Relation between Electric and magnetic fields

While I think your question may be problematic to some because they are very weary of the "why" question, because physics can only go so deep, I recognize that it is hard to just accept a causal relationship between things that seem arbitrarily related, so lets try to look deeper.

A magnetic field is caused by a moving electric charge correct? The moving electric charge causes an increase in the electric field in front of it and a decrease in the electric field in back of it, and these changes create a magnetic field, but let's go back to the charge.

Let's imagine that this charge is moving extremely fast, at relativistic speeds even. Next to it and parallel to its motion is an infinitely long wire with current flowing through it, a lot of current too. Let's say that the electrons in this wire are moving just as fast as the electron, and in the same direction. We could even imagine them with race helmets on, racing each other off to infinity.

Now this wire is electrically neutral, for every electron in the wire there is a proton, so the electron traveling alongside should feel no pull towards the wire or a push away. However, this is all from our perspective. To us the electrons are moving fast, but what about to them?

According to relativity, they have every right to say that they are not moving. What looked like racing hats to us were actually top hats, and they were sitting down having some tea while we zoomed by at nearly the speed of light. Now we would look sort of funny, because the effects of relativity cause us to look squashed. This is important, because we would not be the only things zooming past the electrons.

The protons in the wire as well would be zooming past them as well. The same relativistic squashing happens with them, but this time it's more important. The relativistic length contraction not only squishes the protons, but because it is a whole column of protons moving past them, the column squishes as well, increasing the positive charge density of the wire. The electron feels the effect of this increased charge density as a pull inwards and so it drifts closer to the wire.

We see this in our frame as well and are perplexed, why would the electron feel a pull from the wire? We see no excess charge in the wire, so we ascribe this effect to a different force, the magnetic force. However, from the electrons reference frame, this behavior is perfectly normal, the protons moving past him are closer together than the electrons standing still and the electric force from the wire pull him closer.

This is kind of what magnetism is, electricity's compensation for relativity. For if magnetism didn't exist, we would see the electron attracted to the wire, for no explainable reason. Magnetism is sort of the relativistic form of electricity.

As for them interacting and causing each other, this must happen or else other laws of physics could possibly be broken (or you would have a meaningless thing like a force from nothing). This is a comforting example of how a part of physics holds itself up by itself.

;tldr: You're right about the properties, but the fields don't cause or influence each other in so direct away as a first glimpse of Maxwell's equations might suggest.

Special Relativity is "hidden" in Maxwell's equations. An understanding of their relationship might be useful.

Maxwell's Equations are:

$$\nabla\cdot\vec{E}=\frac{\rho}{\epsilon_0}$$ $$\nabla\times \vec{E}=-\frac{\partial \vec{B}}{\partial t}$$ $$\nabla \cdot \vec{B}=0$$ $$\nabla \times\vec{B}=\mu_0\vec{J}+\frac{1}{c^2}\frac{\partial \vec{E}}{\partial t}$$

Take the curl of the second equation and rearrange.

$$\nabla(\nabla\cdot\vec{E})-\nabla^2\vec{E}=-\frac{\partial}{\partial t}(\nabla \times\vec{B})=-\mu_0\frac{\partial\vec{J}}{\partial t}-\frac{1}{c^2}\frac{\partial \vec{E}}{\partial t}$$

Then rearrange: $$\nabla^2\vec{E}-\frac{1}{c^2}\frac{\partial^2\vec{E}}{\partial t^2}=\mu_0\frac{\partial \vec{J}}{\partial t}+\nabla{(\rho/\epsilon_0)}$$

Taking the curl of the 4th equation and rearranging:

$$\nabla^2\vec{B}-\frac{1}{c^2}\frac{\partial^2\vec{B}}{\partial t^2}=-\mu_0\nabla\times\vec{J}$$

In free space the charge density, $\rho$, and the current density ,$\vec{J}$, are 0

So theses two equations end up with a 0 for the right hand side and become homogeneous wave equations.

Solutions are linear combinations of individual solutions having the form $\vec{E}=\vec{E_0}e^{i(\vec{k}\cdot \vec{x}-\omega t)}$

$k$ is the wave number, $\omega$ is the frequency of the wave.

We can check against Maxwell's equations.

$$\nabla\cdot\vec{E}=i\vec{k}\cdot\vec{E_0}e^{i(\vec{k}\cdot\vec{x}-\omega t)}=0$$

$\vec{k}$ is in the direction of propagation, and since its dot product with $E_0$(and any component $E_0$ in a linear combination), then we can conclude that the electric field is perpendicular to its direction of propagation.

$\nabla \times \vec{E}=i\vec{k}\times\vec{E_0}e^{i(\vec{k}\cdot\vec{x}-\omega t)}=-\frac{\partial \vec{B}}{\partial t} \implies \vec{B}=-\frac{\vec{k}}{\omega}\times\vec{E_0}e^{i(\vec{k}\cdot\vec{x}-\omega t)}$

So the magnetic field is perpendicular to both the direction of propagation and the electric field. It's also worth noting $|\vec{k}/\omega|=1/c$.

