I know that $\frac{1}{\sqrt{\mu_0\varepsilon_0}}$ is equal to the speed of light but is this prediction accurate? I mean is it 100 percent accurate?

  • 2
    $\begingroup$ What is meant by "is it 100% accurate" ?? Do you mean is it experimentally perfectly precise? Accuracy implies measurement, no? $\endgroup$ – N. Steinle Nov 25 '18 at 14:06
  • $\begingroup$ @N.Steinle i mean does it agrees with measurement 100% $\endgroup$ – user210956 Nov 25 '18 at 14:08
  • $\begingroup$ Except possibly for some things that are counted, no measurement agrees with theory with 100% accuracy. There is no such thing as 100% accuracy when it comes to measurements. $\endgroup$ – David Hammen Nov 25 '18 at 17:25
  • 17
    $\begingroup$ This question may not be "answerable" after one realizes that the speed of light has been defined, and the meter has been stated in terms of how far light can travel in a very short amount of time. $\endgroup$ – David White Nov 25 '18 at 18:50

I'm not sure exactly what question you are asking, but it seems to me that the answers that referred to units and definitions of distance and time (e.g. in the SI system) may be missing what the question is.

In the back-up question mentioned as a comment, you ask "does it agree with experiment 100%?" Well no experiment gets 100% precision. The most one can ever say is "it is consistent with all experiments that have been performed, and the best precision attained so far is ...".

If your question is: "is it the case that electromagnetic waves are fully and correctly described by classical electromagnetism?" then the answer is no, because classical electromagnetism is not quite the same as our best understanding of these fields. Our best understanding is offered by quantum field theory and general relativity, and even those are probably not the whole story.

Let me flesh this out a little. In classical electromagnetism, we can measure $\epsilon_0$ by doing static experiments with things like capacitors, and we can measure $\mu_0$ by doing experiments with things like electrical inductors. If someone says those constants have defined values, then such experiments serve as a way of relating field measurements to other properties such as current and charge. Having done all that, we can then measure the speed of electromagnetic waves in vacuum.

Within the theoretical model offered by classical electromagnetism, the answer to your question is that, yes, electromagnetic waves in empty space do propagate at exactly $1/\sqrt{\epsilon_0 \mu_0}$, and you can, if you like, regard this as a good way to define $\epsilon_0$ once $c$ and $\mu_0$ have been given their defined values. However, the physical world does not behave exactly as this theoretical model suggests. Among other effects that complicate the picture are that bright enough light waves will interact with one another in complex ways, owing to their interaction with the Dirac field which describes electrons and positrons. Also, they will cause spacetime curvature. Even the field between the plates of a static capacitor is not nearly as simple as the classical theory suggests (though the corrections are very small in ordinary circumstances).

In the modern understanding based on quantum field theory and general relativity, light in otherwise empty space does move at the special maximum speed, but the fields are no longer fully specified by equations involving just $\epsilon_0$ and $\mu_0$ (in addition to the fields themselves), so the answer to your question is something like "well, $\epsilon_0$ and $\mu_0$ don't exactly capture the physics of these fields, so the answer has to be 'no'".


In the current SI system $\epsilon_0$ is defined as $\epsilon_0=1/c^2 \mu_0$ so in the current SI system $c=1/\sqrt{\epsilon_0 \mu_0}$ is clearly exact. Furthermore, $c$, $\epsilon_0$, and $\mu_0$ are themselves all exact defined quantities with no uncertainty individually.

In the new SI system starting next year we will still have $\epsilon_0=1/c^2 \mu_0$ so in the new system $c=1/\sqrt{\epsilon_0 \mu_0}$ will still be exactly true. However, $\epsilon_0$ and $\mu_0$ will now themselves each be uncertain quantities. Under the new system, $\epsilon_0=e^2/2hc\alpha$ and $\mu_0=2h\alpha/ce^2$. All of these quantities are exact with the exception of the fine structure constant, $\alpha$, which has an experimental uncertainty of 0.23 parts per billion. Note that the contribution of the uncertainty in $\alpha$ is such that the uncertainties cancel out and while $\mu_0$ and $\epsilon_0$ are each individually uncertain their product is exact.

