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Many people have been using very confusing and sometimes contradictory language when describing electromagnetic fields and electromagnetic radiation. It's going to be hard to word this question so everyone understands so I'm going to use some examples.

I want to know if there is simply one large electromagnetic field that is affected by magnets, electricity, and electromagnetic radiation. If not, does that mean that each magnet, each electric flow, and each photon of electromagnetic radiation creates a separate electromagnetic field, all of which overlap but do not interfere with each other?

My understanding is that if electromagnetic fields or waves can interfere with each other, then they are really all just manipulations of one large electromagnetic field that fills all of space much like the Higgs field or spacetime does. In that understanding, would photons (electromagnetic waves) simply be waves in that field?


Here is an experiment example. Say I did a variation of the double-slit experiment. In this variation, instead of having one source that emits electromagnetic radiation from behind the two slits, there are two sources, one at each slit. In essence, I'm performing two single-slit experiments right next to each other projecting onto the same surface. Instead of firing only one photon at a time, each source fires a single photon at a time, both sources firing at the same time.

I have two predictions for what would happen. One for if there is one large electromagnetic field that fills all of space, and one for if each electromagnetic field is separate.

The top image is my prediction of what the result would look like if each field is separate, and by implication the waves do not interfere. The bottom image is my prediction if there is a single electromagnetic field and the waves in the field can interfere with each other.

Predictions


I guess a good change in vocabulary could help. I imagine an "electromagnetic continuum" much like the spacetime continuum. In the spacetime continuum, a gravitational field would simply be a bending/warping of the spacetime continuum. In that sense, an electromagnetic field would be a "bending" or "warping" of the "electromagnetic continuum", and oscillations of the "electromagnetic continuum" would be electromagnetic waves/radiation.

A good way of phrasing my question with that use of vocabulary would be: "Does an electromagnetic continuum exist, or is it all just separate electromagnetic fields and oscillations?"


I hope these examples give enough information for you to understand my question(s) and correct any misconceptions I may have.

Thanks!

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    $\begingroup$ "field that fills all of space much like the Higgs field or spacetime does' - spacetime fills all of space? $\endgroup$ Commented Oct 22, 2015 at 1:19
  • $\begingroup$ It would be good to know how mathematical your understanding of the EM field is. Do you know the math behind it all, or just a vague notion of field? Do this question and its answer make sense? Regardless of the answer, I would suggest to forget entirely about photons until you've mastered the classical concept of a field. $\endgroup$
    – Javier
    Commented Oct 22, 2015 at 1:46
  • $\begingroup$ @Javier - Yes, I understand the math behind the concept of fields. I also understand that any generated electromagnetic field reaches into infinity. My question is if all generated fields are really just bendings of one giant "continuum" or if they are separate and do not interact or interfere with each other. $\endgroup$ Commented Oct 22, 2015 at 1:54
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    $\begingroup$ Well, put in a simple way, the em field is everywhere. It represents the force a charge particle will suffer in a given point in space. The influence of each moving charge that generates this effect adds up (vector math). The double slit experiment is way over this simplification.. You may find interesting to take a look at Maxwell and get a firm grasp of the em force $\endgroup$
    – Alvaro
    Commented Oct 22, 2015 at 5:22

2 Answers 2

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Typically we describe the electromagnetic field using the electromagnetic four-potential, which is represented as $\mathbf{A}$. From the four potential we derive the electric and magnetic fields using:

$$ \mathbf{E} = -\nabla \phi - \frac{\partial\mathbf{A}}{dt} $$

$$ \mathbf{B} = \nabla\times \mathbf{A} $$

where $\phi$ is the electric potential. It is very important to appreciate that the values we get for $\mathbf{E}$ and $\mathbf{B}$ are coordinate dependant, by which I mean that observers in different frames will measure different values for $\mathbf{E}$ and $\mathbf{B}$. You can see this very easily. Suppose in my frame I have a charge at rest, in which case that charge generates a static electric field and no magnetic field. If you are moving with respect to me then you observe a moving charge, i.e. a current, and currents generate magnetic fields.

So if you are considering an all encompassing field this would have to be the four potential. But the four-potential is not a physical object. It is a function of spacetime. If you feed a spacetime point into the function it will return a vector that describes the electric and magnetic potentials at that point.

In principle the four-potential is a function of the position and velocity of every charge present anywhere in the universe. In practice we usually find we can ignore distant charges so the four potential is a function of only finitely many charges:

$$ \mathbf{A} = f_A(q_0, v_0, q_1, v_1, ... q_n, v_n) $$

But the function $f_A$ can be broken up into a sum of separate functions for each charge:

$$ \mathbf{A} = f_0(q_0, v_0) + f_1(q_1, v_1) + ... + f_n(q_n, v_n) $$

and we could write this as:

$$ \mathbf{A} = \mathbf{A}_0 + \mathbf{A}_1 + ... + \mathbf{A}_n $$

where you can regard $\mathbf{A}_0$ as the four-potential of charge 0 and so on.

Now back to your question:

At every point in spacetime there is just one value for $\mathbf{A}$. The four potential cannot simultaneously have more than one value. However we can write that value as a sum of the four-potential of every charge.

