# Relation of electric permittivity $ε_{0}$ and magnetic permeability $μ_{0}$ of vacuum space with capacitance $C$ and inductance $L$ of matter?

I am asking if the permittivity and permeability constants of vacuum space controlling propagation speed of electromagnetic energy, light, through vacuum space:

$$c=\frac{1}{\sqrt{\varepsilon_{0} \mu_{0}}}$$

are somehow correlated or are analogue to the characteristic, capacitance $$C_{0}$$ and inductance $$L_{0}$$ values (i.e. values expressed as per-unit length of a transmission line) controlling the signal velocity propagating through a medium (i.e. matter) modeled as a transmission line?

$$V_{s}=\frac{1}{\sqrt{C_{0} L_{0} }}$$

Both $$V_{s}$$ and $$c$$ refer to group velocities (i.e. in vacuum space, light phase velocity is the same as its group velocity). $$V_{s}$$ in $$m/s$$ units. Characteristic capacitance $$C_{0}$$ of a specific medium modeled as a transmission line is in $$F/m$$ units and characteristic inductance of the medium modeled as a transmission line, is $$L_{0}$$ in $$H/m$$ units. All units are in SI.

• It must be noted that $\epsilon_0$ and $\mu_0$ are artefacts of the way the SI system of units is constructed - these do not appear in many other systems of units, such as, e.g., cgs. Dec 2, 2021 at 9:08

Yes. If the transmission line is surrounded by air then $$\frac{1}{\sqrt{\mu_0\epsilon_0}}= \frac{1}{\sqrt{LC}}.$$ You can compoute the value of a $$L$$ and $$C$$ for a pair of coaxial cylinders, for example, and see that this relation holds. If the transmision line is embedded in a dielectric then the same relation holds, but with $$\mu_0$$ and $$\epsilon_0$$ on the LHS replaced by the appropriate magnetic permeability and dielectric constants $$\mu$$ and $$\epsilon$$.

• So if the line is insulted with a dielectric material this will slow down the propagation of the signal compared to a naked line? But what about the μ and ε values of the copper conductor in the wire? Will these also affect the propagation speed of the signal assuming the signal is low frequency and there is very small skin effect? Dec 2, 2021 at 18:35
• Yes. Insulated line transmit signals at less that $c$. We usually treat the copper is a perfect conductor, so $\mu$ $\epsilon$ are not relevent to the conductor. The resistance of the wire, and thand leakance between the two conductors mostly affect the frequency response/ See: en.wikipedia.org/wiki/Heaviside_condition Dec 2, 2021 at 22:14

Building upon the previous answer of mike stone.

So, if a specific material medium modeled as a signal transmission line, insulated with a dielectric and assuming all the signal is on the surface of the line (i.e. high frequency signal skin effect) has $$ε$$ permittivity (i.e. also known dielectric constant) and permeability $$μ$$ specific values and also $$C_{0}$$ characteristic capacitance and $$L_{0}$$ characteristic inductance values for this medium,

then the following equation must hold for a medium in SI units:

$$C_{0}L_{0}=εμ$$

This is very interesting result because it could be applied when modelling vacuum space as a transmission line where:

$$C_{0}=\varepsilon_{0}=8.8541878128(13) \times 10^{-12} \mathrm{~F} \cdot \mathrm{m}^{-1}$$

and

$$L_{0}=\mu_{0} \approx 4 \pi \times 10^{-7} \mathrm{~H} \cdot \mathrm{m}^{-1}$$

Therefore, permittivity of vacuum space $$ε_{0}$$, is the characteristic capacitance $$C_{0}$$ value of vacuum space and permeability of vacuum space $$μ_{0}$$, is the characteristic inductance value $$L_{0}$$ value of vacuum space.

Also the characteristic impedance $$Z_{0}$$ of vacuum space is given by the equation:

$$Z_{0}=\sqrt{\mu_{0} / \epsilon_{0}}=\sqrt{L_{0} / C_{0}}\approx 377 \Omega$$

From the above analysis and values shown, is it not logical to ask if "empty" vacuum space can be treated after all as an "exotic" matter medium where the information (i.e. light) propagates at the c speed limit controlled by its characteristic capacitance and inductance values therefore also controlled by the impedance value $$Z_{0}$$ of vacuum space?

$$c=\frac{1}{\sqrt{C_{0} L_{0}}}=\frac{1}{\sqrt{\varepsilon_{0} \mu_{0}}}=\frac{1}{\varepsilon_{0} Z_{0}}=\frac{Ζ_{0}}{\mu_{0}}$$