The vacuum is empty, yet it seems to have 2 properties: it's absolute permeability and absolute permittivity, which have specific, finite, non-zero values. Why?

Why are the vacuum permeability and permittivity non-zero and non-infinite?

What would a universe in which the vacuum permeability and permittivity were zero look like?

What would a universe in which the vacuum permeability and permittivity were infinite look like?

  • 4
    $\begingroup$ Indeed, these constants are now called the electric constant and the magnetic constant (of the SI). physics.nist.gov/cgi-bin/cuu/Value?ep0 $\endgroup$
    – user137289
    Nov 10, 2018 at 0:38
  • 2
    $\begingroup$ Permeability is just a number. Permittivity is then fixed by the defined speed of light. Isn't the question just asking what would happen if the speed of light were zero or infinite? $\endgroup$
    – ProfRob
    Nov 10, 2018 at 8:40
  • 3
    $\begingroup$ Perhaps not under the soon-to-be adopted unit changes. $\endgroup$
    – ProfRob
    Nov 10, 2018 at 10:28
  • $\begingroup$ One insight I've had recently is that the vacuum is not defined as being empty, but as being empty of matter. It can still contain energy, quantum fields, and other non-matter entities, and be considered a perfect vacuum. $\endgroup$
    – cowlinator
    Feb 26, 2021 at 21:45

7 Answers 7


The constants $\epsilon_0$ and $\mu_0$ are what a physicist calls "dimensional constants". This means that they are constants whose values contain units. The usual values given are in SI, or the metric system, units, e.g.

$$\epsilon_0 \approx 8.854 \times 10^{-12}\ \mathrm{F/m}$$


$$\mu_0 \approx 1.256 \times 10^{-7}\ \mathrm{H/m}$$

The trick is, when a constant contains a unit, its value thus depends on the units used to measure it. In a different system of units, there may be a different value. In particular, if one looks at the units above, one sees they involve three different physical measurement dimensions, all of which, it turns out, are unrelated enough you can use separate units for all three: capacitance, inductance, and length. It is entirely possible to choose units in whichever way you want, and have these come to be any value you want. For example, I could choose the length unit to not be the metre, but instead to be the Smoot, a quasi-humorous unit equal to exactly 5 imperial feet and 7 imperial inches, or 1.7018 m, exactly. This unit was named after a perhaps-but-perhaps-not-so famous, depending on who you know, professor who spent his college years studying at MIT, the Massachusetts Institute of Technology, and as part of a frat pledge made him use his own body as a measuring device to measure the length of a bridge outside the campus, apparently by flipping him over and over again until the entire bridge - measured at "364.4 Smoots, 'plus or minus an ear'".(*) If we measure the constants in Smoots, but keep the other units (technically thus producing a very bastardized unit system) we get instead

$$\epsilon_0 \approx 5.203 \times 10^{-12}\ \mathrm{F/Smoot}$$ $$\mu_0 \approx 7.380 \times 10^{-8}\ \mathrm{H/Smoot}$$

In fact, with a suitable choice of units, it is entirely possible to make these come to any value at all. From the viewpoint of physical theory, thus, these values are not fundamental. We could even take them to be $1$, and the ability to do so with a suitable choice of units is one of the things that is very useful in theoretical work for simplifying equations.

And this applies to any dimensional constant. The constants that cannot be so changed are those which are formed from such dimensional constants in such a way that all their units cancel, leaving a pure number: these are called dimensionless constants. These dimensionless constants are the ones which are usually considered as having more physical meaning for this reason, as more essentially reflecting properties of the operating principles of the Universe, than upon reflecting what basically amounts to the relationship between those principles and an emergent system - humans - that resulted from their operation in a specific (the only? or not?) instance at a specific place and time therein.

