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Electric Field Propogation in LIGHT

  1. Oct 13, 2013 #1

    san203

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    Hello. I started studying wave optics a while back and here is my first question

    From what i understand, electric field is a region where charged particles feel the coulomb forces.
    And light is basically time varying magnetic and electric waves being propagated in some medium or vacuum.

    What i don't understand is how can something like an electric field be propagated in the first place?
    How can it move. Isnt it supposed to be a mathematical description of Force?
    Or is it something like introducing electric field at points in space with varying magnitude?
    I hope i have made my question as clear as possible. Thanks.
     
  2. jcsd
  3. Oct 13, 2013 #2

    Drakkith

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    The FIELD isn't moving. The CHANGE in the field is.
     
  4. Oct 13, 2013 #3
    The fields propagate by a mutual interaction of the time derivative terms in Faraday's law and Ampere-Maxwell's law. If that sentence didn't make sense, wait until you study Maxwell's equations, because they really are a fundamental prerequisite to the understanding of propagating EM waves.
     
  5. Oct 13, 2013 #4

    UltrafastPED

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    If you were to suddenly unshield an electric charge its field lines would propagate at the speed of light; if the source charge is an electron then the field lines will expand as radial lines from that point.

    Now jiggle the electron, and watch what happens to the field lines: the lines will now originate from a new central point, but if you observe them at distance d=ct you will see the new old field lines "kink" as the charge moves to the new position - and then settle into the new configuration.

    If you have an oscillating charge (perhaps a current running up and down an antenna at a fixed rate then the oscillating charge will generate oscillating kinks in the electric field lines ... and the changing electric fields generate changing magnetic field lines in accordance with Faraday's law. These oscillating, coupled electric and magnetic field lines constitute electromagnetic radiation - perhaps your favorite AM radio station!

    See http://maxwell.ucdavis.edu/~electro/faraday/basic_laws.html

    You can find other visualizations by searching for "radiation kink", etc.
     
  6. Oct 14, 2013 #5

    san203

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    It makes sense when i read it from textbook or from your answers, but what troubles me is imagining electric fields moving.Can i think of it as being constantly generated from the source and from the changing electric and magnetic field which in turn induce each other?

    If we have a varying magnetic field in a isolated region, then will a electric field just come out of nowhere, without any other source?
     
    Last edited: Oct 14, 2013
  7. Oct 14, 2013 #6

    UltrafastPED

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    Yes ... this is correct.
     
  8. Oct 14, 2013 #7

    vanhees71

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    It doesn't make much sense to say the magnetic field induces an electric field and vice versa. There is one and only one electromagnetic field which we describe, after defining a reference frame, in terms of electric and magnetic field components.
     
  9. Oct 14, 2013 #8

    UltrafastPED

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    I think the field structure of light in vacuo (transverse electric and magnetic fields which are coupled) remains the same for any relativistic observer ... after all, they all see it travelling at c!
     
  10. Oct 14, 2013 #9

    WannabeNewton

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    ^That is certainly true PED because ##E\cdot B## is a relativistic invariant so a radiation field propagating in vacuo will maintain the orthogonality of the ##E## and ##B## fields in every inertial frame.
     
  11. Oct 14, 2013 #10
    But why would you make things more complicated than they need to be when explaining the basics of the fields? No need to being electrodynamics into it when the goal is to explain the basics of a propagating field.
     
  12. Oct 15, 2013 #11

    vanhees71

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    If you have a "propagating field" for me this implies you have electromagnetic waves, and it is way less complicated to learn the correct physical meaning of the electromagnetic field than to wrongly think about it as if electric and magnetic field components were something independent of each other. The great achievement of Faraday and Maxwell in the 19th century has been to combine these fields to the one electromagnetic field, although this becomes clear in full beauty only when discussed as a (paradigmatic!) case of a relativistic classical field theory.

    The problem is that in many textbooks the electromagnetic field is introduced for the special cases of the static and stationary field, where the electric and magnetic field components separate (if the constitutive relatitions are approximated in the non-relativistic approximation, which is of course justified almost always for practical situations).
     
  13. Oct 15, 2013 #12

    UltrafastPED

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    I've not seen light described in terms of the Maxwell tensor -ever! Can you point me to something that does?

    My research area was nonlinear optics and ultrafast lasers ... everything I've seen is in terms of E, B.
     
