Does light propagate in a medium of any kind?

[geordief]

I understand it has been said that light does not require a medium to propagate .

 

However I have also come across the idea that it propagates as a disturbance in the em field.

 

Are these two interpretations complimentary  or is there any contradiction between them ?(apart from possibly semantic)

 

Can the em field be condsdered a medium of any kind or is it simply an ordered set of measurements?

[studiot]

17 minutes ago, geordief said:

I understand it has been said that light does not require a medium to propagate .

That is true.

17 minutes ago, geordief said:

 

However I have also come across the idea that it propagates as a disturbance in the em field.

Sort of.

Remember that two (or more) EM fields add together at every point to create only one resultant at that point.
IOW there can be only one (resultant) value of the field at any point.

So light itself (classically, not photons) 'carries' its own electric field wherever it goes.
This means that there is no field at a point before (until) the light arrives there.
So if it goes through empty space then that is all (the em field) there is.

But if there is a preexisting field then the two fields add to form a 'disturbed' field where the light becomes a disturbance of the em field.

Equally the disturbed field can be separated into 'components'. This is what an aerial does.

Remember the way that the fields add can be quite complicated.

[MigL]

That is, as you say, classically, Studiot.

Quantum Field Theory has the field everywhere ( even where zero-valued ), and associated particles simply 'excitations' of that field.
The field in a Quantum Gravity theory would be the geometry of space-time itself.

[swansont]

A field is not a medium. Not a substance.

[geordief]

23 minutes ago, swansont said:

A field is not a medium. Not a substance.

A representation of the likelihood of a physical event (of an electromagnetic character) taking place?

[studiot]

1 hour ago, MigL said:

That is, as you say, classically, Studiot.

Quantum Field Theory has the field everywhere ( even where zero-valued ), and associated particles simply 'excitations' of that field.
The field in a Quantum Gravity theory would be the geometry of space-time itself.

"The Field" in QFT is not (just) and EM field as far as I am aware.

Geordief specifically asked about EM fields and reinforced this just now. EM fields are essentially classical in nature, whether he realises this or not.

[swansont]

1 hour ago, geordief said:

A representation of the likelihood of a physical event (of an electromagnetic character) taking place?

An electric field tells you the force per unit charge at a given point, should you place a charge there.

[quiet]

Let's go slowly.

1. Classical electrodynamics, as far as it has been developed, does not cover quantum behavior.

2. The phenomena included in the part that has been developed have been clearly explained and the theoretical results have agreed perfectly with the practice.

3. Has everything that classic electrodynamics can give has been developed? Or are some developments still missing?

4. If the theme is the propagation of light in a vacuum, we are thinking of developments that involve the wave equation in a vacuum, with all the solutions it admits.

5. How many wave functions can we formulate, referring to the electromagnetic propagation of the vacuum? Let's start with the two most remanded wave functions, corresponding to the electric field and the magnetic field. There is more ? Yes, one for each field and one for each density that the electromagnetic theory formulates. Vector potential wave, energy density wave, linear momentum density wave, are some examples ... And one that corresponds to what Maxwell was forced to take into account and analyze to establish a complete, coherent and consistent theory . That field is the electric displacement. Then we have an electric displacement wave, waiting for it to be formulated and analyzed explicitly. That is to say that we formulate and analyze something fundamental, whose omission dismantles the system and detracts from it.

6. In the simplest case, the electromagnetic wave equation in vacuum admits two solutions in terms of real numbers and a complex solution of exponential type. We have devoted a lot of attention to the two real solutions. Have we done the same with the complex solution?

7. Let's write the complex solution of the wave equation for electric displacement.

[math]D = \hat{D} \ e^{i \left( \omega t - kx  \right)}[/math]

[math]\hat{D} \ \ \ \rightarrow[/math] module of [math] \vec{D} [/math]

Let's write the identity of De Moivre for the case that concerns us.

[math]e^{i \left(\omega t - kx \right)} = cos\left(\omega t - kx \right) + i \ sin\left(\omega t - kx \right)[/math]

Applying that identity we have the following.

[math]D = \hat{D} \ \left[ cos\left(\omega t - kx \right) + i \ sin\left(\omega t - kx \right) \right] [/math]

8. The vector field [math] \vec{D} [/math], which has two components and module [math] \hat{D} [/math], corresponds to a plane electromagnetic wave propagating in the direction of the axis [math] x [/math]. If those components were not mutually perpendicular, they could not correspond to a complex number. They are mutually perpendicular and correspond to different axes of the coordinate system. Which axes?

9. Let's write the vector expression of the electric displacement.

[math]\vec{D}= \vec{P} + \varepsilon \vec{E} [/math]

In the vacuum is [math] \varepsilon = \varepsilon_o [/math]. We apply it.

[math]\vec{D}= \vec{P} + \varepsilon_o \vec{E} [/math]

The components [math] \vec{P} [/math] and [math] \varepsilon_o \vec{E} [/math] are mutually perpendicular. Can they be both cross-sectional? Let's reason. In terms of local results, polarization is a field with colinear symmetry that does not alter the electrical neutrality. That means that, within a finite length segment, there is a pair of equal and opposite vectors, resulting from all local contributions. In the case we are dealing with, could polarization be transversal? Impossible, because two transverse vectors that correspond to different values of [math] x [/math] are not collinear. Two longitudinal vectors corresponding to different [math] x [/math] values are collinear, because both vectors have the [math] x [/math] axis  direction.

10. What does the vacuum do when a wave propagates? Is it inert or participate in any way? The velocity of propagation is determined only by two properties of the vacuum, which are the permeability [math] \mu_o [/math] and the permitivity [math] \varepsilon_o [/math]. That leaves no doubt. The vacuum participates. How do it participate? The expression of displacement leaves no doubt. Participate polarizing. In that way it set the speed [math] C [/math] of propagation. That means that the displacement has a transversal component [math] \varepsilon_o \vec{E} [/math] and a longitudinal component [math] \vec{P} [/math].

[math]\vec{D}= \vec{x} P + \vec{y} \varepsilon_o E[/math]

In terms of the wave function we have the following.

[math] \vec{D} = \vec{x} \hat{D} \ cos\left(\omega t - kx \right) + \vec{y} \hat{D} \ sin\left(\omega t - kx \right) [/math]

[math] \vec{D} [/math] has finite divergence, corresponding to the charge density of the polarization. That divergence has the form of a wave function.

[math]\nabla \cdot \vec{D} = \hat{D} \ k \ sin\left( \omega t - kx \right)[/math] 

Does that mean that some charge travels in a vacuum when the wave propagates? No charge needs to travel to produce that divergence. In the cities there are giant illuminated signs, which are panels populated by thousands of luminous cells, controlled by a programmable device. A program can achieve that the brightness of each cell varies sinusoidally, in the form corresponding to a wave function. You can program two colors, say blue and red. The first cell is initially dark. Then the blue light grows sinusoidally, reaches the maximum and decreases sinusoidally, until the cell becomes dark. Follow the sinusoidal stage of the red light, which does the same. All the cells are immobile on the board, but the program manages to see alternate blue and red areas traveling along the board. The effect is equivalent to colors in movement. At each point of the vacuum, the charge density varies sinusoidally. The signs of the charge do the same as the colors. The effect is equivalent to alternating zones with opposite charges traveling in the direction of propagation, although no infinitesimal or finite charge is actually moving.

Edited August 23, 2018 by quite