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Is the magnetic field a feature of our universe, or is it a side effect of the electric field? In other words: if we were to simulate in a computer a system of moving charged particles, taking in account only equations for the electric field, would we see magnetic effects?

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Related: – Qmechanic Feb 8 '12 at 6:40
up vote 2 down vote accepted

Electric and magnetic fields are fundamentally the same thing. They behave slightly differently, but that's because of the lack of monopoles and consequently "magnetic current". Check out the Maxwell's laws (I'll refer to them as M1,M2,M3,M4). M1 and M2 are nearly the same, except for the zero in M2. This is because of the lack of monopoles. Similarly, M3 and M4 are nearly the same (with some extra constants floating around), except for the $J$ term in M4. This term is not there in M3 due to the lack of "magnetic current". If we discovered magnetic monopoles and magnetic current, the forms of the equations would become exactly the same.

OK. To drive in the fact that both fields are the same thing, imagine a a pair of point charges($q_1,q_2$), lying side by side. You give them both an equal and constant velocity $\vec{v}$ perpendicular to the line joining them. Now, $q_1$ is generating a magnetic field, as it is a moving charge. This magnetic field will interact with $q_2$, and push/pull it in the direction joining their common centers (To check this, imagine $q_1$ to be a wire carrying current in the direction of its motion). Our electrostatic "Coulomb force" will be $\frac{kq_1q_2}{r^2}$, as usual. Looks fine till now.

Now, you hop in your car and start following the charges. You go with the same velocity as them. Now, from your car, you see something different. You see two stationary charges, exerting a force $\frac{kq_1q_2}{r^2}$ onto each other. So what happened to the force caused by the magnetic field? After all, the net force acting on a body should be the same from any inertial, that is non-accelerating, frame (Otherwise you would disagree on their accelerations etc).

A similar scenario is when the charges at at rest, and you first measure them from rest and then you go into motion.

So here we have a bit of a paradox. It's solved pretty easily when you consider the (intentional) flaw in my argument. The Coulomb force only works in electro*static* conditions. Its a tiny bit less/more when the charges are in apparent motion, which recompensates for the magnetic field.

Thus, from one frame (the car), you see two charges at rest, attracting each other electrostatically. Thus you only see an electric field (As you can play about with a third charge in your car, and you will only notice electric effects). From the ground, you calculate two types of attractions, and you can measure two types of fields. One can say that the electric field changed form into the magnetic field. Or you could say the opposite (depending on where you made your measurements first).

So what does this entail? Electric and magnetic fields are the actually the same thing, and they change form whenever you go in relative motion. Note that there is no preferred reference frame for relative motion; so one cannot choose the frame with only the electric field and say "this frame has only $\vec{E}$, ergo $\vec{E}$ is fundamental and $\vec{B}$ is just a side effect.

Another way of looking at the way light propagates. A light wave travels mutually perpendicular with oscillating electric and magnetic fields. Since E/M fields are transmitted/mediated by light (the E/M interaction is due to exchange of photons), this is a good place to visualise it. From this, one can see more clearly how they are equivalent, there is no way to pick one field over the other here; they both have similar characteristics with respect to the light wave.

About your second question involving computer simulations: Well, it depends upon the simulation. If you only use the Coulombic electrostatic force, you should have no problem as long as you neglect magnetic fields, and the particles have uniformish velocity. There will still be aberrations because you are not considering relativity (relativity can be partially deduced from inconsistencies in these equations). Really, it depends upon the accuracy of the simulation, as when dealing with charges only (no currents), the force due to magnetic fields is $\approx c^2$ times smaller than the electric field. If the particles accelerate, then we have the added issue of transmission of fields, as changes to the position of the particle propagate (at the speed of light, not instantaneously) to the field nearby. This is where the different electric field from moving charges comes from. Play with the velocity slider here to see how the field changes. Basically, if a point is at a distance r from a charge, the field at that point will be $\frac{kq_1q_2}{r^2}$ after $\frac{r}{c}$ seconds.

If you incorporate the correct, "field travels at the speed of light" concept, then your system will behave abnormally, and you will notice the lack of magnetic fields (assuming enough precision).

So the answer depends on what you mean by "taking into account only electric fields".

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I'd prefer not to call it a "side effect of the electric field", but rather to recognize that the "true" object is the electromagnetic field tensor $F_{\mu\nu}$, which contains both electric and magnetic effects. What you see in any instance depends upon what measurements you do.

If you try to simulate "just" the electric effects in a system of moving charged particles, then you end up simulating "just" a system of particles moving under the inverse square law, like gravity but with the possibility of repulsive interactions.

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