  14. Oct 15, 2013 #13

    WannabeNewton

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    Well what you'll mostly find if you're after covariant treatments of electromagnetism/radiation are treatments in terms of the 4-potential, from which you can easily get the EM field tensor anyways. For example for a vacuum radiation field in the Lorenz gauge, one has ##\partial^{\gamma}\partial_{\gamma}A_{\mu} = 0## and a simple solution is a wave with constant amplitude: ##A_{\mu} = C_{\mu}e^{i\varphi}##; ##\varphi## is the phase of the wave as usual and level sets of ##\varphi## define null hypersurfaces orthogonal to the wave 4-vector i.e. ##k^{\mu} = \partial^{\mu}\varphi##. For plane waves in particular we have ##\varphi = k_{\mu}x^{\mu}## where ##\{x^{\mu}\}## is a set of global inertial coordinates.

    You could check out section 12.11 of Jackson if you're interested but I personally don't know of any text that does e.g. waveguides, waves in matter, and scattering all in a covariant framework; that would be extremely impractical so it isn't surprising.
     
  15. Oct 15, 2013 #14

    vanhees71

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    I have not said that you should write everything always manifestly covariant. That indeed would be pretty inconvenient for practical purposes and also not very intuitive. Here the three-dimensional formalism is surely better suited.

    However, I'd nevertheless carefully choose the names of objects, e.g., I'd call [itex](\vec{E},\vec{B})[/itex] the components of the electromagnetic field with the electric components [itex]\vec{E}[/itex] and the magnetic components [itex]\vec{B}[/itex] (wrt. to a given reference frame!), when introducing the fields. Further you can say that [itex](\vec{E},\vec{B})[/itex] is an abbreviation of the six independent components [itex]F_{\mu \nu}=-F_{\nu \mu}[/itex] of Faraday tensor (not Maxwell tensor, which name is usually reserved for the spatial part of the energy-momentum, i.e., the stress tensor of the em. field) in the reference frame under consideration.
     
  16. Oct 15, 2013 #15

    UltrafastPED

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    I take it all back: I have seen this before. I even did a bunch of homework problems years ago. I suppose that it is convenient for theoretical physicists who work with particles and other high energy problems.

    But it doesn't seem to be used much when people are just talking about light, which is ordinarily described as an electromagnetic wave. If it comes up in the future I will be sure to call it the electromagnetic tensor.
     
  17. Oct 15, 2013 #16
    I don't agree that it's a problem. Teaching a student electromagnetism and starting with the field tensor would be a worse approach in my opinion. Students have a hard enough time conceptualising the various cross products in quasistatics to worry about how they'd interpret the full relativistic approach.

    For example I have never seen the tensor forms being used in waveguide analysis. So I could say that this approach is equally restrictive (if not more so) to certain people.
     
  18. Oct 15, 2013 #17

    Dale

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    I agree, I would not try to associate a velocity with the fields themselves. As Drakkith said in the first response, it is the change in the field which moves. Similarly, a pressure wave in air travels at mach 1, even though there is no bulk movement of the air at those speeds. So I would think of the velocity of the wave (c) as being different from the velocity of the fields (not even defined).

    Yes, and Yes, except that as others point out, the E and B fields are not independent, so it isn't "out of nowhere".
     
  19. Oct 17, 2013 #18
    OR to simplify couldn't we just say that the E field creates a perpendicular B field which recreates the H field ad infinitum until the energy is transferred. And you could explain the propagation in terms of an analogy to sound You would not need that much math to explain it to him or am I completely wrong?
     
  20. Oct 18, 2013 #19

    vanhees71

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    I don't know, what it should help to "explain" the wave propagation in this way. There is an electromagnetic field, consisting of the electric and magnetic components [itex]\vec{E}[/itex] and [itex]\vec{B}[/itex], obeying the free vacuum (for simplicity) Maxwell equations (in Heaviside-Lorentz units). It doesn't make physical sense to separate them in the sense that one field is the cause of the other.

