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I couldn't help but notice that the expression for the magnetic component of the Lorentz force,

$$\mathbf F = q\,\mathbf v \times \mathbf B\,,$$

is very similar in its mathematical form to the Coriolis force,

$$\mathbf F = 2m\mathbf v \times \mathbf ω\,,$$

providing that we replace electric charge with mass, and angular velocity with the magnetic induction.

Even though I am aware of the physical differences between those two forces (Coriolis is a fictitious force, which acts on objects that are in motion relative to a rotating frame of reference, whereas the magnetic force is caused by a magnetic field), I do remember reading that magnetism is a "relativistic effect of electricity" (Feynman lectures), and wonder whether this analogy is pure coincidence or could obey to a deeper connection. Could it have something to do with Lorentz transformations?

On a more general level, could the magnetic force be viewed as "fictitious", and may this have some relation with the apparent non-existence of magnetic monopoles?

Edit:

I would like to point out that the analogy can be extended to the two other inertial forces, the centrifugal force and the Euler force, as is shown here and here.

My question could then be restated as:

Why is there an analogy between inertial and electromagnetic forces?

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    $\begingroup$ Related: physics.stackexchange.com/q/155093/2451 $\endgroup$ – Qmechanic Mar 9 '16 at 20:23
  • $\begingroup$ I apologize, I didn't notice this had been asked before. I am still curious about the last part of my question, though. $\endgroup$ – David Herrero Martí Mar 9 '16 at 20:55
  • $\begingroup$ @David. I was wondering, did my answer address your questions? $\endgroup$ – Giorgio Comitini Mar 15 '16 at 14:53
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    $\begingroup$ I was worried that the formalism might have hidden the intuition behind the argument I was advancing. Glad I was able to help. Feel free to ask for clarification, if needed. $\endgroup$ – Giorgio Comitini Mar 15 '16 at 16:41
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    $\begingroup$ My question did not cause this kind of a response, I know why...because this is much nicer one. Great question and great answers too. Just...great. $\endgroup$ – Žarko Tomičić Mar 16 '16 at 19:49
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As nobody has done this yet, let's try to give an answer to your question in the right framework, i.e. through the formalism of differential geometry (and of action principles, as far as the physics is concerned). This formalism has the advantage of allowing for the use of arbitrary coordinate systems, so that the problem of the arising of "fictitious" force terms such as the Coriolis force can be addressed in a rigorous way. Moreover, it allows to generalize the standard description of electromagnetism in such a way that magnetic monopoles indeed are permitted to exist. I will show and motivate why in my opinion there is no connection between the Coriolis force and the magnetic force, explain what it means for the magnetic force to be a relativistic effect of the electric force and show how to introduce monopole fields in the magnetic field. I'll try to make myself as clear as possible, as I understand that you are not familiar with the formalism.

First of all, the Lagrangian for a point-like, massive particle in an arbitrarily curved spacetime, subject to the electromagnetic field, can be expressed as

$$\mathscr{L}=-m\sqrt{g_{\mu\nu}\frac{dx^{\mu}}{ds}\frac{dx^{\nu}}{ds}}-e\,A_{\mu}(x^{\mu})\frac{dx^{\mu}}{ds}$$

(factors of $c$ missing). Here $g_{\mu\nu}$ is the spacetime metric, the object which encodes the spacetime curvature, $A_{\mu}$ is the covariant form of the electromagnetic four-potential, $A_{\mu}=(\phi,-\vec{A})$, $s$ is an arbitrary parameter, $m$ and $e$ the mass and charge of the particle, $x^{\mu}(s)$, with $\mu=0,1,2,3$, the trajectory of the particle in spacetime. We will be interested in flat spacetimes, i.e. Minkowski spacetime, but one needs the full generalization to extract useful results from the formalism. One finds the equations of motion for the particle by minimizing the action integral $S$, that is

$$ S[x]=\int_{a}^{b}\mathscr{L}(x^{\mu}(s),\dot{x}^{\mu}(s))\ ds $$ where the dot denotes a derivation with respect to the parameter $s$. Finding a minimum for $S$ is completely equivalent to the procedure shown in Frédéric's answer: the curve which minimizes the action is the curve which solves the Euler-Lagrange equations, or equivalently the Hamilton equations (those in the cited answer). Notice that the action

$$ S=\int_{a}^{b}\bigg\{-m\sqrt{g_{\mu\nu}\frac{dx^{\mu}}{ds}\frac{dx^{\nu}}{ds}}-e\,A_{\mu}\frac{dx^{\mu}}{ds}\bigg\}\ ds $$ is invariant with respect to three different kinds of transformation. The first one is a monotone, increasing change of parametrization $s\to s'$ (i.e. one with $ds'/ds>0$), as the transformation gets absorbed in the measure of integration $ds$. The second one is an arbitrary change of coordinates: every time you see two indices contracted, as the two objects involved in the contraction transform in opposite ways (this is symbolically expressed by the positioning of the indices), the overall object remains invariant under a change of coordinates. The last one is the transformation

$$ A_{\mu}\to A_{\mu}+\partial_{\mu}\chi $$ where $\chi$ is an arbitrary function of the variables $x^{\mu}$, known as a gauge transformation of the electromagnetic potential. Under such a transformation, the action gains the additional term

