# Transmission Line EM Wave vs EM Wave in Free Space

• I
• LarryS
In summary: I understand the modern approach in QFT in general and QED in particular is to set c = 1, thus forcing ε0 and μ0 each to be 1 also. It's a very convenient approach. But setting constants to 1 does not make them go away. Their still there. You've just done a change of scale. The constants no longer need to be visible in the equations. That's all.Setting constants to 1 does not make them go away. Their still there. You've just done a change of scale.
LarryS
Gold Member
TL;DR Summary
Do capacitance and inductance have meaning for just an EM Wave in free space, in the absence of physical wires and currents?
According to Maxwell’s Equations, the speed an EM plane wave in free space, far from its source, is determined by the electric constant, ε0, and the magnetic constant, μ0, such that c = 1/√( ε0 μ0).

The units of ε0 are capacitance per unit length and the units of μ0 are inductance per unit length. Those two quantities have real meaning for a transmission line consisting, say, of two parallel perfectly conducting wires. Together they define the impedance of such a transmission line. That impedance seems to explain the finiteness of the transmission speed, the speed of light.

Do the quantities of capacitance and inductance have any meaning for an EM wave in free space far from its source? Can their definitions be generalized beyond physical wires and currents?

berkeman
WernerQH, vanhees71, LarryS and 1 other person
But the question is why does a vacuum, which classically contains nothing, impose that ratio upon the fields?

tech99 said:
But the question is why does a vacuum, which classically contains nothing, impose that ratio upon the fields?
Space is not empty, it is full of dark matter and dark energy. No matter where you go, you will see photons, while being ignored by neutrinos.

It is the EM character of the photon that gives space an intrinsic impedance. The propagation of energy along the Poynting vector requires there be the cross product of an E and an M field, neither can be zero.

tech99 said:
But the question is why does a vacuum, which classically contains nothing, impose that ratio upon the fields?
Makes one yearn for some aether does it not?

davenn, DaveE, LarryS and 1 other person
tech99 said:
But the question is why does a vacuum, which classically contains nothing, impose that ratio upon the fields?
That's a bit the old-fashioned idea that there must be a medium whose motion is what's manifests to our senses as light/electromagnetic waves. This has been overcome for at least 120 years now. According to the modern picture the fundamental descriptions are the fields themselves. The various constants like ##\epsilon_0## and ##\mu_0## in elctrodynamics are just arbitrarily choosen conversion factors needed to adopt the physical laws to our arbitrary choice of (SI-)units. There's nothing deep in their numerical values, which are simply as they are by historical chances.

vanhees71 said:
There's nothing deep in their numerical values, which are simply as they are by historical chances.
Then why does the reciprocal of the square root of their product just happen to be the speed of light, one of the most important constants in our universe?

That's because the em. field is a massless field. The most natural description is to use "natural units", i.e., you set all the fundamental constants of nature to 1. In such units, the Heaviside-Lorentz units with ##c=1##, the Maxwell equations read
$$\vec{\nabla} \times \vec{E}+\partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{E}=0, \quad \vec{\nabla} \times \vec{B} -\partial_t \vec{E}=\vec{j}, \quad \vec{\nabla} \cdot \vec{E}=\rho.$$
In the usual way you get for free fields, i.e., for ##\rho=0## and ##\vec{j}=0##
$$(\partial_t^2-\vec{\nabla}^2) \vec{E}, \quad (\partial_t^2-\vec{\nabla}^2) \vec{B}=0,$$
and this describes waves with phase velocity ##1##, i.e., with the limiting speed of the relativistic spacetime model. That this is describing electromagnetic phenomena correctly is an empirical fact.

LarryS
LarryS said:
Then why does the reciprocal of the square root of their product just happen to be the speed of light, one of the most important constants in our universe?
They're artifacts of using SI units. In a sense they're just unit conversion factors, and you can make them go away by making a "natural" choice of units (although such choices are often impractically large or small for day-to-day use). That their product is ##c^2## in that unit system is inevitable for a consistent set of units.

I understand the modern approach in QFT in general and QED in particular is to set c = 1, thus forcing ε0 and μ0 each to be 1 also. It's a very convenient approach. But setting constants to 1 does not make them go away. Their still there. You've just done a change of scale. The constants no longer need to be visible in the equations. That's all.

These values of these two constants were originally determined by careful experiments done by Coulomb and Ampere (I think). The fact that together they determine the precise speed of light was a complete surprise to Maxwell.

LarryS said:
But setting constants to 1 does not make them go away. Their still there. You've just done a change of scale.
That depends. Consider ##F=ma##. I want to measure distance in inches instead of meters but otherwise use SI units; thus I need to modify Newton's second law to ##F=kma## where ##k## is a constant with value ##0.0254\ \mathrm{N\ s^2\ kg^{-1}\ in^{-1}}##. By your argument, Newton's second law in normal SI units is actually ##F=kma## where ##k## takes the value ##1\ \mathrm{N\ s^2\ kg^{-1}\ m^{-1}}## in this unit system, and it is remiss of us not to teach it like that. Or you can just absorb the constants into the unit definitions and say ##F=ma## - but then why object to doing the same with ##\epsilon_0##?

Ibix said:
That depends. Consider ##F=ma##. I want to measure distance in inches instead of meters but otherwise use SI units; thus I need to modify Newton's second law to ##F=kma## where ##k## is a constant with value ##0.0254\ \mathrm{N\ s^2\ kg^{-1}\ in^{-1}}##. By your argument, Newton's second law in normal SI units is actually ##F=kma## where ##k## takes the value ##1\ \mathrm{N\ s^2\ kg^{-1}\ m^{-1}}## in this unit system, and it is remiss of us not to teach it like that. Or you can just absorb the constants into the unit definitions and say ##F=ma## - but then why object to doing the same with ##\epsilon_0##?
All that makes perfect sense. But let's suppose, extremely hypothetically, that you then noticed, much to your surprise, that k had the same value as Newton's Gravitational Constant (analogous to the electric constant and magnetic constant's relation to c). You might then decide that k deserved a second look.

