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JP

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What are evanescent waves?
« on: 13/04/2010 09:57:35 »
This came up out of a recent discussion on why gravity bends light.  The questions are these: With regards to light/electromagnetism, what are evanescent waves?  Why are they important?  How can we explain them classically and/or quantum mechanically?

JP

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What are evanescent waves?
« Reply #1 on: 13/04/2010 09:59:06 »
By the way rereading evanescent waves. The explanation for how they come to be seems to bite its own tail? " In optics and acoustics, evanescent waves are formed when waves traveling in a medium undergo total internal reflection at its boundary because they strike it at an angle greater than the so-called critical angle "

Okay, makes sense, maybe?
Whenever did a wave get a critical angle? :)

But.

" The physical explanation for the existence of the evanescent wave is that the electric and magnetic fields (or pressure gradients, in the case of acoustical waves) cannot be discontinuous at a boundary, as would be the case if there were no evanescent wave-field." Wait a Minute, do we explain a evanescent wave with the idea of a evanescent wave-field. that then have some undefined angle of approach that our just before absolutely normal waves hits.

So, because we have a evanescent wave-field we will suddenly find our selves with evanescent waves and the evanescent wave field we have is due to the existence of?

Evanescent waves?

Nah it has to be that angle??
And that has to be of a reflection or refraction.

" When light crosses a boundary between materials with different refractive indices, the light beam will be partially refracted at the boundary surface, and partially reflected. However, if the angle of incidence is greater (i.e. the ray is closer to being parallel to the boundary) than the critical angle – the angle of incidence at which light is refracted such that it travels along the boundary – then the light will stop crossing the boundary altogether and instead be totally reflected back internally.

This can only occur where light travels from a medium with a higher [n1=higher refractive index] to one with a lower refractive index[n2=lower refractive index]. For example, it will occur when passing from glass to air, but not when passing from air to glass."

Okay now I know how we can get a standing wave from only one wave, if I'm getting it right? It bumps and meet itself :) But why would it refract, reflect? It leaves your antenna f.ex and bumps into what? Air, but that was what it meet directly it left the antenna anyway? So either that 'near field' have to be very near indeed or there have to be something more to it? and in this case your antenna must have the higher refractive index than air? Or is it the other way around? Not a transmitter but a receiver? Then the air have the higher refractive index and your antenna have the lower which to me seems to make more sense as your antenna more or less leads waves, and lightening :)And where would now that place that evanescent wave-field? At the place where it starts to bump and that must then have to be directly at the antenna, right? So a transmitter won't have a evanescent wave-field?

Awhhh.
Better avoid this :)

Farsight

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What are evanescent waves?
« Reply #2 on: 13/04/2010 12:53:22 »
The evanescent wave is a standing wave, essentially a static stress-energy or "pressure" gradient. Also see http://en.wikipedia.org/wiki/Evanescent_wave where it says evanescent waves are found in the nearfield region within one-third wavelength of any radio antenna.

IMHO the best way to understand it is to think in terms of waves in water. A boat makes waves, see http://en.wikipedia.org/wiki/Wake. Here in Poole there's a place near the harbour mouth where people sit in the sun watching the boats go by. When a big one goes past fast, it sends a noticeable wave crashing onto the beach. This wave is of course travelling. However if you were standing on the boat looking down at the water around the bow, you'd see a standing wave.

There's something similar going on with a radio transmitter. It's making electromagnetic waves, and whilst the transmitter itself isn't moving, electrons are. If you were standing on the transmitter looking down at the space around the aerial, you'd "see" a standing wave. See http://en.wikipedia.org/wiki/Near_and_far_field and note the Quantum field theory view where it says far field effects are manifestations of real photons, while near field effects are due to a mixture of real and virtual photons.

Also see Evanescent modes are virtual photons by Stahlhofen and Nimtz. Andrew Meulenberg told me about this.

Edit: Ah, I see you've covered this already in the other thread.

