Planck Length 1 Detector? Emitter?

[Silly Questions]

Do we have the ability to detect the gamma ray of Planck Length 1? Is there any known phenomenon, natural or man-made, that could emit on Channel 1? How dangerous to man and machine would Channel 1 be / at what density (is that the right term?) would it become dangerous?

Would there be any theoretical advantage to using Channel 1 to send and receive data (i.e. synchronous encoding of bits onto a stream of Planck Lengths, having a clean channel nothing else in the universe uses, etc)? Would there ever be "static" on Channel 1? Could any signal broadcast on Channel 2 (like synchronization pulses) affect / crosstalk / interfere with Channel 1?

Must all Channel 1 broadcasts universe-wide be synchronized with each other? To put this another way can any emission on Channel 1 ever take place less than 1 Planck Time away from any other?

Lastly what do you think about calling the One Planck Wonder "Channel One"? I feel like that's what Douglas Adams would've called it ...

Your source for silly questions,
Silly Questions

 

[hilbert2]

A photon with wavelength similar to Planck length would have energy measured in gigajoules, quite a lot for one particle.

 

[PeroK]

A photon with wavelength similar to Planck length would have energy measured in gigajoules, quite a lot for one particle.

And quite a lot for one detector!

 

[Vanadium 50]

1. There is nothing magic about Planck units.
2. In some reference frame, any photon is above the Planck energy.
3. I have no idea what this "Channel 1" is, even after your message. Please use conventional terminology; it promotes clear communication.

 

[phinds]

I have no idea what this "Channel 1" is

He wants radio stations to use Plank Lengths in place of frequency, so "channel 1" is a transmission with a wavelength of 1 Plank Length. Great idea, huh.

Just for grins, I did a quick calculation and found that my favorite rock station, at 90.5Mhz would be come station (approximately)
1,981,187,500,000,000,000,000,000,000,000,000,000,000

I think I like 90.5 better.

 

[Silly Questions]

I'd like to learn some proper terminology.

If all valid EM waves were enumerated from smallest to longest what would the first wave be named? Does it have a special name or is it just "h=1"?

Does "h" itself have a special name besides "h", like "Planck Coefficient?"

 

[Nugatory]

I'd like to learn some proper terminology.

If all valid EM waves were enumerated from smallest to longest what would the first wave be named? Does it have a special name or is it just "h=1"?

Does "h" itself have a special name besides "h", like "Planck Coefficient?"

What you're trying to do is like trying to enumerate all the real numbers from smallest to largest - there's no such thing.
Electromagnetic waves only appear in classical electrodynamics, where both frequency and wavelength are continuous variables. Thus there is no smallest or longest; no matter how short the wavelength there could be a wave with shorter wavelength, no longer how long the wavelength there could be a wave with longer wavelength, and given two waves of different wavelength there can always be another wave with wavelength in between.

 

[phinds]

I'd like to learn some proper terminology.

If all valid EM waves were enumerated from smallest to longest what would the first wave be named? Does it have a special name or is it just "h=1"?

Does "h" itself have a special name besides "h", like "Planck Coefficient?"

You have succumbed to the common pop-sci fabrication that says nothing can be smaller than a Plank length. Not true. Plank length is a man-made construct just like the foot and the meter and nature doesn't care.

 

[PeterDonis]

You have succumbed to the common pop-sci fabrication that says nothing can be smaller than a Plank length. Not true.

More precisely, not known to be true. But it's not known to be false, either. The smallest length scale we can make measurements of is, IIRC, currently about 17 or 18 orders of magnitude larger than the Planck scale, so we have no evidence either way about what actually happens at that scale.

 

[PeterDonis]

More precisely, not known to be true. But it's not known to be false, either.

A more cogent criticism of the pop sci claims would be that they are not based on any actual theoretical model. There is work being done on theoretical models of this general type, but it is far too early to say whether any of that work will end up leading to something useful. AFAIK none of the models being worked on are as simple as "nothing can be smaller than a Planck length".

 

[Silly Questions]

What is the shortest valid EM wave? If they're quantized and bounded there must be a "shortest"?

