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

[Rhizomorph]

Can someone explain to me why all models of an EM wave (or photon) are always shown in-phase? Isn't the total energy of a photon stored in the two oscillating fields? And isn't the total energy of a photon constant (according to conservation of energy)? If so then an in-phase model of perpendicular fields describes a wave in which, at the beginning of every new cycle, the strength of E+B = 0, meaning the total energy at that instant in time is 0. See http://www.astronomynotes.com/light/emanim.gif

It would seem to me that in order for the strength of E+B, and thus the total energy, to be constant the two fields would have to be out of phase by a factor of Pi/2. Am I incorrect in my understanding of the relationship between the two fields E & B and the total energy?

 

[Galileo]

E and B are mutually in phase because they must obey Maxwell's equations.
It's a direct consequence of that.

For a monochromatic plane wave the energy density of the wave is indeed a function of time, but the frequencies for visible light is $10^4$ to$10^5$ Hz. In most cases a time average of the energy density delivered is all that is required.

Light consists indeed of photons, but they do not fit into the classical picture of light as an electromagnetic wave (as you see in the picture of the E and B-fields).

The intensity (or irradiance) of the light (average power per unit area) is different for the wave description and the particle description.
For waves you take the average value of the magnitude of the Poynting vector, and for photons you take the number of photons times their energy ($\hbar\omega$) per unit time per unit area.

The photon description is correct. The wave picture from electromagnetism is an approximation when dealing with a large number of photons.

 

[Tyger]

Hold on a minute here. The E and B components are 90^o out of phase, both in space and in time. So that the square of the magnetic and electric field intensities, which represents the energy density, is constant and energy is conserved. The fact that the E and B components and the direction of motion are othogonal to each other is called tranversality, and is an important property of free photons or electromagnetic fields.

 

[Rhizomorph]

Tyger, that's whay I'm saying. But I keep finding illustrations where E and B are orthogonal but IN-phase, like the one in the link. And I'm not sure the explanation that it "must obey Maxwell's equations" fully justifies an in phase wave model. If they are out of phase by 90deg, the equation could remain unchanged by replacing sin with cos (whose results are naturally 90deg out of phase). Then everything else still follows from those equations but the apparent violation of conservation of energy during propagation is resolved. Any thoughts?

 

[Tyger]

I just looked at the picture in the link you gave and it isn't very realistic at all. The E and B components should be 90^o out of phase. There must be a more accurate representation somewhere on the web.

 

[Eye_in_the_Sky]

---------------------------------------------

(1/c)∂E/∂t = curl B ... (originally I missed out the "1/c")

---------------------------------------------

let:

E = e_yE_ocos(kx – ωt)

B = e_zB(x,t)

---------------------------------------------

then:

∂E/∂t = e_yωE_osin(kx – ωt)

curl B = -e_y∂B(x,t)/∂x

---------------------------------------------

so:

∂B(x,t)/∂x = -(ω/c)E_osin(kx – ωt)

B(x,t) = (ω/ck)E_ocos(kx – ωt) ... [throw away const]

= E_ocos(kx – ωt)

---------------------------------------------

therefore:

E = e_yE_ocos(kx – ωt)

B = e_zE_ocos(kx – ωt)

E and B are mutually in phase because they must obey Maxwell's equations.
It's a direct consequence of that.

 

[Eye_in_the_Sky]

... the apparent violation of conservation of energy during propagation ...

The energy didn't disappear, it just moved over.

 

[Sammywu]

It's so fun seeing so many people having the same issue I had.

I had an old thread about it:

Does E field and B field has a 90 degree phase difference in EM wave ?

The differential EQ. does give us an in-phase solution.

But I guess everyone intuitively feel that the 90-degree-out-of-phase sounds to give us an image of energy and momentum conservation.

Regards

 

[Rhizomorph]

Then perhaps someone can illuminate for me how the strengths of the fields E and B relate to the total energy. Because according to the in-phase interpretation, it would seem that at 2 points (instantaneous moments) in every cycle of an EM wave both E*B and E+B yield a value of 0. So if the total energy of a photon is in any way related to the strengths of these two fields, then there are 2 moments in every cycle of a photon that look a bit troubling.

That said, it would seem that the only way to avoid a potentional violation of conservation of evergy with an in-phase interpretation is to come to one of the following conclusions:

Either...
A) The strengths of the electrical and magnetic fields have nothing to do with total energy of the wave when analysed instantaneously.
Or...
B) Energy can only exist over an interval of time, and cannot be properly measured instantaneously.

