where the electron is at rest will not expect any force acting on the compass? It turns out that, if you model the compass as a magnetic dipole of moment $${\bf m}$$ acquires an electric dipole moment $${\bf p}$$.

This acquired electric dipole moment ensures that the person who observes the charge in motion will detect no force on the compass.

See, for example the following:

The Electric Dipole Moment of a Moving Magnetic Dipole George P. Fisher American Journal of Physics 39, 1528 (1971);

Classical electromagnetism and relativity: A moving magnetic dipole W. G. V. Rosser American Journal of Physics 61, 371 (1993);

Magnetic dipole moment of a moving electric dipole V. Hnizdo American Journal of Physics 80, 645 (2012);

Suppose you have a point charge at rest in the laboratory. Within the lab, particles only experience an Electric field. A moving observer will experience a Magnetic Field as well.

Any moving charge, however small, produces a magnetic field.

There are multiple ways to find the strength of this field.

1) You can Lorentz transform the electrostatic potential of the lab frame using the Potential 4-Vector. Basically, that's a regular vector with a time dimension as well as the three spacial dimensions. The transform is Matrix Multiplication.

2) You can integrate over the charge distribution taking into account the Retarded Time. Objects don't react to current charge and current distributions, but how they were some time in the past since it takes time for EM signals to reach a destination.

3) A geometric analysis might expose what is going on in Magnetism. As a charge moves away from a region of space, the electric field drops off as the inverse square of the distance. The direction of relative motion establishes a privileged direction regarding the time variance of the magnetic field. The change of the charges causes the change in the fields at some point in the future of that changing charge distribution.