Why are E/B fields of photon in-phase?

[vanesch]

Mymymy.

There are quite some people here trying to discuss advanced concepts who lack some basic mathematics of wave propagation such as the one on a cord (used to be called string but that is doomed to make a conceptual link with advanced stuff it has nothing to do with)

To try to put some things in order, let us first consider a piano cord, but of a very very big piano. Doing some elementary Newtonian mechanics, you can find the wave equation (d^2 u/dx^2) - 1/v^2 (d^2 u/dt^2) = 0
Here, u(x,t) is the sideways displacement of the cord from its rest position, and v is a constant which is calculated from the tension strength on the cord and the mass density of the cord.

If this is not clear to you, honestly, I think the whole discussion on EM waves, photons and so on is absolutely not accessible.

The above wave equation is a partial differential equation which has as a general solution: u(x,t) = f1(x-vt) + f2(x+vt), where f1 and f2 are two completely arbitrary functions. In this one-dimensional case we're lucky that we can write explicitly down the general solution ; in most of the cases this is not true, and we have to resort to some special techniques ; one of those techniques is called Fourier transforms.
If you calculate the Fourier transform of u(x,t) and write it U(k,w), then the equation becomes:

k^2 U(k,w) - 1/v^2 w^2 U(k,w) = 0

The nice thing is that we changed the partial differential equation into an algebraic one, so we have:

(k^2 - w^2/v^2) U(k,w) = 0

This essentially tells you that for a point {k0,w0} in the k-w plane where U(k,w) is defined, to have U(k0,w0) not equal 0, you have to have that:
k0^2 - w0^2/v^2 = 0 or that v * k0 = +/- w0. For other combinations of k and w, U(k,w) = 0
The curve in the k,w plane where U is allowed to be different from 0 is called a dispersion relation, in our case:

w(k) = +/- v * k

As the differential equation is linear, we do not have to work with a general solution, it is sufficient to work with a basis set of solutions which we can then superpose. A way of doing so is by considering ONE SINGLE POINT on the dispersion curve, namely a single couple (w0, k0) satisfying the dispersion relation, where we consider U(k,w) to be different from 0.

This gives us then a HARMONIC WAVE in the x-t domain:

u(x,t) = Exp(i (k0 x + w0 t)) by inverse Fourier transformation.

We usually work with these harmonic waves but it is understood that a more general solution is a SUPERPOSITION of these harmonic solutions.
In fact, we already have to combine u(x,t) with its conjugate (the point at -w0, -k0) because u(x,t) is to be a real number, not a complex number, so we have as real solutions sin(k0x + w0 t) and cos(k0 x + w0 t).

We now come to the concept of a "wave packet". This is nothing else but such a superposition of harmonic waves, that has as a peculiar property that it is "lumped" as well in the (x,t) domain as in the (u,k) domain - this is a property of the Fourier transform.
So in order to make a wave packet, you take a U(k,w) which is non-zero over a certain "lump" of k-w space (but of course only on the dispersion curve), instead of in one single point. If you choose the amplitudes and phases of U(k,w) right, the inverse transform will give you a function u(x,t) which, for a given t-value, is lumped in x.

The funny thing is that such a wave packet propagates with speed v: the center of the wave packet, x_c, at time t is a function of t as follows:

x_c(t) = x_c(0) +/- v t

This is the underlying idea of a wave propagation, but when you have done this a few times, you already see this when looking at the harmonic solution, so in most texts, people don't bother by going through this mantra again and again.

Now consider energy. The only energy we have is kinetic energy ; you could be tempted by thinking there is a potential energy associated with a displacement u(x,t), but this is not true, as can be easily seen:
Imagine that at one side of the very long chord, you apply a "step function": u(x=0, t) = h(t). Clearly, you only do a modest amount of work when moving the point x=0 from u=0 to u=1. The solution of this will be that at time t, u = 1 for x < v*t, and u = 0 for x > v*t with a propagating step front. If having u = 1 amounts to storing energy, there would clearly be a conservation of energy violation, because you only introduced one small amount when you moved the point at x=0 at time t=0, and the part of the chord where u = 1 grows with t.
So you see that the energy (kinetic energy) of a wave motion u(x,t) is concentrated in those parts of the chord that are moving:
the energy density E(x,t) is given by 1/2 rho ( d u/dt )^2.
If you calculate the energy density of a harmonic wave sin(k0 x - w0 t) you will find that the energy density is not constant but "moves" with a speed v.