$$ \delta S=\int_{a}^{b}-e\ \partial_{\mu}\chi\ \frac{dx^{\mu}}{ds}\ ds=\int_{a}^{b}-e\ \frac{d\chi}{ds}\ ds=-e\ \bigg[\chi(b)-\chi(a)\bigg] $$ which is a constant. So the action may not actually be invariant under such a transformation, but as $S$ only gets shifted by a constant amount, its minima are preserved by the transformation. The property of $S$ being invariant under such transformations has one important consequence: the general form of the dynamical equations one gets from the minimization of $S$ is valid with respect to any parametrization of the curve of the particle, which in turn can be expressed in any coordinate system you like, and the equations are not changed by a gauge transformation. To give the equations a simpler look, I will use the parametrization in which

$$ \sqrt{g_{\mu\nu}\frac{dx^{\mu}}{ds}\frac{dx^{\nu}}{ds}}=1 $$

Please notice that this condition does not affect the choice of the coordinates through which you express the dynamical curve: the condition itself is unaffected by a change of coordinates. From the above action $S$ and given the former condition on $s$, it can be shown that the dynamical equations have the following form:

$$ \frac{d^{2} x^{\mu}}{ds^{2}}=-\Gamma^{\mu}_{\nu\tau}\ \frac{dx^{\nu}}{ds}\frac{dx^{\tau}}{ds}+\frac{e}{m}\ F^{\mu}_{\ \nu}\ \frac{dx^{\nu}}{ds} $$

Here $$ F^{\mu}_{\ \nu}=g^{\mu\sigma}(\partial_{\sigma}A_{\nu}-\partial_{\nu} A_{\sigma}) $$ with $g^{\mu\sigma}$ the inverse of the matrix $g_{\mu\sigma}$, is the electromagnetic field tensor and the functions $$ \Gamma^{\mu}_{\nu\tau}=\frac{1}{2}g^{\mu\sigma}\ \big[\partial_{\nu}g_{\sigma\tau}+\partial_{\tau}g_{\sigma\nu}-\partial_{\sigma}g_{\nu\tau}\big] $$ are called the "Christoffel symbols" related to the metric $g$ in the $x^{\mu}$ coordinate system. The Christoffel symbols encode two kinds of information: first of all, if the metric $g$ describes a curved spacetime, they encode the effect of curvature on the particle, i.e. they encode the gravitational field; second of all, they encode the fictitious forces due to the choice of a specific coordinate system. It can be shown that, when spacetime is not curved, there exist coordinates with respect to which every $\Gamma$ is zero. These are the (in)famous intertial frames, in which $\eta_{\mu\nu}=\text{diag}(1,-1,-1,-1)$ and the equations take the form

$$ \frac{d^{2} x^{\mu}}{ds^{2}}=\frac{e}{m}\ F^{\mu}_{\ \nu}\ \frac{dx^{\nu}}{ds} $$ $$ F^{i}_{\ 0}=-\vec{\nabla}\phi-\partial_{0}\vec{A}=\vec{E} $$ $$ F^{i}_{\ j}=-\partial_{i}A_{j}+\partial_{j}A_{i}=\epsilon_{ijk}(\vec{B})_{k} $$ If we substitute $ds$ with $dt ds/dt$, where $t$ is the time coordinate of the inertial observer, i.e. $t=x^{0}$, we find for the $\mu=1,2,3$ equations $$ \frac{d}{dt}\bigg(\frac{dt}{ds}\frac{d\vec{x}}{dt}\bigg)=\frac{e}{m}\ \left(\vec{E}+\vec{v}\times\vec{B}\right) $$ As $ds/dt$ turns out to be $\sqrt{1-v^{2}/c^{2}}$, the former is exactly the Lorentz equation for a relativistic particle. Now let's work in more general coordinates. Let's call these coordinates $y^{\mu}$, with equations of motion $$ \frac{d^{2} y^{\mu}}{ds^{2}}=-\Gamma^{\mu}_{\nu\tau}\ \frac{dy^{\nu}}{ds}\frac{dy^{\tau}}{ds}+\frac{e}{m}\ F^{\mu}_{\ \nu}\ \frac{dy^{\nu}}{ds} $$ It can be shown that the general relation between the Christoffel symbols wrt the $y$ and the $x$ coordinates is

$$ \Gamma^{\mu (y)}_{\nu\tau}=\frac{\partial x^{\sigma}}{\partial y^{\nu}}\frac{\partial x^{\lambda}}{\partial y^{\tau}}\Gamma^{\alpha\ (x)}_{\sigma\lambda}\frac{\partial y^{\sigma}}{\partial x^{\alpha}}+\frac{\partial y^{\mu}}{\partial x^{\alpha}}\frac{\partial^{2} x^{\alpha}}{\partial y^{\nu}\partial y^{\tau}} $$ where $\partial y/\partial x$ and its inverse are the matrices of the change of coordinates. In our specific case ($\Gamma^{\alpha\ (x)}_{\sigma\lambda}=0$), $$ \Gamma^{\mu}_{\nu\tau}=\frac{\partial y^{\mu}}{\partial x^{\alpha}}\frac{\partial^{2} x^{\alpha}}{\partial y^{\nu}\partial y^{\tau}} $$ So the dynamical equations can be written as $$ \frac{d^{2} y^{\mu}}{ds^{2}}=-\frac{\partial y^{\mu}}{\partial x^{\alpha}}\frac{\partial^{2} x^{\alpha}}{\partial y^{\nu}\partial y^{\tau}}\ \frac{dy^{\nu}}{ds}\frac{dy^{\tau}}{ds}+\frac{e}{m}\ F^{\mu\ (y)}_{\ \nu}\ \frac{dy^{\nu}}{ds} $$ Now let's focus on each term of the equation. $\frac{d^{2} y^{\mu}}{ds^{2}}$ plays the role of an acceleration wrt to the parameter $s$ (which does not need to be time). $\frac{e}{m}\ F^{\mu (y)}_{\ \nu}\ \frac{dy^{\nu}}{ds}$ is the ordinary electromagnetic acceleration, expressed in an arbitrary coordinate system and wrt to the parameter $s$. But what about the functions $ -\frac{\partial x^{\alpha}}{\partial y^{\nu}\partial y^{\tau}}\ \frac{dy^{\nu}}{ds}\frac{dy^{\tau}}{ds}$? It is useful to notice that these functions depend on second derivatives, i.e. they do not appear if the relation between the coordinates $y$ and $x$ is linear, of the form