I hightly doubt the first conclusion. The second *might* work, but this means that the conservation of energy law needs a footnote - energy is conserved over a complete EM cycle, but NOT conserved for time intervals smaller than that required for an EM wave to complete a full cycle.

Traditionally, (as I understand it) we express a plane electromagnetic wave that propagates in the +z direction of a rectangular coordinate system as:

E = ˆx*E0 sin(kz - wt)
B = ˆy*E0 sin(kz - wt)

If this is wrong then perhaps I am misguided. But if this is correct, then we can phase shift the magnetic field by 90deg and replace the sin with cos since sin(x) = cos(x-90deg). Then all the mathmatical results based upon the original relationship still hold, but we can use the sin^2(x)+cos^2(x)=1 identity to establish a constant value for E throughout the entire propagation of the wave, and can talk about instantaneous E and not just average E.

Any thoughts?

 

[turin]

The energy is in flux. That's the whole "point" of propagation. Maxwell's equations, in fact, do give a 90^o phase shift between the E and H fields. However, there is a wave solution that is (almost fundamentally) different than the non-propagating solution. The energy is necessarilly moving through space in an EM wave. So, just because some point along an EM wave has zero EM energy does not mean that there is a problem; it just means that the energy that used to be there has moved.

 

[Rhizomorph]

"The energy is in flux"

But if the energy of a photon is fluxuating, then it is not constant. And if it is not constant, then violation of conservation of energy has occured.

"it just means that the energy that used to be there has moved."

To where? I'm looking at the photon now, not later. If I'm analysing a specific moment in time, and the energy is 0 at this moment, then for this moment a violation has occured. There is no physical location that the energy could have "moved to" to make up for this.

Here's an example:

Let's shoot a single photon with frequency 5x10^14.
The period for this photon = 1/f = 2x10^-15 seconds.
So in 2x10^-15 seconds there are 2 points where E+B and ExB are both 0.
Now, over the entire 2x10^-15 second span (i.e. one complete cycle), energy may be conserved. But if I look at any time interval smaller than that, an in-phase model gives moments of 0 energy. To say "that's ok, it will average out by the end of the cycle" is not an acceptable explanation for the temporary violation of one of the most fundamental laws of physics.

It seems that either the law of conservation of energy needs to be revised (to account for sufficiently small time intervals within which the rule does not apply), or the interpreted phase of EM waves needs to be revised. As it is now, they seem to stand in opposition to one another.

 

[Eye_in_the_Sky]

A question.

... Maxwell's equations, in fact, do give a 90^o phase shift between the E and H fields. However, there is a wave solution that is (almost fundamentally) different than the non-propagating solution.

Can you be more specific about "non-propagating" solutions for which E and H are out of phase?

 

[Galileo]

A wave has a fixed shape and moves with constant velocity. Actually, the E and B fields are not fluctuating at all. They are constant in time, but moving thourgh space.
Point is, if you look at a fixed point, then the field is oscillating because the wave is passing by.
So the energy (which is proportional to E^2) at any point is traveling along with the wave. If it is zero at a given point then t seconds later it will be zero a distance ct further in the direction of propagation. The energy 'moves along' with the wave.
So there is no violation of any conservative law.

For example. Let's make a very short wave. Light of one frequency traveling in the +z-direction with angular frequency $\omega$.
Let: $\vec E(z,t) = E_0sin(kz-\omega t)\hat x$.
And let the wave extend from $z=-\pi$ to $z=\pi$ at t=0.
At any moment in time the wave profile (shape) stays the same (one oscillation). So the whole energy density distribution of the wave has not changed, but moved with the wave. The formula for the energy density is:
$$u=\epsilon_0 E_0^2sin^2(kz-\omega t)$$
which is actually a wave itself. The energy density has fixed profile, but moves along the z-axis at the same velocity as E.

It tried to clear some things up, but I hope I didn't make it seem more complicated than it actually is.

 

[Eye_in_the_Sky]

The reason why E and B have the same phase

Traditionally, (as I understand it) we express a plane electromagnetic wave that propagates in the +z direction of a rectangular coordinate system as:

E = ˆx*E0 sin(kz - wt)
B = ˆy*E0 sin(kz - wt)

If this is wrong then perhaps I am misguided.

This is not wrong.

However, your subsequent statements are problematic.

But if this is correct, then we can phase shift the magnetic field by 90deg and replace the sin with cos since sin(x) = cos(x-90deg) ... we can use the sin^2(x)+cos^2(x)=1 identity ...

