Do Electric field lines propagate by themselves away from a charge?

[renormalize]

While the charge does have KE, that is not the energy I am talking about. I am talking about the EM field energy density: $u=(E^2+B^2)/2$ with whatever constants are needed to make the units work

Yes, but doesn't the EM field around a moving charge contribute to its total kinetic energy due to the so-called "electromagnetic mass": https://www.feynmanlectures.caltech.edu/II_28.html ?

 

[Dale]

Yes, but doesn't the EM field around a moving charge contribute to its total kinetic energy due to the so-called "electromagnetic mass": https://www.feynmanlectures.caltech.edu/II_28.html ?

Yes, definitely. But although the mass from the field energy contributes to the kinetic energy, it is much larger than the kinetic energy. So again, you are not wrong that there is KE, but I was specifically referring to the EM field energy.

I am not correcting you, I am clarifying my own statement.

 

[vanhees71]

Except that the energy transfer does not happen through the electrons. The energy goes from the battery to the light bulb in the fields, not the electrons.

The electrons are involved in changing energy from chemical to electromagnetic at the battery and in changing the energy from electromagnetic to thermal at the light bulb. They are not involved in moving the energy from the battery to the light bulb, except for the fact that the current produces the B field.

If you do not understand $\vec S = \frac{1}{\mu_0} \vec E \times \vec B$ then you do not understand the energy flow in a DC circuit. Once you do understand the energy flow in a DC circuit then you also understand it for an EM wave. They are the same.

This is not true. Static E and B fields, such as those in a simple DC circuit as I mentioned above, have this same transfer. Anywhere that S is nonzero, energy is transferred, regardless of whether the field is static or not.

That is correct. The difference is where the energy goes. In a constant speed charge the energy goes forward and stays close to the charge. In an EM wave the energy goes out away from the charge. Similarly with a DC circuit. There is no wave because the energy stays close to the wires at all times.

That's very important and easy to observe when switchin on the light in your room. There are some meters distance between the switch and the light source, and the light turns on very quickly. As it turns out when you analyze what happens using Maxwell's equations is that the signal velocity is the speed of light. You can calculate this for simple geometries (e.g., a coax cable) even using the simplifying telegrapher's equation.

If it were the electrons transporting the energy from the source to the light bulb it would take minutes until you get light after switching it on, because the velocity of the electrons in the cable is a few mm/s (millimeters per second!).

 

[gionole]

Thanks so much everyone. I will study couple of things to get ready to better understand all this and will reply as soon as possible.