  • $\begingroup$ Since $\alpha$ enters into the definitions now, does that mean that both $\varepsilon_0$ and $\mu_0$ change as a function of the energy at which a process happens, since $\alpha$ is a running coupling constant? $\endgroup$ – Graipher Nov 25 '18 at 18:01
  • 2
    $\begingroup$ My understanding is that $\alpha$ is specifically defined as the low energy limit of the coupling, so at higher energies $\alpha$ still remains constant but the coupling is some larger multiple of $\alpha$. However, I freely admit that I am not very familiar with the higher energy equations, so please take my comment for what it is: a minimally informed opinion. $\endgroup$ – Dale Nov 25 '18 at 18:04
  • 1
    $\begingroup$ It's irrelevant for this question, though, since $c$ will stay constant since it involves a product of the two. $\endgroup$ – Graipher Nov 25 '18 at 18:05

The wikipedia article has some great info. on the historical progression of this question.

Using Maxwell's equations, one recovers the wave equation for electric and magnetic fields. From these equations, Maxwell postulated that light can be thought of as an electromagnetic wave since the electric and magnetic fields solve the wave equation with a phase velocity of $c$. This theoretical value of the phase velocity of light waves is exactly $299,792,458 \frac{m}{s}$.

I mean is it 100 percent accurate?

Nothing is ever 100% accurate - that's an idealization of the human mind since there is always systematic errors in the measurement process (even if perfect humans are conducting the experiments). So, no, it is not 100% accurate. It's worth noting that there's a difference between accuracy and precision.

As the wiki article shows, the precision and the accuracy of the measurements of the speed of light have improved greatly with time: as better measurement techniques are used, the measurements agree with each other better and better, and any one of the measurements agree with the theoretical value of $c$ better and better.

So, although we cannot have 100% accuracy in principle, we can have such small uncertainties in the measurement that we might as well consider the value to be 100% accurate. From the wiki article,

After centuries of increasingly precise measurements, in 1975 the speed of light was known to be $299792458 m/s$ with a measurement uncertainty of 4 parts per billion.

A measurement uncertainty of 4 parts per billion is very small, i.e. 0.0000001% of uncertainty: it's like saying for every million years you live you only fudge 30 seconds of it.

Lastly, in 1983, the meter was redefined in the International System of Units (SI) as the distance traveled by light in vacuum in 1/299792458 of a second. This is justified because, as I said above, the precision of measurements of $c$ are so precise that we might as well just define $c$ to be exactly the value that they all (within certainty) agree upon - this value as it turns out is also what Maxwell's equations predict.

Thus, yes the value of $c$ is defined to be exactly that given by Maxwell's equations, but importantly this is justified because the measurements are very very very accurate.