Maxwell's equations in the vacuum do indeed lead to two (separate) wave equations, one for E and one for B. For our case of a linearly-polarized plane-wave, which we can write (more generally) as

[0] E = n₁E(z,t) and B = n₂B(z,t) , with n₁,n₂ ┴ e_z

those two wave equations reduce to

[1] ∂²E(z,t)/∂z² = (1/c²) ∂²E(z,t)/∂t² ,

[2] ∂²B(z,t)/∂z² = (1/c²) ∂²B(z,t)/∂t² .

These two equations give the "appearance" that, for any given frequency component, the amplitudes for E and B can be independently specified, and likewise, the phases for E and B can be independently specified, and, similarly, the angle between the unit vectors n₁ and n₂ can be at anything we choose. But equations [1] and [2] are not the "whole story", because Maxwell's equations tell us more than just [1] and [2]. In particular, for the case at hand, they also tell us that

[3] n₂ = e_z x n₁ ,

and

[4] (1/c) ∂E(z,t)/∂t = - ∂B(z,t)/∂z ,

[5] (1/c) ∂B(z,t)/∂t = - ∂E(z,t)/∂z .

Now, equation [3] shows us that if we set n₁=e_x, then we have, by implication, n₂=e_y.

Furthermore, equations [4] and [5] show us that for a given frequency component we must have E and B with the same amplitude and the same phase. You can easily verify that this is so. Just write

E(z,t) = E_o cos(kz - ωt + Φ₁) ,

B(z,t) = B_o cos(kz - ωt + Φ₂) ,

and plug them into either one of [4] or [5]. Upon doing so, you will find that

E_o = B_o and Φ₁ = Φ₂ .

---------------------------------------------
Rhizomorph, are you now convinced that E and B must have the same phase? If not, is it because you are not convinced that [4] and [5] are implied by Maxwell's equations? (... or is it that the presentation I have offered is too advanced? (... or could there be another reason?))

 

[reilly]

There's another way to show that the E and B phases are the same:

If A is the vector potential, obeys the homogeneous wave eq., then

E=-dA/dt and B=curlA. The fields will thus be spatially perpendicular, and in phase temporarly.
Regards,
Reilly Atkinson

 

[turin]

"The energy is in flux"

But if the energy of a photon is fluxuating, then it is not constant. And if it is not constant, then violation of conservation of energy has occured.

You may have misunderstood me with this one. I made no mention of photons, and no mention of fluxuating photon energy. Electromagnetic radiation is characterised by electromagnetic energy flux through space. In this regard, I don't understand how you can have a problem with the energy changing value at only one point in space. All that shows is that the energy moves, which is exactly why we say that it propagates.

"it just means that the energy that used to be there has moved."

To where?

To the next spot over. The energy moves. That's what a wave is: propagation of energy.

If I'm analysing a specific moment in time, and the energy is 0 at this moment, then for this moment a violation has occured.

Violation of what? If you restrict yourself to only one moment of time, then no physical process whatsoever occurs. What other kind of violation are you talking about? If you see zero energy at a particular moment in time, then no energy arrives at your detector at that one moment. This can be specificed as the absence of a photon, if you so desire. This is, indeed, a possible feature of the particle nature of light. Assuming a coherent beam, the electromagnetic energy could be monitored to demonstrate a periodic detection of maximum and vanishing energy. There is no violation of physical laws in this case either, as there is no rule (AFAIK) that says energy must move, and therefore subsequently arrive at its desitination, in a steady, uniform, unaltering flow. It can, and does, arrive at the deterctor in waves.

Let's shoot a single photon with frequency 5x10^14.
The period for this photon = 1/f = 2x10^-15 seconds.
So in 2x10^-15 seconds there are 2 points where E+B and ExB are both 0.

You're inappropriately overlapping the particle and wave behaviors. A photon of electromagnetic radiation at 500 Thz is a quantum of energy at 2 eV. It does not exist at a particular point in space or time unless it is observed (it is an energy eigenstate, not a position eigenstate). If it is observed, it is observed to have its characteristic energy ... period. If it is not observed, that is just how the quantum coin toss turns out. An electromagnetic wave at 500 THz is a pattern of variations in the electric and magnetic field that moves through space at the speed c (and has periodic occurences of concurrently zero field strength). In order to interact with this wave (i.e. detect it), only discrete chunks of energy = 2 eV can be extracted. These extractions are the photons.

... over the entire 2x10^-15 second span (i.e. one complete cycle), energy may be conserved. But if I look at any time interval smaller than that, an in-phase model gives moments of 0 energy.
...
It seems that either the law of conservation of energy needs to be revised (to account for sufficiently small time intervals within which the rule does not apply) ...

