Author Topic: How does the electron make a photon in an antenna?  (Read 29643 times)

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Offline HP-ILnerd

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Re: How does the electron make a photon in an antenna?
« Reply #25 on: February 12, 2017, 11:24:26 am »
OK but there must be different mechanism to radiate sub Infrared (longer) photons. When a wire generates a radiowaves no electrons are jumping from a higher orbital to a lower one as is the case to make IR light UV and Xrays (balmer series etc). The longer the jump the higer energy/shorter wave length photon. IR seems to emitted by phonons in matter as way to get rid of thermal energy. But what happens at lower energies?   

Ah!  I think I see the problem.  The wavelength of a photon is not its physical length, but a property of how it propagates through space.  Like all particles in the Standard Model, the photon is dimensionless.  Consider the following experiment:
Say you have a hole that you can block in a tiny fraction of a second.  Through this, you try to shoot a photon where the period of it's wavelength is twice the time it takes to close the hole.  Can you close the hole to "slice" the photon in half to catch half a photon?  No.  The photon made it through in its entirety or it did not.  The E=hv equation has to be used in integral amounts, i.e., 1(hv) or 2(hv)...n(hv) but never 1/2(hv) or any other fraction.

I think it's less confusing if you consider fields rather than particles or waves (notions which have their uses).  Certain phenomena seem less magical then.

Fun Sean Carroll lecture on particles and fields: 
 

Offline Rick Law

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Re: How does the electron make a photon in an antenna?
« Reply #26 on: February 12, 2017, 10:17:57 pm »
Photon and EM wave are the same thing.  To say EM wave make photons is rather like saying H2O makes water.  Photon is a packet of EM wave energy.  EM wave traveling is a bunch of photons traveling.

When energy is released in an atom, such as when an electron falls from a higher energy state to a lower energy state, it will emit a photon - that is the same as saying it will emit EM wave.  That released EM wave (photon) contains the energy it released.

Mass and energy are the same thing.  Photon (mass-less) carries momentum.

Photon and EM wave relationship is a different concept from wave-particle duality.  Wave-particle duality is the concept that all particles exhibits wave properties and the reverse is also true.  You can pick any subatomic particle, be it photon, electron, alpha particle, or for that matter, any particle.  When you treat it as a particle, you can measure it's particle properties.  When you treat it as a wave, you can measure its wave properties.  You will find alpha particles doing crazy things like being at two places at the same time when you are treating it as a wave.
work are being done.  We don't know all there is to know yet.

OK but there must be different mechanism to radiate sub Infrared (longer) photons. When a wire generates a radiowaves no electrons are jumping from a higher orbital to a lower one as is the case to make IR light UV and Xrays (balmer series etc). The longer the jump the higer energy/shorter wave length photon. IR seems to emitted by phonons in matter as way to get rid of thermal energy. But what happens at lower energies?   

Exact same thing - you got a photon with less energy.  That it was less energy doesn't matter, it just turns into a lesser photon.

Eventually, the energy can get so low that uncertainty principal comes into play.  Then you don't know what energy that photon has, or where that photon is - or if that photon even exist.

Now you are into philosophy more than you are in Physics.

Even ignoring inflation and the expansion of the universe, how do you measure a photon with frequency< (1 / (13.7 billion years)), or in other words a photon with wave length>13.7 billion light years?  That photon if created at the birth of the universe has not completed one cycle of oscillation yet.  You have to wait another 0.1 billion years for the first oscillation cycle to complete.

We think that uncertainty principal appears valid.  So below a certain point, it can just vanish perhaps to reappear at a later time in another form.  What happen to that photon for the moment it vanished or does it exist?  That I don't think you can find the answer in Physics yet.
 

Offline T3sl4co1l

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Re: How does the electron make a photon in an antenna?
« Reply #27 on: February 12, 2017, 10:25:28 pm »
Even ignoring inflation and the expansion of the universe, how do you measure a photon with frequency< (1 / (13.7 billion years)), or in other words a photon with wave length>13.7 billion light years?  That photon if created at the birth of the universe has not completed one cycle of oscillation yet.  You have to wait another 0.1 billion years for the first oscillation cycle to complete.

We think that uncertainty principal appears valid.  So below a certain point, it can just vanish perhaps to reappear at a later time in another form.  What happen to that photon for the moment it vanished or does it exist?  That I don't think you can find the answer in Physics yet.

Misapplying Fourier analysis doesn't invalidate what the universe does. ;)

Consider, for example, the case of the Sun-Earth system, as a quantum "atom":

- What is the quantum number of this system?
- If the quantum number were to decrease by one, what is the wavelength of "photon" emitted?  (It's a gravitational rather than electromagnetic system, so it would actually be a graviton, as such.  For the problem, a boson is a boson, so that's fine.)
- Based on what you know about the Earth-Sun system, what is interesting about this wavelength?

(Borrowed from Griffiths' Quantum Mechanics.  My favorite problem in the book.)

Tim
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Offline Rick Law

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Re: How does the electron make a photon in an antenna?
« Reply #28 on: February 13, 2017, 12:11:24 am »
Even ignoring inflation and the expansion of the universe, how do you measure a photon with frequency< (1 / (13.7 billion years)), or in other words a photon with wave length>13.7 billion light years?  That photon if created at the birth of the universe has not completed one cycle of oscillation yet.  You have to wait another 0.1 billion years for the first oscillation cycle to complete.

We think that uncertainty principal appears valid.  So below a certain point, it can just vanish perhaps to reappear at a later time in another form.  What happen to that photon for the moment it vanished or does it exist?  That I don't think you can find the answer in Physics yet.

Misapplying Fourier analysis doesn't invalidate what the universe does. ;)

Consider, for example, the case of the Sun-Earth system, as a quantum "atom":

- What is the quantum number of this system?
- If the quantum number were to decrease by one, what is the wavelength of "photon" emitted?  (It's a gravitational rather than electromagnetic system, so it would actually be a graviton, as such.  For the problem, a boson is a boson, so that's fine.)
- Based on what you know about the Earth-Sun system, what is interesting about this wavelength?