$$ x^{\mu}=\Lambda^{\mu}_{\ \nu}y^{\nu}+a^{\mu} $$ It is easy to realize that the former is the correct relation between two inertial coordinate systems. In a special-relativistic setting, we choose $\Lambda$ to be a Lorentz transformation, in order to keep the metric $g_{\mu\nu}=\text{diag}(1,-1,-1,-1)$, and in turn the $g^{\mu\nu}$ in the definition of tensor $F^{\mu}_{\nu}$, invariant. Following a Lorentz transformation then, the equations of motion do not change, as advised by the theory of special relativity. But if we were to make different coordinate changes, the equations would indeed be different. In particular, new terms proportional to the velocities $dy^{\mu}/ds$ would arise. This is how Coriolis and fictitious forces arise: $\vec{v}\times\vec{\omega}$ is none other than the product between a velocity and a reference frame parameter $\vec{\omega}$, which you can see in the general equation given above. The other velocity disappears when you choose $s$ to be time.

Now that we have the needed machinery and conceptual rigour, let's go back to your questions. First of all, what does it mean for the magnetic force to be a relativistic effect of the electric force? The electric and magnetic fields are related by coordinate transformations through the equations

$$ F_{\mu\nu}^{(y)}=\frac{\partial x^{\sigma}}{\partial y^{\mu}}\frac{\partial x^{\lambda}}{\partial y^{\nu}}\ F_{\sigma\lambda}^{(x)} $$ On the right side of the equation, the electric and magnetic fields get mixed due to the change of coordinates. Are there coordinate systems in which the field is entirely electric or magnetic? Yes (and in principle they can even be inertial frames). This is because if the electric source is static in some reference frame, then in that frame the field is entirely electric. Thus, for example, in any frame related to this frame by a space rotation or translation the field will remain totally electric (Lorentz boosts, on the other hand, mix the two). On the other hand, imagine setting the source into motion. Then the field produced by the force in that frame is both electric and magnetic, but the source itself hasn't changed a bit! This means that the splitting of the EM field between an electric component and a magnetic component is not an intrinsic property of the source, but rather an artifact of the choice of a coordinate system, relative to the state of motion of the source itself. This is seen by the very definition of the E and B fields, which are deduced from the tensor $F_{\mu\nu}$, which in turn depends on the choice of coordinates. So in this case "relativistic effect" means "relative wrt the state of motion of the source, wrt to a chosen coordinate system". (The question of magnetic fields due to spin currents, on the other hand, must be addressed in a different, quantum-mechanical, setting).