The revision that I think you're looking for is called the uncertainty principle. Check this out:

From basic definitions:
E_photon = hf
T = f^-1

From fundamental quantum uncertainty:
ΔEΔt >= h/2

Combining:
ΔE = the minimum "energy swing" of the wave = E_photon = hf
Δt = time between "problematic features" = T/2 = f^-1/2
=>
ΔEΔt = (hf)(f^-1/2) = h/2

This is a bit "hand-wavy," but interesting.

To say "that's ok, it will average out by the end of the cycle" is not an acceptable explanation for the temporary violation of one of the most fundamental laws of physics.

I am not at all trying to say that the problem "averages out." My position is that there is no problem.

Can you be more specific about "non-propagating" solutions for which E and H are out of phase?

Just take a reactive electronic component as an example:

V = L(dI/dt)
or
I = C(dV/dt).

Then:

E = dV/dx
and
H is somehow directly related to I.

When a circuit operates at some frequency, the electromagnetic energy gets stored in these reactive components (it does not propagate). A sinusoidal voltage produces a sinusoidal current (or vice versa) 90^o out of phase!

Faraday's Law and the Ampere-Maxwell law are to blame.

... are you now convinced that E and B must have the same phase? If not, is it because you are not convinced that [4] and [5] are implied by Maxwell's equations? (... or is it that the presentation I have offered is too advanced? (... or could there be another reason?))

You forgot about the invalidity of Maxwell's equations (just kidding).

 

[Sammywu]

I hope this might help.

Just looking this from purely classical point-of-view, a wave spans a range of space at an exact moment of time. At one point of the space at the time, the E and B fields might be zero, but E/B fields on other space spanned by the wave at the same moment do have magnititude. So, at any point of time the energy is conserved by the range of space the wave spans.

For example, we usually create an E/M wave by accelerate some electric charge from rest to max. speed and back to rest. At the initial moment right from start, E/B field right around it is zero. With the acceleration of the charge, the wave is created and start spreading out through space. Said, after t seconds of the initial moment, the wave already span on a range of the space as of t*c. the wave front has zero E/B field, but the other area of the wave have non-zero E/B fields.

 

[Sammywu]

By the way, the formula of E & B as a sin function of z and t are published all over the texbooks.

But from a purely classical point of view, that formula is not wrong but flawed. -- My opinion.

If you only creates a wave packet, the formula shall only have values in the range of the space it spans for that moment and it shall be zeros elsewhere.

If you put a condition something like a*t < z < b*t, then that would be perfect, where a and b are a simple linear function of w - the frequnency. You ashall be able to work that out by yourself.

Simplly put, if I send a signal ( only one signal ) from here toward the moon, after 10 seconds, the signal is somewhere between me and the moon and not detectable by me any more.

 

[turin]

Sammywu,
Look at the theory of retarded potentials. It is targeted at precisely the "flaw" you mention.

 

[Sammywu]

Hi Turin,

Where does this "theory of reatrded potential" usually appear, in classical ED or wave , QM or QFT?

Do you have a quick place you can point me to?

By the wording of the theory, does it mean it is really reagrded as vanishing of E/B fields something like cancelling effect?

Thanks

 

[turin]

Sammywu
Someone else should probably answer that as I do not have a formal physics background. The first place that I saw it was in first semester graduate E&M as a foreshadowing of second semester graduate E&M (but I never took the second semester). I recall seeing it in an undergraduate senior level "mathematical methods for physicists" text by Arfken. Another place I saw it used (though not explained) was in one of those compilations made of Einstein's relativity theory, but I don't remember which one.

 

[robphy]

links for "retarded potentials"

http://www.astro.umd.edu/~miller/teaching/astr601/lecture11.ps (you need a postscript viewer)

http://farside.ph.utexas.edu/teaching/em1/lectures/node45.html

 

[Rhizomorph]

Thanks to all who've replied so far. I still have to ponder this (massless) matter for a while and look into this "retarded potential". I started this thread because I'm determined to be able to understand the fundamental aspects of the physical universe in a manner that can be visualized and understood intuitively (as opposed to an abstract collection of equations). And sometimes a mathmatical equation can compute correct yet make "no sense" (like imaginary numbers). I want to make sure my fundamental understanding of EM isn't undermined by the models that describe it.

 

[Sammywu]

Robphy,

Thank you for the info.

I browsed it for a few minutes.
I guess it's named as "retarded" because it uses the concept of retarded time.