(Borrowed from Griffiths' Quantum Mechanics.  My favorite problem in the book.)

Tim

You are thinking along a line that I am not in sync with.  I am unsure how Fourier Analysis comes to play in your line of thinking.   Along the line of energy being a function of wave-length (or frequency, same thing just inverse),  Fourier Analysis doesn't come into play.  So I am having problem following your line of thought here.

I chose very long wave length is to illustrate the point of the philosophical difficulty in scaling mathematics to real-life.  A frequency of 1/13.7 BillionYears is just a number for mathematics, you can plug that into any equation.  But the universe is just 13.6 billion years old.  So can you have an oscillation cycle that lasts 13.7 billion years?  So whether such photon can exist or not is philosophical.  (Again, forgoing the complexity of inflation and universe expansion.)

I am also unsure of your example Earth-Sun system analogy in reference to the very low energy discussion.

Do you scale Planck's constant with it?  The issue is not the absolute size of the quantum, rather, the issue is how close is it to Planck's constant.  The closer to Planck's constant, the bigger the uncertainty.  At the size of the solar system, uncertainty due to the uncertainty principle is not even in the scale of rounding errors.  So even if you are talking about a single particle of graviton, you are talking a huge amount of energy far exceed the scale of uncertainly.  There would be no chance of it hiding within the grey area covered by the uncertainty principle.

That said, much much much bigger "borrowing from uncertainty" came into play before - namely the big bang.
 

Offline T3sl4co1l

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Re: How does the electron make a photon in an antenna?
« Reply #29 on: February 13, 2017, 05:06:31 am »
You are thinking along a line that I am not in sync with.  I am unsure how Fourier Analysis comes to play in your line of thinking.   Along the line of energy being a function of wave-length (or frequency, same thing just inverse),  Fourier Analysis doesn't come into play.  So I am having problem following your line of thought here.

Simply highlighting a fallacy of analysis -- at the simplest level, wave-particle duality is identical to time-frequency duality, and the Heisenberg uncertainty principle is simply the relationship between time-domain bounds and frequency-domain bounds.

That is to say -- the universe is more than happy to allow conditions which, within the scope of a Fourier analysis, should count as a photon (say) with a frequency lower than the age of the universe -- that you consider it as such, is merely your fault of applying an analysis that can only resolve things in terms of frequency, and not in other, more suitable terms. :)  (What those terms are, is an exercise for the student, naturally...)

Now, I'm not sure under what conditions you could ever observe such a thing -- :-DD -- but the takeaway point is, use what analysis is most suitable; Fourier analysis falls apart at "DC", where "DC" is merely however long you wish to look at a signal.

Remember also that Fourier analysis (of our simplest, most favorite functions) is symmetrical: the transform must exist for all time, including all negative time and all positive time.  Neither condition of which can be properly met in a finite-time universe!

Fortunately, Fourier transforms fail softly, so we can dirty up our graphs by bounding them within realistic windows.  We remind ourselves of the extents and limitations of our experiments, and perhaps we choose to exclude that pesky DC term from our subsequent analysis because it's an artifact of the transformation, or measurement.  But remembering, also, that we should contemplate its origin, in case it's really there (the universe has a net charge..?!). :)

In QM, Fourier isn't quite right, because QM isn't pure signal analysis.  But the duality phenomenon is common to all wave systems, and so we should naturally expect to see similar concerns arise in all wave systems.

Basically, for QM, you might find it's better to use a time-domain analysis than a frequency-domain analysis, in such a case.  The frequency-domain (or momentum, or..) results arise from eigenvalues of the solved equation; the eigenfunctions give their spacial distribution.  This works nicely when the problem is static (like the energy levels of a particle in a box, or the hydrogen atom), for which you expect a frequency analysis to work nicely (because it's not otherwise changing over time!).

The choice of analysis, is a convenience to the solver -- consider solving for the time-domain waveforms of an RLC circuit (analytically, not with SPICE ;) ), versus with Fourier analysis.  Once you've trudged through all the awful integrals and found your series of exponential functions, you still can't do much with it because if you want to change the input signal, you have to integrate the damn thing again (output signal = convolution of input signal with impulse response).

AC steady state analysis is doing the whole thing in the Fourier domain, though they don't often tell you that that's what you're doing (hey, it's only the second course in the average EE curriculum).

If nothing's changing over time, of course the two approaches converge; we don't even bother writing the integrals nor the reactances, and the whole thing reduces to DC resistor networks: EE101. ;)

Conversely, Fourier analysis won't help you much with a switching supply circuit -- it's bad enough if the duty cycle is varying over time, but if the frequency is varying as well, you're pretty much screwed. :P  Combined with the nonlinear parameters in a real semiconductor circuit, you're better off using energy arguments and events in time. 

Quote
I am also unsure of your example Earth-Sun system analogy in reference to the very low energy discussion.

Do you scale Planck's constant with it?

Nope, as stock.  Basically take the already-solved hydrogen atom equations, and plop in the correct potential (gravitational vs. Coulomb) and masses.

Quote
The issue is not the absolute size of the quantum, rather, the issue is how close is it to Planck's constant.  The closer to Planck's constant, the bigger the uncertainty.  At the size of the solar system, uncertainty due to the uncertainty principle is not even in the scale of rounding errors.  So even if you are talking about a single particle of graviton, you are talking a huge amount of energy far exceed the scale of uncertainly.  There would be no chance of it hiding within the grey area covered by the uncertainty principle.

That said, much much much bigger "borrowing from uncertainty" came into play before - namely the big bang.