This is perhaps more clearly illustrated with a constant which is more simply relatable to things we humans experience in everyday life, and that is not these somewhat more specialized constants that only physicists and engineers typically bump into, but one that at least a fair percentage of the general population has perhaps an inkling of, and that is the speed of light, $c$, typically given as:

$$c = 299\ 792\ 458\ \mathrm{m/s}$$

Clearly, sincce it has a unit, it is a dimensional constant. Naively, you might think this means "light goes really, really fast". But actually, on second thought, that is not quite so. As you may know, it takes enormous amounts of time for light to travel across the Universe, so is light "actually" fast, or actually is it very slow? The speed is relative to us, humans. And indeed, we could take a unit system that makes the speed very slow, if we wanted to:

$$c = 0.000\ 002\ 997\ 924\ 58\ \mathrm{Pm/s}$$

where we now have used the distance unit as petameters (Pm), a metric-system unit that is suitable for measuring astronomical scales. We could even dink around with the time unit, too:

$$c = 0.000\ 002\ 997\ 924\ 58\ \mathrm{m/fs}$$

where we have now exchanged it for a timescale suited to the atomic realm. The point is here that the speed looks dramatically different when viewed from scales different from the human, and thus what it really is telling us is not "how fast light is", but rather "where we are in relation to the Universe's own scales". To further elaborate this we should first note in more detail how the units we used - the meter and second - relate to us: we are about 1.70 m tall on an international, demographically-weighted (i.e. not eurocentric, focused on peoples of color) average at least to this author's best-guess research (so the Smoot is rather close to a fully "average" human, from a world point of view) given such a figure is hard to turn up directly, only lists of the values for separate countries because there is fair variability, and moreover, one second is roughly the temporal scale we operate on - our individual thoughts take up about a second and our heart beats at 1 or 2 times per second depending on if we're resting or active (at least for a healthy enough human). Thus we see these units very roughly encapsulate "human scale", and there, sitting right on the right in the unit symbol, is "human scale".

Thus, from one interpretation, what it says is that light is at the order of magnitude of 300 000 000 times higher than the speeds that are important at typical scales of human movement. However, there's also a much more interesting interpretation and that deals with how that this constant, which we are calling here the "speed of light", actually isn't perhaps best thought of as a speed from a more physically fundamental point of view. Instead, what it "really" is is the factor which interrelates space and time - the "exchange rate" that tells us how much space we need to exchange for a given amount of time, and vice versa, because as Albert Einstein helped at least those in the western world to realize(**) and moreover set the grounds for confirming as a very accurate picture, that space and time are two parts of the same continuum. Thus it could also just as well be interpreted as telling us how that the human scale is scaled within the space-time continuum, that is, "$c$" isn't just a "speed" but rather our dimensions (as in our physical measurements like measuring a box with rulers) in space-time are in a ratio of roughly the order of 300 000 000:1 of time to space, that is, that we are a heck of a lot longer in time than we are in space! In fact, in more exact terms, we are much, much longer indeed: our typical lifespans are about 2.2 Gs - that's gigaseconds, get used to 'em - for the international average (Wealthy nations can exceed 2.6 Gs of lifespan and sadly, there are many nations whose people cannot expect to reach into their third gig.). Using the speed of light, we then easily see that amounts to 660 000 000 Gm, or 660 Pm - petameters, the unit we mentioned before - of space. That's like 400 quadrillion times longer temporally than we are at our maximal spatial extent! In fact, if you took one of us and laid us out along our TRULY longest axis in space, we would reach to some stars of considerable distance - note that the nearest other star to the Sun, Proxima, no doubt familiar to many, is only 46 Pm away, and Spock's fictive homeworld, Vulcan, imagined as being in orbit of the star Keid, which many fewer may have heard of, would still only be 150 Pm away were it real! Think about that: in some real sense, YOU are as long as interstellar space! THAT is what $c$ is "really" telling you!

Likewise, the values $\epsilon_0$ and $\mu_0$ are thus telling us how we, humans, relate electromagnetically to the Universe, not a fundamental property of the Universe itself. For that, we should choose a system of units that is more in line with bringing out those properties to the best ways we understand them, and we can do that by choosing units in which the right dimensional constants go to $1$, and thus drop out of our equations altogether. In particular, if we take

$$\hbar = c = G = k_B = e = 1$$

we have a good candidate for "the scale of the Universe itself". There's still some arbitration in just which constants we so set, but it turns out this choice is particularly illuminating for the case of electromagnetic theory. It is close to, but not quite, the same as the "Planck units", with the crucial difference being here that we took the natural unit of electric charge, $e$, to be our unit, while Planck units massage $4\pi \epsilon_0$ to be $1$. Arguably, I find this choice to be a more pleasing and intuitive one, compared to the Planck unit system, and we are about to see why.

