6
$\begingroup$

From my later school days on, the formula that impresses me the most is

$$c=\frac{1}{\sqrt{\varepsilon_0\mu_0}}$$

I know how this can be derived, both from Maxwell Equations and more intuitively. I can read up on wikipedia what $\varepsilon_0$ (permittivity) and $\mu_0$ (permeability) are. I roughly know what Ampere, Volt, Ohm and Watt are and that already concludes my knowledge of electromagnetism, everything else from school I have forgotten already.

When I have speed, I know that $1\frac{m}{s}$ means one meter per second change in distance. When I have acceleration, I know that $1\frac{m}{s^2}$ means one $\frac{m}{s}$ per second change in acceleration. Acceleration is still intuitive from driving in cars. I know $1Pa=1\frac{F}{m^2}$ can be visualized as the pressure coming from (roughly) having 100g of chocolate rasped and spread out over a square meter, while $1bar$ is about a pack of sugar (1kg) being hold up in the air by my thumb (1cm²). Giving $Pa$ as $\frac{kg}{ms^2}$ bears no meaning I know of, but is only the short form of $\frac{kg}{m^2}\cdot\frac{m}{s^2}$ where the second factor basically just carries the factor 10 so I get from 100 gram to 1 Newton. I know that $1kcal$ is about the energy needed to heat water at normal pressure by one degree Celsius. My intuitive understanding of electromagnetic units is rather lacking.

So now I'm basically given $v=(\sqrt{\varepsilon\mu})^{-1}$ and I want an intuitive understanding why the units work out. I understand the equation in that way that e.g. if I were to take four times the permittivity and leave the permeability, I would get half the speed, just in terms of the equation. I'm aware I can't simply do that with the light speed equation above because all numbers involved are constants. But why do the units work out? I'm not asking if they do, I can see that. I'm asking why. And please no "There are no fundamental units" philosophy, I'm asking for an intuitive grasp of the units making sense just like my pressure example was graspable and made sense.

$\endgroup$

6 Answers 6

3
$\begingroup$

A quite different interpretation is as follows.

$ \epsilon = \frac 1{zc} \qquad \mu = \frac zc$

In a modern interpretation, $c$ represents the space-time parameter. Massless particles travel at this speed.

$z$ is a conversion from flux (i.e. transport of effect) to field (force per charge). The actual relation here is the photon continuity equation.

$E = cB = zH = zcD$, whence $\epsilon = 1/cz, \ \ \mu=z/c$.

Maxwell actually compared the speed of light to the Weber-Kohlrausch value. The latter represents the ratio of charge accumulated in a condenser, in esu, to the current delivering it, in emu. From this near equality, Maxwell concluded that light travels in the same medium as EM waves.

It is interesting that if gravitons are also massless, there is a graviton continuity equation as above. Oliver Heaviside explored this in 1893, from a conclusion, that if gravity travels at a finite speed, a co-gravitational force must exist too.

$\endgroup$
5
  • $\begingroup$ This approach seems promising, just like I wouldn't try to find meaning of time in $t=\square\frac{a}{s}$. Unfortunatly, I currently can't make sense of your $B,H,D$, I assume $E$ is energy. I'm also not used to flux and field. Again I can read some information up, but if you could add some examples like I did in my question, that would probably make things easier to understand in relation with the question. $\endgroup$
    – SK19
    Mar 16, 2018 at 11:02
  • $\begingroup$ I like this answer, and the Weber-Kohlrauch experiment is a great intuitive experimental illustration of what you are saying, but what do you mean by the "photon continuity equation"? I haven't heard of this before. I usually think of the fact of the units of $E/B$ being ones of velocity as coming from their presence in the Faraday tensor, as in my answer. $\endgroup$ Mar 16, 2018 at 12:12
  • $\begingroup$ @SK19 It's a standard upper-level electromagnetism notation. $E$ is the electric field. $D$ is the electric field with modifications from polarized materials. $B$ is the magnetic field, including modifications from polarized materials, and $H$ is the part of the magnetic field due to free electric currents. Any E&M textbook gives more details. $\endgroup$
    – rob
    Mar 16, 2018 at 12:38
  • $\begingroup$ If you suppose a single photon does not break up, then the photon continuity must be true: that is, the fields of transport (D,H) and of force (E,B), must be related as such. $\endgroup$ Mar 17, 2018 at 6:42
  • $\begingroup$ I shouldn't call that anything needfully to do with photons. I'd be more inclined to say that the link is a property of both classical and quantum fields: $[E] = [c]\,[B]$ comes from the local geometry (Lorentzian) appearing through membership of $E$ and $B$ of the Faraday tensor and $D\propto E$, $B\propto H$ is, as you say, a flux to field proportionality that is fixed by the unit of charge. Also convenient when we deal with materials, since often polarization / magnetization can be modelled by simply squidging the electric and magnetic constants. But BTW, your answer is still the best here $\endgroup$ Mar 19, 2018 at 0:28
2
$\begingroup$