« Last Edit: 13/04/2010 12:56:32 by Farsight »

yor_on

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What are evanescent waves?
« Reply #3 on: 14/04/2010 17:18:08 »
Well JP, thanks.
So, can a transmitter antenna have a evanescent wave-field?
==

Ah well, that one was up the walls then :)

"In 2006, Marin Soljačić and other researchers at the Massachusetts Institute of Technology discovered a new way to wirelessly transfer power using non-radiative electromagnetic energy resonant tunneling. Their theoretical analysis showed that by sending electromagnetic waves around in a highly angular waveguide, evanescent waves are produced which carry no energy. If a proper resonant waveguide is brought near the transmitter, the evanescent waves can allow the energy to tunnel (specifically evanescent wave coupling, the electromagnetic equivalent of tunneling) to the power drawing waveguide, where they can be rectified into DC power. Since the electromagnetic waves would tunnel, they would not propagate through the air to be absorbed or dissipated, and would not disrupt electronic devices or cause physical injury like microwave or radio wave transmission might. Researchers anticipate up to 5 meters of range for the initial device, and are currently working on a functional prototype."

So, not only evanescent but also 'tunneling'?
« Last Edit: 16/04/2010 18:46:53 by yor_on »

yor_on

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What are evanescent waves?
« Reply #4 on: 16/04/2010 20:13:59 »
I think i did knew this one, well sort of :)

It's what's used in Tesla coils but there called resonant energy transfer through what's also called Evanescent wave coupling. And ""Evanescent" means "tending to vanish", which is appropriate because the intensity of evanescent waves decays exponentially (rather than sinusoidally) with distance from the interface at which they are formed.

Evanescent waves are formed when sinusoidal waves are (internally) reflected off an interface at an angle greater than the critical angle so that total internal reflection occurs. "

And " 'the critical angle' total internal reflection is an optical phenomenon that occurs when a ray of light strikes a medium boundary at an angle larger than a particular critical angle with respect to the normal to the surface. If the refractive index is lower on the other side of the boundary, no light can pass through and all of the light is reflected. The critical angle is the angle of incidence above which the total internal reflection occurs." Total internal reflection

So what do we have? A standing wave right? Made from Sin-waves 'bouncing' due to that they meet "different refractive indices, where the light beam will be partially refracted at the boundary surface, and partially reflected. However, if the angle of incidence is greater than the critical angle – the angle of incidence at which light is refracted such that it travels along the boundary – then the light will stop crossing the boundary altogether and instead be totally reflected back internally."

Well, it's a beginning :) But how can they be said to be energy less? That's a mighty strange statement about a wave, isn't it? "Their theoretical analysis showed that by sending electromagnetic waves around in a highly angular waveguide, evanescent waves are produced which carry no energy."
===

It might be that they take out each others 'vectors' or direction then and so also the momentum? And depending on how you look at such a statement you then can accept that they truly have 'no energy'. It comes down to how 'thingies' interact then I guess. In form of waves you could look at it as water-waves meeting each other neutralizing one another. But in forms of energy? ah well..
===

Here is a nice one explaining it in form of light. Evanescent Waves
« Last Edit: 16/04/2010 20:39:54 by yor_on »

JP

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What are evanescent waves?
« Reply #5 on: 17/04/2010 04:33:01 »
Evanescent waves and quantum mechanical tunneling follow the same mathematics.  Basically, the allowable states of your wave/particle are either sinusoidal or exponentially decaying.  An evanescent wave is not a standing sinusoidal wave or combination of sinusoidal waves, however.  I'm not as sure how to describe it easily in optics, so let's go to the quantum mechanical tunneling waves.

In QM, your particle's sine wave is basically wiggling with a frequency that matches it's energy.  You determine its energy from its kinetic energy minus the potential energy it's had to overcome to get to where it is.  As long as that kinetic energy is greater than the potential energy, there's no issue, as it still has energy left.  If its kinetic energy is less than the potential energy, then it wouldn't classically be able to travel there--not enough energy.  Quantum mechanically, the energy in that region is negative, which corresponds mathematically to an exponential decaying wave.

If you're decent with equations, the sinusoidal wave is given by
sin(ωx),
where ω is frequency of the wave and x is position.  The exponentially decaying wave is given by
exp(-ωx),
where ω is now the rate of decay of the wave.

The mathematics works out similarly in evanescent light waves, but I don't know how to explain it as easily, unless you're comfortable with negative numbers.

Evanescent waves are also cool because they can beat the rule that light only interacts with things of its wavelength.  If you can get information out of the evanescent field, you can beat the usual limit for the smallest things you can image with that light.

yor_on

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What are evanescent waves?
« Reply #6 on: 17/04/2010 18:17:34 »
Very esoteric to me JP?