    Starting from the homogeneous Maxwell equations
    [tex]\frac{1}{c} \partial_t \vec{B}+\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{B}=0,[/tex]
    you find that the electromagnetic field can be described by a scalar and a vector potential
    [tex]\vec{E}=-\frac{1}{c} \partial_t \vec{A}-\vec{\nabla} \Phi, \quad \vec{B}=\vec{\nabla} \times \vec{A}.[/tex]
    These potentials are only determined up to a scalar field, i.e., changing the potentials to
    [tex]\vec{A}'=\vec{A}-\vec{\nabla} \chi, \quad \Phi'=\Phi+\frac{1}{c} \partial_t \chi,[/tex]
    leads to the same physical field components [itex]\vec{E}[/itex] and [itex]\vec{B}[/itex]. Thus one can give one constraint, the gauge-fixing condition. One example is the Lorenz-gauge condition
    [tex]\frac{1}{c} \partial_t \Phi+\vec{\nabla} \cdot \vec{A}=0.[/tex]
    The free fields, i.e., if there are no charges and currents, then obey the wave equation
    [tex]\Box \Phi=0, \quad \Box \vec{A}=0,[/tex]
    with the d'Alembert operator
    [tex]\Box=\frac{1}{c^2} \partial_t^2 - \Delta.[/tex]
    For the free-field case the Lorenz-gauge condition is not sufficient for a unique solution, because one can still use a gauge transformation with a scalar field [itex]\chi[/itex] that fulfills [itex]\Box \chi=0[/itex]. Thus you can give one more constraint. The most simple one is the "radiation gauge",
    [tex]\Phi=0, \quad \vec{\nabla} \cdot \vec{A}=0.[/tex]
    Then the Lorenz condition is still fulfilled, and you have unique solutions for [itex]\vec{A}[/itex] in terms of, e.g., a plane-wave mode decomposition.

    To that end you make the ansatz
    [tex]\vec{A}=\vec{A}_0(\vec{k}) \exp[-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x}].[/tex]
    The gauge condition tells you
    [tex]\vec{k} \cdot \vec{A}_0=0,[/tex]
    and the wave equation for [itex]\vec{A}[/itex] gives the dispersion relation
    [tex]\omega = c |\vec{k}|.[/tex]
    Thus the general solution reads
    [tex]\vec{A}_{\lambda,\vec{k}}=A_0(\vec{k},\lambda) \vec{\epsilon}(\lambda,\vec{k}) \exp[-\mathrm{i} c t |\vec{k}|+\mathrm{i} \vec{k} \cdot \vec{x}] + \mathrm{c.c.}[/tex]
    Here the vectors [itex]\vec{\epsilon}(\lambda,\vec{k})[/itex] for [itex]\lambda \in \{1,2 \}[/itex] are two arbitrary unit vectors perpendicular to [itex]\vec{k}[/itex] (conveniently chosen perpendicular to each other).

    Thus electromagnetic waves are always transverse, and there are two polarization states for each wave number [itex]\vec{k}[/itex].

    As you see, there is no physical possibility here to say the electric field is caused by a time-changing magnetic field or vice versa.

    The "cause" of an electromagnetic field are of course always charge-current distributions. The corresponding solutions are given by the corresponding retarded potentials.
     
  21. Oct 19, 2013 #20

    Claude Bile

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    Think of a wave on a string.

    The atoms in the string aren't propagating forward in the direction of the wave; instead it is the change in displacement is propagating forward.

    In the same sense, it is not the electric field propagating forward, it is the change in electric field that is propagating (as originally pointed out by Drakkith).

    If you think about it, it makes no sense to say that the electric field propagates, since it is a quantity that has a value as a function of space, rather than having a location in space (as a chunk of matter would).

    Claude.
     
  22. Oct 19, 2013 #21

    Born2bwire

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    This is commonly used as an introductory description of the EM field but I've seen a growing opposition, with which I agree, to using it. The idea that a changing electric field creates the changing magnetic field which in turn creates... implies a causal relationship between the two fields. But this is not true, there is only the electromagnetic field, not electric and magnetic fields. That is, the two changing fields simultaneously exist (which can be clearly seen from Jefimenko's Equations) or not at all. It may be better to explain the concept of a force field and what the electric and magnetic fields represent. Classically, the electric and magnetic fields represent the force experienced by a test charge due to the presence of stationary and moving charge densities. In this manner, the idea of propagating fields and such become more understandable upon realizing that it is an intermediary to describe the retarded force between two bodies.

    In a way, Jefimenko's Equations help in this because they allow one to express the uncoupled electric and magnetic fields as a function of the sources alone. One could then substitute these into the Lorentz force equation to completely remove the fields from the picture (though I do not think that such a representation is useful for actual calculations).
     
    Last edited: Oct 19, 2013
  23. Oct 20, 2013 #22

    vanhees71

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    To say it very clearly again. It is wrong to say "that the E field creates a perpendicular B field which recreates the H field". It doesn't make sense! There is only one electromagnetic field and E and B are components of it. This is most easily seen that by choosing another frame of reference the electric and magnetic components are different.