I'm not sure the meaning you are seeking is in this relationship. Something like Wendy Krieger's answer is the nearest one can come, I believe to an answer to this question. I'd answer something similar in a slightly different way. The reason I think you are doomed in your quest is that the electric and magnetic constants are actually artifacts of certain unit systems, and one can spirit them away by convention.

Firstly, the relationship $c^2\,\mu_0\,\epsilon_0=1$ is not universal: it depends on the unit system one is using. The general relationship is $c^2\,\mu_0\,\epsilon_0=\kappa^2$, where one chooses one's "rationalization" constant $\kappa$ for convenience. For example, both Gaussian and Lorentz Heaviside units have $\kappa=c$ and $\epsilon_0=\mu_0=1$. Let's look further at what is going on here.

The fundamental physical quantity in the relationship, and in Maxwell's equations is $c$: its meaning is clear from special relativity as a cause-effect speed bound, and so are the reasons for its units.

Maxwell's equations come from special relativity, e.g. in the way I describe in this answer here and the fundamental Lorentz four force $q\,F^\mu_\nu\,v^\mu$, where $q$ is the charge of the Lorentz-forced particle, $v$ its four velocity and $F$ the Faraday tensor. One can see here that, given units of force and velocity, one's definition of the unit charge affects the units of the components of the Faraday tensor. The fundamental velocity constant $c$ then fixes the relationship between the "electric" (time-space, symmetric parts of $F$) and "magnetic" (space-space, skew-symmetric parts of $F$). In other words, our unit of charge wholly sets one of the electric/ magnetic constant and $c$ sets the other.

So, like what Wendy says, the electric and magnetic constants come from what field one ascribes to what flux / charge. They are force per flux and flux per force constants.

$\endgroup$
2
$\begingroup$

It comes down to comparisons of dels of one vector to time derivatives of another. From $\boldsymbol{\nabla}\cdot\mathbf{E}=-\dot{\mathbf{B}}$, $\mathbf{E}$ has the dimensions of $v\mathbf{B}$ for speed $v$. Thus in $\boldsymbol{\nabla}\times\mathbf{B}=\mu_0\mathbf{J}+\mu_0\varepsilon_0\dot{\mathbf{E}}$ the coefficient $\mu_0\varepsilon_0$ must have the dimension of $v^{-2}$.

$\endgroup$
2
$\begingroup$

Let me try and expand on the answer given by Philip Wood.

There is to some extend an analogy between deriving the speed of sound from first principles (first done by Newton) and deriving the speed of electromagnetic waves from first principles.

In order to have propagation of a mechanical wave you need two properties: the wave carrying medium must have inertia, and it must have elasticity. Mechanical oscillation is an oscillation between a state where all the energy is potential energy, and a state where all the energy is kinetic energy. In the case of a vibrating string: at the point of maximal deflection all the energy is potential energy, at the point of moving through the point of zero deflection all the energy is kinetic energy.

Let me try and phrase that very generally:
To have propagation of a wave you need a restoring effect, acting to restore back to zero state, and you need momentum (or something analogous to that) so that if the state is moving it will overshoot the zero state.

The wave equation is designed to be populated with those two properties. That is, by design the wave equation will describe a propagating wave when populated with that combination of properties.

In the case of derivation of the speed of sound in air:
The restoring effect enters the equation in the form of elasticity: how air pressure and air density relate to each other. The overshooting effect enters the equation in the form of air density; weight per unit of volume. Populate the wave equation correctly and the units work out.