"... both frequency and wavelength are continuous variables ..."
That seems to contradict Planck's solution to the "Ultraviolet Catastrophe"?

"Nothing can be shorter ..."
I'm referring specifically and only to EM waves. What else might or might not be valid for what other natural phenomena I have not speculated upon.

 

[PeterDonis]

If they're quantized and bounded

First, "quantized" is not the same thing as "bounded".

Second, as @Nugatory pointed out in post #7, the concept of "EM wave" is a classical concept, not a quantum concept.

 

[PeterDonis]

That seems to contradict Planck's solution to the "Ultraviolet Catastrophe"?

Planck's solution to the ultraviolet catastrophe was not to quantize radiation itself, it was to quantize the emission of radiation by atoms. His solution still assumes a continuous infinity of possible frequencies and wavelengths of radiation; there is no "shortest wavelength" and the possible wavelengths are not discrete. It just puts different rules on how likely a given wavelength is to be emitted by the atoms of the emitting substance, based on the Planck relation between wavelength (or frequency) and the energy required to emit it.

 

[Silly Questions]

"Planck's solution to the ultraviolet catastrophe was not to quantize radiation itself ..."

Whoa. Question answered. Holy smokes. Now it all makes sense.

Of course that brings a new question: is it possible to broadcast a "Planck-invalid" EM wave? Is there any way to emit an EM wave that is impossible for atoms to radiate?

 

[PeterDonis]

is it possible to broadcast a "Planck-invalid" EM wave?

At the very least, not unless you have an emitter that can put out a radiated power, for a very brief time, many orders of magnitude beyond what our current or foreseeable future technology is capable of.

If we ever get to that point, we will be able to test whether this is actually possible. I would recommend not holding your breath.

Is there any way to emit an EM wave that is impossible for atoms to radiate?

What do you think such an EM wave would look like? In other words, what do you think are the limitations on what it is possible for atoms to radiate? (Note carefully what I said about the possible wavelengths in my previous post.)

 

[Silly Questions]

"What would it look like?" Between two consecutive quantized-anythings lies a value invalid for the quantization parameters.

Take "valid emission of radiation x[1] by atom y" and "next valid emission of radiation x[2] by atom y", add them together, divide by 2, set the transmitter to that wavelength, and throw the switch. I'm expecting you to say, "No transmitter is that precise," but if it isn't would it still be broadcasting on that wavelength along with others? That still satisfies the question's truth condition.

 

[vanhees71]

"Planck's solution to the ultraviolet catastrophe was not to quantize radiation itself ..."

Whoa. Question answered. Holy smokes. Now it all makes sense.

Of course that brings a new question: is it possible to broadcast a "Planck-invalid" EM wave? Is there any way to emit an EM wave that is impossible for atoms to radiate?

According to the known physics there's neither a longest nor a shortest possible wave length for electromagnetic waves. The wave length (and thus momentum and energy) of electromagnetic waves are continuous also in the quantum theory (quantum electrodynamics).

The solution to the ultraviolet catastrophe does not need a quantization of the energy of the em. field but the quantization of the possible amounts of energy being exchanged with matter (i.e., charged particles) for each wave mode of definite frequency.

It's much easier to understand in the modern quantum-field theoretical formulation than via the very tricky way Planck used to derive his famous formula.

First start with a finite volume. For simplicity we use a cube of length $L$ and assume periodic boundary conditions, i.e., the vector potential in the radiation gauge (which uniquely describes free em. fields) should obey the conditions $\vec{A}(t,\vec{x}+L \vec{n})=\vec{E}(t,\vec{x})$ . Then all you need to know about the quantized electromagnetic field is that it is equivalent to an infinite set of harmonic oscillators.