And thus we are now in a position to elucidate the true contours of electromagnetism. Coulomb's law now goes from this

$$F_E = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$$

to the rather more profound form

$$F_E = \alpha \frac{q_1 q_2}{r^2}$$

Here, the constant that has popped out in front is now in fact not dimensional - this $\alpha$ is a constant which is not some simple happenstance of our unit choices, but rather one which better represents a true parameter of the Universe, or at least one that goes considerably deeper into our understanding thereof than the others did - that is, we have now gone down the rabbit hole and into Wonderland. This is the so-called "fine-structure constant", sometimes known as Sommerfeld's constant for those with an (in my view unhealthy) obsession with eponyms (something of which I am every-so-slightly leery). It has the famous value $\alpha \approx \frac{1}{137}$ and its importance is that it is considered the "master dial" which "describes the strength of the electromagnetic force". This is hidden with our previous choice of units, which seem to suggest it is related to $\epsilon_0$ and $\mu_0$, but that is because actually the origin of electromagnetics is quantum-mechanical, and by introducing the quantum of charge we have brought this deeper level into view. In this system, we can see then $\alpha$ as literally describing the exact amount by which a charge produces a force: if we increased $\alpha$ somehow, then $F_E$ would get stronger in proportion. From the viewpoint of our other unit systems, we would interpret this as increasing the charge on the electron, which seems kind of odd given the way the constant is usually described which is as a force gauge, not a battery rating.

Moreover, going further the full Maxwell equations are

$$\nabla \cdot \mathbf{E} = 4 \pi \alpha \rho$$ $$\nabla \cdot \mathbf{B} = \mathbf{0}$$ $$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$$ $$\nabla \times \mathbf{B} = \left(4 \pi \alpha \mathbf{J} + \frac{\partial \mathbf{E}}{\partial t}\right)$$

And as you can see, here the constant $\alpha$ is directly controlling the relationship now of the electric charges - the sources of electromagnetic forces - to the electromagnetic fields they generate, as it appears only on the charge-related terms $\rho$ (charge density) and $\mathbf{J}$ (current density).

Thus $\alpha$ is the real "magic constant" behind electromagnetism, and its direct interpretation here is "how much electromagnetic field a charge pumps out". The larger $\alpha$, the more EMF that a charge of any given size will produce, and the smaller, the less. Finally, we see that your question really should be not "why does the vacuum have these values $\epsilon_0$ and $\mu_0$", but "why does $\alpha$ have the value it does?" And this is, my friend, a real puzzle in physics. Solve it to the bottom, and you will win yourself a Nobel Prize!

ADD: I notice that some commenters below asked a question as to what this has to do with the constants being zero or not. And I admit that the answer above was mostly focused on the "why do they have the values they do?" aspect of the question, and also I did not notice the original question also asked why they are nonzero. And in fact, this is an important point and moreover it is distinct from the above, because while it is true that the specific value a dimensional constant has is (modulo the constraints that arise from the need for the dimesionless ratios of suitable constant combinations to be what they are in terms of "truly" physically meaningful parameters of our Universe) effectively a product of our measuring artifices, whether a constant is zero or nonzero is, on the other hand, actually a different matter and could indeed be argued to be physically meaningful, since it is also independent of the unit system: any mathematically sensible choice of units will leave a zero constant zero or a nonzero constant nonzero, you cannot find one that will result in a nonzero dimensional constant becoming zero or vice versa.

There are, thus, a number of ways to look at this. One is from the viewpoint of the system of fundamental units we have above. In this view, we can work out that $\epsilon_0 = \frac{1}{4\pi \alpha}$, and $\mu_0$ is its reciprocal (the way to do this is to first note that the Coulomb force constant is just $\alpha$, and then solve $\alpha = \frac{1}{4\pi \epsilon_0}$, and then also note that $1 = c = \frac{1}{\sqrt{\epsilon_0 \mu_0}}$ so $\epsilon_0 \mu_0 = 1$). As reciprocals, then, we see that both cannot be zero simultaneously, since $0 \cdot 0 = 0$, but we have $\epsilon_0 \mu_0 = 1$. Moreover, if only one were zero, then the product giving $c$ would be indeterminate, since the other would have to be infinite. The laws of physics would not be happy with that, and a Universe built using such laws wouldn't make any sense. So they have to be something nonzero, given the corpus of laws on which we best understand it as being based upon.