The nice thing about this example is, it illustrates the problem of limited analysis over a much more human time scale than the Big Bang.  Since, as you say, the math doesn't care, you can simply plug in any number -- why not ask the same questions of an atom 2 A.U. across, or 100pm across?  :)

For the Earth-Sun system, some pertinent questions are:
- It is well studied that the Earth's orbit affects the orbits of its neighbors.  If it radiates so little energy, how can this be?  (Even given the Earth has been in its orbit for 4.5 billion years.)  Surely, so little energy cannot distort space-time enough to do that!
- If radiation is given off at the rate it seems to be given off at (and, well, why wouldn't it?), then how is it that we can seemingly measure its effect on a far shorter time scale?  (The answer to this, on the truly quantum scale, is a field of active development, actually -- the mechanism is analogous to using parametric sensors to measure the presence of a signal, altering the signal slightly in the process but not absorbing it whole.)

If you don't know/remember the QM to work it out by hand (it's a good study to work through, I encourage you to give it a try if you can!), the answer is here.
SPOILER: http://www.physicspages.com/2013/01/15/earth-sun-system-as-a-quantum-atom/

Tim
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Bringing a project to life?  Send me a message!
 

Offline Rick Law

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Re: How does the electron make a photon in an antenna?
« Reply #30 on: February 13, 2017, 06:44:38 am »
...
That is to say -- the universe is more than happy to allow conditions which, within the scope of a Fourier analysis, should count as a photon (say) with a frequency lower than the age of the universe -- that you consider it as such, is merely your fault of applying an analysis that can only resolve things in terms of frequency, and not in other, more suitable terms.
...
Quote
I am also unsure of your example Earth-Sun system analogy in reference to the very low energy discussion.

Do you scale Planck's constant with it?

Nope, as stock.  Basically take the already-solved hydrogen atom equations, and plop in the correct potential (gravitational vs. Coulomb) and masses.

Quote
The issue is not the absolute size of the quantum, rather, the issue is how close is it to Planck's constant.  The closer to Planck's constant, the bigger the uncertainty.  At the size of the solar system, uncertainty due to the uncertainty principle is not even in the scale of rounding errors.  So even if you are talking about a single particle of graviton, you are talking a huge amount of energy far exceed the scale of uncertainly.  There would be no chance of it hiding within the grey area covered by the uncertainty principle.

That said, much much much bigger "borrowing from uncertainty" came into play before - namely the big bang.

The nice thing about this example is, it illustrates the problem of limited analysis over a much more human time scale than the Big Bang.  Since, as you say, the math doesn't care, you can simply plug in any number -- why not ask the same questions of an atom 2 A.U. across, or 100pm across?  :)
...
...
Tim

[This was going to be a quick reply, but to be clear I had to add more and more words.  Hope I don't lose you half way.]

I see where we are out of sync.  You are talking about adopting a different (in your word, "more appropriate") methodology of doing analysis.

Whereas, I am simply describing the inability of scaling math to the limit (not in the mathematical sense of limit but in the practical sense) and consider that as applicable to the physical world.

Let me ask you a question (and it is not a trick question, question just to discern your philosophy): This is similar to but is not the silly question.  Let me dispatch with the silly question first so you know is not this: "if a tree falls in a forest and no one is around, does it make a sound."   The forest is a known condition, the physical laws that govern are known.  The tree will impact something and air molecule will vibrate.  Such vibration is called sound so as a result, yes a sound will be made.

So however similar sound the question, this is a real non-silly question: If a particle is mathematically evaluated not to decade in 10E40 years, and I am not talking half-life, I am talking non-probabilistic hard number of 10E40 years.  The question is: does it decade?   Based on current rate of acceleration of expansion, in 10E30 years or so our universe is energy-dead.   At 10E40 years, the particle has a life longer than the expected life of the universe.  At 10E30 years, there would be no energy to use, no matter visible, no matter interaction since there is no matter left within reachable distance (rate of expansion>c)...

The universe in 10E40 years is an unknown condition.  We don't know what physical laws will apply.  We also know that the act of measurement itself affects what we are measuring...

I would not consider the question "does it decade" a question of physics but instead it is a philosophical question.  In other words, it is outside the domain I think of as Physics since it is in a domain where I don't know if our physical laws apply.

We can hypothesize that our known physical laws still apply at the extremes, but we don't know.  In fact, we don't know even today how our physical laws apply beyond near our own solar system.  96% of the universe is dark matter and dark energy.  We don't even know what they are let alone form laws that may apply to that extend.  So, I tend to think physics as what we can have a prayer of forming laws that we can describe and perhaps prove directly or indirectly.

So, bottom line (in my view), you can choose whatever way you choose to look at it, all we can say is "well, we think it does, but we really don't know for sure."

If you say the particle WILL decade in 10E40 years, I can accept that too.  You have more faith in the robustness of our physical laws than I do.
 

Offline John Heath

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Re: How does the electron make a photon in an antenna?
« Reply #31 on: February 13, 2017, 07:25:53 am »
Great thread guys , enjoying it. Especially the old Feynman flicks. Would like to add that there is an advantage to using the photon model vs EM wave model for RF energy leaving an antenna. With a photon you can model the exact energy and shape of each individual photon. For a 100 MHz transmitter the energy of each photon is E=hv or energy in joules = Planck's constant times frequency for each photon. Too small to measure as h = 6.2 * 10^-34 , ouch . However the diameter of the photon is also set by frequency making a 100 MHz RF photon a 3 foot beach ball. This is useful information as it tells one ahead of time the limit of resolution of a 100MHz photon in a intuitive way by just thinking of it as a 3 foot beach ball. You can tell if there is a building by bouncing it off it but the doors and windows would be hard to make out armed only with a 3 foot beach ball. 10 GHz on the other hand is a gulf ball photon so the windows and doors can be resolved. In short the photon is a nice way to think of RF as it paints an intuitive picture of how the RF will act when it reflects of objects.
 

Offline RoGeorge

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Re: How does the electron make a photon in an antenna?
« Reply #32 on: February 13, 2017, 08:27:01 am »
Quote
How does the electron make a photon in an antenna?