There is

In EM (classical EM), we have exactly the same equation, but it is 3-dimensional, instead of 1-dimensional. A wave packet is then a "pulse of light" which has a certain spread in frequency, and has a certain localisation in space. This "pulse of light" propagates at speed c.

Now to the "mystery" of E and B fields. First of all, you could work with the electromagnetic potential, A, and then there is only ONE potential, and you wouldn't be bothered by two fields oscillating "in phase", there would only be one field. Concerning the energy density in space, yes, it is not constant for a HARMONIC solution. So what ?
If a car is moving over a road, then its "mass density" is not constant: when the car is at point A, its mass is at point A and no mass is at point B ; while a bit later when the car is at B, the mass density at A is gone, and it is now at point B. So the same happens with a classical EM wave, at least the harmonic wave, which is a special solution.

This has nothing to do with photons ! We are working purely with classical waves here.

cheers,
Patrick.

 

[Sammywu]

Patrick,

Do you mean that without this dispersion, the wave will not spread thru space but rather stand vibrating in the same space ?

Thanks

 

[vanesch]

Patrick,

Do you mean that without this dispersion, the wave will not spread thru space but rather stand vibrating in the same space ?

Thanks

If you mean that w(k) = 0, this comes down to saying that v ->0 and that the wave equation reduces to d^2 u / dt^2 = 0, so the general solution will be u(x,t) = f1(x) + t f2(x). You have a structure in space which can stand there (not vibrating) or can grow linearly in time. If you replace that by w(k) = constant, it can indeed vibrate with a constant frequency (given by the constant in w(k) = constant), but cannot "move" in space.

cheers,
Patrick.

 

[Olias]

If you mean that w(k) = 0, this comes down to saying that v ->0 and that the wave equation reduces to d^2 u / dt^2 = 0, so the general solution will be u(x,t) = f1(x) + t f2(x). You have a structure in space which can stand there (not vibrating) or can grow linearly in time. If you replace that by w(k) = constant, it can indeed vibrate with a constant frequency (given by the constant in w(k) = constant), but cannot "move" in space.

cheers,
Patrick.

So does this explain why standing waves can be seen as having mass, without speed, and consequentely photons (particles) can be seen to have speed without Mass/wave?

 

[Sammywu]

Patrick,
I reread your answer and compare that to this section of wave packet. Now I am certain that is what it was saying. Without a certain dispersion of k and w under the constraint w=ck, the wave solution is just a standing wave.

The e^ikx+wt can only represents a standing wave when k and w are constant. My original answer by simply restricting only value appearing in the wave function in the space lump is flawed because it will not satisfy the wave Eq. at the two boundary pints. -- I knew there were somethings wrong -- So, Fourier transformation has to be brought into rescue the situation and gave a general solution to represnt a lump-type of wave. And it's also shown that in this case, the solution actually moves with time.

By this, what I see is a light pulse has to be represented by a wave packet, because it moves thru space. A radiation might be able to be represented as standing waves because it bounces around us. I already knew that wave packet failed to represent a particle. Roughly I knew that was because it was proved that wave packet will spead out thru space and particles don't. I still need to read the detail.

Thank you again. I have read this wave packet things somewhere. Now I truly have a better grasp of it.

 

[turin]

The e^ikx+wt can only represents a standing wave when k and w are constant.

This expression does not represent a standing wave, it represents a complex valued propagating plane wave. If you happen to have the conjugate plane wave to this traveling in the same direction, then you have a physical (classical) wave. If you happen to have one of these plane waves traveling in the opposite direction, then you have a standing wave. The standing wave condition is met when w is a constant wrt a change in k at a particular value of w and k, not when both w and k are constant. What would that mean? Since this is a plane wave, the w and k are constant by definition in this expression. Otherwise, the expression is arbitrary, and you might as sell have written f(k,x,w,t).

... it's also shown that in this case, the solution actually moves with time.

A wave packet is not a requirement for this condition. All of the individual basis functions e^{i(&omega;t-kx)} themselves "have motion." A "non-moving" particular solution to the diff. eq. (a standing wave) will only arise if certain boundary conditions exist that restrict the basis functions to pairs that cancel each other's motion. In other words, standing waves occur at the edge of the Brillioun (sp?) zone where &omega;(k) = constant => &part;&omega;/&part;k = 0 (zero group velocity).