The Poynting Vector representing the direction of energy flux of the EM field is : $$\vec{S}=\frac{1}{\mu_0}\vec{E}\times\vec{B}=\frac{-1}{\mu_0}e^{2i(\vec{k}\cdot\vec{x}-\omega t)}\vec{E_0}\times(\frac{\vec{k}}{\omega}\times \vec{E_0})= \frac{-1}{\mu_0}e^{2i(\vec{k}\cdot\vec{x}-\omega t)}(\frac{\vec{k}}{\omega}E_0^2-\vec{E_0}(\frac{\vec{k}}{\omega}\cdot\vec{E_0}))$$

Since $\vec{E}$ is perpendicular to $\vec{k}$:

$$\vec{S}=\frac{-1}{\omega\mu_0}e^{2i(\vec{k}\cdot \vec{x}-\omega t)}E_0^2\vec{k}$$

in agreement with $\vec{k}$ as the direction of propagation.

So Maxwell's equations demonstrate certain properties of electro magnetic radiation without making any explicit reference to special relativity. Some questions arise though.

In the wave equation, the coefficient of the second time derivative is the reciprocal square of the velocity of the wave. So Maxwell's equations imply that electro magnetic waves travel at the speed of light. But with respect to what frame of reference? How can radiation be traveling one speed to a stationary observer on earth and the same speed relative to a spaceship traveling in its direction at half the speed of light? Since it always takes some amount of time for a a cause to manifest an effect, how can the time variation of a magnetic field instantly influence the electric field as Maxwell's Equations seem to imply?

This issue might best be approached through an analysis of electric and magnetic potentials.

Given that $\nabla \cdot \vec{B}=0$, we have $\vec{B}=\nabla \times \vec{A}$ for some vector $\vec{A}$ by a theorem from vector calculus, a vector with zero divergence can be expressed as the curl of some other vector. This Vector is the magnetic potential or the Vector Potential.

So we also have $\nabla \times \vec{E}=-\frac{\partial}{\partial t}(\nabla \times \vec{A})\implies \nabla\times(\vec{E}+\frac{\partial \vec{A}}{\partial t})=0$

Another theorem from vector calculus tells us that if the curl of a vector is zero, it can be expressed as the gradient of some scalar.

So: $$\nabla \times (\vec{E}+\frac{\partial \vec{A}}{\partial t})=0 \implies \vec{E}=-\nabla \phi - \frac{\partial \vec{A}}{\partial t}$$

From this we have:

$$-\nabla^2\phi-\frac{\partial(\nabla \cdot \vec{A})}{\partial t}=\frac{\rho}{\epsilon_0}$$

Taking the curl of the expression of $\vec{B}$ in terms of $\vec{A}$ we have

$$\nabla(\nabla \cdot \vec{A})-\nabla^2\vec{A}=\mu_0\vec{J}+\frac{1}{c^2}\frac{\partial}{\partial t} (-\nabla \phi - \frac{\partial \vec{A}}{\partial t})$$

Implying:

$$\nabla^2 \vec{A}-\frac{1}{c^2}\frac{\partial^2 \vec{A}}{\partial t^2}=-\mu_0\vec{J}-\nabla(\nabla \cdot \vec{A}+\frac{1}{c^2}\frac{\partial \phi}{\partial t})$$

Now for any $\vec{v}$, vector $\vec{u}$, and scalar $g$,

$$\nabla \times (\vec{v}+\nabla g)=\nabla \times \vec{v}$$

$$ \nabla \cdot \vec{v}=\nabla \cdot (\vec{v}+\nabla \times \vec{u})$$

Since our fields only depend on the derivatives of potentials, we are allowed to change them so long as the original vector derivatives remain constant.

So:

$$\vec{A'}=\vec{A}+\nabla g$$ $$ \phi'=\phi-\frac{\partial g}{\partial t}$$

So modifying with some scalar $g$ will change the potential perahps to a more convenient form without changing the vector derviatives and hence the fields. This Gauge Freedom allows us to assert that: $$\nabla \cdot \vec{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0$$

That is called the Lorentz Gauge.

It simplifies the equations for our potentials. They become:

$$\nabla^2 \vec{A}-\frac{1}{c^2}\frac{\partial ^2 \vec{A}}{\partial t^2}=-\mu_0\vec{J}$$ $$\nabla^2\phi-\frac{1}{c^2}\frac{\partial ^2 \phi}{\partial t^2}=\frac{- \rho}{\epsilon_0}$$

We can use Green's Method to find integral equation solutions to the potentials from which the fields can be derived.

These solutions give us time dependent versions of the Biot-Savart Law and Coulomb's Law. In particular, the solutions necessarily have a dependence not on the charge and current distribution at the present time, but the distributions as they were in the past. In other words we have (time) retarded potentials.

Their derivatives produce the Jefimenko's Equations for electric and magnetic fields relating the current feild values to past charge and current distribution values. They resolve our questions above by implying that electromagnetic information travels at light speed and the various relations deduced above regarding the electric and magnetic fields do not actually violate causation. They don't influence each other instantaenously, the relationship between their changes is due to changes in the charge and current distributions in the past.