(I think it's quite likely that in the history of physics this has been used as a heuristic: if you're not sure how to populate the wave equation, and have to resort to guesses, then only try inputs for which the units work out!)

Maxwell recognised that electric effects and magnetic effects combined are acting as a unified electromagnetic phenomenon, and that this electromagnetism features the two properties required for wave propagation: restoring effect and overshooting-when-in-motion effect. Using the appropriate wave equation Maxwell populated the equation and arrived at the speed of propagation of electromagnetic waves.

LATER EDIT:
I noticed that I goofed when I talked about 'populating the wave equation'. Stackexchange contributor Mark Eichenlaub wrote an anwer demonstrating how a wave equation can be obtained from Maxwell's equations though substitution and rearranging.
As we know, historically Maxwell didn't have Maxwell's equations (those were introduced by Oliver Heavyside). Still, Maxwell, expecting the existence of electromagnetic waves, was able to find a way from the electromagnetic equations he had to a wave equation form. That in itself must have been confirmation to Maxwell that he was indeed on to something.

If Maxwell's equations can be rearranged into the general format of wave equation then implictly Maxwell's equations are a wave equation.

$\endgroup$
3
  • $\begingroup$ @ Cleonis "As we know, historically Maxwell didn't have Maxwell's equations (those were introduced by Oliver Heavyside)." Heaviside was one of the pioneers of vector calculus, enabling Maxwell's equations to be written in their modern form as four vector equations. Maxwell had four $sets$ of equations, essentially the vector equations in component form. $\endgroup$ Mar 16, 2018 at 16:24
  • $\begingroup$ @PhilipWood About Maxwell's equations: after looking up information about the history of 'Maxwell's equations' it is clear to me now that it is best to take 'Maxwell's equations' to refer to all historical variations of them. Heaviside's version went beyond introduction of vector notation. As introduced by Maxwell the set was 20 equations with 20 variables. As I understand it, Heaviside showed that for the purposes of the time 8 of the 20 variables were redundant. Heaviside's formulation is sufficient to demonstrate wave propagation. $\endgroup$
    – Cleonis
    Mar 16, 2018 at 20:35
  • 1
    $\begingroup$ Thank you. You've inspired me to dig deeper into Heaviside's role. A bit of an unsung hero? $\endgroup$ Mar 18, 2018 at 19:11
2
$\begingroup$

The way we derive this now is to write down four (so-called vector partial differential) equations (Maxwell's equations) that relate magnetic and electric fields to each other and to their sources, charges. We find that for accelerating sources the equations show that the electric and magnetic fields move away from the source as waves travelling (in a vacuum) at speed $c=\frac{1}{\sqrt{\mu_{0}\epsilon_{0}}}$.

I don't suppose this has been very helpful, but I'm afraid that to gain more insight one has to spend time getting to understand Maxwell's equations. A somewhat offbeat way of getting a feel for what's going on is to learn about how Maxwell conceived of his equations in the first place…

Maxwell thought of a mechanical medium which, if it filled all space, would account for the phenomena of electromagnetism. The medium had tiny cells of semi-elastic fluid separated by small spherical 'idlers'. Magnetic fields consisted of the cells spinning. [Because of the idlers, cells next to each other would spin in the same sense.] Electric fields consisted of stress experienced by the medium. An accelerating charge would set the cells spinning and these, via the idlers, would set cells next to them spinning, and so on, so the fields would propagate outwards away from the source. What would determine the speed of propagation? (1) The greater the density of the fluid the harder it would be to set the cells spinning, so the smaller the wave speed. Maxwell showed that fluid density in his model was linked to $\mu_0$ in order for his model to fit with the known facts of electromagnetism. (2) The stiffer the fluid of the cells the faster the waves would travel. Maxwell showed that the stiffness had to be inversely correlated with $\epsilon_0$ in order to account for the phenomena of electrostatics.

