What makes you think light would bend in a magnetic field? Even if treated as a photon (particle) it has no charge. Charged particles curve when moving through a magnetic field (that's also what exerts a force on a wire with current flowing through a magnetic field) but an uncharged particle will move straight through a magnetic field.

Treated as a wave, the fields sum linearly, so even if you have a strong magnetic field, and that field happened to line up with the magnetic component of the wave, it doesn't actually alter the wave amplitude at all. Think of a wave on the surface of water. Changing the surface level of the water (assuming there's no "bottom" near that surface so that the wave isn't interacting with anything else) does not alter the magnitude of the oscillations of the wave about the surface level. Likewise, an EM wave passes unimpeded and unaltered through a strong magnetic OR electric field.

Note that's all very different than what happens when an EM wave passes through a dielectric. There the fields of the wave interact with charged particles, doing work on them (making them move) and thereby altering the wave. The energy induced in the charges can be lost to scattering in the material (become heat) or can re-radiate after an (essentially mechanical) delay. This is why light (appears to) "move" slower in a dielectric than in free space.

@ Michael: Your arguments sound convincing, if the wavelength is too short to check with realizable metallic waveguides. But do they apply to much larger wavelengths, which exceed the dimension of conventional waveguide? Take a straight, round waveguide made of copper with radius = 0.05 m and length = 10 m, filled with air or vacuum.

It is known that photons of a very short wavelength (1 μm or similar, ie light) can pass through the waveguide unhindered. But it does not work with photons of wavelength 1 m. How so? Are the photons thicker than the tube? Why does this change when the copper pipe is slotted so that certain current directions are interrupted? How is it possible that the uncharged photons can be influenced by electric current?

I am happy to see that my questions are getting more questions and more answers.

It might. What's the basis for ascertaining straight line motion in a magnetic field? What is the straight line measured relative to? Light is a propagated disturbance in a magnetic field. It could be propagating along a magnetic field line or at right angles to it, or at any angle in between. The magnetic field is non-isotropic and so there is reason to believe that it may have a steering effect on the path of light.

We also have to consider that when we switch on an electric circuit, a magnetic field forms around it. If the AC frequency is high, much of the EM radiation escapes never to return. But in a DC circuit, when the power is switched off, the magnetic field collapses back into the circuit current again. This implies that low energy EM waves have been following the curved solenoidal path of the magnetic lines of force in the near field. High energy radiation however escapes the solenoidal near field, more like in the case of a body which has exceeded a planet's escape velocity and flies by with a hyperbolic trajectory.

Hi Herbert:

You mention a waveguide, which as the name would suggest means you need to be talking about the wave nature of light/EM waves, and not the particle nature. Propagation modes in a waveguide are not in free space, so you have to take into account the interactions with matter (a conductor in this case) and are not talking about simple wave propagation through a uniform medium. The wave interacts with the metal walls of the waveguide and causes current to flow. The change in the induced current likewise can re-radiate another EM wave. This is why we talk about a "PEC" (perfect electric conductor) as reflecting all EM waves.

When we talk about propagation in a waveguide, waves larger than the lowest mode of the waveguide simply don't "fit" and thus don't propagate very far. In simplest terms that's because the conductor of the waveguide walls is "shorting out" the wave and preventing it from being able to create a peak and a null that propagates forward without loss. By the time you're on a scale where you're talking about photons, you're typically no longer constrained by the waveguide (wavelength much much smaller than the dimension of the waveguide). If you imagine shining light down an RF waveguide, the light that never hits the walls will propagate unimpeded, but light that hits the walls will scatter and some will be lost as heat in the conductor. If you imagine polishing the waveguide to a mirror finish so that any light hitting the walls bounces its way down the waveguide between mirrors will very little loss, then you have a picture of what's happening in higher order modes within a waveguide.

At any rate, this discussion really has nothing to do with the original post of does an external magnetic field impact a wave that propagates through it. It doesn't.

Thanks,

Michael

Hi Frederick:

If you're asking how to define a straight line, then you're far outside the topic at hand. We aren't talking special relativity and curvature of space (due to MASS not magnetic field) where this concept might have some meaning (but not in our Euclidean reference frame). So I'll just refer back to the traditional definition of "the shortest distance between two points" and leave it at that.