How do you define 'potential energy' there?
And how do one lift out 'information' from a field of possibly 'no energy'?
Assuming that the idea of energy and information needs each other to exist.

If they don't then the world becomes 'spooky' indeed. Then information needs only 'nothing' to be 'readable' it seems? the potential of 'existing' perhaps? And what exactly would decide that potential? Your choice of 'system'?

Awhh :)

JP

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What are evanescent waves?
« Reply #7 on: 18/04/2010 04:17:00 »
I'm not sure I 100% follow you, so let me know if I'm making sense here.

Potential energy is the same here as it is classically.  It's the energy of a particle by virtue of its position.  (I'm working with a model that doesn't consider the fields as quantum objects here.)  For example, when you lift a ball off the ground, you give it potential (or stored) energy, since when you let it go, it will accelerate.

If something is classically trapped between two "peaks" of potential energy, it can't escape unless it gets enough kinetic energy to over come the potential energy.  An example is that you need to get a high enough velocity (kinetic energy) to escape earth's gravitational field if you want to get to outer space.  Classically an object would either escape or not escape from the potential energy barrier.  An easier model to deal with it just to put a "wall" of potential energy in the way.  The only way your object will pass by the wall is to have more energy than the height of the wall.  Physically, this wall would feel like a force field--you'd be moving along and at some point there would be a strong force pushing you back.

Quantum mechanics is all governed by probabilities, so it turns out if you put a big potential energy barrier in the way that the particle has a small probability of getting through it even if you wouldn't classically expect it to.  Here's an example of the particle's wave function as it passes through the barrier (from Wikipedia): .  On the other side of the wall, your particle starts moving like normal again, but with a much reduced probability of being there if you try to measure it.

I'm not sure I follow the questions about information.  You shouldn't be losing any information, since the probability of detecting the particle anywhere should still be 100%, I believe.

yor_on

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What are evanescent waves?
« Reply #8 on: 18/04/2010 22:13:12 »
Yes, you are right in this :)

Quantum mechanically all seems to fall down to probability. I'm afraid I misread you there.  "You determine its energy from its kinetic energy minus the potential energy it's had to overcome to get to where it is.  As long as that kinetic energy is greater than the potential energy, there's no issue, as it still has energy left.  If its kinetic energy is less than the potential energy, then it wouldn't classically be able to travel there--not enough energy."

It's not that it's seen as having no energy. it's just that its probability of 'existing' there becomes negative, when comparing the potential energy it should have been expected 'classically' to expend, as compared to the energy it's expected to have, does this way of looking at it make sense to you JP?

I did the mistake of understanding you to say that it had no, or negative, energy. But rereading you I see that you meant that this was in comparison with energy spent. So a Evanescent field do have an energy localized where it exists. And my later thought was just how I saw information, as needing 'energy' to exist. but in this case our 'potential' energy is defined from where we expect this wave to have come right?

And yes, the probability must be there (100% all together) as long as you haven't measured it somewhere. So what exactly do you define a Evanescent wave from quantum mechanically? That it has a 'negative energy'? It all seems to go back to how we see waves classically then, propagating? As that seems to be what defines those 'fields', or am I misunderstanding it?

JP

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What are evanescent waves?
« Reply #9 on: 19/04/2010 01:15:13 »
Well, I was using the QM analogy since the mathematics is basically the same.  Both the QM particle and an EM wave look like wiggly traveling sine waves in empty space.  When they hit an appropriate surface, they both have this exponential decay that isn't wiggly.  If that surface ends, a tiny bit of the wave can continue on (where its amplitude is decreased by a lot due to that exponential decay).  In QM this is called tunneling.  In EM propagation, it's due to evanescent waves (the exponential decay wave).  Evanescent waves are a property of waves, as the name suggests, so if you didn't describe these processes using waves, they would be hard to explain.  As it is, they're still weird.

Quote
It's not that it's seen as having no energy. it's just that its probability of 'existing' there becomes negative, when comparing the potential energy it should have been expected 'classically' to expend, as compared to the energy it's expected to have, does this way of looking at it make sense to you JP?
The probability of existing is always positive.  Negative probabilities wouldn't have physical meaning:  what would it mean to say "there's a negative 50% probability of seeing this particle"?