So there is a sense in which there is a single electromagnetic four-potential, but there is also a sense in which there are lots of individual electromagnetic four-potentials that add up to give a single value at every point.

I think which of these interpretations you prefer is philosophy rather than physics. My view is that there is a single four-potential and it can be mathematically decomposed into individual components.

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    $\begingroup$ I would more strongly emphasize that there are lots of sources (i.e., charges and currents), but one field. The way one describes the field, as you correctly stated, can be as one single field or a superposition of multiple fields. This is no different than saying $1+0=1$ is equivalent to $1=1$ or $0.5+0.5=1$ (please be kind number theorists... it's a loose[poor?] analogy), all add up to 1. $\endgroup$ Commented Oct 22, 2015 at 10:54
  • $\begingroup$ Now that is a good answer. Thank you. I now have a question to see if I can understand it. Would there be any mathematical differences in the effects of electromagnetic four-potentials if they are considered as separate yet additive instead of as affecting the same space? For example would two magnets bend each others' fields or would they add their field on top of each other? Would those results be identical? And finally, this is the same field that describes electromagnetic radiation, correct? $\endgroup$ Commented Oct 22, 2015 at 15:54
  • $\begingroup$ Also, the Lorentz transformations correct for the different measurements of the different observers do they not? $\endgroup$ Commented Oct 22, 2015 at 16:04
  • $\begingroup$ @Danegraphics: No. You can derive $\mathbf{E}$ and $\mathbf{B}$ from the total $\mathbf{A}$, or you can derive separate $\mathbf{E_i}$ and $\mathbf{B_i}$ from all the component $\mathbf{A_i}$s then add them (vector addition remember) and the result would be exactly the same. To use your example, the fields of two magnets add together (using vector addition). The $\mathbf{A}$ field does include propagating EM waves. Remember that it is a function of spacetime i.e. it varies in time as well as space. An finally, yes, the Lorentz transformations transform the fields between different observers. $\endgroup$ Commented Oct 22, 2015 at 16:11
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The answer by W.Barnett is a handwaving one in the framework of quantum electrodynamics.

John Rennie's answer is in the framework of classical electrodynamics.

First let us define what a field is in physics:

In physics, a field is a physical quantity that has a value for each point in space and time.

.......

A field can be classified as a scalar field, a vector field, a spinor field or a tensor field according to whether the represented physical quantity is a scalar, a vector, a spinor or a tensor, respectively. A field has a unique tensorial character in every point where it is defined: i.e. a field cannot be a scalar field somewhere and a vector field somewhere else.

.....

Moreover, within each category (scalar, vector, tensor), a field can be either a classical field or a quantum field, depending on whether it is characterized by numbers or quantum operators respectively.

So one has to keep oranges and apples clearly in one's mind when talkign of fields.

The classical framework is simple and the single potential field can be supposed to be the underlying field. In John's answer the electromagnetic four-potential ( a four vector) is the field that is defined at every (x,y,z,t) point in space. This can be zero, no energy at that point, or have some values because a charge exists in the neighborhood, or an electromagnetic wave is passing ( classical electromagnetic radiation).

When one goes down to the quantum mechanical microcosm of particles , one is dealing with quantum field theory, and in this framework the fields are creation and annihilation operators, defined over all (x,y,z,t) . These operators operating on the wavefunction at the point (x,y,z,t) will generate a photon, if they are photon creation and annihilation operators, or an electron, or a muon, or... according to which field creation ( annihilation) operators they are. That is the meaning of saying the whole space is permeated by fields of creation/annihilation operators.

Coming to your question:

A good way of phrasing my question with that use of vocabulary would be: "Does an electromagnetic continuum exist, or is it all just separate electromagnetic fields and oscillations?"

It depends on which framework, classical or quantum dynamical the viewpoint rests. In the classical formulation John's answer is sufficient, a four vector field can be defined, it is not "electromagnetic" but both electricity and magnetism and radiation can be built up by this field.

In the quantum mechanical formulation with creation and annihilation operators there will be a field of operators, with creation and annihilation of photons which build up the electrical and magnetic fields. Also radiation itself, being the photon , continually is created (ahead) and destroyed ( behind) as it rushes off on the geodesic towards infinity or interaction.

So something more complicated than an electromagnetic continuum represents reality as we have observed and measured it.

I must add that the classical framework emerges from the quantum electrodynamical framework smoothly, as Lubos Mottl has described in his blog. The classical fields emerge from a confluence of innumerable photons.

Now your thought experiment with the two slits is irrelevant to the above discussion, as it will depend on the boundary conditions of the experiment, whether the quantum mechanical solution which gives the probability of finding the photon on the (x,y) of a screen at z will show an interference. Two independent laser beams can show interferences, depending on the boundary conditions.

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  • $\begingroup$ Thank you so much for the information. I really appreciate the multiple perspectives of classical and quantum. $\endgroup$ Commented Oct 22, 2015 at 15:58
  • $\begingroup$ Also, the chosen answer has been changed because of the new answers that I have received. You might want to edit your answer to refelct that. $\endgroup$ Commented Oct 22, 2015 at 16:10

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