Coriolis forces are too an effect of the choice of a specific coordinate system. As seen, they arise from second derivatives of the coordinate transformation from inertial coordinates. Their "coordinate dependence" status, however, is quite different from that of the E/M splitting. First of all, they do not depend upon the state of motion of any kind of source relative to a specific coordinate system (here we don't regard spacetime itself as one). Second of all, they only appear in non-inertial frames, whereas the E/M splitting is an issue even in inertial frames. Last but not least, Coriolis forces depend upon the mass of the particle, whereas the magnetic force does not. This means that particles with different masses whose motion is described in the same coordinate system will experience different Coriolis forces, but they will experience the very same E/M splitting (the opposite happens with respect to the accelerations). So magnetic forces and Coriolis forces should not be compared to one another: they are two very different objects, and as such no deep connection can exist between them. You noticed, though, that their mathematical form is similar. The mathematical form of a dynamical equation (apart from general covariant equations such as those I wrote above, but this is not the case), though, depends very much on the choice a coordinate system, so one must be careful when comparing force terms of dynamical equations, especially when one is going from one coordinate system to another. The choice of coordinates is unphysical, in the sense that it needn't be connected to underlying physical principles, nor it affects the physics of the system. In this case, though, a comparison can be made on a solid basis and turns out to be useful to answer your question. Now, a Coriolis force term of the form $2m\vec{v}\times\vec{\omega}$ appears in coordinate systems (rotating coordinates wrt a given inertial frame) where the magnetic force can as well not be of the form $e\,\vec{v}\times \vec{B}$. Consider a non-relativistic charged particle in a uniform, constant magnetic field (description given in an inertial frame). The particle will spin around some axis parallel to the magnetic field with frequency $\omega=\frac{eB}{m}$ (factors of $c$ missing, depending on convention on the definition of the magnetic field). If you go to a frame which uniformly rotates around that axis with angular frequency $\omega=\frac{eB}{m}$, there will seem to be no magnetic field at all acting on the particle. This is because the frame is moving together with the component of the motion of the particle which changes due to the magnetic field, thus no motion induced by the magnetic field can be observed in that frame. This does not signal a deep connection between the Coriolis and the magnetic force, it only confirms that there exist specific frames in which specific forces do not appear to act on specific systems. This, of course, is valid for any kind of force, if you don't restrain yourself to simple inertial frames. In this case, magnetic fields make electrically charged particles spin, so it is obvious that the coordinate systems in which the magnetic field does not appear must be a rotating coordinate system. Let us see this for the case of interest. I'll use the same formulas as in Wikipedia's "Rotating reference frame" article (with $\vec{\omega}$ taken in the opposite direction). The acceleration for a charged particle in a uniformly rotating frame about the center of the trajectory, subject to a uniform constant magnetic field, takes the form: $$ \vec{a}_{r}=-\vec{\omega}\times\vec{\omega}\times \vec{r}_{i}+2\vec{\omega}\times\vec{v}_{r}+\frac{e}{m}\ \vec{v}_{i}\times \vec{B} $$ where the subscript $r$ denotes a quantity in the rotating reference frame, while the subscript $i$ denotes the same quantity in an inertial frame. We have: $$ \vec{v}_{i}=\vec{v}_{r}-\vec{\omega}\times\vec{r}_{i} $$ so that $$ \frac{e}{m}\ \vec{v}_{i}\times \vec{B}=\frac{e}{m}\ \vec{v}_{r}\times \vec{B}-\frac{e}{m}\ \vec{\omega}\times\vec{r}_{i}\times \vec{B} $$ As you can see, the mathematical form of the magnetic force changes in a uniformly rotating reference frame. Moreover, as $$ -\vec{\omega}\times\vec{\omega}\times \vec{r}_{i}=-\vec{\omega}(\vec{\omega}\cdot\vec{r}_{i})+\vec{r}_{i}(\vec{\omega}\cdot\vec{\omega})=\frac{e^{2}B^{2}}{m^{2}}\ \vec{r}_{i} $$ where the last identity follows from

$$ \vec{\omega}=\frac{e}{m}\ \vec{B}\ ,\qquad\quad \vec{\omega}\cdot\vec{r}_{i}=0 $$ and $$ -\frac{e}{m}\ \vec{\omega}\times\vec{r}_{i}\times \vec{B}=-\frac{e}{m}\ \vec{r}_{i}(\vec{\omega}\cdot\vec{B})+\frac{e}{m}\ \vec{B}(\omega\cdot \vec{r}_{i})=-\frac{e^{2}B^{2}}{m^{2}}\ \vec{r}_{i} $$ we have $$ \vec{a}_{r}=-2\vec{v}_{r}\times \vec{\omega}+\frac{e}{m}\ \vec{v}_{r}\times\vec{B}=-2\vec{v}_{r}\times \vec{\omega}+\vec{v}_{r}\times\vec{\omega}=-\vec{v}_{r}\times \vec{\omega} $$ Again, $$ \vec{a}_{r}=-\vec{v}_{r}\times \vec{\omega}=-\vec{v}_{i}\times\vec{\omega}-\vec{\omega}\times\vec{r}_{i}\times\vec{\omega} $$ which is zero due to the fact that, having solved the equation in the inertial system, $\vec{v}_{i}=\vec{r}_{i}\times\vec{\omega}$. Hence no uniform constant magnetic field nor Coriolis forces appear in the equations expressed in the uniformly rotating reference frame. This is why the form of the magnetic force term in the inertial frame coincides with the form of the Coriolis force term in the rotating frame: one must compensate the other (together with the centrifugal force) in the transition between the two coordinate systems, given the right rotation frequency.

The same can be done for the electric force: it can be made to disappear in simple accelerating coordinate systems, as the force term has the form of a simple acceleration (in contrast with the $\vec{v}\times$ form of the magnetic force).

A magnetic force is not fictitious in the following sense. Magnetic and electric forces have very different effects on the motion of point-like charges. This statement is supported by the very form of the Lorentz equation. Thus, as previously stated, one may find a coordinate system such that the magnetic force does not appear in the equations, but one cannot remove the effect of the magnetic field on the trajectories of charged particles: the trajectories themselves do not depend on the description you give of them. The curve $x^{\mu}(s)$ is an actual collection of points in spacetime, and the location of these points does not depend on the way in which you parametrize it. The equations will change in such ways as to produce the very same effect on the dynamical trajectories, only through a different coordinate description of the dynamics. This is the statement of the coordinate-transformation invariance of the action $S$. It is the description which changes, not the physics. Thus a magnetic field will always have an influence on the system, whether you call it magnetic or (in a different description, i.e. in a different coordinate system) not. One should though keep in mind that magnetic fields arise from the perturbation of the EM field by non static (wrt to some inertial frame) electric charges, with emphasis on the electric (i.e. non-magnetic) nature of the charges. (Again, we are totally underlooking the role of spin currents on the production of magnetic fields).