Quote
I did the mistake of understanding you to say that it had no, or negative, energy. But rereading you I see that you meant that this was in comparison with energy spent. So a Evanescent field do have an energy localized where it exists. And my later thought was just how I saw information, as needing 'energy' to exist. but in this case our 'potential' energy is defined from where we expect this wave to have come right?
Exactly.  The particle/wave has a certain amount of energy in a vacuum.  Classically this wouldn't be enough to get through a wall of potential energy, or exist within it, but because it acts like a wave it can.  It still has its own kinetic energy, though.  The wave/particle coming out the other side of the wall will have the same kinetic energy, just a very reduced probability of being detected there.

Quote
And yes, the probability must be there (100% all together) as long as you haven't measured it somewhere. So what exactly do you define a Evanescent wave from quantum mechanically? That it has a 'negative energy'? It all seems to go back to how we see waves classically then, propagating? As that seems to be what defines those 'fields', or am I misunderstanding it?
I think I mostly addressed  the first part of your question earlier, so I'll be brief: the wave is still a probability of seeing the particle when you try to observe it.

To answer the second part, waves are usually disturbance in space/time that carry energy.  They obey wave equations.  The kind of QM we're talking about here is where you can describe particles as waves as well as tiny "bits" of something.

Fields are things that exist over all space and time and which can be assigned a value at every point.  They generally are associated with forces.  There are classical fields.  For example, classical gravity or electromagnetism can be described as gravitational or electromagnetic fields.  Fields can classically give rise to waves and quantum mechanically they can also be described as particles.

Tunneling is a property of waves, so that quantum wave/particles and classical electromagnetic waves can both tunnel.

yor_on

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What are evanescent waves?
« Reply #10 on: 19/04/2010 14:13:51 »
"The probability of existing is always positive.  Negative probabilities wouldn't have physical meaning:  what would it mean to say "there's a negative 50% probability of seeing this particle"?"

Can't argue with that one, as for now, that is.. :)
But just you wait and see ::))

"The wave/particle coming out the other side of the wall will have the same kinetic energy, just a very reduced probability of being detected there."

Now that is an altogether fascinating statement :)
On one side 'classical'. On the other making 'quantum jumps'.

"The kind of QM we're talking about here is where you can describe particles as waves as well as tiny "bits" of something."

But not simultaneously, right? We are talking about wave particle duality, but depending on the measurement. Or are you seeing this as an 'artificial' division forced by the 'instrument measuring'?

As for fields :) I kind'a like that idea. I think Einstein did too, even though he hesitated in stating it as a 'truth'? Fields seems to have that advantage that you can stop discussing 'speed' or 'velocity' And depending on your view it also can eliminate the discussion of 'distances'. That is, depending on how you look at our universe, in a classical way, or not.

Tunneling is a very interesting phenomena. But I'm still not sure how to see this evanescent wave(s) ? You wrote "exponential decay that isn't wiggly". Not wiggly :) would that be something, ah, standing? And what exactly would that 'surface' they hit represent here? The Antenna? But in a transmitter then?

As you can see I'm totally with you, yep :)
Step by step, a mile or two behind..

« Last Edit: 19/04/2010 14:15:37 by yor_on »

JP

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What are evanescent waves?
« Reply #11 on: 20/04/2010 01:08:58 »
"The kind of QM we're talking about here is where you can describe particles as waves as well as tiny "bits" of something."

But not simultaneously, right? We are talking about wave particle duality, but depending on the measurement. Or are you seeing this as an 'artificial' division forced by the 'instrument measuring'?
Not simultaneously.  The probability of finding a particle somewhere is described by the waves.

Quote
Tunneling is a very interesting phenomena. But I'm still not sure how to see this evanescent wave(s) ? You wrote "exponential decay that isn't wiggly". Not wiggly :) would that be something, ah, standing? And what exactly would that 'surface' they hit represent here? The Antenna? But in a transmitter then?

Well, the mathematics is the same.  (See the picture above.)  Waves that carry energy or probability far away from the source are traveling waves.  These are generally "wiggly" sinusoidal waves, like those in the picture.  With the right physical situation, evanescent waves can exist, which decay exponentially as you go away from the source.  They don't carry energy far away.  Mathematically, this exponential decay is just like the exponential decay in tunneling: they both arise in situations where the waves die out quickly and both have the same exponential decay.

