# Why does vacuum have a nonzero characteristic impedance towards electromagnetic radiation?

On Wikipedia, the impedance of free space $$Z_0$$ is defined as square root of the ratio of the permeability of free space $$\mu_0$$ to the permittivity of free space $$\epsilon_0$$, i.e. $$Z_0 = \sqrt{\mu_0 / \epsilon_0} \, .$$ The value is approximately 377 Ohms.

Now impedance is described as an impeding effect to flow of something, it makes more sense for electric current travelling in a wire where the characteristic impedance of the line (as the line consists of capacitance and inductance per unit length) prevents the flow of AC/DC flow.

Why in the world does free space have a characteristic impedance? That makes no sense to me. Wires makes a lot of sense, but free space having 377 ohm of impedance is too much and not clear why such a value exists.

• Your 2nd paragraph makes no sense at all. I suspect that your understanding of transmission line theory and, in particular, characteristic impedance, isn't quite right. – Alfred Centauri Oct 2 '13 at 15:50
• What I meant to say is that, in a wire the atoms hinder the flow of electrons. This hindrance is what creates impedance. Why does impedance have to exist in free space and that also to EM radiation, this is what makes no sense. I have never been taught about this at all. – quantum231 Oct 3 '13 at 9:40
• Because free space has capacitance (permittivity) and inductance (permeability). Those are intrinsic properties of space. Capacitance and inductance when excited by time varying source (like EM wave) they behave as if they were resistors (impedance). The only difference is that those resistors do not dissipate power as a real resistor does. That is why a plane wave is not attenuated while it is traveling in free space. Does it make sense now? – Gotaquestion Oct 3 '13 at 22:06
• @Gotaquestion I like your writing style, and invite you to think about this: how does space happen to have these "intrinsic properties" as you call them, without some kind of medium in it ?? Am I going in a direction which is "beyond the standard model" with this question ?? You bet !! – PERFESSER CREEK-WATER Mar 2 '18 at 16:42

Numerically, it is a ratio between two numbers. Intuitively or physically, you can think of free space impedance as a limiting factor of the rate of change in electric/magnetic field. The following relation arises naturally from Maxwell equations:

$$\eta = \frac{E}{H}$$

where $$\eta$$ is the characteristic impedance of free space which is 377 Ohm, $$E$$ is the electric field of wave and $$H$$ is the magnetic field of the wave. You can compare that to Ohm’s law. If $$\eta$$ was set to zero, that means one of two things must be true:

1. Either the magnetic field ($$H$$) is infinity with finite electric field ($$E$$).

2. Or the electric field ($$E$$) is zero with non-zero magnetic field ($$H$$).

The first scenario describes a wave (none of the fields is zero) but it is wrong because it requires infinite $$H$$ which is nonphysical. Accordingly, having a non-zero value of $$\eta$$ limits the required magnetic field such that both $$E$$ and $$H$$ become non-zero and finite. The second scenario doesn't describe electromagnetic wave. It describes a magnetostaic situation.

The free space permittivity has a unit of $$\mathrm{F/m}$$ while free space permeability has a unit of $$\mathrm{H/m}$$. If you think of it you can model the propagation of EM wave in vacuum with an infinitely long circuit composed of inductance and capacitance. This circuit is drawn per unit of length that is why the units are given per meter. See the next figure

From a circuit perspective, the capacitor doesn’t allow sudden changes in the voltage, because that requires an infinite current (which is non-physical obviously) according to: $$i_C = C \frac{\mathrm{d}v_C}{\mathrm{d}t}$$

In the same sense, inductance doesn't allow sudden changes in the current because that requires voltage (which is non-physical obviously) according to: $$v_L = L \frac{\mathrm{d}i_L}{\mathrm{d}t}$$

The current and the voltage are equivalent to magnetic and electric fields respectively. Capacitance and inductance are equivalent to permittivity and permeability respectively. In such a circuit, the resonance frequency is equivalent to the speed of EM wave: $$\omega = \frac{1}{\sqrt{LC}}\\c = \frac{1}{\sqrt{\epsilon_0 \mu_0}}$$

Studying waves using circuit is a fully developed field; circuits used to study EM wave propagation not only in space but on any other media including transmission lines are called transmission lines circuits.