As a bonus, here is the potential for a monopole magnetic field. Define a coordinate system that covers the entire $\Bbb{R}^{4}$ spacetime except for the non-positive $z$ axis and take $\vec{A}$ to be

$$ \vec{A}^{N}=\left(g\ \frac{y}{r(r+z)},-g\ \frac{x}{r(r+z)},0\right) $$ Then define a coordinate system that covers $\Bbb{R}^{4}$ except for the non-negative $z$ axis and take $\vec{A}$ to be $$ \vec{A}^{S}=\left(-g\ \frac{y}{r(r-z)},g\ \frac{x}{r(r-z)},0\right) $$

$g$ is the magnetic coupling constant, and the magnetic field which corresponds to $\vec{A}^{N}$ and $\vec{A}^{S}$ is the same; it equals

$$ \vec{B}=\vec{\nabla}\times\vec{A}=-g\frac{\vec{r}}{r^{3}} $$ and it is a monopole field. Differential geometry teaches us that it is ok to choose local coordinate systems, i.e. coordinate systems which cover spacetime only in part. It also teaches us that, thanks to the gauge invariance of the action, we can choose potentials to be different in those local coordinate systems, as long as they are related by a gauge transformation in the overlapping regions. In this case, we have

$$ \vec{A}^{N}=\vec{A}^{S}+2g\vec{\nabla} \arctan\frac{y}{x} $$

so that an overall gauge potential is well defined, and it correctly reproduces a monopole field. So no, magnetic monopoles are not forbidden, not even in a classical setting.

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    $\begingroup$ (1/2) Thank you for this thoughtful response- for what's it's worth, I enjoyed reading it quite a bit! I appreciate your point about not mixing up artifacts of the coordinates with physical phenomena, but I have a doubt about this argument in this instance. If you will permit me to generalize the OP's question slightly, it is a coordinate-independent statement that if one takes an interferometer and sets it into circular motion, a phase shift results. This is the Sagnac effect. For charged particles, an analogous phase shift occurs in a uniform magnetic field: the Aharonov-Bohm effect. $\endgroup$ – Rococo Mar 16 '16 at 7:52
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    $\begingroup$ (2/2) The relation between these two effects involves the same substitution of terms as in the OP’s question (see, for example, PRD 21, 2993), and I think it is reasonable to claim that both are manifestations of the same physical analogy. So one might reformulate the OP’s question as: what is the underlying reason for the resemblance between the Aharonov-Bohm and Sagnac effects? And it is not clear to me that this question is an unphysical artifact of a choice of coordinates. $\endgroup$ – Rococo Mar 16 '16 at 7:52
  • $\begingroup$ Sure. I addressed this point, but I may not have given it the emphasis it deserved. I will modify that paragraph later. Now, first of all, there is a problem with wave and quantum mechanics in that the dynamical equations for wave and quantum systems (Maxwell, Schrödinger, Heisenberg, path integral kernel of the evolution operator) are quite different from those of geometric mechanics (i.e. particle mechanics). Specifically, the Aharonov-Bohm effect cannot be observed in geometric mechanics, while it is in quantum mechanics. $\endgroup$ – Giorgio Comitini Mar 16 '16 at 10:37
  • $\begingroup$ I was not aware about the Sagnac effect, but the same applies in this case, as it is manifestly a wave-like phenomenon. Moreover, I don't think that comparison with wave/quantum mechanics clarifies the problem. Geometric mechanics is a self-consistent theory and thus need not be given a foundation/interpretation in terms of a wave/quantum theory (though it is possible to do so, of course, especially when one starts from an action principle). $\endgroup$ – Giorgio Comitini Mar 16 '16 at 10:42
  • $\begingroup$ The reason why the Coriolis force and the magnetic force for a uniform, constant magnetic field have a similar mathematical form in two different, specific coordinate systems (the rotating frame and the inertial frame), as I wrote in the third-last paragraph, is that a uniform constant magnetic field has the effect of producing a uniform circular motion on the particles subject to it. This, of course, is a coordinate-invariant (i.e. physical) statement. One can even measure the geometrical features of the trajectory, with no reference to a specific coordinate system. $\endgroup$ – Giorgio Comitini Mar 16 '16 at 10:48
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I came across this wonderful paper on arXiv.org which exactly answers what you asked for.

"Why is the magnetic force similar to a Coriolis force?" - Antoine Royer

Abstract from the paper:

In this paper, it is pointed out that the underlying reason why the magnetic force is similar to a Coriolis force is that it is caused by Thomas rotations, induced by successions of non-collinear Lorentz boosts. The magnetic force may even be viewed as a kind of Coriolis force (making perhaps more acceptable the apparent non-existence of magnetic monopoles). We also show that under a change of inertial frames, Faraday lines of force Lorentz contract as if ‘etched’ in space, while ‘Coriolis’ terms get added on.

Please read the full paper. Something I noticed was that the question you asked and the description you gave, like the lecture by Feynman, is also described in this paper.

Read it here.

PS:

"What led me more or less directly to the special theory of relativity was the conviction that the electromotive force acting on a body in motion in a magnetic field was nothing else but an electric field." - A. Einstein

"This magnetic force has a strange directional character […] Magnetism is in reality a relativistic effect of electricity." - R. P. Feynman

The paper describes that magnetic forces have their strange Coriolis-like character because they enact Thomas rotations, induced by successions of non-collinear boosts. So if one views magnetism as a “relativistic effect of electricity” (Feynman), as was apparently also the initial idea of Einstein, then magnetic forces are a kind of Coriolis force. More generally, any Newtonian force (i.e. depending on the positions, but not on the velocities of particles) in some inertial frame, is necessarily accompanied by a ‘Coriolis’ force in another inertial frame.

