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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
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?
"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'?
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?
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.
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?
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.