Hopefully that helped!

• Nice explanation on why it has to have a value. Once I saw why it can't be zero, things make much more sense now. – zeta-band Oct 1 '18 at 21:34

The 377.Ohm impedance of empty space "makes no sense" intuitively, basically because it has no physical meaning. That is, the value of 377.Ohms arises solely from the arbitrarily defined (and totally un-needed) MKS "constants" of $\mu_0$ and $\epsilon_0$ . These constants were introduced into SI for engineering convenience (to give human-scale magnitudes); but they also include "units" (Farads and Henrys) which greatly confuse the underlying physics.

The Maxwell-Lorentz equations, which describe ${E,B}$ fields and forces on charges, require only 1) the defined units of length and time (e.g. meters and seconds) with the resultant speed of light $c$; and 2) a defined unit for electron charge $e$. The equations say that charges generate $E$, moving charges generate $B$, and ${E,B}$ couple through their time variations. Once generated by charges, this coupled time variation propagates through empty space as a "photon", carrying energy and momentum. The only "conductance" is the speed $c$, and the only "impedance" is $1/c$.

From a "units" perspective, CGS says that a charge $e$ creates an electric field $E$ at a distance $r$ given by $E = e / r^2$, whereas MKS says $E = e / 4 \pi \epsilon_0 r^2$. The MKS constant $\epsilon_0 = 8.85\times 10^{-12}$ (Farads/meter) defines the permittivity of empty space to be a specific number, and also gives it units. In CGS, the permittivity of empty space might be expressed as "yes, it permits $E$ fields", representing the "1" in the equation above for $E$. The MKS permeability $\mu_0 = 4 \pi \times 10^{-7}$ (Henrys/m) is similarly arbitrary, connected to physical reality only by the requirement that $\epsilon_0\times\mu_0 = 1 / c^2$ .

A further confusion is that resistance (as in electrical circuits) is dissipative, turning coherent work or power into entropic heat; whereas the general term "impedance" includes non-dissipative out-of-phase reactances, such as arise with capacitors and inductors. Whatever its numerical value, the impedance of empty space for photons is non-dissipative, like an LCR tank circuit with $R=0$.

Since a finite fluctuating electric field creates a finite fluctuating magnetic field (and vice versa), their ratio must also be a finite value. In other worlds, that being a finite quantity is only a consequence of the existence of light in vacuum.

Now that value itself is merely a consequence of the definition of units you choose, so it being $376.730... \Omega$ isn't something very interesting by itself.

However, the fact that it doesn't depend on frequency is very important (and visible to the naked eye) : that is because both $\epsilon_0$ and $\mu_0$ are fundamental constants. In other words, vacuum is a non-dispersive medium ; that is why we see all colors from an object arriving at the same time.

Be aware that if a vacuum does not have matter, i.e., it is an empty space, how can we define a permittivity and permeability for it? On the other hand, the free space (atmosphere) is not an empty space, therefore it has a specific permittivity and permeability, which was defined with a subscript zero, notch, to identify as free space and not a vacuum, as usually done. The measurement of the permittivity in the vacuum it is said to be approximately equal to that of the free space. It is usually a misundertanding on the definition. For example, do we have any kind of matter at the interstellar space? The electromagnetic waves, as one can determine its speed as presented in the discussion above, need matter to propagate. Then, if it is true, how does an eletromagnetic wave propagate in the vacuum? On the other hand, if it propagates in the vacuum is because it is a light particle and if not is because it is pure electromagnetic wave that requires medium to propagate.

• This is way off. EM waves work just fine in a vacuum. – Navin Feb 16 at 20:02

## protected by AccidentalFourierTransformOct 1 '18 at 20:38

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