It is recommended to check the "conclusion" part of the paper.

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    $\begingroup$ This does look like an interesting argument. It would be nice if you could provide more description than just reposting the abstract though. $\endgroup$ – Rococo Mar 14 '16 at 4:57
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A comment about this answer: as I now think about it, this answer really gives a motivation, starting from some gauge theory assumptions, for why a uniform magnetic field should cause a moving charged particle to rotate in a circle. The answer of Giorgio, on the other hand, explains quite nicely why any force that causes circular motion (in an inertial frame) must have a form analogous to the Coriolis force (in a rotating frame).


I have wondered about this myself (and I don't regard the answer to the linked question as particularly helpful). This analogy appears strikingly in, for example, the comparison between the response of a super-fluid to rotation and of a superconductor to a uniform magnetic field, as I mentioned in a recent answer.

I don't see myself as very qualified to give a definitive answer about this, but I will put something down in the hope of encouraging discussion. I think the root of this comparison lies not in Lorentz transformations, but in the structure and meaning of gauge theory. This will require a deep dive into quantum physics, so if you haven't studied that I apologize in advance...

One way that the laws of electromagnetism can be derived is by starting with the regular expression for the Lagrangian of a charged particle, then demanding that it be invariant under a local gauge transformation. This ends up requiring that you introduce the vector potential $A_{\mu}$ and modify the kinetic energy term in the minimal coupling way: $\partial_{\mu}\rightarrow(\partial_{\mu}-\frac{ie}{\hbar c}A_{\mu})$.

One way to think about this is that $A_{\mu}$ describes how the phase of the particle's wavefunction changes as one moves from point to point. Indeed, Chen-Ning Yang, among others, has remarked that a 'gauge field' really ought to be called a 'phase field' instead. If you're not familiar with this set of ideas, relevant phrases to start looking at are 'fiber bundle,' '(gauge) covariant derivative,' 'parallel transport,' and 'connection.'

Okay, so to recap: when we look at the evolution of the wavefunction of a charged particle in a region with some EM field, taking a spinless particle for simplicity, the Hamiltonian has a term like: $(\partial^{\mu}+\frac{ie}{\hbar c}A^{\mu})(\partial_{\mu}-\frac{ie}{\hbar c}A_{\mu})\phi$ . We can view the terms with $A$ as telling us how the phase changes in space, in much the same way that a curvature tensor could tell us how to transport a vector along a curved surface. But, in practice, we invariably instead think of the phase of the wavefunction as being connected in the same way it would be in the absence of an EM field, and then add a new influence to the wavefunction (aka the electromagnetic field) to 'correct' the behavior. Therefore, I suspect (and I say suspect because I haven't quite seen anyone else say this explicitly) that this process of ignoring the non-trivial phase field, then adding in some term to reproduce the same behavior, is fundamentally in the same spirit as ignoring that you are really rotating and putting in a fictitious Coriolis force to make things work out.

For example, in a uniform magnetic field along the $\hat{z}$ direction, one may write $A$ as: $A=\frac{1}{2}(Bx\hat{y}-By\hat{x})$. Now think of $(\hat{r}\cdot\vec{A}(x,y))$ as the rate of change of the phase at (x,y) for an infinitesimal step along $\hat{r}$, which is in fact (up to some constants) exactly what it means. One can see that the phase is in a sense spiralling around in a circle, so it makes sense that when we ignore this spiralling and then try to add some force to duplicate its effect, we end up getting something that closely resembles a fictitious force for uniform circular motion.

One obvious complication is that the gauge of $A$ is not unique- for example, we can just as easily say that $A=Bx\hat{y}$. But although this no longer looks like a whorling pattern, it is still true, of course, that it has the same curl with value $\vec{B}$.

Although I know even less about this, my understanding is that Kaluza-Klein theory was an attempt to unify EM and gravity around the '30s by adding an extra tiny 5th dimension at every point in space. This tiny space would basically be a physical place for the phase (or, in other words, the U(1) symmetry) to be manifest. I therefore suspect (although I would appreciate any comments from more knowledgeable people on this subject) that in Kaluza-Klein the magnetic field really does literally correspond to the fictitious force that results from pretending that you aren't spinning around in the little 5th dimension when you actually are. Sadly, this didn't quite work out, so at least in the Standard Model one should regard all these phases of the field as a purely internal property.

I think it is fair to say that, in this picture, the difference between a fictitious and non-fictitious force becomes somewhat fuzzy, as is the case in GR. I don't see any connection with monopoles, and since this formulation takes care of both electricity and magnetism, I don't think it is getting at some deep difference between the two of the type you have in mind.


Edit: in response to the comments of ACuriousMind, I will go into some more depth.