The other view though is to use what we were just saying. Keep in mind that when we normalized the constants to $1$, we were effectively implicitly assuming all of them are nonzero. But as we've said, whether a dimensional constant is zero or nonzero is fundamentally physically meaningful unlike its specific nonzero value, and thus we should also consider all such ways in which such constants could be of such qualitatively different "kinds" of values and what they might mean. Namely, we have in arbitrary units that

$$\epsilon_0 = \frac{e^2}{2\alpha \hbar c},\ \epsilon_0 \mu_0 = \frac{1}{c^2}$$

We note that if somehow $\epsilon_0$ and $\mu_0$ were both zero, the latter would imply that $c$ would have to be $\infty$ - which also is another unit-independent fact. $0$ and $\infty$ are special points, the latter not even being a usual real number, and behaving in some ways even more "exceptionally" than $0$ does. If $c = \infty$ we effectively have no special relativity. If there is no special relativity, however, then there are no electromagnetic waves, and even better it may be (though my chops aren't enough to know) the whole usual structure of the quantum field theories either does not work or becomes very degenerate. The Universe becomes rather sterile and lifeless, I'd think, if the laws still keep making sense.

In general, whether one of the dimensional constants $\hbar$, $c$, or $G$ is or is not zero basically corresponds to whether or not we have a universe that is or is not (a binary choice) involving quantum mechanics, special relativity, or general relativity (gravitation), respectively.

(*) In fact, the actual length is 387.72 Smoots, so they actually didn't do so good estimating their uncertainty! Nonetheless, the real uncertainty - under 10% - is rather impressive, and I can't imagine what it must have felt like to be rolled over head over heels a little more than 364 times. I can imagine vomit, however, as a plausible sight at least somewhere during the process, and that one's stomach would, afterward, probably hurt like death - the Chinese way of saying "hurts really f***ing bad". (Also, the author of this post is interestingly almost exactly one smoot tall - within 1 cm!)

(**) Concepts similar to our modern spacetime, while only possible to test observationally recently, were thought of before by at least some Indigenous peoples of the Andes region (who still exist, by the way), in particular, in Peru and Bolivia. These peoples also have some other interesting ways by which, at least in their traditional understandings of language, they relate to time.

  • 5
    $\begingroup$ Uh, ok, very interesting, but how does this even answer the question? The constants are nonzero, no matter which unit we choose. $\endgroup$
    – user212613
    Nov 11, 2018 at 14:38
  • 1
    $\begingroup$ Awesome answer! Unfortunately the system only allows me to do +1 $\endgroup$
    – Dale
    Nov 17, 2018 at 4:41
  • $\begingroup$ @Dale that's ok, in the right unit system, your +1 can amount to anything ;-) $\endgroup$
    – Alexis
    Oct 3, 2019 at 11:41
  • 1
    $\begingroup$ This more answers a question like "why does the value of the vacuum permeability contain the numeral sequence it does?" than it does answer more fundamental questions like "why is the vacuum permeability non-zero?" or "why isn't the vacuum permeability twice as strong as it is?" $\endgroup$
    – cowlinator
    Feb 26, 2021 at 21:55
  • $\begingroup$ Could you provide some reading on the indigenous people of the Andes coming up with a concept similar to spacetime? It sounds interesting $\endgroup$
    – Siupa
    Jul 15, 2021 at 11:49

I will assume throughout this answer that we fix the value of $c$ independently of $\varepsilon_0$ or $\mu_0$. The vacuum permittivity and permeability are related to one another by $\varepsilon_0\mu_0 = 1/c^2$, so they're not independent constants — as we should expect given that electricity and magnetism are both manifestations of the same fundamental force.

The permittivity is related to the dimensionless fine structure constant $\alpha$ by $\alpha = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{\hbar c}$. The fine structure constant determines the strength of the coupling of charges to the electromagnetic field. Since it's dimensionless, it doesn't depend on a choice of units and in this sense is more fundamental than $\varepsilon_0$.

If we take $\alpha \rightarrow 0$ ($\varepsilon_0 \rightarrow \infty$), charges aren't affected by EM fields at all, and there's no electromagnetic interaction between charges. There would be no atoms, so no macroscopic matter as we know it. If we take $\alpha \rightarrow \infty$ ($\varepsilon_0 \rightarrow 0$), then the EM coupling between charges is infinitely strong. I don't really have good intuition for what happens in this case.