    The causal source of the em. field are electric charge-current distributions. Jefimenko's equations (I never understood why they are named after Jefimenko, because they are nothing else than the retarded solution of the Maxwell equations for a given charge-current distribution and thus for sure more than 100 years old) indeed clearly show this, as is nicely explained in Wikipedia

    http://en.wikipedia.org/wiki/Jefimenko_equations
     
  24. Oct 20, 2013 #23
    I still am not getting how it doesn't make sense. It just seems like you disagree with splitting it into two fields on grounds of aesthetics. To me it seems almost misleading to say only charges and currents produce fields, because then a student will clearly ask "then what does the displacement current do?" These time derivative terms are not just used to create light, they're used in transformers, inductors, non-plane wave solutions, etc... where people would definitely say that a changing E field creates a rotational B field and vice versa.

    I can't say anything factually wrong with saying this.
     
  25. Oct 20, 2013 #24

    Jano L.

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    You can explain displacement current with the great example Feynman gives in his Lectures: charging a capacitor and asking what is the magnetic field around it.

    The displacement current is the term ##\mu_0 \epsilon_0 \partial\mathbf E/\partial t## introduced as a supplement to the conduction current ##\mathbf j## flowing in wires in the Ampere law to make sure the value of magnetic field obtained from the Ampere law is the same for every auxiliary integration surface.

    Formally, the displacement current looks like some forgotten contribution to total electric current. But physically, we do not think of it as real electric current. It is just important term in the Maxwell equations that makes them consistent.

    In engineering practice, if the frequencies are not too high so the electric field is almost gradient, the magnetic field is given as a function of the conduction charge density by the Biot-Savart law as

    $$
    \mathbf{B} (\mathbf{x}, t) = \frac{\mu_0}{4\pi}\int \frac{\mathbf{j}(\mathbf{x}', t) \times \mathbf{R}}{R^3}~d^3 \mathbf{x}',
    $$
    where
    $$
    \mathbf R = \mathbf x - \mathbf x'
    $$
    and this already gives

    $$
    \nabla\times \mathbf B = \mu_0 \mathbf j + \mu_0 \epsilon_0 \frac{\partial \mathbf E}{\partial t}.
    $$

    There is no displacement "current" in the Biot-Savart law.


    If the frequencies are high enough, the proper formula is more complicated (Jefimenko's formulae based on the retarded solution of the wave equation), but again the magnetic field can be written as a function of the conduction current ##\mathbf j## only. That's why it is more appropriate, if cause-effect distinction is useful at all, to assign the role of cause to the electric current, not to the change of electric field.

    This paper explains in more detail:

    A. P. French and Jack R. Tessman, Displacement Currents and Magnetic Fields, Am. J. Phys. 31, 201 (1963)
    http://dx.doi.org/10.1119/1.1969359

    Yes, the cause-effect relation is sometimes invoked to help explain the basic laws in non-mathematical language, which people find helpful when learning. But it is often flawed by the fact that the alleged cause and effect happen at the same time, thus making the very distinction between cause and effect doubtful.
     
  26. Oct 20, 2013 #25
    We know current J is part of the cause. But in an equation with three terms, it makes no sense to have only one cause and two effects, because the apportioning of effect between the two terms is underconstrained.

    So you can't look at Ampere's law and say "current causes dD/dt and curl(H)". One of these other field terms terms must also be part of the cause of the other one. So do we say a current and/or a curl of H cause dD/dt, or do we say current and/or dD/dt cause a curl of H?

    I suppose your point is that the two choices are physically equivalent and therefore the choice is irrelevant? But it has always helped me to think about it in one way rather than the other.




    What makes me think I'm right is that the electric field would never have a curl if it weren't for Faraday's law. Static charges alone cannot create a curl(E), but they can create a curl(E) if they vary over time (with nonzero 2nd derivatives), because a d^2D/dt^2 causes a dcurl(H)/dt which causes a curl(E).

    So charges are still the origin of the curl(E), but Faraday's law is an essential part of the logical chain of events, and I see no reason to say that this chain of events has no cause nor effect. Faraday's law takes a dH/dt and gives a curl(E), in that order, in this particular story. Can you come up with an equivalent story where a curl(E) is the cause of a change in charge density? I don't think so, and the symmetry is broken.
     
    Last edited: Oct 20, 2013
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