You're also incorrect that light is a "disturbance" in a magnetic field. It's been proven that there is no "ether" that a wave propagates through (although again quantum physicists have reintroduced something similar in the idea of a"quantum foam"). An EM wave is its own freely propagating structure that carries energy through empty space. On the other hand, a static magnetic field does not carry any energy. It holds/stores energy in the form of inductance, but it does no work in an of itself. When a static magnetic field and an EM wave happen to intersect, during the period that they occupy the same space, the net magnetic field is simply the sum of the two components. There's nothing magical about that and no interaction beyond what might happen to a charged particle moving through the same region. There's a lot more I could talk about here, but again it doesn't really affect the original discussion.

Regarding your mention of DC vs. AC, you're again missing some points. The main one is that when you turn off the DC, it's no longer DC. There's a discontinuity that has a tremendous spectrum of frequencies associated with it. During that transition, the circuit can radiate those frequencies. Turn on an AM radio and flip a light switch on and off nearby. If there's enough inductive load, you'll definitely hear it on the radio. Heck, that's how radio was discovered! They were inducing sparks in loops of wire at a distance and eventually learned that it wasn't just a mutual inductance thing that was being observed.

And BTW, if what you say about EM escaping if the frequency is high were as simple as that, we wouldn't have cable TV or wired internet. The frequency doesn't determine whether or not something radiates.

Sincerely,

Dr. Michael D. Foegelle

Michael, You don't add an EM wave to a steady state magnetic field. One is an oscillation in the other. As for cable transmission, that is a separate topic.

Hi Frederick:

The two magnetic field vectors are linear independent quantities. They simply add. The total magnetic field is the vectorial sum of the static magnetic field vector and the magnetic field vector of the wave. Neither is a function of the other, which is why there can be no interaction that alters one due to the presence of the other. The static field can be as high as you want and the net field at a given location will still just be the sum of the static field and the EM field. When the wave propagates away from the vicinity of the magnetic field (assuming free space plane wave propagation), it has the same magnetic components it had prior to encountering the magnetic field. Likewise the static magnetic field that the wave propagated through is unaltered after the wave passes (e.g. once the signal is disabled) and has the same value as before the wave arrived. It's only the TOTAL field that changed during the period that the two intersected, not the individual components. Each exists without the other.

Sincerely,

Dr. Michael D. Foegelle

Michael, An EM wave does not own its own magnetic field. It is a propagated disturbance in an already existing magnetic field. There has to be a magnetic field present for the magnetic field to oscillate. It's not like as if the oscillating field is an entity in its own right that moves along at speed 'c'. It's the disturbance in the already existing magnetic field that moves along at 'c'.

Now I don't know whether or not the curvature in a magnetic field can steer an EM wave off a straight line path, but I suspect that if the wavelength is in the same order of magnitude as the arc of the curve of a magnetic field pattern, that it may be able to do so noticeably. As such I doubt if a laboratory magnet would have much of a noticeable effect on the path of a beam of light.

Hi Frederick:

Where are you getting your information? Please cite a reference.

Thanks,

Michael

@ Michael: Disagree. We live within a waveguide: the inner electrode is earth / sea, the outer electrode is the ionosphere. The inferior conductivity (in comparison with copper) is not a counter-argument, because the Schumann resonance (7.8 Hz) is proven. So all experiments on earth are done inside a waveguide. So photons could not be detected in any experiment.

How do you explain the following contradiction: When a photon (lambda = 1 μm) flies through a copper tube, you speak of a propagation in the free space. But when a photon (lambda = 1 m) flies through the same copper tube, suddenly there is no free space left. Suddenly, electrical currents are argued. Are there different-sized photons? What do electrically neutral photons have to do with currents?

Michael, This goes right back to when Maxwell first showed that EM radiation is a propagated oscillation in a magnetic field.

You seem to be under the impression that an EM wave consists of oscillating magnetic fields as an entity in themselves, divorced from the background magnetic field. Every magnetic field has a source electric current. That is a fact expressed in Ampère's Circuital Law. So as you see it, where is the source electric current in the oscillating magnetic field in the section of a ray of light that is entering our solar system from a distance star?

Hi Herbert:

Free space is commonly used to imply a relative permeability and permittivity of 1.0. Air is very close to that, so we generally ignore it. However, in your example of the ionosphere, the permittivity is obviously not 1.0, since it's an ionic conductor. Thus, you have the same types of interactions that I referred to regarding the impact of a conductor (moving charges) on an EM wave. This is primarily and electric field reaction, but nonetheless, even though air can obviously bend light, it's not doing so because of an applied magnetic field, it's doing so because of an interaction between the charges of the atoms (and their component electrons and protons) making up the atmosphere and the fields of the EM wave itself.

