PhysicsMan68 said:
What causes electromagnetic waves to first radiate from its source?
In addition to what others wrote ...
Definition of the
nabla operator:
##\vec \nabla := ({\partial \over \partial x},{\partial \over \partial y},{\partial \over \partial z })##
In vacuum, in the absence of sources, Maxwell's equations are
##\vec \nabla \cdot \vec E = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)##
## \vec \nabla \times \vec B = \epsilon_0\mu_0{\partial \vec E \over \partial t}\ \ (2)##
##\vec \nabla \cdot \vec B = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)##
## \vec \nabla \times \vec E = -{\partial \vec B \over \partial t}\ \ \ \ \ (4)##
An identity of vector calculus is
##\vec\nabla \times (\vec\nabla \times \vec v ) = \vec\nabla (\vec\nabla \cdot \vec v) - \vec\nabla^2 \vec v##.
This identity applied to the electric field and using equation (1):
##\vec\nabla \times (\vec\nabla \times \vec E ) = 0 - \vec\nabla^2 \vec E##
It follows, using equation (4):
##\vec\nabla \times (-{\partial \vec B \over \partial t} ) = - \vec\nabla^2 \vec E##
##{\partial \over \partial t}(\vec \nabla \times \vec B) = \vec\nabla^2 \vec E##
It follows, using equation (2):
##{\partial \over \partial t}(\epsilon_0\mu_0{\partial \vec E \over \partial t}) = \vec\nabla^2 \vec E##
Using ##c^2={1 \over\epsilon_0\mu_0 }##, the result is a 2
nd order linear differential equation:
##\vec\nabla^2 \vec E - {1 \over c^2}{\partial^2 \vec E \over \partial t^2} = 0##
In an analog way, one can derive the same for the magnetic field:
##\vec\nabla^2 \vec B - {1 \over c^2}{\partial^2 \vec B \over \partial t^2} = 0##
These differential equations are solved by a function, that describes the propagation of an electromagnetic wave with speed ##c## in vacuum.
Source: Book "Covariant Electrodynamics" by John M. Charap, chapters 2.1 and 3.4
https://www.amazon.com/Covariant-Electrodynamics-John-M-Charap/dp/1421400146?tag=pfamazon01-20