By this, what I see is a light pulse has to be represented by a wave packet, because it moves thru space.

Not because it moves through space, but because it is a pulse. The complex valued plane waves move through space. Two conjugates traveling together give a physical (classical) plane wave, not a wave packet, that travels through space. In order for the wave to exist in only a finite region of space, the destructive interference of the plane waves must eliminate the displacement outside of this region.

A radiation might be able to be represented as standing waves because it bounces around us.

Can you clarify?

I already knew that wave packet failed to represent a particle. Roughly I knew that was because it was proved that wave packet will spead out thru space and particles don't.

I don't think I agree with this justification. From what I understand, the Schroedinger equation predicts a spreading of a wavefunction for any particle whose state evolves with time.

 

[Sammywu]

Turin,

Let me try to understand you on the first question.

This is my point:

If w and k are both conatant and there is no other conditions, the EQ. e^i(kx+wt) will have exact amplitude at every positions apart of kx along x-axis at the same moment of t. The whole universe will be vibrating in a constant rate, both in amplitude and frequency.

I can't see any propagating here. The whole universe needs to be vibrating since beginning of the time to the end of it.

Regards

 

[turin]

The magnitude of this expression never changes throughout time or space, however, the amplitude does change. The amplitude at any given point will vibrate, so the entire universe would indeed be vibrating, but these vibrations do not all occur in phase. In fact, it is their phase relationship that gives rise to the propagation. Consider the point x = 0 @ t = 0. The amplitude is unity. Then, if you just consider that one point in space, you will see the amplitude "vibrate" (it will repeatedly circle around the origin of the complex plane). At any point in space, you will see this same behavior.

Now, consider a set of points from x = 0 to x = 2π/k (a range of 1 wavelength). At t = 0, the amplitude of the expression over this range looks like a helix that completes exactly one period. Now, if you let time progress, this helix will rotate. It will still complete exactly one period within this range, but the point of the helix that has a value of unity will slide along the axis (in the -x direction). If you extend the domain of consideration to include all values of x from - to + infinitiy, then you simply stack identical helices together like the one just discussed. Letting time progress simply allows this stack to rotate. The point along the stack that has unit amplitude occurs once every 2π/k, and all of these points slide along the axis. This is propagation.

The peculiar thing about the plane wave idealization is that there is no source. It is just there. It is not considered to be strictly physical, as far as I know, but the set of all possible plane waves provides a complete basis to describe any physical situation (such as a wave packet).

 

[Sammywu]

Turin,

I am very sorry.

I can't see your helix in this plain formula. All I see in this plain formula is E and B fields are always on the same axis , orthogonal to each other, their amplitudes go up and down in phase at every point since the beginning of time. I don't see anything rotating either.

As long as I look from the real part.

Somehow we are not on the same page.

I think you are looking at the complex part as one. That seems to fit this picture. In that case, you will have to assume two more extra dimensions at every points similar to hyperspaces on the 3-dim space. Your world will be a 5-dim space then. The problem is what do these two additional dimention reprent.

Maybe you want to lead to super string theory?

Another possibility -

If the light is polarized in a 90 degree phase , ( as I remember, you need to interwine two light pulses that interfere with each other in 90 degrees. ), that might fit a rotation picture to this. I am not so sure that fits this equation though.

Regards

 

[turin]

Sammywu,
I'm not developing a theory here; this isn't just my crazy idea (except for calling it a helix, though I still think that's a good way to picture it). If you're going to ignore the complex part, then why are you expressing it as a complex valued expression. We have expressions for the complex-shy people; they're called sine and cosine. Why not use one of those? For instance, if you just want to consider the real part of this expression:

cos(kx+ωt)

Where in e^i(kx+ωt) do you see E and B fields? I don't see them. This is an expression for a generic plane wave.

I think this last thing you said was in reference to circular polarization, correct? That is not at all what I'm talking about, though I can see how someone may interpret it that way. I appologize for confusing anyone with my helix picture. Let me re-emphasize that the expression:

e^i(kx+ωt)

is NOT considered to be physical. Only a superposition of complex-valued plane waves with different k's and ω's is considered physical (observable).

 

[Sammywu]

Turin,

OK. Now I see where you are coming from.

As I understand, the complex number representation here is more like conveniency. Actually, all we look is either real part or imaginary part.

You can actually just write it in sin or cos. I think that's a classical view.