As far as the "size" of a photon, again, it's nonsensical to talk about photons (particles) when talking about EM wave mechanics. But yes, since the wavelengths are of different sizes, the photons are ALSO of different sizes. A common description of a photon is as of a wave "packet" which can't possibly be shorter than at least half a wavelength. So yes, if you want to talk about a waveguide in particle physics, below cutoff, the photon is bigger than the waveguide (as is the wavelength) and won't fit. However, it really doesn't make much sense.

Hi Frederick:

So I think perhaps the problem is in your statement that an EM wave is an oscillation "in" a magnetic field. It's not. It IS an oscillating/propagating electromagnetic field. There is no magnetic field required before the wave and no magnetic field required after the wave. The magnetic field in the wave also does not exist without the associated electric field. The energy that is stored in a wave is propagating through space in the form of orthogonal electric and magnetic fields.

I shouldn't have to do this here, but go look at the Wikipedia page on EM radiation (https://en.wikipedia.org/wiki/Electromagnetic_radiation) to learn more about what an EM wave really is, but I'll try to help simplify it here.

If you look at the picture of an EM wave you'll see the blue and red E and B fields (more often E and H, but I digress). You'll notice of course that those sinusoidal field curves go through zero and the magnitude oscillates around zero. An observer at a fixed point in space sees that oscillation going by, but the net field (magnetic or electric) averages over time are zero unless there is a static field applied in addition to the EM wave. However, that static field has nothing to do with the EM wave and is not required.

Now let's talk about what the wave actually is and how energy is carried. Take a ride along with the wave (i.e. travel at the speed of light) and you'll see that the wave is fixed, frozen in time and space relative to the moving observer. At some points relative to the observer, it's zero (no energy) and at some points both E and H are at their maximums. Basically you have this one sinusoid of E and H going from zero to max and back to zero that's just moving along at the speed of light. That single packet from zero to zero could be considered a photon. For a continuous wave, that is followed by another zero to zero packet with the fields in the opposite direction and so on. However, as mentioned, the field isn't oscillating at all. It's static in the light speed reference frame but that is moving through the physical (static) reference frame at the speed of light. The field in a given direction was created at the source in that direction and has stayed that way forever, simply propagating away from the source at the speed of light. Only if it encounters a charge somewhere (free floating or in a conductor) will it change as it loses (or gains) energy.

And since someone is bound to bring up the speed of light and relativity, put yourself in any other arbitrary reference frame moving relative to the source. What happens to the frequency and wavelength of that "packet" as it passes a point that's moving relative to the original source. By definition, the speed of light is a constant, so it's constant in the source reference frame and constant in the moving observer reference frame. But in the moving frame, the oscillations in the EM wave are changing faster or slower than they were in the reference frame (depending on the direction of the relative velocity). Thus, the frequency and energy in the wave increases (blue shift) or decreases (red shift) and the wavelength decreases or increases correspondingly. That's relativity in a nutshell. Everything's a wave so waves moving past each other faster or slower change frequency, not speed.

Sincerely,

Dr. Michael D. Foegelle

Michael, The EM wave equation is derived from Faraday's law and Ampère's Circuital Law. Those laws are rooted in the inter--relationship between an electric current and a magnetic field. You seem to be under the impression that an EM wave has been cut off from its roots in a magnetic field and taken off on its own. How could that possibly happen? How can you have a propagating magnetic oscillation as an entity in its own right? Where has the electric current whose existence would be implied by Ampère's Circuital Law which is used in the derivation of the wave equation? You cannot divorce an EM wave from its electromagnetic foundations.

Hi Frederick:

I'm not under any impression. That's exactly what's happened. If it hadn't, then what you're saying is the photons we receive now from a star at the edge of universe somehow has something to do with the current state of the star now and the fields that it may be producing. Sounds pretty nonsensical to me.