The question is misleading. First, because it suggest an answer, as in the electron being the one to be blamed for the produced photons.
AFAIK, an electron does not "make" or "release" photons. What we call a photon is, in fact, a pack of energy with some specific behaviors, or properties.

Let's try to visualize a photon:
- There are fields and particles.
- What we call a particle is just a ripple, a perturbation in a field. A photon would be a ripple in the electromagnetic field.

To picture this, imagine the surface of a still lake. The surface of the lake is "The Field". Now, let's make "A Particle". That would be to make some ripples, or waves, on that perfect mirror of the still lake.

See that group of ripples moving away from the central drop? That whole group of ripples is a pack of energy. Let's call it a Riplon:




Can a drop of water make a Riplon? Apparently it can. Look, it's a fact:




But it's the drop of water who's making the Riplon? Or it's the rain?
Can an insect make a Riplon? Yes, it can:




Now, who's making the Riplon? It's the insect, or it is the surface tension of the water, or it is just the gravity? Causality seems to be an empirical concept and a rabbit whole, so it's hard to say "who's fault" is this or that. I guess the safest one can say is that some energy was transferred to the surface of the lake in the form of a Riplon. We will end here our analogy between photons and water waves. Analogies might help to "visualize" abstract concepts, but this can be dangerous in the long run. A mathematical representation serves better.

In conclusion, electrons "jumping from one energy state to another" (whatever the hell that could mean) it's not the only way to "produce" photons. Ripples in the electromagnetic field (AKA photons) can be made in many ways, and causality (AKA who's making the photons in an antenna) is more of a philosophical concept, because in a mathematical equation, there is no such thing as "variable x is the cause of variable y". It's just an equality, where the x and y terms can be rearranged upon wish, as long as we don't brake the math.

So, who's making a photon in an antenna? One electron jumping from a higher to a lower energy state? No.
I'd rather say it's the "dance of a gazillion of electrons", all moving in the rhythm of the variable electric potential produced by the radio transmitter. That dance produces ripples in the electromagnetic field, and those ripples are what we like to call photons.
« Last Edit: February 13, 2017, 09:40:42 am by RoGeorge »
 
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Offline John Heath

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Re: How does the electron make a photon in an antenna?
« Reply #33 on: February 13, 2017, 03:57:11 pm »
Quote
How does the electron make a photon in an antenna?

The question is misleading. First, because it suggest an answer, as in the electron being the one to be blamed for the produced photons.
AFAIK, an electron does not "make" or "release" photons. What we call a photon is, in fact, a pack of energy with some specific behaviors, or properties.

Let's try to visualize a photon:
- There are fields and particles.
- What we call a particle is just a ripple, a perturbation in a field. A photon would be a ripple in the electromagnetic field.


To picture this, imagine the surface of a still lake. The surface of the lake is "The Field". Now, let's make "A Particle". That would be to make some ripples, or waves, on that perfect mirror of the still lake.

See that group of ripples moving away from the central drop? That whole group of ripples is a pack of energy. Let's call it a Riplon:




Can a drop of water make a Riplon? Apparently it can. Look, it's a fact:




But it's the drop of water who's making the Riplon? Or it's the rain?
Can an insect make a Riplon? Yes, it can:




Now, who's making the Riplon? It's the insect, or it is the surface tension of the water, or it is just the gravity? Causality seems to be an empirical concept and a rabbit whole, so it's hard to say "who's fault" is this or that. I guess the safest one can say is that some energy was transferred to the surface of the lake in the form of a Riplon. We will end here our analogy between photons and water waves. Analogies might help to "visualize" abstract concepts, but this can be dangerous in the long run. A mathematical representation serves better.

In conclusion, electrons "jumping from one energy state to another" (whatever the hell that could mean) it's not the only way to "produce" photons. Ripples in the electromagnetic field (AKA photons) can be made in many ways, and causality (AKA who's making the photons in an antenna) is more of a philosophical concept, because in a mathematical equation, there is no such thing as "variable x is the cause of variable y". It's just an equality, where the x and y terms can be rearranged upon wish, as long as we don't brake the math.

So, who's making a photon in an antenna? One electron jumping from a higher to a lower energy state? No.
I'd rather say it's the "dance of a gazillion of electrons", all moving in the rhythm of the variable electric potential produced by the radio transmitter. That dance produces ripples in the electromagnetic field, and those ripples are what we like to call photons.
-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
The second picture with blue water shows the drop moving up not down. This is a nice way to demonstrate Heavyside's reflection caused by the degree of impedance difference between falling through air and falling through water. The drop of water experience an impedance difference leading to a reflection as seen with a portion of the water drop being reflected back up. In a RF coaxes cable it is only 1 dimension however with the water drop the reflective action can be seen in full 3 dimensions. A full 3 dimensional demonstration of the dynamics of a sudden change in impedance. That is cool.

I would add that waves leaving of the surface of the water outward become smaller and smaller through distance but photons do not as seen in the photoelectric effect. Green light will force an electron off metal but red light no matter how strong will not. The waves are quantified into energy packets of E=hv for reasons unknown.
 
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Offline RoGeorge

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Re: How does the electron make a photon in an antenna?
« Reply #34 on: February 13, 2017, 05:01:03 pm »
...A full 3 dimensional demonstration of the dynamics of a sudden change in impedance. That is cool.

You, Sir, have made my day!  :-+

Offline T3sl4co1l

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Re: How does the electron make a photon in an antenna?
« Reply #35 on: February 13, 2017, 07:31:31 pm »
[This was going to be a quick reply, but to be clear I had to add more and more words.  Hope I don't lose you half way.]

I appreciate the long form discussion!  So many lengthy posts go unchallenged, or probably even unread.  Few people seem to appreciate that 140 characters simply cannot communicate the complexity, detail and subtlety that encompasses... most everything!

Quote
I see where we are out of sync.  You are talking about adopting a different (in your word, "more appropriate") methodology of doing analysis.