  • $\begingroup$ Are the values of $\epsilon_0$ and $\mu_0$ also fixed and what fixes them in the current SI? If they are not, do people do measurements comparing $(\epsilon\mu)^{-1/2}$ to the defined value of $c$? $\endgroup$ – jacob1729 Nov 25 '18 at 14:45
  • 1
    $\begingroup$ @jacob1729. As N. Steinle writes, at least in the SI system of units, the speed of light, $c,$ is given a fixed value. So has $\mu_0$ for several years ($4\pi \times 10^{-7}\ \rm{H/m}$). And $\epsilon_0$ has then been defined as $1/(c^2 \mu_0).$ $\endgroup$ – md2perpe Nov 25 '18 at 14:45
  • 1
    $\begingroup$ md2perpe is correct in that we define the value of $\mu_{o}$. In terms of measuring the value of $\epsilon_{o}$, the measurement uncertainty is likewise very tiny en.wikipedia.org/wiki/Permittivity#Measurement $\endgroup$ – N. Steinle Nov 25 '18 at 14:51
  • $\begingroup$ You are using accuracy and precision in your latter paragraphs in a manner that is sure to confuse readers. The 299792458 value is not a privileged value, as the length of the meter was arbitrary. The theoretical velocity of light is only that number based on experimental measurements of other physical properties, it's not some mathematical constant like pi or e. Then leading to the redefinition of the meter, that was done because the precision of many measurements was so tight, everyone agrees on a value. $\endgroup$ – whatsisname Nov 26 '18 at 5:41
  • $\begingroup$ @whatsisname Thank you very much for pointing this out. After thinking about it, I agree with you: I should have said "precision" instead of "accuracy" when talking about why we define $c$ to be exact in the SI system - it certainly is because all the various measurements agree with each other not because they agree with a theoretical value. Thank you again! I have made the appropriate edit. $\endgroup$ – N. Steinle Nov 27 '18 at 13:27

The answer is unequivocally no: as the quantities are currently officially defined in SI, $c=\frac{1}{\sqrt{\mu_0\varepsilon_0}}$ is absolutely not a prediction of the speed of light. The speed of light is defined to have a certain, exact value. That is what forms the basis for the definition of the meter. If someone came along and made a more accurate measurement of the speed of light, it would actually not change the speed of light: it would change the meter.

In theory, this would mean that all the meter sticks in the world would need to be thrown out. But in practice, any change would be orders of magnitude smaller than the precision of a meter stick, or any other common way of measuring distance, so it would actually not lead to real world changes at all. It's still worth emphasizing that the effect of a more accurate measurement of the speed of light would not be to change the numerical value assigned to that speed, or to any other constant, but rather to introduce a tiny (in almost all cases, too tiny to matter) bias into all the measurements which have ever been made that depend on the definition of the meter, which include not just distance, but also quantities whose units include meters in their definition, such as force, etc.

The same goes for $\mu_0$ and $\varepsilon_0$: both have defined values, and if increasing experimental accuracy were to lead to a discrepancy, say in the amount of force measured between charges or currents (doubt this is the most sensitive way to do such experiments, but it's just a hypothetical), this discrepancy would not be remedied by changing the values of $\mu_0$ and $\varepsilon_0$ (or any other official value). It would instead be done by tuning the calibration of the apparatus used to make the measurement so that it gave the desired result. The apparatus would then presumably (eventually) become part of the new standard method for most accurately experimentally reproducing the meter, the kilogram, the ampere, or whatever.

Addition in response to comment by @garyp

As of May 2019, SI will be redefined. $c$ will still have a defined value, but the charge on the electron will change from being subject to measurement to being a defined value, in coulombs. The meter, as we already said, is and will continue to be defined by $c$, the second will continue to be defined by cesium radiation, and the kg will now be defined by specifying a value for Planck's constant. The units of $\varepsilon_0$ are $\frac{C^2 s^2}{m^3 kg}$, so are already completely defined. Thus it will no longer be logically consistent to assign a defined value. So starting next May, the values assigned to $\mu_0$ and $\varepsilon_0$ will be subject to change on the basis of more accurate experiments. If one is changed, the other will need to change as well (along with other values, including the fine structure constant), to maintain $c=\frac{1}{\sqrt{\mu_0\varepsilon_0}}$.