As mentioned above, one way to think about the effect of the magnetic field (or, more generally, a non-trivial vector potential) is that it modifies the phase accumulated by a charged particle. The minimum set of observables needed to describe the effects of this field, as mentioned above, are the so-called non-integrable phase factors:

$$I(C)=e^{\frac{ie}{\hbar c} \oint_C A_{\mu} dx^{\mu}}$$

Note that these are gauge-invariant quantities. In a uniform magnetic field this expression simplifies to (setting $\hbar$,$c$, and $e$ to one):

$$I(C)=e^{i BA}$$

where $A$ is the area enclosed by the path. These phase factors contain all the information about the dynamics in the field. In particular, they imply the Lorentz force. Instead of giving a derivation, I will give a heuristic argument that will be useful later: by definition, these phase factors mean that a charged particle transported in a loop gets an extra phase of $BA$. As a result, the free particle phase of $(\vec{p}\cdot \vec{x}-\omega t)$ must be modified, and since this phase is only dependent on the spatial path taken we must modify $\vec{p}\cdot \vec{x}$. There are various equivalent choices of how to do this. For example, if the path taken is a circle with radius $r$, one may modify the momentum: $\vec{p} '=\vec{p}-xB\hat{y}$, and this will get the phase right. Of course, this is a special case of the minimal coupling expression given above. This modification of the momentum necessarily implies a velocity-dependent force of the Lorentz form, as is shown in any course covering QM in a magnetic field.

We normally think about the magnetic field (or, more precisely, the vector potential) as modifying the expression for the momentum, but one could also interpret it as modifying the $\vec{x}$ term in $\vec{p}\cdot \vec{x}$ instead. In this interpretation, the magnetic field effectively changes the path length for a given charged particle. This is an unusual perspective to take, but it is completely identical: it results in the same phase factors, which completely describe the effects of the magnetic field. In this perspective, the path length for the same circular trajectory becomes longer by $\Delta x=B \pi r^2/p$.

With all this in place, the comparison with a rotating frame is straightforward. An object traveling along a rotating disc (as in the picture) really does travel a longer path than one measures in the frame that rotates along with the disc. In particular, something that travels in a circle at radius $r$ and velocity $v$, in the same direction as the overall rotation, travels an extra distance of $\Delta x=2\pi d^2 \omega/v=2 m \omega \pi r^2/p$. As expected, this is the same expression with the substitution $B\rightarrow2m\omega$.

enter image description here

To summarize this argument: the phase factor formulation of electromagnetism implies that a non-zero vector potential may just as easily be interpreted as modifying the spatial structure that a given charged particle travels through as it may be interpreted as modifying the momentum in the usual prescription. Upon taking this perspective, one sees that the expression for the change in path length in this system is analogous to that of a system in uniform rotation. Therefore, if one attempts to correct for the dynamics caused by this path length change by introducing a force, the force in either case has the same form. The fact that it is called a fictitious force in one case and not the other is, in my view, somewhat arbitrary.

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    $\begingroup$ Thank you for your answer! Although I unfortunately can't follow the entire explanation –I'm still an undergraduate and haven't studied those subjects yet–, I think I somehow managed to grasp some of the ideas. At least, it is helpful to have some clues on where to keep looking for further information. $\endgroup$ – David Herrero Martí Mar 10 '16 at 23:19
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    $\begingroup$ I glad you liked it. I think it would be nice for this to get a more definitive answer than mine, though, so I'm going to put a bounty on the question and see what kind of competition I can attract. $\endgroup$ – Rococo Mar 11 '16 at 23:55
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    $\begingroup$ I believe you are taking the "geometrical" analogy too far here. The geometrical space in which a gauge theory acts is not a physical space, the gauge choice acts purely on unphysical degrees of freedom, while a spatial rotation is manifestly a physical choice. Also, the Lorentz force involves only gauge invariant quantities, so I don't see the analogy you claim. Newton's law actually changes under non-inertial coordinate changes, but the Lorentz law does not change under the change of gauge. How do you think the Lorentz law corresponds to a fictitious force in the sense of Coriolis? $\endgroup$ – ACuriousMind Mar 12 '16 at 0:26
  • $\begingroup$ Please see my edit on the original question, as it may be relevant to the discussion. $\endgroup$ – David Herrero Martí Mar 13 '16 at 13:38
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    $\begingroup$ Some random remarks: 1. The phrase "non-integrable phase factor" sounds archaic to me, it's just the parallel transport operator associated to the connection that is the gauge field. 2. The modification of $x$ instead of $p$ works for a single particle, but fails to generalize to a more general gauge theory. 3. The answer seems to freely switch between quantum viewpoint ("free particle phase") and classical viewpoint ("rotating disk"). In particular, the argument for the classical Lorentz force being a fictitious force relying on the quantum notion of particle phase seems a bit strange. $\endgroup$ – ACuriousMind Mar 14 '16 at 10:34
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The formal identity in the two expression is maybe just due to the fact that it is the easiest way to add a velocity dependent force in a Lagrangian. Let’s look at this more formally, in the framework of Lagrangian and Hamiltonian mechanics.

If one want a velocity-dependent force (like Coriolis of Lorentz forces) coming from a Lagrangian, the simplest way is to add a term which varies linearly on the velocity $\dot{\mathbf r}$. Such a term can always take the form ${\mathbf a}(\mathbf r)⋅\dot{\mathbf r}$. That would give the following Lagrangian for a single particle $$\mathcal L=\frac12 m\dot{\mathbf r}² - V(\mathbf r) + {\mathbf a}(\mathbf r)⋅\dot{\mathbf r}.$$

If you are not familiar with Lagrangian mechanics,

  • The independence of this Lagrangian with respect to $t$ ensures energy conservation
  • The first two terms correspond to the kinetic energy and the potential energy of the other forces present in this sytem
  • As shown in the appendix, the last term creates a Coriolis/magnetic-like force, and ${\mathbf a}$ plays a role similar to the vector potential, and $m\mathbf Ω$ or $q \mathbf B$ is given by $∇×\mathbf a$.

