- #1
- 5,695
- 2,477
Starting from the microscopic form of Maxwell's equations and following standard mathematical procedure outlined in
Inhomogeneous electromagnetic wave equation - Wikipedia
we can have as end result the following equations:
$$(\nabla^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2})\mathbf{E}=\frac{\nabla\rho}{\epsilon_0}+\mu_0\frac{\partial \mathbf{J}}{\partial t}$$
$$(\nabla^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2})\mathbf{B}=\mu_0\nabla\times\mathbf{J}$$
These two equations seem to imply that the speed of the waves is always c (due to the ##\frac{1}{c^2}## term appearing in front of the second time derivative in the left hand side). On the right hand side of course the ##\rho## and ##\mathbf{J}## are not the free charge and current density but rather the total ##\rho=\rho_{free}+\rho_{bound}## and ##\mathbf{J}=\mathbf{J_{free}}+\mathbf{J_{bound}}##. But what's the catch here, how can the exact nature of the right hand side (that depends on the materials used) can affect the left hand side and the speed of the waves?
Inhomogeneous electromagnetic wave equation - Wikipedia
we can have as end result the following equations:
$$(\nabla^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2})\mathbf{E}=\frac{\nabla\rho}{\epsilon_0}+\mu_0\frac{\partial \mathbf{J}}{\partial t}$$
$$(\nabla^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2})\mathbf{B}=\mu_0\nabla\times\mathbf{J}$$
These two equations seem to imply that the speed of the waves is always c (due to the ##\frac{1}{c^2}## term appearing in front of the second time derivative in the left hand side). On the right hand side of course the ##\rho## and ##\mathbf{J}## are not the free charge and current density but rather the total ##\rho=\rho_{free}+\rho_{bound}## and ##\mathbf{J}=\mathbf{J_{free}}+\mathbf{J_{bound}}##. But what's the catch here, how can the exact nature of the right hand side (that depends on the materials used) can affect the left hand side and the speed of the waves?