It's very analogous to the gravitational radiation's model I read in GR. I don't see any quantum concept here. So I guess it's a part of classical theory.

I will read it more thoroughly to see anything I missed.

Regards

 

[turin]

It is most definitely within the classical realm. It is a solution to the inhomogeneous wave equation with a 4-D Dirac-delta function as the driving function (it is the space-time Green function of the D'lambertian). To think of it intuitively, it describes what the electromagnetic field does if a point charge suddenly appears and then immediately disappears at some point in space (thus the 4-D Dirac-delta function). It is a very straightforward application of simple PDE.

 

[Sammywu]

Robphy,

The website you posted is a great site. I roughly toured it and found it contains lots of clear analyzation of different aspects of classical E/M theory.

Turin,

I am not sure why it has any things to do with Dirac 4-D delta function though. Is it because the Green's function is a derived concept from Dirac 4-D delta function?

Actually, I just read something related to "wave packet" of light quantum , the minimum "band width" and "definite" Eigen value, etc ...

I found I am not so sure here. Are we able to get a "definite" energy eigen value for a light quantum or we always need to take some uncertainty on the wave frequency of a light quantum?

Thanks

 

[turin]

You could say that the Green Function is "derived" from the Dirac-delta function, but to clarify:

The Green function is a characteristic function of an operator. It is the function on which the operator must operate in order to yield a Dirac-delta.

I far as I know about wave packets, they are a suggested improvement to the picture of a photon (its nature and behavior).

 

[Sammywu]

Turin,

I recalled I encountered the "wave packet" in three places. 1. It's described in how a TV or radio formulates ( or modulates ) its radio signals. 2. In this old book "The Quantum Theoty" by David Bohm, it mentioned an early attempts in making electron model, the "wave packet" was suggested, whcih came from the concept of a light's wave packet always has a very small deviation of the wave frequency, from the distribution function of the "momentum", by Fourier transformation we can make the wave function of the light in the "position" for the light quantum. 3. This "wave" packet concept was used to interprete Schrondinger EQ. but eventually abandoned by modern QM people. I think the operator theory thrilled.

Doesn't this look familiar to you?

This section seems to suggest that all light quanta have a certain uncertainty of its momentum which coresponds to its wave length and so frequency.

While I have an impression that we shall be able to prepare a light quantum to a definite eigenvalue.

Does this seem to be a conflict?

I just tried to see whether you or anyones have a quicker answer for this.

Is there a way that we can prepare a light quantum to a definite momentum? Or there would be always a certain degree of uncertainty to the momentum of a light quantum?

I haven't finished my readings. Somewhere there might be an answer.

Regards

 

[turin]

I don't know very much behind the notion of a wave packet. I am skeptical of the first point you mentioned about the radio (perhaps this is not the same "wave packet" as what I mean, and what I think you mean). The need for the wave packet, as opposed to a perfectly well defined momentum eigenfunction for EM radiation, arises when one incorporates special relativity. I am not strong on the relativistic considerations of the quantum world (and, therefore, shame on me for even trying to talk about photons). It seems as though there is a fundamental uncertainty in the energy of a photon, which is (IMO unsatisfactorily) explained as a time-energy uncertainty relationship:

&Delta;E&Delta;t > h/2

In order for the photon to have perfectly defined energy, the atom would have to have an infinite uncertainty in emission time. With an ensemble of millions of atoms, this would lead to complete uncertainty in energy flux anyway.

 

[StealthC]

know what I don't know that math very well but it's starting to make a tinsy bit of sense...
But think about it, drop the math and use what we know. That photon is ALWAYS going to have the same amount of energy unless it is losing it and that energy that it loses is long gone. You are not creating nor destroying energy by having a photon, it is ALWAYS there and it is ALWAYS made of the exact same things.
Photons are not energy in flux, photons ARE energy. A photon can never have 0 energy so therefore those calculations giving you that answer are flawed, and I believe they are flawed, even though I don't know all that math, I can tell you that the fact that the wave concept is a generalization and the particle approach you are dealing with individual objects then extending into a generalization (because you aren't going to care what 1 billion photons are doing individually at once, you are going to care more about the big picture). That's why those calculations exist, because otherwise they are an abomination designed only to make your life easier; get a computer instead of using bad generalizing calculations. Sure they sort of fit but they aren't accurate at all.
I think what it looks like from that model and that FLAWED calculation is that a photon is a one handed ping pong player playing against himself with his feet glued to only one side of the table. I guess that would be a little confusing wouldn't it?

 

[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.