  • 3
    $\begingroup$ It's my understanding that with the 2019 redefinition of SI base units the status of $\epsilon_0$ will change from being exact to being subject to uncertainty. $\endgroup$ – garyp Nov 25 '18 at 15:37
  • 4
    $\begingroup$ This is historically incorrect -- in Maxwell's time, it certainly did predict the speed of light. We subsequently adjusted the SI definitions because our uncertainty in defining a meter was greater than that in measuring the speed of light. $\endgroup$ – Jerry Schirmer Nov 25 '18 at 17:11
  • 2
    $\begingroup$ @JerrySchirmer That is the point. Question is asked in present tense, not about history. As things are currently defined, and as they will be defined for the foreseeable future, there is no experiment that will change the numerical value of the speed of light. $\endgroup$ – Ben51 Nov 25 '18 at 17:14
  • 1
    $\begingroup$ The new SI defines both the vacuum permittivity $\varepsilon_0$ and vacuum permeability $mu_0$ in terms of the Planck constant, the electron charge, the speed of light, and the fine structure constant. The product of the two remains exactly equal to $1/c^2$. $\endgroup$ – David Hammen Nov 25 '18 at 17:56
  • 4
    $\begingroup$ This is a comment on the philosophy of metrology, and it is not wrong in that context. But when people comment on this "prediction" they usually mean to draw attention to the fact that Maxwell's equations predict a relationship between a purely dynamic quantity ($c$) and some quantities ($\epsilon_0$ and $\mu_0$) that can be obtained from purely static measurements. In that sense there is a prediction here and it is a prediction and is accurate to experimental tolerance. Quite a success. $\endgroup$ – dmckee --- ex-moderator kitten Nov 25 '18 at 22:45

$\let\eps=\varepsilon \def\qy#1#2{#1\,\mathrm{#2}}$ I always wonder seeing how few people really understand that units systems and physical dimensions are entirely conventional, and that as a consequence many "universal constants" are matter of convention as well.

So that e.g. "measuring" $\eps_0$ may get quite different meanings according things are defined. You can't measure $\eps_0$ unless a unit of charge is independently defined. Otherwise, how can you measure a capacity?

In SI as it still defined (before the new definition enters in force next year) $\eps_0$ is a fixed constant: $\eps_0 = 1/(\mu_0 c^2)$, with $\mu_0 = \qy{4\pi\cdot10^{-7}}{H/m}$, $c=\qy{299792458}{m/s}$. (More exactly, the value of $\mu_0$ enters definition of unit of current, $c$ in the one of metre.)

In the new SI there is a complete change, as now the coulomb (and therefore the ampere) are defined in terms of the electron charge. Then it will make sense to speak of a measurement of $\eps_0$ or $\mu_0$. This is better explained by Ben51.

Some words about @Andrew Steane's answer. He's certainly right in saying that our present view of electromagnetic field is much more complex than Maxwell's theory could encompass (how could it not be so, given 150 years in between?) After all, the same can be said for the whole of physics...

The main difference is that Maxwell's theory is a linear one. A consequence is that if two light beams intersect in the same region of space, they pass each other unhindered. (Interference is no counterexample, but I don't want to dwell explaining why.) Today we know that it is not exactly so: e.g. photon-photon scattering exists. But its probability is so faint that AFAIK there is only indirect evidence of such process, no direct experimental proof.

"well, $\eps_0$ and $\mu_0$ don't exactly capture the physics of these fields, so the answer has to be 'no'".

Nobody could disagree, but... IMHO we should be very cautious with statements like this when speaking to non-specialists. To understand its actual reach it's necessary to have an adequate understanding of how physics works. A simple "no", even though strictly right, is at risk of being interpreted too literally. After all Andrew also writes

Our best understanding is offered by quantum field theory and general relativity, and even those are probably not the whole story.

It will never be the whole story, in the whole of physics. It's the adverb "exactly" in question's title which had to be criticized. Nothing in physics is "exactly" true, but at the same time an overwhelming part of our knowledge does have wide fields where it can be safely and confidently applied. As to Maxwell's electromagnetism, think of how many electromagnetic devices are in general use today, both between domestic walls and in sophisticated scientific laboratories and space undertakings. It would take a post much longer than this just to touch, as an example, upon how complex and demanding is GPS system, how deeply it relies on our thorough understanding of electromagnetic fields and their propagation. Unfortunately the very success of projects like that drives us to take them for granted.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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