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

1. Jul 26, 2004

### 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?

Last edited: Jul 26, 2004
2. Jul 26, 2004

### 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.

3. Jul 26, 2004

### Tyger

Hold on a minute here. The E and B components are 90o 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.

4. Jul 27, 2004

### 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?

5. Jul 27, 2004

### 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 90o out of phase. There must be a more accurate representation somewhere on the web.

6. Jul 27, 2004

### Eye_in_the_Sky

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

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

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

let:

E = eyEocos(kx – ωt)

B = ezB(x,t)

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

then:

E/∂t = eyωEosin(kx – ωt)

curl B = -ey∂B(x,t)/∂x

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

so:

∂B(x,t)/∂x = -(ω/c)Eosin(kx – ωt)

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

= Eocos(kx – ωt)

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

therefore:

E = eyEocos(kx – ωt)

B = ezEocos(kx – ωt)

Last edited: Jul 27, 2004
7. Jul 27, 2004

### Eye_in_the_Sky

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

8. Jul 27, 2004

### Sammywu

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

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

9. Jul 27, 2004

### 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?

10. Jul 27, 2004

### turin

The energy is in flux. That's the whole "point" of propagation. Maxwell's equations, in fact, do give a 90o 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.

11. Jul 27, 2004

### 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.

12. Jul 27, 2004

### Eye_in_the_Sky

A question.

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

13. Jul 28, 2004

### 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 travelling 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 travelling 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.

Last edited: Jul 28, 2004
14. Jul 28, 2004

### Eye_in_the_Sky

The reason why E and B have the same phase

This is not wrong.

However, your subsequent statements are problematic.

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,n2ez

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 = ez 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=ex, then we have, by implication, n2=ey.

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) = Eo cos(kz - ωt + Φ1) ,

B(z,t) = Bo cos(kz - ωt + Φ2) ,

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

Eo = Bo 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?))

Last edited: Jul 28, 2004
15. Jul 28, 2004

### 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

16. Jul 28, 2004

### turin

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.

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

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.

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.

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

From basic definitions:
Ephoton = hf
T = f-1

From fundamental quantum uncertainty:
ΔEΔt >= h/2

Combining:
ΔE = the minimum "energy swing" of the wave = Ephoton = 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.

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

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) 90o out of phase!

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

You forgot about the invalidity of Maxwell's equations :tongue2: (just kidding).

Last edited: Jul 28, 2004
17. Jul 29, 2004

### 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.

18. Jul 29, 2004

### 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.

19. Jul 29, 2004

### turin

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

20. Jul 30, 2004

### 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