LarryS said:
But let's suppose, extremely hypothetically, that you then noticed, much to your surprise, that k had the same value as Newton's Gravitational Constant (analogous to the electric constant and magnetic constant's relation to c). You might then decide that k deserved a second look.
##k=G## in what unit system? You can set ##G=1## (or any other value) by an appropriate choice of units.

LarryS
LarryS said:
These values of these two constants were originally determined by careful experiments done by Coulomb and Ampere (I think). The fact that together they determine the precise speed of light was a complete surprise to Maxwell.
Right. You cannot set all physical constants to 1. In this case they are constrained by the relation ## \epsilon_0 \mu_0 c^2 = 1 ##. The impedance of free space is ## \sqrt{\mu_0/\epsilon_0} \approx 377~\Omega ##. The mismatch of impedance becomes apparent (for example) when light hits a window pane and is partially reflected, glass having different material parameters ## \epsilon ## and ## \mu ##. You may find it helpful to look at chapter 5 on reflection in volume 3 ("Waves") of the Berkeley physics course. I had been completely unaware of the term "spacecloth" and the magic value ## 377~\Omega ##.

LarryS

The c that appears in c-1/\sqrt{\epsilon_0 \mu_0 i s the same one that appears in the Lorentz force law F=q(E+(v/c) \cross B) in any consistent set of units,

It comes about because of how we choose to split the electromagnetic field into "electric" and "magnetic" components.

The c that appears in c-1/\sqrt{\epsilon_0 \mu_0 i s the same one that appears in the Lorentz force law F=q(E+(v/c) \cross B) in any consistent set of units,
No, there's no ##c## in the Lorentz force law when you use SI units.
It comes about because of how we choose to split the electromagnetic field into "electric" and "magnetic" components.
When the same reference frame is adopted, physicists should always agree on which components are electric or magnetic. But what some people refer to as ##\mathbf{B}##, is what others write as ## c \mathbf{B} ## (when using SI units).

LarryS
LarryS said:
I understand the modern approach in QFT in general and QED in particular is to set c = 1, thus forcing ε0 and μ0 each to be 1 also. It's a very convenient approach. But setting constants to 1 does not make them go away. Their still there. You've just done a change of scale. The constants no longer need to be visible in the equations. That's all.

These values of these two constants were originally determined by careful experiments done by Coulomb and Ampere (I think). The fact that together they determine the precise speed of light was a complete surprise to Maxwell.
They are not there anymore. It's just using a "natural system of units".

Of course, to represent the units in practice you need to find measurement procedures, and indeed historically that's how Maxwell discovered that light very probably might be an electromagnetic wave as predicted by his equations. At this time they used electrostatic and magnetostatic units, and the experimentum crucis in connection with the speed of light was the experiment by Kohlrausch and Weber, determining the relation between the electrostatic and the magnetostatic units of charge.

LarryS
LarryS said:
TL;DR Summary:
Do the quantities of capacitance and inductance have any meaning for an EM wave in free space far from its source? Can their definitions be generalized beyond physical wires and currents?
I think in classical terms we can only describe the properties of free space by observing the behaviour of "massless" charges subjected to an EM wave. Such charges have the mechanical properties of springiness (due to their attraction) and inertia (due to the creation of a magnetic field when they move, which stores energy). Charges in a wire have different mechanical properties, due to their proximity, which tell us the characteristics of the line. A similar situation occurs with particles in a sound wave, where pressure and velocity (instead of E and B) tell us the characteristics of the medium.

LarryS
To the contrary: Massless charges are even more problematic than massive ones, because the radiation-reaction problem gets even worse for massless charges.

hutchphd

## What is the primary difference between an EM wave in a transmission line and an EM wave in free space?

The primary difference is the medium through which the EM wave propagates. In a transmission line, the wave is guided by conductors, such as coaxial cables or waveguides, which confine and direct the energy. In free space, the wave propagates through the vacuum or air without any guiding structure, spreading out in all directions.

## How do the propagation speeds of EM waves in transmission lines compare to those in free space?

EM waves in free space travel at the speed of light (approximately 299,792,458 meters per second). In transmission lines, the propagation speed is slower due to the dielectric material surrounding the conductors and the inductive and capacitive properties of the line. The speed in a transmission line is typically a fraction of the speed of light, determined by the line's velocity factor.

## What are the typical losses associated with EM waves in transmission lines versus free space?

In transmission lines, losses occur due to the resistance of the conductors (ohmic losses), dielectric losses in the insulating material, and radiation losses. In free space, the primary loss mechanism is free-space path loss, which increases with distance as the wave spreads out. Transmission lines generally have higher losses over long distances compared to free space propagation.

## How do impedance characteristics differ between transmission lines and free space?

In free space, the intrinsic impedance is a constant value of approximately 377 ohms, derived from the permittivity and permeability of free space. In transmission lines, the characteristic impedance depends on the geometry and materials of the line, such as 50 ohms for coaxial cables or 75 ohms for certain types of television cables. Matching the impedance of the transmission line to the load is crucial to minimize reflections and power loss.

## What are the typical applications for EM waves in transmission lines versus free space?

EM waves in transmission lines are commonly used in telecommunications, data transmission, and power distribution, where guided propagation is necessary for efficient and reliable signal delivery. Examples include coaxial cables, fiber optics, and waveguides. EM waves in free space are used in wireless communication, radar, and satellite systems, where the ability to transmit over long distances without physical connections is advantageous.

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