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B EM wave medium

  1. Dec 14, 2016 #1
    While studying the difference between sound and light, I am puzzled by the following statement in this blog:
    Is this just another way of saying it doesn't need any medium?
    The Transmission medium Wikipedia page makes me think the vacuum of space is its "medium".
  2. jcsd
  3. Dec 14, 2016 #2


    Staff: Mentor

    Basically, yes. Or you could say it is its own "medium".

    Looking at SI units might also lead one to think that, since they assign a "permittivity" and "permeability" to the vacuum. But I would say it's better to view the vacuum as the absence of a medium, and to view EM radiation as either not needing a medium or being its own medium.

    The real question is what function the concept of a "medium" serves. In the case of sound, the properties of the wave, such as its speed and dispersion (dependence of speed on frequency), can be derived from properties of the medium, such as pressure, temperature, density, modulus of elasticity, etc., that obviously exist independently of sound waves, since they are involved in many other things. In the case of light in a vacuum, saying the vacuum is its "medium", or that it doesn't need a medium, or that it is its own medium, don't tell you anything you didn't already know, and don't help you to figure out properties of the wave that you don't already know. In other words, the "medium" concept doesn't really do any work in the case of EM waves, so why bother with it?
  4. Dec 14, 2016 #3
    Well said, thanks Peter.
  5. Dec 15, 2016 #4


    User Avatar
    Science Advisor

    Yes. The reason it doesn't need a medium comes down to Maxwell's Equations. In the absence of any electric charges, those equations are:

    [tex]\nabla \cdot \vec{E} = 0[/tex]
    [tex]\nabla \cdot \vec{B} = 0[/tex]
    [tex]\nabla \times \vec{E} = -{\partial \vec{B} \over \partial t}[/tex]
    [tex]\nabla \times \vec{B} = {1 \over c^2}{\partial \vec{E} \over \partial t}[/tex]

    What you see here is that a changing electric field creates a magnetic field, and a changing magnetic field creates an electric field. If you solve the above equations, you find that E and B follow a wave equation with speed c.
  6. Dec 17, 2016 #5
    Although Maxwell equations can be used to calculate the characteristics of the EM waves, they do not explain the medium or empty space in which EM waves travel. That is the realm of quantum and particle physics. If anything the equations imply some sort of "medium" since there are the mu and epsilon terms. Ultimately Maxwell equations are not fundamental with respects to the nature of "empty space". The more I've read about it, it seems "empty space" is not so empty.
    Although I need to do a bit more research on the actual meaning of mu and epsilon, they do not make sense at the subatomic level. For example, how would you define mu and epsilon between the distance of the electrons and the nucleus around which the electrons orbit? Mu and epsilon are more like measurement at the macro level.
    To make things a bit more complicated, the magnetic and electric fields are basically both sides of the same coin and they could come from a more fundamental force although some might argue it's a matter of semantic. Richard Feynman in his lecture going so far to say that the fundamental force that gives rise to electric/magnetic field should be properly referred to as "potential vector" which determines, base on the given frame of reference, how to distribute how much electric field and magnetic field should be subjected to that given frame of reference. In other words, different observers in different frame of reference actually will disagree on where the observed force is electric or magnetic in nature. That is observer A might see the force as electric whereas observer B might see the force as magnetic. Or it is entirely possible that observer A may see 10% of the force as electric and 90% as magnetic while observer B will see the opposite means 10% as magnetic and 90% electric. Hence the term electromagnetic field as it could be both at the same time although different in composition.
    As a personal experience, for a long time after college, I always knew what is electric field but I always had a hard time to understand where magnetic field comes from. I mean I could not find anything physical that gives rise to magnetic field until I had some times to research that magnetic field actually is the result of relativistic effect.
    To think a bit further, at least to me, it complicates what we think of electrons that they are supposed to give rise to "electric field" as if electric field is fundamental, but if you accept that magnetic and electric field are basically both side of the same coin, then the field that is given off by electrons is not fundamental. Again some people may say it's a matter of semantic but I hope others may chime in and the thought of such already gives me a headache.
    Last edited: Dec 17, 2016
  7. Dec 17, 2016 #6


    Staff: Mentor

    These terms are only present in particular units (SI units); they go away if you use different units. In "natural" units there are no dimensionful constants at all in Maxwell's Equations, just pure numbers like ##4 \pi##.
  8. Dec 17, 2016 #7
    I think the term 1/c2 actually comes from 1/square root of (mu*ep). Actually I think the original Maxwell equation did not have the 1/c2 term, only later when he realized that the speed of light happens to come from those two terms. In his time, mu and epsilon were actually measured in the lab independent of light speed (c). And to be accurate, Maxwell equations are a collection of equations previously discovered by Gauss, Ampere, Lorentz but Maxwell did make his own contributions so he did not come up with the equations all by himself. And if you look at the equations by Gauss and others, the term mu and epsilon were there.
  9. Dec 18, 2016 #8


    Staff: Mentor

    In natural units, ##c = 1##, so the ##1 / c^2## factor is just ##1## and doesn't even need to be written down.

    More precisely, the equations of electrodynamics before Maxwell did not contain the equivalent of the displacement current, ##(1/c^2) dE / dt##. Those equations weren't written in the form Maxwell wrote them at that time.

    Maxwell realized that, if the displacement current term were added to the equations of electrodynamics, they would be more symmetrical, so he added the term and investigated the consequences. One consequence was that there could now be electromagnetic waves in a vacuum, and the speed of the waves turned out to be the speed of light.

    None of this has anything to do with the units in which the equations can be written.

    Only in SI units. There are other possible systems of units, including one called "natural units" in which ##c = 1## and there are no dimensionful constants at all in Maxwell's Equations. You continue to ignore this crucial point.
  10. Dec 18, 2016 #9


    Staff: Mentor

    This topic was already briefly addressed above, but I will address it directly. ##\mu_0## and ##\epsilon_0## are artifacts of the SI system of units. As long as you are using SI units then they apply the same at all scales. They are not properties of space or time or EM waves, they are results of a series of decisions by a committee, the BIPM.

    Edit: I see that it was addressed more directly. Just to add my voice of support. In the current SI, both are defined constants with exact values. How could they be exact values if they were a property of the universe instead of the units?
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