Magnetic induction as a reletivistic effect too?

Question

Magnetic field doesn't really exist. Magnetic attraction of two conducting wires can be explained as a relativistic effect of moving electrons with respect of stationary nuclei. Can magnetic induction be explained as a relativistic effect too, without the existence of magnetic field?

Answer 1

I think the first statement is incorrect. With two identical current carrying wires, the force is entirely magnetic (since both wires are neutral in the rest frame of the wire):

$$ \rho_ + = -\rho_- $$ $$ \rho = \rho_+ + \rho_- = 0 $$

So there is no electric effect, and the attraction is due to the current.

If you boost to a frame in which the electrons are stationary, the charge densities become:

$$ \rho'_+ = \gamma\rho_+ $$ $$ \rho'_- = \frac{\rho_-}{\gamma} $$ $$ \rho' = \rho(\gamma - \frac 1{\gamma}) = \gamma\beta^2\rho$$

so the electric field repels, and if you work out the magnetic field, it provides a stronger attraction.

Meanwhile, Maxwell's equations are Lorentz covariant, so attribution various nuances to relativity or not is really not productive.

There is a causal equivalent formulation called: https://en.wikipedia.org/wiki/Jefimenko%27s_equations .

Answer 2

Magnetic field doesn't really exist.

A magnetic field is as real as an electric field. While a macroscopic electric field can be induced by separating negatively (electron) from positively (proton) charged particles, this is done for a macroscopic magnetic field by aligning the magnetic dipoles of electrons and protons. At the subatomic particle level, both fields - electric and magnetic - are absolutely equal (see the NIST constants for electron magnetic moment and electron elementary charge).

Can magnetic induction be explained as a relativistic effect too, without the existence of magnetic field?

A counter-example is already enough to refute an assertion. Two permanent magnets - consisting of the aligned magnetic dipoles of the subatomic particles of the magnets - attract each other without a particle flow and thus without relativistic effects.

Answer 3

Let us consider a charged cloud on the sky moving above us. We observe a vertical electric field. There is also a hot air balloon inside the cloud, with a string hanging out vertically. Oh yes there are many watches tied on the string, uniformly spread out along the string.

OK so now we start moving vertically upwards. Relativity says that there occurs a change of the relative phases of those aforementioned clocks, when we start moving. The proceeding of different string segments are also clocks whose relative phases change as we start moving. Which means that the string becomes tilted, according to us, when we start moving.

Now it happens to be so that the string is still aligned with the electric field lines, according to us, after we started moving.

A test charge that we hold, experiences a different force after we start moving, and the difference is the motion induced electric field.

Answer 4

The $\vec E$ field and $\vec B$ field are both components electromagnetic field tensor $F_{\alpha \beta}$. The components may change with different reference frames, but $F_{\alpha \beta}$ does not.

Saying "magnetic field does not exist" is incorrect, and it's like saying in the $xy$ plane that the $y$ direction does not exist because you can always define a direction $x'$ that points tangential (locally) to the trajectory of an object. Regardless of that, the plane is still two dimensional.

Answer 5

Magnetic field doesn't really exist …

Whether or not some physical concept “really exists” is theory dependent. There could be multiple theories explaining the same set of phenomena and a concept fundamental to one theory could be relegated to an auxiliary role in another and be entirely absent from the third.

The role the concept of magnetic field would play depends on the postulates you are putting in the theory intended to explain electromagnetic phenomena. One such possible postulate is the principle of locality. If we try to construct a theory satisfying this principle (such theory would be called local) then in order to explain some electromagnetic phenomenon at a spacetime point $p$ we must use only properties associated to that point.

And it turns out that two vectors $(\mathbf{E}, \mathbf{B}) $ at each spacetime point (in addition to data needed to specify charges and currents) are enough to explain all the (classical) electromagnetic phenomena. So we might as well say that those two vectors (or rather vector fields) “really exist” within local theory of electromagnetism.

… Magnetic attraction of two conducting wires can be explained as a relativistic effect of moving electrons with respect of stationary nuclei.

I am assuming this refers to the demonstration that in the reference frame of the charge moving near a current carrying wire the force on the charge arises as the Coulomb force from the nonzero charge density induced in the wire by the frame change due to different velocities of charges of opposite signs.