In this way he arrived at $c=\frac{1}{\sqrt{\mu_{0}\epsilon_{0}}}$. He was wise enough not to think that his mechanical medium actually existed. Instead he kept just the equations that described how it behaved, in terms of the fields and charges that the medium was supposed to be modelling. These are Maxwell's equations. $\mu_0$ and $\epsilon_0$ had already (by the 1860s) been measured, and Maxwell saw that $\frac{1}{\sqrt{\mu_{0}\epsilon_{0}}}$ was equal to the speed of light, which had been measured directly. The conclusion, he famously said, was inescapable: light was an e-m wave.

$\endgroup$
1
$\begingroup$

From my later school days on, the formula that impresses me the most is $c=\frac{1}{\sqrt{\varepsilon_0\mu_0}}$

IMHO it's nothing special. All it's saying is that the speed of an electromagnetic wave depends on the permittivity and permeability of space.

I know how this can be derived, both from Maxwell Equations and more intuitively. I can read up on wikipedia what $\varepsilon_0$ (permittivity) and $\mu_0$ (permeability) are. I roughly know what Ampere, Volt, Ohm and Watt are and that already concludes my knowledge of electromagnetism, everything else from school I have forgotten already.

I'd say some of the things you read about electromagnetism don't get it across. My pet hate is the way people say the electric wave generates the magnetic wave which generates the electric wave and so on. It isn't like that. It's an electromagnetic wave.

When I have speed, I know that $1\frac{m}{s}$ means one meter per second change in distance. When I have acceleration, I know that $1\frac{m}{s^2}$ means one $\frac{m}{s}$ per second change in acceleration. Acceleration is still intuitive from driving in cars. I know $1Pa=1\frac{F}{m^2}$ can be visualized as the pressure coming from (roughly) having 100g of chocolate rasped and spread out over a square meter, while $1bar$ is about a pack of sugar (1kg) being hold up in the air by my thumb (1cm²). Giving $Pa$ as $\frac{kg}{ms^2}$ bears no meaning I know of, but is only the short form of $\frac{kg}{m^2}\cdot\frac{m}{s^2}$ where the second factor basically just carries the factor 10 so I get from 100 gram to 1 Newton. I know that $1kcal$ is about the energy needed to heat water at normal pressure by one degree Celsius. My intuitive understanding of electromagnetic units is rather lacking.

You are not alone. But let's not forget that $c=\frac{1}{\sqrt{\varepsilon_0\mu_0}}$ is just a wave speed and focus on that.

So now I'm basically given $v=(\sqrt{\varepsilon\mu})^{-1}$ and I want an intuitive understanding why the units work out. I understand the equation in that way that e.g. if I were to take four times the permittivity and leave the permeability, I would get half the speed, just in terms of the equation.

The equation for the speed of a transverse seismic wave is given as csolid,s = √(G/ρ), where G is the shear modulus and ρ is the density. Increase the shear modulus or "strength" and the speed is faster. Increase the density and the speed is slower. The equation for the speed of light in vacuo takes the same form. It's c = 1/√(ε0μ0) where ε0 is vacuum permittivity and μ0 is vacuum permeability. There’s a reciprocal because permittivity is a “how easy” measure rather than a “how hard” measure. Permittivity is effectively the inverse shear modulus for an electromagnetic wave in space, whilst permeability is effectively the density.

I'm aware I can't simply do that with the light speed equation above because all numbers involved are constants.

They aren't. That's a myth I'm afraid. I know you can find "reliable" sources that say this, but check out the fine structure constant $\alpha =\frac {e^{2}}{4\pi \varepsilon _{0}\hbar c}$. See NIST and note that it's a "running constant". It varies with the energy scale. Now look at the terms in the expression, remember conservation of charge, and look at this.

But why do the units work out? I'm not asking if they do, I can see that. I'm asking why. And please no "There are no fundamental units" philosophy, I'm asking for an intuitive grasp of the units making sense just like my pressure example was graspable and made sense.

For this I would refer you to Maxwell: “light consists of transverse undulations in the same medium that is the cause of electric and magnetic phenomena”. This is often considered to be archaic, but don't forget LIGO, see this, and check out what Percy Hammond said in the 1999 Compumag: "We conclude that the field describes the curvature that characterizes the electromagnetic interaction". IMHO the bottom line is this: when an ocean wave moves through the sea, the sea waves. When a seismic wave moves through the ground, the ground waves. When a light wave moves through space, space waves.

$\endgroup$

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