For evanescent waves in total internal reflection, the analogy with tunneling is pretty good.  You expect all the wave's energy to be reflected, and therefore the tiny bit that gets into the reflecting material is evanescent and dies off quickly.  If you had a very thin layer of reflecting material, some of that energy would "tunnel" through just as in QM, so that you'd get a little bit of light coming out the other side with most of it reflected.  Near an antenna or anything else that's emitting radiation, the evanescent waves are a little more tricky.  They have the same mathematical form, but they're associated with the fact that the antenna is emitting all kinds of waves, including these evanescent ones.

yor_on

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What are evanescent waves?
« Reply #12 on: 20/04/2010 12:40:48 »
Now I remember JP why I took you too for defining it as 'no energy'. If you look at the post where I cite 'Marin Soljačić and other researchers at the Massachusetts Institute of Technology' they specifically seem to state that "Their theoretical analysis showed that by sending electromagnetic waves around in a highly angular waveguide, evanescent waves are produced which carry no energy. If a proper resonant waveguide is brought near the transmitter, the evanescent waves can allow the energy to tunnel (specifically evanescent wave coupling, the electromagnetic equivalent of tunneling) to the power drawing waveguide, where they can be rectified into DC power."

Now? How can that be? How can you ever have a wave without energy? That's one of the most confusing statements I've seen. So what do one then expect those 'waves' to consist of. Probability only? :)
« Last Edit: 20/04/2010 12:42:30 by yor_on »

JP

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What are evanescent waves?
« Reply #13 on: 20/04/2010 16:06:16 »
In the waveguide example you site, here's what's going on.  The waves reflect around inside of the waveguide completely, so no wave should be leaking out.  In fact, there is a tiny evanescent field leaking through the waveguide, but if you look more than a few wavelengths from the waveguide, you see no energy.  This is why evanescent fields are said to carry no energy: they don't radiate it to large distances like usual EM radiation.  Meanwhile, inside the waveguide, the EM radiation is bouncing around like crazy.

However, if you bring another appropriate waveguide very close to the first one (so close that the evanescent field is still strong), the evanescent field CAN transfer energy to the second wave guide since it hasn't died out by that point.  It will then induce a traveling EM field in the second wave guide.

What these EM waves and evanescent fields represent are energy, so this all has to do with energy transfer.

------------------
The mathematics in QM is the same, so you could devise a similar experiment.  You could have an electron bouncing around in a box with really thick walls (made of a force field--such as a magnetic trap, rather than matter, but don't worry too much about that in my explanation).  Outside the walls is air.  The electron can't "travel" in the walls, so its wave function (which represents the probability of finding it there) dies off quickly within the walls rather than traveling (mathematically analogous to evanescent fields).  If the walls are thick enough, there is virtually no probability of it escaping, since the wave function has basically died to zero by the time the electron would get into the air.  If you cut one wall so it's very thin, then the electron's probability isn't nearly zero by the time it touches the air.  At that point, the evanescent wave turns into a traveling wave again and the electron has a probability of being seen outside the box.

yor_on

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What are evanescent waves?
« Reply #14 on: 14/04/2011 04:12:41 »
I'm hereby solemnly and formally calling this thread reopened.
ahem.

Both because it's so good, in its own right, (as I see it:), as well as because I suspect there may be more to be said? I've recently read a friend argue that the reason why a photon has a 'recoil' comes to be out of it being in the in the so called 'near field', involving 'virtual photons'´? And I'm not really comfortable with the idea of 'action and reaction' when it comes to a light quanta leaving. So I thought I needed more opinions on this.

To me it somehow seems to assume a acceleration, and if I do so we're only a small step from stating that 'photons' have a invariant mass. And then we suddenly have both acceleration and mass and so also 'time'. From there we can degrade a photon to a particle just as a electron. Then we either start to define 'virtual photons' the same way, or assume that what Einstein 'really' talked about was 'virtual photons' that now get new and different properties from the 'real' ones. All of this giving me a immense headache as I'm a ardent 'believer' of Einsteins vision of relativity.

