5
$\begingroup$

As in electrical impedance, Causes:

Resistance - collision of electrons with atoms and other electrons,

Reactance - Capacitive and inductive effects.

Likewise, what offers opposition to a wave traveling in a medium?

$\endgroup$
  • $\begingroup$ As a starting point, the lumped element parameters (resistance, capacitance, inductance, reactance) are simplified models for the electromagnetic behavior. The underlying effects are the same, but building devices like resistors and inductive coils controls what's going on to the point where we can understand a circuit using a simplified model. $\endgroup$ – The Photon May 30 '18 at 16:13
3
$\begingroup$

The "opposition" to the wave or wave impedance or impedance of a medium to a wave is caused by characteristics of the medium analogous to the resistance, capacitance and inductance. While resistance is a pretty generic term, applicable to different types of waves, the energy storing characteristics, capacitance and inductance, could be generalized as compliance or stress and inertia or motion.

When a wave is propagated, it energizes the medium and the speed of the propagation is reduced or opposed by the resistance of the medium and by its ability to store energy. So both high capacitance or compliance and high inductance or inertia of the medium act to slow down the wave or, we can say, it takes more time and energy to energize a medium with high capacitance and inductance. This for instance, is reflected in a formula for the wave propagation speed in an ideal transmission line, $V_p=\frac 1 {\sqrt {LC}}$, where both capacitance and inductance contribute symmetrically to oppose or slow down the wave.

The impedance, on the other hand, characterizes the tendency of a medium to oppose the motion component of the wave at a given stress level or, in electrical domain, the tendency to oppose the current or the magnetic field at a given level of voltage or electric field. This is reflected in a formula for the characteristic impedance of an ideal transmission line, $Z_0=\sqrt \frac L C$. Here, capacitance and inductance are not contributing symmetrically: high capacitance encourages the current flow, while high inductance impedes it.

For EM wave in space, the formulas for the propagation speed and impedance, $V=\frac 1 {\sqrt {\mu \epsilon}}$ and $Z=\sqrt \frac {\mu} {\epsilon}$, have similar meaning and underling mechanisms. The propagation is opposed or slowed down by both greater magnetic permeability (inductance) and electric permittivity (capacitance) of the medium. On the other hand, the impedance (to the motion component of the wave) is increased with the magnetic permeability and decreased with its electrical permittivity.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I didn't understand the concept of wave energizing the medium. If by that you mean wave is loosing energy, then like in case of resistance the collision of electrons with atoms is the cause of loosing energy, here why exactly wave is loosing energy in the medium? I wanted to know what is happening physically. $\endgroup$ – shul Jun 1 '18 at 10:23
  • $\begingroup$ No, I am not talking about losses. When a wave is passing through a medium, it sets up oscillations in the medium, which have energy associated with them. See details in this post: physics.stackexchange.com/questions/397467/…. If the medium has high inertia (inductance) and high compliance (capacitance), it would require more energy to energize it and, given limited power of the source, more time. Hence the prop. speed will be lower. Impedance considerations are similar. $\endgroup$ – V.F. Jun 1 '18 at 19:05
0
$\begingroup$

assuming we are discussing longitudinal waves (as opposed to gravity or capillary waves), the two factors influencing the movement of waves through a medium are its density and its compliance.

on the other hand, if you are talking about electromagnetic waves, the determinants of their speed (c) in a vacuum are the electric constant and the magnetic permeability of free space.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ From the tags, OP seems to be asking about EM waves. $\endgroup$ – The Photon May 30 '18 at 16:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.