Whereas, I am simply describing the inability of scaling math to the limit (not in the mathematical sense of limit but in the practical sense) and consider that as applicable to the physical world.

So what you're getting at is, not just the simple matter of analysis (which I see more often, hence my focus on that), but more fundamental.

Quote
So however similar sound the question, this is a real non-silly question: If a particle is mathematically evaluated not to decade in 10E40 years, and I am not talking half-life, I am talking non-probabilistic hard number of 10E40 years.  The question is: does it decade?   Based on current rate of acceleration of expansion, in 10E30 years or so our universe is energy-dead.   At 10E40 years, the particle has a life longer than the expected life of the universe.  At 10E30 years, there would be no energy to use, no matter visible, no matter interaction since there is no matter left within reachable distance (rate of expansion>c)...

NB: decay?

Some theories suppose the proton will decay, in some unimaginably long time scale; still, for the vast number of them in the universe, perhaps one could stand the chance of observing such an event, within that time scale.

But that's a probabilistic decay; you mean a fixed, known decay time?  There is no mechanism to conceive of such a thing, within the Standard Model as we know it.  This supposed particle is very alien indeed, and making any hard statements about, say, its mass or energy or internal state, is quite undefined!

So, proposing a simple thought experiment about one property, really strains the understanding of many more properties within present theory!  I'm not sure that this was the intended effect...

Quote
The universe in 10E40 years is an unknown condition.  We don't know what physical laws will apply.  We also know that the act of measurement itself affects what we are measuring...

I would not consider the question "does it decade" a question of physics but instead it is a philosophical question.  In other words, it is outside the domain I think of as Physics since it is in a domain where I don't know if our physical laws apply.

Yeah, physics is a very practical science: if you can measure it, if it comes out of the math of established (read: measurement-supported) theory, there's something there.  If you can't measure it, maybe it's there, maybe not, it doesn't really matter in that case, does it?

(This is sort of the problem that string theory has: my limited understanding of it is, it's a superset of earlier theory, so it suffers from the problem of being too general -- indeed, that many of the new variables can't be measured.  More of a framework than a physical theory.  When we finally obtain new experimental data, we have plenty of choice of framework to fit them into -- but for now, it's kind of just a plaything.)

So, given that it's unlikely the study of physics (at least, as we know it) will persist for 10^30 years, say -- I would be more than happy to answer "no, it doesn't decay*".  The implication is: *for all intents and purposes.

But if you have reason to believe it decays (such as the theories which suggest proton decay, if proven), then who knows, you might measure it some day, and you can say "yeah sure, why not?" instead.

Or, if you've somehow determined that a particle will go off like a time bomb at exactly 1e40, you probably have a very good reason to believe that, in which case the "Yes!" is as resounding as the strength of the theory behind that result, and the sigmas of its supporting data!

And, since these things always cut both ways -- whatever mechanism you've discovered, that gave rise to this timebomb-ino, as it were -- likely has profound effects on other particles or relationships within the theory!  If this one particle is deterministic, there must be a companion that's also deterministic?  Or is there even less determinism (more chaos, or randomness) in other events?


Besides proton decay, a more practical example might be double-beta decay: it was hypothesized for some time, and expected to be very unlikely, but it has in fact been observed in a number of quite long half-life nuclides.  This is, of course, a probabilistic decay, not a time bomb, which makes it practical to measure, despite the half-life being on the order of (age of universe)^2.

Is it useful to know?  Yes -- it validates our theories about the weak force.  Is it practical, in and of itself?  Probably not; anything with a half-life that long is pretty well "stable" for all intents and purposes, outside maybe a very sensitive detector (don't accidentally get any 100Mo in your neutrino detector tank!).

Quote
We can hypothesize that our known physical laws still apply at the extremes, but we don't know.  In fact, we don't know even today how our physical laws apply beyond near our own solar system.  96% of the universe is dark matter and dark energy.  We don't even know what they are let alone form laws that may apply to that extend.  So, I tend to think physics as what we can have a prayer of forming laws that we can describe and perhaps prove directly or indirectly.

So, bottom line (in my view), you can choose whatever way you choose to look at it, all we can say is "well, we think it does, but we really don't know for sure."

If you say the particle WILL decade in 10E40 years, I can accept that too.  You have more faith in the robustness of our physical laws than I do.

So to sum up, it seems like we're in agreement.  Physics, as a field of study, is only as certain of its results as it can measure.  No need for faith, just soft fuzzy error bars!

Tim
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Offline T3sl4co1l

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Re: How does the electron make a photon in an antenna?
« Reply #36 on: February 13, 2017, 08:03:26 pm »
So, who's making a photon in an antenna? One electron jumping from a higher to a lower energy state? No.

BTW, note that the antenna is only a guide for the EM field.  The EM field extends from within the cable (or waveguide, or whatever), through the radiating structure, out into space.  If you wish to use photons in your reasoning, then they can be present in all these locations.

Ultimate photon interactions (creation and absorption) -- not just scattering, occurs at sources and sinks.  Really, anything with resistance, or equivalent resistance.

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Offline T3sl4co1l

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Re: How does the electron make a photon in an antenna?
« Reply #37 on: February 13, 2017, 10:03:18 pm »
I would add that waves leaving of the surface of the water outward become smaller and smaller through distance but photons do not as seen in the photoelectric effect. Green light will force an electron off metal but red light no matter how strong will not. The waves are quantified into energy packets of E=hv for reasons unknown.

Honestly: the photoelectric effect is far more complex (surface physics, yay!..) than its traditional origin story suggests.  It's remarkable to me that any research on the subject could be eligible for a Nobel prize!  Though to be fair, it's generally been said that Einstein was honored, not so much for that particular paper (On The Photoelectric Effect), but more generally, because, well, Einstein.

Familiar examples are phototubes, with sensitivity curves seemingly much longer in wavelength than the materials used would suggest.  (But, they use compounds, selected for wideband sensitivity, which might not obey the same rules as pure metals.)