We can see a little physics of both of these limits from Coulomb's law, \begin{equation} F = \frac{1}{4\pi\varepsilon_0} \frac{q q'}{r^2}, \end{equation} where the former limit gives $F\rightarrow 0$ and the latter $F \rightarrow \infty$ for finite charges and distances.

  • 3
    $\begingroup$ Is it not the case the the permeability and permittivity are independent, and it's the speed of light that's determined by them? (This may not be a meaningful question.) $\endgroup$ Nov 10, 2018 at 4:18
  • 3
    $\begingroup$ The way I am thinking about it, $c$ is the speed of massless particles — in particular, gravity propagates at $c$. I think that we should be able to change $\varepsilon_0$ or $\mu_0$ without changing anything about how gravity behaves. So in my picture, $c$ is god-given and we vary the EM coupling. But I don't know what the "real" answer is. $\endgroup$
    – d_b
    Nov 10, 2018 at 4:40
  • 2
    $\begingroup$ The product of $\epsilon_0 \mu_0$ remains fixed. $\endgroup$
    – ProfRob
    Nov 10, 2018 at 10:27
  • 4
    $\begingroup$ @Alexis we're not talking about how the BIPM defines things, we're talking about how, theoretically, things fundamentally are. And one (maybe) reasonable way to look at it is that the speed of light in vacuum is a consequence of the dielectric properties of vacuum, same as other materials. It probably doesn't work that great, but I think that's what BallpointBen is after :) $\endgroup$
    – hobbs
    Nov 10, 2018 at 16:41
  • 6
    $\begingroup$ @hobbs : common understanding of these things is that special relativity is more fundamental than e&m, and the value of $c$ is a parameter from special relativtity that tells you the unit conversion between time and space. That leaves you with only one independent constant out of $\mu_{0}$ and $\epsilon_{0}$, which we typically take to be the latter. $\endgroup$ Nov 10, 2018 at 17:42

If they were zero, you recover Newtonian mechanics (the speed of light would become infinite). If they were infinite, there is no recognizable universe, and you would have frozen light (zero speed) and thus neither mass nor reference frames could exist.

  • 1
    $\begingroup$ Could you explain how infinite speed of light means light is frozen? $\endgroup$
    – Schwern
    Nov 10, 2018 at 1:26
  • $\begingroup$ No, I said infinite permeability means zero speed of light $\endgroup$
    – user65081
    Nov 10, 2018 at 1:54
  • 3
    $\begingroup$ Could you explain that inference? $\endgroup$
    – Schwern
    Nov 10, 2018 at 2:45
  • 2
    $\begingroup$ the limit of 1/x as x->+inf = zero $\endgroup$
    – cowlinator
    Nov 11, 2018 at 3:58
  • 1
    $\begingroup$ @Wolphramjonny I'm not trying to contradict, but understand. I think I found my answer and again in layman's terms. $\endgroup$
    – Schwern
    Nov 11, 2018 at 7:23

Both are meaningless values that go away when you use Gaussian units. It is meaningful that neither is zero and neither is infinite, but their particular finite values don’t have physical significance, just as how the values of other dimensional constants don’t have significance.

  • $\begingroup$ What are Gaussian units? $\endgroup$ Nov 10, 2018 at 7:05
  • $\begingroup$ They are explained at en.wikipedia.org/wiki/Gaussian_units. Look at how much nicer they are! There are no ugly factors like $\frac{1}{4\pi\epsilon_0}$ or $\frac{\mu_0}{4\pi}$, only factors of $\frac{1}{c}$, and the electric and magnetic fields have the same units which seems like a good idea since they are just different components of an electromagnetic field tensor. $\endgroup$
    – G. Smith
    Nov 10, 2018 at 8:03
  • $\begingroup$ The SI system of electrodynamical units has two degrees of freedom: the magnitudes of the interacting charges, $q$, and the coupling strength defined by $\alpha \hbar c = 4\pi\epsilon_0 e^2$. The fact that all fundamental charges have the same magnitude $e$ is a nontrivial experimental discovery which has always felt hidden in Gaussian units: it's lovely to do algebra with $F=q_1q_2/r^2$, but getting numerical results requires me to know that the proton's charge is 480.3204 pico-esu, as if there might exist some other particle in nature with charge 481.2340 pico-esu. There isn't, and won't be. $\endgroup$
    – rob
    Nov 10, 2018 at 18:18
  • $\begingroup$ To get numerical results in SI units you have to know the proton’s charge in Coulombs. Plus you have to know the value of $\epsilon_0$. And I don’t see how charge quantization is more obvious in one system than the other. $\endgroup$
    – G. Smith
    Nov 10, 2018 at 18:26
  • $\begingroup$ All fundamental charges do not have magnitude $e$. Some have magnitude $e/3$ and some have magnitude $2e/3$. And this is not obvious in either SI or Gaussian units. $\endgroup$
    – G. Smith
    Nov 10, 2018 at 18:30

Why are the vacuum permeability and permittivity non-zero and non-infinite?