Since a linear dependence in $\dot{\mathbf r}$ is the simplest, the linearity can arise for many reasons, and the common cause is that both force velocity dependent Lagrangian forces. Of course, to have a velocity dependence, one need a “non-Galilean” reference frame change (non-inertial for Coriolis, relativistic for $\mathbf B$, since magnetism is explained by electrostatic + relativity.) In the latter case, the linearity might be explained by a first order approximation in $\dot{\mathbf r}/c$.

Appendix: Formal derivation of the Coriolis/magnetic forces from $\mathcal L$

Here, I derive the magnetic/Coriolis force expression from the Lagrangian introduced above. It is just straightforward and standard Hamiltonian mechanics, and it does not need to be read if you accept that a simple linear term added in the Hamiltonian creates the desired force.

For this $\mathcal L$, the generalised momentum $\mathbf p$ is $ {\mathbf p}:=\frac{∂\mathcal L}{∂\dot{\mathbf r}}=m\dot{\mathbf r}+\mathbf a $. The Hamiltonian $\mathcal H$ is then given by \begin{align} \mathcal H &:=\dot{\mathbf r}⋅\mathbf p-\mathcal L=\frac{(\mathbf p - \mathbf a({\mathbf r}))^2}{2m} + V({\mathbf r}). \end{align}

the expression of the momentum allows us to compute its change rate $$\dot{\mathbf p}=m\ddot{\mathbf r}+\dot{\mathbf r} ⋅\frac{∂\mathbf a}{∂{\mathbf r}} =m\ddot{\mathbf r}+ \begin{bmatrix} \dot x \frac{∂a_x}{∂x}+\dot y \frac{∂a_x}{∂y}+\dot z\frac{∂a_x}{∂z}\\ \dot x \frac{∂a_y}{∂x}+\dot y \frac{∂a_y}{∂y}+\dot z\frac{∂a_y}{∂z}\\ \dot x \frac{∂a_z}{∂x}+\dot y \frac{∂a_z}{∂y}+\dot z\frac{∂a_z}{∂z} \end{bmatrix}.$$

On the other hand, Hamilton’s equations give us $$\dot{\mathbf p} = -\frac{∂\mathcal H}{∂{\mathbf r}} = -\frac{∂}{∂{\mathbf r}}\frac{(\mathbf p - \mathbf a)^2}{2m} -\frac{∂V}{∂{\mathbf r}} =\begin{bmatrix} \dot x \frac{∂a_x}{∂x}+\dot y \frac{∂a_y}{∂x}+\dot z\frac{∂a_z}{∂x}\\ \dot x \frac{∂a_x}{∂y}+\dot y \frac{∂a_y}{∂y}+\dot z\frac{∂a_z}{∂y}\\ \dot x \frac{∂a_x}{∂z}+\dot y \frac{∂a_y}{∂z}+\dot z\frac{∂a_z}{∂z} \end{bmatrix}-\frac{∂V}{∂{\mathbf r}}. $$ Puting the equations together, we obtain the following expression for the acceleration (and the forces): \begin{align}\ddot{\mathbf r} &=\begin{bmatrix} \dot x \frac{∂a_x}{∂x}+\dot y \frac{∂a_y}{∂x}+\dot z\frac{∂a_z}{∂x}\\ \dot x \frac{∂a_x}{∂y}+\dot y \frac{∂a_y}{∂y}+\dot z\frac{∂a_z}{∂y}\\ \dot x \frac{∂a_x}{∂z}+\dot y \frac{∂a_y}{∂z}+\dot z\frac{∂a_z}{∂z} \end{bmatrix}-\frac{∂V}{∂{\mathbf r}} - \begin{bmatrix} \dot x \frac{∂a_x}{∂x}+\dot y \frac{∂a_x}{∂y}+\dot z\frac{∂a_x}{∂z}\\ \dot x \frac{∂a_y}{∂x}+\dot y \frac{∂a_y}{∂y}+\dot z\frac{∂a_y}{∂z}\\ \dot x \frac{∂a_z}{∂x}+\dot y \frac{∂a_z}{∂y}+\dot z\frac{∂a_z}{∂z} \end{bmatrix}\\ &=\begin{bmatrix} \dot y (\frac{∂a_y}{∂x}-\frac{∂a_x}{∂y})+\dot z(\frac{∂a_z}{∂x}-\frac{∂a_x}{∂z})\\ \cdots\\ \cdots \end{bmatrix}-\frac{∂V}{∂{\mathbf r}}\\ &=\begin{bmatrix} \dot y (∇×\mathbf a)_z-\dot z(∇×\mathbf a)_y\\ \cdots\\ \cdots \end{bmatrix}-\frac{∂V}{∂{\mathbf r}}\\ &=\dot{\mathbf r}×(∇×\mathbf a) -\frac{∂V}{∂{\mathbf r}} \end{align} The first term is the magnetic/Coriolis force, and the second the usual potential derived forces.

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  • $\begingroup$ Thank you for your answer! I find this approach interesting, too. However, please see my recent edit on the original question: could the two other analogies between inertial and electromagnetic forces also be explained by this approach? $\endgroup$ – David Herrero Martí Mar 15 '16 at 21:40

protected by Qmechanic Mar 11 '16 at 16:01

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