But also of note -- EVERYTHING is nonlinear, to some degree.  Some crystals exhibit multi-photon mixing (essentially: upconversion due to a locally-quadratic field response).  Such phenomena are uncommon, both because of material properties (only specialized crystals are chosen for applications), and because of small probabilities (the likelihood of two photons being in the same location, phase and polarization, is small; however, laser light is quite luminous and coherent, so the effect can be usefully applied).

A more banal example: simply heating something, with any radiation source sufficiently intense.  Induction heating achieves up-conversion from 60Hz (~coherent) to ~400THz ("white" noise -- oh, wait, white indeed!), and that's just in ordinary industrial steelmaking! :)

In the extreme, space itself is nonlinear: the kugelblitz is a black hole, created from a sufficient density of pure light at a point.  The mass of a black hole, summoned purely from energy!  Of course, you need a truly astronomical amount of light, converging on an exact point, which might not be possible given optical limitations; but there's nothing prohibiting it on the lowest physical levels, that we know of.

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Offline John Heath

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Re: How does the electron make a photon in an antenna?
« Reply #38 on: February 14, 2017, 01:04:18 pm »
I would add that waves leaving of the surface of the water outward become smaller and smaller through distance but photons do not as seen in the photoelectric effect. Green light will force an electron off metal but red light no matter how strong will not. The waves are quantified into energy packets of E=hv for reasons unknown.

Honestly: the photoelectric effect is far more complex (surface physics, yay!..) than its traditional origin story suggests.  It's remarkable to me that any research on the subject could be eligible for a Nobel prize!  Though to be fair, it's generally been said that Einstein was honored, not so much for that particular paper (On The Photoelectric Effect), but more generally, because, well, Einstein.

Familiar examples are phototubes, with sensitivity curves seemingly much longer in wavelength than the materials used would suggest.  (But, they use compounds, selected for wideband sensitivity, which might not obey the same rules as pure metals.)

But also of note -- EVERYTHING is nonlinear, to some degree.  Some crystals exhibit multi-photon mixing (essentially: upconversion due to a locally-quadratic field response).  Such phenomena are uncommon, both because of material properties (only specialized crystals are chosen for applications), and because of small probabilities (the likelihood of two photons being in the same location, phase and polarization, is small; however, laser light is quite luminous and coherent, so the effect can be usefully applied).

A more banal example: simply heating something, with any radiation source sufficiently intense.  Induction heating achieves up-conversion from 60Hz (~coherent) to ~400THz ("white" noise -- oh, wait, white indeed!), and that's just in ordinary industrial steelmaking! :)

In the extreme, space itself is nonlinear: the kugelblitz is a black hole, created from a sufficient density of pure light at a point.  The mass of a black hole, summoned purely from energy!  Of course, you need a truly astronomical amount of light, converging on an exact point, which might not be possible given optical limitations; but there's nothing prohibiting it on the lowest physical levels, that we know of.

Tim

Your response makes you the qualified to do a photoelectric test as you have the understanding that the real world is not the elegant picture painted by arm chair physics. Why would you assume the minds of the men and women 100 years ago to be different. The test was done with care to control third variables.
 

Offline aetherist

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Re: How does the electron make a photon in an antenna?
« Reply #39 on: January 31, 2023, 11:38:00 am »
I was taught, and I just did some google searching because its been so long, is that the actual radio wave propagation (In the common communication ranges) is purely electromagnetic and electrostatic based on Maxwells equations. They propagate independent of photons in a manner dictated by QED. The photons are just an emitted byproduct generated by the intrinsic energy of the signal itself. In other words an antenna is producing the EM field, but any photons that are being released are not the principal emission and nowhere near the frequency of the base band, or in other words the antenna does not emit electrons or photons as a primary mode, only fields. It emits no more photons than any other piece of metal with that amount of energy going on about it. Things get a bit more complicated as you go higher up in frequency though. Via QED more "Loss" of energy is expressed via photons until you have an infrared light source.  Somebody please correct me. Like I said. It has been a very long time.
Wow - i just then read this -- i thort that this forum (& modern electricity & radio) was populated with idiots -- but calexanian is a breath of fresh air -- yes, radio waves are not photons -- radio waves are em radiation.
And, re the OP....
 ....................We know how an electron makes a photon in an LED. It jump from a higher orbital to a lower one emitting a photon. But in an antenna its occupying the same valance just with a different nucleus each hop.  How is the energy transferred to a radio frequency photon?.................
.... i reckon that electricity aint due to the movement of electrons -- neither in the Cu nor on the Cu.
« Last Edit: January 31, 2023, 09:21:22 pm by aetherist »
 

Offline CatalinaWOW

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Re: How does the electron make a photon in an antenna?
« Reply #40 on: January 31, 2023, 04:30:17 pm »
Another example that may make the apparent contradiction less jarring.  Take a simple silicon diode detector.  When a photon  exceeds the band gap energy it can boost an electron across the junction, resulting in current.  The band gap is large enough in silicon to require photons of visible wavelength or shorter to generate these currents.   Other semiconductors, InSb for example allow the lower energy photons of roughly 5 micrometer infrared radiation to boost electrons across the band gap.  But at room temperature thermal electrons also cross this bandgap in copious amounts making it impossible to notice the photon induced current.  So these detectors are cooled to cryogenic temperatures typically 150 K or less to make it possible to detect the photocurrent.  When wavelength is increased to the far infrared further cooling is required, down to a few tens of Kelvin's.  Radio photons have energies 100,000 times smaller than LWIR photons, and even if a material with appropriate band gap existed would require cooling to tiny fractions of a Kelvin to avoid thermal noise.