"Wave packet" is actually a classical view that tried to envision a particle existing as a wave packet moving thru the space.
It did lead to QM eventually, but since there is no way to tell what is the real part and imaginary part in a classical way, that's why probability interpretation and opeartor theory came to rescue. So, now, we don't need to actually picture it as a wave but just a "blurred" state.

This is not a quantum wave function that you have to inlucde both real and imaginary part -- my understanding. Maybe I am wrong.

 

[Sammywu]

My impression from this chapter of "wave packet" is:

1. wave "propagation" and "spread" are not the same.

2. wave propagation is on a factor of \partial w / \partial k.

3. wave spread is on a factor of \partial^2 w / \partial k^2.

 

[turin]

As I understand, the complex number representation here is more like conveniency.

I suppose you could say it that way, but it's a little deeper than that. For one thing, the complex-valued plane wave is a momentum eigenstate!

"Wave packet" is actually a classical view that tried to envision a particle existing as a wave packet moving thru the space.

My own humble, basic understanding would agree with this.

It did lead to QM eventually, ...

I don't know. The apparent wave-particle duality was pivotal in the development of QM, but I'm not sure that wave packets were a necessary nudge in that direction.

... since there is no way to tell what is the real part and imaginary part in a classical way, that's why probability interpretation and opeartor theory came to rescue.

Maybe. I wouldn't have ever thought of it like that. I have an acquaintance that is working on the connection between the intimate and unavoidable use in QM as apposed to CM, but I haven't heard anything from him in a coon's age (maybe he figured out how to build a worm hole in the process ).

So, now, we don't need to actually picture it as a wave but just a "blurred" state.

I think that if you're picturing it as a wave-packet, you are unavoidably considering the wave nature, specifically superposition. That's why they call it the "wave function," even when it is describing a "particle."

1. wave "propagation" and "spread" are not the same.

Agree.

2. wave propagation is on a factor of \partial w / \partial k.

Agree.

3. wave spread is on a factor of \partial^2 w / \partial k^2.

Off the top of my head I'm not sure, but it seems reasonable.

 

[Sammywu]

Turin, Now back to your helix picture. Actually I am not totally against it. The opinions I have said basically are what was said in QM and how it was interpreted. Myself have imagined your picture in the past for a long time and I originally thought that the real part is E field and the imaginary part is B field . When I found out E/B fields are in phase, I realized that thought failed. Well, but remember we have voltage and current in a 90-degree phase difference in a capacitor resonance. So, if the real and imaginary parts are actualy coresponding to certian voltage and current and the electron works similar to a capacitor, this might work. Actually, there is a guy who has this site proposing this solution. Can't remember his name. Something like Warren Betty. He is in the forum, of course as a figure of anti-mainstreamer. When I finished my QM readings, I might want to see what he says in his site. At this point, I am still trying to understand QM. On one side, I can accept the theory of the abstarct states, but on the other side, I still want to see whether there is any possibility that a true Physical picture is behind the veil and was left out completely. Another issue is electron spin. I am studying it for quite a while. I always think this intrinsic spin might have a physics picture. Actually from what I have learn from QM baout it now -- the way how Caley-Klein parameter and Pauli matrix are put together to represent a spin or angular momentum --, I don't see it excluding the possibility of the electron might work as an empty shell surounding the proton and keep chuning in a manner similar to a mobius strap. Best Regards.

 

[turin]

... I originally thought that the real part is E field and the imaginary part is B field . When I found out E/B fields are in phase, I realized that thought failed. Well, but remember we have voltage and current in a 90-degree phase difference in a capacitor resonance. So, if the real and imaginary parts are actualy coresponding to certian voltage and current and the electron works similar to a capacitor, this might work.

This represents a dangerously confusing similarity between the EE approach to circuit analysis and the Physics approach to wave mechanics. The issue is even further confounded by the allusion to the same basic phenomenon: electromagnetism. However, two key differences to point out:

1) Electromagnetic waves don't propagate in a circuit in "normal" operation. (That is, in considering the behavior of capacitors and inductors, one makes assumptions that are contrary to the support of wave propagation.) If a capacitor supported an electromagnetic wave, it would not be functioning correctly, as a basic assumption in linear circuit analysis is that the entire circuit is of a size at least one order of magnitude smaller than c/f_max. Therefore, any electromagnetic coupling among circuit elements is a first order coupling through either Faraday's Law or the Ampere-Maxwell Law, not a second order coupling (through the wave equation which constitutes both). Actually, the similarity attains its highest level of guile in the form of an LC resonance, as it then maintains the characteristic exchange between electric and magnetic energy. However, don't be fooled. There is still the fundamental 90^o phase shift between the current and voltage in such an exchange. In fact, upon careful inspection of this example, you will find the key difference that distinguishes between the two behaviors. In a circuit, there is an exchange from one type of energy to another (between electric and magnetic), with any losses characterised by resistance (that is, impedance that is not purely imaginary). In wave propagation, there is a flux of energy (in the entangled form called electromagnetic energy) from one point in space to another, with any losses characterised by reactance (that is, impedance that is not purely real).

2) The 90^o phase difference is a property of the physcial substrate in the case of a circuit, whereas wave propagation is a fundamentally dissapative phenomenon which has a characteristically purely real-valued impedance. (See "intrinsic impedance" for propagation through free space or a wave guide and "radiation resistance" for the mechanism by which the electromagnetic energy coherently leaves the circuit.) The phase difference in a circuit comes from the phase of the impedance of the physical elements in the circuit. Both the voltage and the current can be (and generally are) independentaly described by complex valued sinusoids. This is the only way a complex valued impedance can make sense. With the definition of impedance being:

Z = V/I,

then, if the driving current and driving voltage are to be measurable and oscillate through a reactive element, they must have complex-valued amplitudes. Of course, in EE we take the real parts, but both voltage and current are individually treated as complex valued during the analysis.

 

[what_are_electrons]

Turin, Now back to your helix picture. Actually I am not totally against it. The opinions I have said basically are what was said in QM and how it was interpreted. Myself have imagined your picture in the past for a long time and I originally thought that the real part is E field and the imaginary part is B field . When I found out E/B fields are in phase, I realized that thought failed. Well, but remember we have voltage and current in a 90-degree phase difference in a capacitor resonance. So, if the real and imaginary parts are actualy coresponding to certian voltage and current and the electron works similar to a capacitor, this might work. Actually, there is a guy who has this site proposing this solution. Can't remember his name. Something like Warren Betty. He is in the forum, of course as a figure of anti-mainstreamer. When I finished my QM readings, I might want to see what he says in his site. At this point, I am still trying to understand QM. On one side, I can accept the theory of the abstarct states, but on the other side, I still want to see whether there is any possibility that a true Physical picture is behind the veil and was left out completely. Another issue is electron spin. I am studying it for quite a while. I always think this intrinsic spin might have a physics picture. Actually from what I have learn from QM baout it now -- the way how Caley-Klein parameter and Pauli matrix are put together to represent a spin or angular momentum --, I don't see it excluding the possibility of the electron might work as an empty shell surounding the proton and keep chuning in a manner similar to a mobius strap. Best Regards.

The helix representation is most definitely an excellent mental image for light (EM energy or EM radiation) and its inherent chiral nature. The helix image is even more useful and more accurate if you envision the double helix of DNA with both strands, the chirality of the strands, the chirality of the inner molecular parts and all the molecular interconnections (hydrogen bonds etc.).

Some 6-8 months back, I spotted the high degree of similarity between the DNA helix and the internal structure of light (EM radiation), electrons (a EM force field based mass) when my friend and I were discussing and theorizing about the chiral nature of electrons (ie +1/2 vs -1/2 spins when in an atom). At that particular moment I happened to look at my drawings of light and electrons and a nicely rendered, shiny metallic 3D image of DNA in Chem Eng News. I put 2 and 2 together and became hooked on the similarity and the symmetry of it. (Side note: My thesis involved chiral organic chemicals, circular dichroism, conformation analysis and the3D imaging of the conformation of molecules, but my current hobby involves understanding the physical nature of electrons and photons.)

The DNA helix inherently has several degrees of chirality and the many chiral nucleic acid molecular strings which represent the many different energy states of the "wave packets" that are all on the same strand. The next level of complexity involves thinking of space filled with endless arrays of DNA strings going in all directions that do nothing until perturbed.

The interconnections between the two strands of the helix represents the permanent inter-connection that exists between electric and magnetic properties of all elementary particles. So, the picture that EM is a single entity and is not a blend of E and B is best.

There is also the factor of time, what it is or is not, does it exist or not, but...

 

[Sammywu]

At first, I thought this is not worthy to think.

But deeper thought said that two electrons working in a helix structure similar to DNA might be true.