Why does vacuum have a nonzero characteristic impedance towards electromagnetic radiation?
Intrinsic impedance given by $\eta=\sqrt{j\omega\mu / (\sigma +j \omega \epsilon)}$. It gives slope of transformation of $\mathbf E$ to $\mathbf H$ and vice versa.
Here $\eta$ is complex.
And in this expression real part is the cause of attenuation and imaginary part is the cause of phase shift.
In case of free space since $\sigma = 0$, we have $\eta = \sqrt{j\omega\mu / j\omega\epsilon} = \sqrt{\mu / \epsilon}$, which is real.
This suggests presence of resistive part in intrinsic impedance which means there should be attenuation. Also curiosity is how free space can offer resistance and however, the expression for electric field in plane wave $\mathbf E = E_0 \exp(wt-\beta z)$ where $\beta =2 {\pi}/{\lambda} $ and $\lambda $: wavelength
suggests constant electric field. How can we reconcile the real impedance of space with the expression for electric field, which has no attenuation?
Best Answer
It's important to make the following distinction: it's not that vacuum "has" an intrinsic impedance. It's that electromagnetic waves IN a vacuum have an intrinsic ratio between their electric field (E) and magnetic field (H), which we call impedance. That impedance is given by Z = E/H, and it is a fundamental constant; it's only when EM waves travel through some medium other than a vacuum that the impedance gets altered. The units are Ohms because E is measured in Volts/meter and H is measured in Amperes/meter, and 1 Volt/Ampere is defined as an Ohm. This does not imply that vacuum "resists" electromagnetic waves and dissipates them like a resistor would.
The specific value of Z(in free space) is related to the speed of light, and to the way we define the Volt and the Ampere. You could think of the "impedance" as being what limits the speed of propagation of the wave, if that is helpful.