Both wave descriptions and photon descriptions are tools that work to describe physical phenomenon.  As far as we can tell they work at all wavelengths.  The photon tool just isn't very helpful at radio wavelengths and energies.  The increment of energy from a single radio photon is swamped by other sources of energy. 
« Last Edit: January 31, 2023, 04:36:33 pm by CatalinaWOW »
 

Offline TimFox

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Re: How does the electron make a photon in an antenna?
« Reply #41 on: January 31, 2023, 06:59:24 pm »
The fundamental reason is that charge radiates when it is accelerated.
In an antenna, the oscillating current accelerates the conducting charge along the wire.
In a synchrotron, charged particles are accelerated centripetally (by appropriated magnetic fields) to follow an approximately circular orbit.
 
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Re: How does the electron make a photon in an antenna?
« Reply #42 on: January 31, 2023, 07:16:38 pm »
Since time doesn't exist, neither does acceleration. So accelerated particles are just for the birds. ;D
 

Offline aetherist

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Re: How does the electron make a photon in an antenna?
« Reply #43 on: January 31, 2023, 09:09:09 pm »
Another example that may make the apparent contradiction less jarring.  Take a simple silicon diode detector.  When a photon  exceeds the band gap energy it can boost an electron across the junction, resulting in current.  The band gap is large enough in silicon to require photons of visible wavelength or shorter to generate these currents.   Other semiconductors, InSb for example allow the lower energy photons of roughly 5 micrometer infrared radiation to boost electrons across the band gap.  But at room temperature thermal electrons also cross this bandgap in copious amounts making it impossible to notice the photon induced current.  So these detectors are cooled to cryogenic temperatures typically 150 K or less to make it possible to detect the photocurrent.  When wavelength is increased to the far infrared further cooling is required, down to a few tens of Kelvin's.  Radio photons have energies 100,000 times smaller than LWIR photons (Long-wave infrared or LWIR) , and even if a material with appropriate band gap existed would require cooling to tiny fractions of a Kelvin to avoid thermal noise.

Both wave descriptions and photon descriptions are tools that work to describe physical phenomenon.  As far as we can tell they work at all wavelengths.  The photon tool just isn't very helpful at radio wavelengths and energies.  The increment of energy from a single radio photon is swamped by other sources of energy.
Interesting. That kind of stuff is over my head. But if u are referring to my idea (that radio waves are not photons) then i am not sure how my idea relates to a silicon diode detector.
1. I guess that a silicon diode detector can detect photons. I don’t know whether a silicon diode detector can detect radio waves (radio waves are not photons). Can they?

2. U mention radio photons. I reckon that radio photons do not exist. I reckon that radio is due to em radiation (not photons), in particular radio is due to a changing em radiation, with a wavelength(s).
Hence if a silicon diode detector can detect radio waves then that detection is due to the action of em radiation not photons.

3. U mention thermal electrons. I reckon that electricity on a wire is actually elekticity &  is due to photons (what i call elektons) propagating along the surface of the Cu, not due to any movement of electrons in or on the Cu.
I reckon that electrons do not orbit the nucleus of an atom – elektrons orbit the nucleus. In other words an atom is elekticity orbiting a nucleus.
However, i do believe that electrons exist – but these electrons are photons that have formed a loop by biting their own tails (or in some cases other tails).
In any case i don’t believe that there is any such thing as a thermal electron. But praps u mean an electron that has been created by a thermal photon (i might be ok with that).
« Last Edit: April 29, 2023, 01:41:26 am by aetherist »
 

Online IanB

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Re: How does the electron make a photon in an antenna?
« Reply #44 on: January 31, 2023, 09:36:29 pm »
2. U mention radio photons. I reckon that radio photons do not exist. I reckon that radio is due to em radiation (not photons), in particular radio is due to a changing em radiation, with a wavelength(s).
Hence if a silicon diode detector can detect radio waves then that detection is due to the action of em radiation not photons.

3. U mention thermal electrons. I reckon that electricity on a wire is due to photons (what i call electons) propagating along the surface of the Cu, not due to any movement of electrons in or on the Cu.
I reckon that electrons do not orbit the nucleus of an atom – electons orbit the nucleus. In other words an atom is electricity orbiting a nucleus.
However, i do believe that electrons exist – but these electrons are photons that have formed a loop by biting their own tails (or in some cases other tails).
In any case i don’t believe that there is any such thing as a thermal electron. But praps u mean an electron that has been created by a thermal photon (i might be ok with that).

Why do you reckon all these things? Have you spent the last 200 or 300 years doing careful scientific experiments, evaluating the results and carefully constructing mathematical equations that show how to predict the results of new experiments? Have you been alive the the last 200 years to be able to do this?

Otherwise, it would seem you are simply talking through your hat.

That kind of stuff is over my head.

If that kind of stuff is over your head, why to you feel qualified to expound upon it?
 
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Online Sal Ammoniac

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Re: How does the electron make a photon in an antenna?
« Reply #45 on: January 31, 2023, 09:41:39 pm »
Keep in mind that QED describes what happens, not how it happens. There's nothing in the theory, or any other theory, that describes how an electron converts excess energy into a photon, just that it does.
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Offline aetherist

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Re: How does the electron make a photon in an antenna?
« Reply #46 on: January 31, 2023, 10:10:25 pm »
I was taught, and I just did some google searching because its been so long, is that the actual radio wave propagation (In the common communication ranges) is purely electromagnetic and electrostatic based on Maxwells equations. They propagate independent of photons in a manner dictated by QED. The photons are just an emitted byproduct generated by the intrinsic energy of the signal itself. In other words an antenna is producing the EM field, but any photons that are being released are not the principal emission and nowhere near the frequency of the base band, or in other words the antenna does not emit electrons or photons as a primary mode, only fields. It emits no more photons than any other piece of metal with that amount of energy going on about it. Things get a bit more complicated as you go higher up in frequency though. Via QED more "Loss" of energy is expressed via photons until you have an infrared light source.  Somebody please correct me. Like I said. It has been a very long time.
Wow - i just then read this -- i thort that this forum (& modern electricity & radio) was populated with idiots -- but calexanian is a breath of fresh air -- yes, radio waves are not photons -- radio waves are em radiation.
:bullshit:

Radio waves are *made* of photons and those photons *do* have wavelengths that match the wavelength of the bulk RF emission. Bulk RF emission (radio waves) are packets of coherent photons. You can filter them down with attenuators until they're at the single-photon-per-second level, and those photons will mostly have wavelengths around the same wavelength as the bulk RF signal.