Since you refer to Faraday and Ampere's laws, let's finish that. You need to be looking at Maxwell's equations, which "finished" those laws. So, away from the presence of a charge, both of Gauss's laws are zero. No charge. Faraday's law relates the electric field to the rate of change of the magnetic field, and since there's no current, the Maxwell-Ampere law looks identical (with the free-space permeability and permittivity scaling factors applied). This is the essence of a propagating EM wave. The electric and magnetic fields are orthogonal and equivalent in magnitude (with the scaling factor) and propagating along the normal relative to them. There's no static field involved, only the propagating field. Note also that the time derivative portion of this is actually related directly to the fact that the field is propagating. If it wasn't propagating, then to the fixed observer there would be no changing field and thus no orthogonal component associated. The whole thing would fall apart and literally would not exist.

You also keep going back to this idea that an EM wave is somehow "rooted" (only) in a magnetic field. It's not, although it potentially can be. The most common radiator is a dipole, which has a current at the feed but none at the tip, and a voltage profile from tip to tip. Thus, it is ideally situated to produce an EM wave from both the electric fields parallel to the dipole and the orthogonal magnetic field around the feed current in the middle. On the other hand, it is possible to make a magnetic loop antenna where (at least at low frequencies) the electric field component generated by the antenna is negligible.

At any rate, it sounds like either you aren't familiar with Maxwell's equations or just aren't understanding what those equations are telling you. If you took the time to read through that wiki link I'd sent you'd see the section on "Behavior of the fields in the absence of charges or currents" which gives the free-space version of Maxwell's equations I alluded to earlier. You'll note that in the absence of charge, fixed or moving, (thus no waveguides, atmospheric or otherwise) the fields in the EM wave are ONLY a function of a changing field, and thus a static field, regardless of the strength (including zero!) does not alter the changing field. That's the correct answer to the original question and remains so. The remainder of this discussion has unfortunately digressed into repeated attempts to explain what Maxwell's equations say about EM waves and if you don't understand them by now, I'm not likely to be able to help you further. I'll leave that to the textbooks and the professors that teach from them. Then again, if that's who convinced you of what you are saying then I would be asking them for my money back!

Sincerely,

Dr. Michael D. Foegelle

It depends on whether you want to go by Maxwell's explanation for displacement current or the textbook's explanation. They are not the same. For Maxwell, displacement current was a real current. The textbook displacement current on the other hand adds an additional term to Ampère's Circuital Law which is not a real current, whereas Maxwell's displacement current emerges from the total current. The textbook displacement current therefore contradicts Ampère's Circuital Law. You can read Maxwell's original views on displacement current in the preamble to Part III of his 1861 paper "On Physical Lines of Force" at this link, beginning at page 39 in the pdf file's system, http://vacuum-physics.com/Maxwell/maxwell_oplf.pdf

You can also look at my analysis of Part III in section I of of this article,

Presentation Radiation Pressure and E = mc²

or to see my commentaries on the original Maxwell's equations take a look at this paper,

Article Maxwell's Original Equations

Berend, Yes you seem to have got the picture. But you are assuming that the current will be zero on the grounds that the negative and positive charges cancel. If you read the theory more carefully you will see that it is proposed that electric current is something more fundamental than a flow of charged particles. There is a section on "electric current" in the article suggesting that charged particles can be carried along with the flow as a side effect.

Berend, The double helix alignment is a magnetic line of force. The principle behind it is that the electrostatic force of attraction that acts between the electrons and positrons in the background dielectric sea is channeled along the double helix. I.e. magnetic attraction is simply electrostatic attraction between electrons and positrons in a particular arrangement. Magnetic repulsion on the other hand acts at right angles from magnetic lines of force due to centrifugal force acting in the equatorial plane between two adjacent rotating dipoles.

In the case of electric current, if positive particles are sources in the fundamental aethereal fluid that comprises electric current, then they will be pushed along with the flow. Negative particles being sinks, will eat their way upstream in the opposite direction. But the primary electric current itself will be the uni-directional aether flow. It will be a hydrodynamical momentum and in many cases it will arise from the velocity field of an electric field. Rotating dipoles are in fact dipolar aether vortices and the circulation current/momentum A nowadays known as the magnetic vector potential is in fact Maxwell's displacement current.Article Displacement Current and the Electrotonic State

Delbrück scattering and photon-photon interactions are quantum mechanical behaviors. While this is an gross oversimplification, they can be considered to relate to the particle nature of photons such that they can collide and rebound in different directions. Extrapolating from that the possibility that a static magnetic field of sufficient strength could bend a photon sounds like pseudo science. There's no basis to even formulate such a hypothesis and the known physics denies the assumption.