Why not?

What would a universe in which the vacuum permeability and permittivity were zero look like?

Because $$c = \frac{1}{\sqrt{\epsilon_0 \mu_0}},$$

you'd have an infinite speed of light as Wolphram jonny pointed out.

What would a universe in which the vacuum permeability and permittivity were infinite look like?

Wolphram jonny answered this and it follows from the equation I gave.

  • $\begingroup$ "Why not?", yes, that's exactly the question. So, why or why not, hmm? $\endgroup$
    – cowlinator
    Mar 13, 2019 at 19:08
  • $\begingroup$ @cowlinator I'm pointing out that there there isn't necessarily a more fundamental reason for the vacuum permeability and permittivity to be non-zero and non-infinite other than that they are. $\endgroup$
    – user113773
    Mar 13, 2019 at 20:31
  • $\begingroup$ @cowlinator In asking "why X?" you're assuming that there's something more fundamental than X that can explain X. You didn't give a reason to believe that there is something more fundamental than the fact that the vacuum permeability and permittivity are non-zero and non-infinite. $\endgroup$
    – user113773
    Mar 13, 2019 at 20:37
  • $\begingroup$ Things are typically assumed to have a cause, but you are stating that you believe that they are causeless. You didn't give a reason for such a belief. $\endgroup$
    – cowlinator
    Mar 13, 2019 at 21:20
  • $\begingroup$ @cowlinator I suggest you watch this Richard Feynman interview: youtube.com/watch?v=36GT2zI8lVA $\endgroup$
    – user113773
    Mar 13, 2019 at 23:01

Permeability of free space describes the curvature of the universe, or alternately, the dissipation of an omnidirectional force with distance.

Standard value: μ0 = 4π×10^−7 H/m

The 10^−7 comes from the particular definition of the Ampere. If we "correct" this definition to not contain the 10^−7 embedded in it, we get μ0 = 4π which is merely the surface area of the unit sphere in three (space) dimensions. This corresponds to the inverse square law.

The two-dimensional universe would have a value of μ0 = 2π and the one-dimensional universe would have a value of μ0 = 1.

A four-dimensional universe would have μ0 = 2π^2.

A zero permeability would result in a singularity in the laws of physics that I cannot transform into any meaningful description.

An infinite permeability implies an infinite number of dimensions, which in turn results in forces having no ranges.

In theory, μ0 could be changed with local curvature and be used as the mechanism to adjust the physical laws for highly-curved regions of spacetime. However I know of nobody who actually does this.


For the record, I was able to determine what the consequences for various values of the vacuum permeability and the vacuum permittivity would be by using the following formulas:

$$c_{0} = \sqrt{\frac{1}{\varepsilon_{0}\mu_{0}}}$$ $$\varepsilon_0\mu_0 = 1/c_{0}^2$$ $$Z_{0} = \sqrt{\frac{\mu_{0}}{\varepsilon_{0}}}$$ $$\alpha = \frac{e^{2}Z_{0}}{2h}$$ $$h = \frac{e^{2}Z_{0}}{2\alpha}$$ $$e = \sqrt{\frac{2\alpha h}{Z_{0}}}$$


  • $\varepsilon_{0}$ is the Vacuum Permittivity (a.k.a. the Electric Constant)
  • $\mu_{0}$ is the Vacuum Permeability (a.k.a. the Magnetic Constant)
  • $c_{0}$ is the Speed of Light in the vacuum
  • $Z_{0}$ is the Vacuum Impedance (a.k.a. Impedance of Free Space)
  • e is the Elementary Charge
  • h is the Planck Constant
  • $\alpha$ is the Fine-structure Constant

Depending on which 2 of (e, h, $\alpha$) you consider to be fixed, it has some interesting results.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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