Also it seems to confine them (virtual photons) to certain localities which doesn't sit quite right with me, as well as with the QM idea of a 'quantum foam' and them being 'force carriers' everywhere, well, as I see it? But it is a interesting angle and? Anyone that want to share his/hers views on this subject? And yes, it's evanescent waves, although from a QM perspective as I understands it.
« Last Edit: 14/04/2011 04:23:22 by yor_on »

yor_on

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« Reply #15 on: 14/04/2011 04:38:46 »
And yes, it do matter (no pun) if a photon has a invariant mass, or any ideal type of 'mass'). At least as I see it. Read this and ponder..

"
'Kay, Let's define why a photon is said to be massless.

1. It has no frame where it is at rest.
2. It has no acceleration
3. Therefore no inertia.

Inertia is a property of invariant mass.

"Bose was interested deriving Planck's radiation formula, which Planck obtained largely by guessing. Using the particle picture of Einstein, Bose was able to derive the radiation formula by systematically developing a statistics of massless particles without the constraint of particle number conservation. He was quite successful, but was not able to publish his work, because no journals in Europe would accept his paper.

"This "Bose-Einstein statistics" described the behavior of a "Bose gas" composed of uniform particles of integer spin (i.e. bosons). When cooled to extremely low temperatures, Bose-Einstein statistics predicts that the particles in a Bose gas will collapse into their lowest accessible quantum state, creating a new form of matter, which is called a superfluid. This is a specific form of condensation which has special properties. "

In 1924, Bose wrote to Einstein (in Germany) explaining his work and enclosed his manuscript written in English. Einstein was so happy with Bose's work that he translated the manuscript into German and arranged its publication in Zeitschrift f. Physik (the most prestigious physics journal at that time). Furthermore, in 1926, Einstein completed the Bose-Einstein statistics by extending Bose's work to the case of massive particles with particle-number conservation."

(And the condensates exist.)

Let us go back to the photon statistics formula derived by Bose. There is a factor "2" sitting on the numerator of this formula. The usual explanation is that it is because photons are massless particles. Then why not 1 or 3 ? Bose argued that the photon can have two degenerate states. This eventually led to the concept of photon spin parallel or anti-parallel to the momentum.

The question of why the photon spin should be only along the direction of momentum has a stormy history. Eugene Wigner (1939) showed that the internal space-time symmetry of massless particles is isomorphic to the symmetry of two-dimensional Euclidean space consisting of one rotation and two translational degrees of freedom. It is not difficult to associate the rotational degree with the photon spin either parallel or anti-parallel to the momentum, but what physics is associated with the translational degrees of freedom.

These translational degrees were later identified as gauge transformations. This does not solve the whole problem because there is one gauge degree of freedom while there are two translational degrees of freedom. How do they collapse into the one gauge degree of freedom? This problem was not completely solved until 1990."
==
« Last Edit: 14/04/2011 04:40:35 by yor_on »

JP

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What are evanescent waves?
« Reply #16 on: 14/04/2011 13:51:27 »
Both because it's so good, in its own right, (as I see it:), as well as because I suspect there may be more to be said? I've recently read a friend argue that the reason why a photon has a 'recoil' comes to be out of it being in the in the so called 'near field', involving 'virtual photons'´? And I'm not really comfortable with the idea of 'action and reaction' when it comes to a light quanta leaving. So I thought I needed more opinions on this.

I don't see why the near-field is needed to explain recoil.  You can send a photon off to the ends of the universe, and when you release it, you'll experience a recoil from conservation of momentum.

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« Reply #17 on: 15/04/2011 00:47:03 »
Let me see if I got this right. Assuming that the system (source/photon) are at 'rest' before the excitation the recoil observed is the direct response to the demand that the the final total momentum (source/photon) must equal zero? So if we assume a photon to have a certain momentum then the recoil have to exist to 'counteract/equalize' this?

JP

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« Reply #18 on: 15/04/2011 00:52:05 »
Let me see if I got this right. Assuming that the system (source/photon) are at 'rest' before the excitation the recoil observed is the direct response to the demand that the the final total momentum (source/photon) must equal zero? So if we assume a photon to have a certain momentum then the recoil have to exist to 'counteract/equalize' this?

Precisely.