The standard explanation of this within the local theory of electromagnetism is that electric and magnetic fields constitute different components of a single four-dimensional object, Faraday tensor, so under the Lorentz transformations between different frames electric field gains contribution from magnetic and vice versa, and examples like this provide us with the law of transformation between reference frames: $(\mathbf{E}, \mathbf{B}) \to (\mathbf{E}^{'}, \mathbf{B}^{'})$. Similarly, charge and current densities combine into a 4-vector and so a neutral wire with current in one frame becomes charged in another.

But if instead we interpret this exercise as demonstrating the auxiliary character of magnetic field we must conclude that electric field also does not really exist because if we repeat the same calculation with another charge located at the same point but having different velocity we could not obtain Coulomb force in the second charge's reference frame just from knowing the force on the first charge. Instead, we anew must calculate the charge distribution in the reference frame of that second charge taking into account position and velocities of all the charges in the wire. What we have is the start of a non-local theory of electromagnetism that does not have either electric or magnetic fields (as fundamental objects) and where the forces acting on a charge depend (potentially) on positions and motions of all the charges in the universe.

The biggest obstacle to building such non-local, yet fully relativistic theory is the existence of electromagnetic radiation: an isolated system of moving charges may lose energy. With local EM theory we say that this energy was carried away by the EM waves and we then can calculate their characteristics. But in the non-local theory without the electric or magnetic fields to account for this lost energy we must include interaction with distant entities that could absorb this energy.

The possible way around this problem is to limit ourselves only to weakly relativistic systems of charges, incorporate only the lowest order relativistic corrections and ignore relativistic corrections of higher orders including those that corresponds to emission of EM radiation. The resulting theory would incorporate interaction between charges with Coulomb forces as well as interactions usually described via magnetic field, but now this could be done using a multi-particle Lagrange function, the Darwin Lagrangian, depending only on the charges positions and velocities: \begin{eqnarray} \mathcal{L} &\mathcal{=}&\sum\limits_{i=1}^{i=N}m_{i}c^{2}\left( -1+\frac{% \mathbf{v}_{i}^{2}}{2c^{2}}+\frac{(\mathbf{v}_{i}^{2})^{2}}{8c^{4}}\right) -% \frac{1}{2}\sum\limits_{i=1}^{i=N}\sum\limits_{j\neq i}\frac{e_{i}e_{j}}{|% \mathbf{r}_{i}-\mathbf{r}_{j}|} + \notag \\ &&+\frac{1}{2}\sum\limits_{i=1}^{i=N}\sum\limits_{j\neq i}\frac{e_{i}e_{j}% }{2c^{2}}\left[ \frac{\mathbf{v}_{i}\cdot \mathbf{v}_{j}}{|\mathbf{r}_{i}-% \mathbf{r}_{j}|}+\frac{\mathbf{v}_{i}\cdot (\mathbf{r}_{i}-\mathbf{r}_{j})% \mathbf{v}_{j}\cdot (\mathbf{r}_{i}-\mathbf{r}_{j})}{|\mathbf{r}_{i}-\mathbf{% r}_{j}|^{3}}\right] .\notag \end{eqnarray} So within that theory the “effects”, such as accelerations of the charges, coming from the terms in the Lagrangian with $c^{-2} $ in it could be termed “relativistic effects”. Note, that electric and magnetic fields can still be calculated within this theory, only those are no longer the fundamental entities but of auxiliary role.

Can magnetic induction be explained as a relativistic effect too, without the existence of magnetic field?

As a matter of fact, yes, within the theory of point charges governed by the Darwin Lagrangian. Here is a paper that does exactly that:

• Boyer, T. H. (2015). Faraday induction and the current carriers in a circuit. American Journal of Physics, 83(3), 263-271, DOI:10.1119/1.4901191, arXiv:1410.1197.

From the abstract (emphasis mine):

In this article, it is pointed out that Faraday induction can be treated from an untraditional, particle-based point of view. The electromagnetic fields of Faraday induction can be calculated explicitly from approximate point-charge fields derived from the Liénard-Wiechert expressions or from the Darwin Lagrangian. Thus the electric fields of electrostatics, the magnetic fields of magnetostatics, and the electric fields of Faraday induction can all be regarded as arising from charged particles.