Barry_G said:
Ok, they are not synonyms. Field is a concept that describes geometrical distribution of charge, and charge just refers to the magnitude. For example, a field can have shape, described by gradients, while charge is a scalar number, a measure of the magnitude at some point in the field.
The point is, when you ask whether there is electric field present at some point in space, it is just about the same as asking whether there is electric charge there, so in that sense they are synonyms, and you can see those two articles use the two terms interchangeably.
After reading through this and the thread you hijacked in the Relativity section, I must be the fifth or sixth person to tell you what I'm about to tell you:
There does not need to be any charges for there to be an electric field. Maxwell's Equations tell us that a changing magnetic field will also produce an electric field. Similarly, you don't need any currents for there to be a magnetic field; a changing electric field will produce a magnetic field. Essentially what happens in an EM wave is that a changing electric field produces a changing magnetic field which produces a changing electric field which... This allows the electric and magnetic fields to propagate without any charges or currents. I'll even go through how this is derived, step by step:
As you've been shown numerous times, Maxwell's Equations in a vacuum are:
\nabla \cdot E=0
\nabla \times E = -\frac{\partial B}{\partial t}
\nabla \cdot B=0
\nabla \times B = \frac{1}{c^2} \frac{\partial E}{\partial t}Now, using the following "curl of the curl" identity: \nabla \times \nabla \times A=\nabla(\nabla\cdot A)-\nabla^2A
\nabla \times \nabla \times E=\nabla(0)-\nabla^2E=-\nabla \times \frac{\partial B}{\partial t}=-\frac{\partial }{\partial t}[\nabla \times B]=-\frac{\partial }{\partial t}[\frac{1}{c^2} \frac{\partial E}{\partial t}]
Simplifying you get:
\nabla^2E=\frac{1}{c^2}\frac{\partial^2 E}{\partial t^2}
You can apply the same identity to the curl of the magnetic field to get the following:
\nabla^2B=\frac{1}{c^2}\frac{\partial^2 B}{\partial t^2}
The solution to both of these differential equations is a sinusoidal wave which moves at a velocity c.
You accuse K^2 of being condescending, yet he's absolutely correct. You have no idea how basic electrodynamics works. What you need to do is stop pretending you know what you're talking about and go pick up a textbook. You're arguing about things you don't understand.