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« Reply #19 on: 15/04/2011 01:17:59 »
It makes a very nice sense to me JP and I agree :) for once, heh..
It's a symmetry and it also builds on our ideas of the universe and its 'forces' as being in a constant equilibrium until acted on. And as another point, it reminds me very much of Newtons ideas of 'action and reaction'.

So what would happen if we found/made up a system where that didn't hold true?  Either we would have to assume that our description breaks down, or that it's opposite 'symmetry' is expressed in a way we don't see, Am I right?

Do we have any such systems? I made me look up CPT symmetry in fact and

"It has been shown, however, that in certain quantum theories of gravity (including string theory and loop quantum gravity) Lorentz and CPT symmetry might be broken on the ultra-short length scale known as the Planck scale. For a long time, the conventional wisdom was that to probe physics at the Planck scale one would have to perform experiments at extremely high energy -- out of reach of any accelerator -- making these experiments completely unfeasible. However, Robert and his collaborators have shown that another way to search for new physics at the Planck scale is by examining extremely low-energy Lorentz and CPT tests but with very high precision."

From CPT.

But then we have the CP violation of course, why there isn't a equal proportion of matter/antimatter in the universe? And yes, anti matter takes up space just as ordinary matter (Pauli exclusion principle), but annihilate itself as well as any 'normal' matter it interacts with. What is antimatter?

Is there any more discrepancies?

JP

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What are evanescent waves?
« Reply #20 on: 15/04/2011 01:24:37 »
Some symmetries don't seem to hold quite true under careful observation, but to my knowledge, no one's found a problem with momentum conservation.  Momentum conservation relates to the symmetry that says physics behaves the same at any position in the universe.  So if it were broken, I guess it would mean that the laws of physics vary from point to point in space...
« Last Edit: 15/04/2011 01:31:05 by JP »

yor_on

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What are evanescent waves?
« Reply #21 on: 15/04/2011 02:30:51 »
Ahh, JP, beautifully argued ('Momentum conservation relates to the symmetry that says physics behaves the same at any position in the universe':)  It would in fact invalidate the theory of relativity. So yes, there must be a symmetry. I really like this thread :)

synxourfuture

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Re: What are evanescent waves?
« Reply #22 on: 24/10/2013 05:38:16 »
In the waveguide example you site, here's what's going on.  The waves reflect around inside of the waveguide completely, so no wave should be leaking out.  In fact, there is a tiny evanescent field leaking through the waveguide, but if you look more than a few wavelengths from the waveguide, you see no energy.  This is why evanescent fields are said to carry no energy: they don't radiate it to large distances like usual EM radiation.  Meanwhile, inside the waveguide, the EM radiation is bouncing around like crazy.

However, if you bring another appropriate waveguide very close to the first one (so close that the evanescent field is still strong), the evanescent field CAN transfer energy to the second wave guide since it hasn't died out by that point.  It will then induce a traveling EM field in the second wave guide.

What these EM waves and evanescent fields represent are energy, so this all has to do with energy transfer.

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The mathematics in QM is the same, so you could devise a similar experiment.  You could have an electron bouncing around in a box with really thick walls (made of a force field--such as a magnetic trap, rather than matter, but don't worry too much about that in my explanation).  Outside the walls is air.  The electron can't "travel" in the walls, so its wave function (which represents the probability of finding it there) dies off quickly within the walls rather than traveling (mathematically analogous to evanescent fields).  If the walls are thick enough, there is virtually no probability of it escaping, since the wave function has basically died to zero by the time the electron would get into the air.  If you cut one wall so it's very thin, then the electron's probability isn't nearly zero by the time it touches the air.  At that point, the evanescent wave turns into a traveling wave again and the electron has a probability of being seen outside the box.
Hi JP,
I am current working on the evanescent waves in elastic waveguide, that is Lamb waves. As you stated in the post, "if we can bring another appropriate waveguide very close to the first one where the intensity of the evanescent wave is big enough, the evanescent waves can transfer energy to the second guide and can convert the evanescent waves to propagating." Can you explain more about the mechanism here for the conversion between evanescent to propagating waves? Or can you recommend some materials that I can read on this?
Your reply will be greatly appreciate!

Thanks!

The Naked Scientists Forum

Re: What are evanescent waves?
« Reply #22 on: 24/10/2013 05:38:16 »

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