To elaborate on the negative: if you want to describe the output of a 2.4 GHz antenna connected to a 2.4 GHz oscillator, it is not "photons with a frequency of 1.3 GHz" or "photons with a frequency of 720 GHz" or something like that. It is photons with a frequency distribution centered around 2.4 GHz.

Otherwise, there would be no point engineering passive nanostructures with length scales around the relevant wavelength as single-photon emitters: https://pubs.acs.org/doi/10.1021/acsphotonics.7b00061

Just because you can knock a hammer (or short a charged capacitor) on a bell tuned for 200 Hz (or an antenna tuned for 1 MHz) and make it ring, for the briefest of moments, at 20 Hz (or 100 kHz) does not mean the bell's (antenna's) normal mode of operation is to emit acoustic waves (photons) of random frequencies. If you let the micro-scale (electron-scale) dynamics settle for a bit, the device will ring at whichever frequency it was tuned for.

Quote
And, re the OP -- electricity aint due to the movement of electrons -- neither in the Cu nor on the Cu.
A statement this vague yet absolutist qualifies as trolling. One could argue either side because the framing is nonsensical. Are you even talking about RF or just bulk electricity? Are you talking about the electron-scale dynamics that we refer to generally as "electricity," or are you talking about sparks and smoke and magic shows? Any way you pick, electrons are moving!

Unless you'd like to show us how to create radio emissions and/or "electricity" from nothing but crystalline structures (say, quartz) with "fixed" positive and negative charges without applying any heat, mechanical shock, acoustic waves, photons, or mechanical stress, and without waiting for their atoms to decay. Good luck  :-+
I accept your challenge. My answer will take me about 1 second of thinking, & a few minutes of typing. It goes like this.
In a crystal atom elektrons (photons) orbit (hug) the nucleus, ie elekticity orbits the nucleus, ie a shock etc can dislodge an elerton, & the elekton can then propagate along (hug) a conductor (eg a Cu wire), in which case we have elekticity.

What is vague about my saying that elekticity on a wire is due to elektons propagating along (hugging) the surface of the Cu?
What is vague about my saying that an atom is elekticity orbiting a nucleus?

Veritasium & others have said that electricity on a wire is not due to the movement of electrons -- & i agree (with that simple statement) – but then Veritasium & Co launch into some krapp about the Poynting Vector or Poynting Field being the carryer of the electrical energy (No).
So, u should firstly have a fight with Veritasium & Co -- & if u win that then come & see me (i have truth & facts in my corner).

But, back to the main topic re radio not being photons.
I reckon that radio is due to em radiation. U say that radio is due to photons.
Photons have a natural frequency. Em radiation duznt have a natural frequency – in radio it has a forced frequency.
Em radiation is emitted by every photon, ie it is emitted by elektons propagating upndown a transmitting antenna.
The em radiation emitted by elektons will mimic the elektons movement upndown theantenna.
A 1MHz antenna emits a 1 MHz em radiation wave, not a 1MHz photon.

In the receiver the em radiation will excite the elektons already on the Cu -- & will create a mini-version of the elekton current on the transmitter.
« Last Edit: April 29, 2023, 01:45:32 am by aetherist »
 

Offline aetherist

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Re: How does the electron make a photon in an antenna?
« Reply #47 on: January 31, 2023, 10:18:11 pm »
Uh-oh. Seems I've taken the opinion of someone who thinks electrons are photons seriously! Shame on me.

If I disagree with aetherist, he gets off on people reading and unavoidably amplifying his fantasy takes. If I ignore his posts, readers have to think and judge whether or not his claims are accurate for themselves, possibly without evidence, possibly spreading his fantasies to others. Never would have thought I'd run into Facebook-grade social dynamics on an engineering forum, but here we are. ::)

I'll just take this opportunity to plug Griffiths' QM instead: https://www.fisica.net/mecanica-quantica/Griffiths%20-%20Introduction%20to%20quantum%20mechanics.pdf
Does Griffiths tell us what electricity is, & what radio is?
 

Offline aetherist

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Re: How does the electron make a photon in an antenna?
« Reply #48 on: January 31, 2023, 10:33:23 pm »
The fundamental reason is that charge radiates when it is accelerated.
In an antenna, the oscillating current accelerates the conducting charge along the wire.
In a synchrotron, charged particles are accelerated centripetally (by appropriated magnetic fields) to follow an approximately circular orbit.
NO. Charge duznt radiate when accelerated. Charge radiates all of the time. But, when a charge is at rest then the radiation is impotent. When charge is accelerated the radiation become potent, ie it can then have effect, ie it can then produce a force in some instances.

NO. In a transmitting antenna there is no conducting charge. In an antenna the oscillating elektons going upndown the Cu have their own charge, an elekton has a negative charge.

NO. Elektons going upndown an antenna do not ever accelerate. They always propagate at the speed of light. They have one speed. The signal going upndown an antenna will have a say sinusoidal form, & this sinusoidal form is due to the numbers of electons flying in formation, it aint due to any acceleration of the elektons.

In a synchrotron i suppose that electrons etc are made to follow a circular path at high speed (i say made, made to follow, by em radiation) – that sounds ok to me – here the electrons are electrons, they are not elektons, & they are not free photons (they are photons that have formed a loop by biting their own tail).
« Last Edit: April 29, 2023, 01:09:50 am by aetherist »
 

Online Sal Ammoniac

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Re: How does the electron make a photon in an antenna?
« Reply #49 on: January 31, 2023, 10:36:37 pm »
Charge radiates all of the time. But, when a charge is at rest then the radiation is impotent.

Please explain what you mean by "impotent" -- this isn't a very descriptive term when applied to charged particles emitting radiation.
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