Each harmonic oscillator describes a field mode with definite momentum $\vec{p}=\hbar \vec{k}$ and definite helicity $\pm 1$ (i.e., right- and left-circular polarized fields). Such a field mode or "photon" has also definite energy $E(\vec{p})=\hbar \omega=\hbar |\vec{k}|c=|\vec{p}| c$. In our periodic-box setup the momenta are indeed "quantized", i.e., allowed are the wave vectors $\vec{k}=\frac{2 \pi}{L} \vec{n}$ with $\vec{n} \in \mathbb{Z}$. For each allowed $\omega(\vec{k})$ the possible energies of the corresponding harmonic oscillator are $E_j=j \hbar \omega$ with $j \in \{0,1,2,\ldots \}$.

Thus the partition sum for the field modes with definite frequency in the cavity with walls kept at given absolute temperature $T$ in thermal equilibrium is
$$Z(\omega,t)=\sum_{n=0}^{\infty} \exp(-\beta j \hbar \omega)=\frac{1}{1-\exp(-\beta \hbar \omega)},$$
where $\beta=1/(k T)$.

The mean energy in this field mode is
$$\langle E(\vec{k}) \rangle = -\frac{1}{Z} \partial_{\beta} Z=\frac{\hbar \omega}{\exp(\beta \hbar \omega)-1}.$$
Now we take the "thermodynamic limit", i.e., we make the volume very large. Then in any volume of momentum space $\mathrm{d}^3 \vec{p}$ we have $\frac{2 \mathrm{d}^3 \vec{p} L^3}{(2 \pi \hbar)^3}$ field modes (the factor of 2 takes account of the two polarization states). Thus the spectral distribution of the energy is given by
$$\mathrm{d} U=\frac{2 \mathrm{d}^3 \vec{p} L^3}{(2 \pi \hbar)^3} \langle E(\vec{k}) \rangle = \frac{2 \mathrm{d}^3 \vec{k} L^3}{(2 \pi)^3} \frac{\hbar \omega}{\exp(\beta \hbar \omega)-1}.$$
This tells us that since each photon with frequency $\omega$ carries an energy $\hbar \omega$ that photons are massless bosons with the corresponding Bose-Einstein distribution.

A more common form of Planck's Law is to refer it to the energy density per frequency rather than wave-vector interval. Since $\omega^2=c^2 k^2$ we have
$$\omega \mathrm{d} \omega=c^2 k \mathrm{d} k\; \Rightarrow \; \mathrm{d} k=\frac{\mathrm{d} \omega}{c}$$
and thus
$$\mathrm{d}^3 \vec{k}=4 \pi k^2 \mathrm{d} k =\frac{4 \pi}{c^3} \omega^2 \mathrm{d} \omega$$
and thus finally Plancks radiation formula,
$$\frac{\mathrm{d} U}{\mathrm{d} \omega}=\frac{\hbar \omega^3 V}{\pi^2 c^3} \frac{1}{\exp(\beta \hbar \omega)-1}.$$

 

[Silly Questions]

Thank you for a very precise answer.

I noticed you put zero in the domain of j. What exactly happens when j = 0? What does that look like?

 

[Nugatory]

"What would it look like?" Between two consecutive quantized-anythings lies a value invalid for the quantization parameters.

That's simply incorrect. Electromagnetic radiation is quantized (in the quantum electrodynamical treatment of the phenomenon) in the sense that all exchanges of energy with the field happens in discrete amounts; but all possible energy values are represented in blackbody radiation.

Take "valid emission of radiation x[1] by atom y" and "next valid emission of radiation x[2] by atom y", add them together, divide by 2, set the transmitter to that wavelength, and throw the switch.

You have been victimized by a common misunderstanding. Individual atoms will emit at discrete frequencies as as electrons change energy levels within the atom, but this is not the mechanism of electromagnetic radiation from a transmitter or blackbody radiation (you've mentioned both in this thread). The latter comes from the bulk movement of unbound or loosely bound charged particles, and the spectrum is continuous.

 

[PeterDonis]

"What would it look like?" Between two consecutive quantized-anythings lies a value invalid for the quantization parameters.

You are not reading other people's responses very carefully. You have already been told that frequency and wavelength of radiation are not "quantized" in the sense you mean: the entire continuum of values for frequency/wavelength is possible. This is true even when you adopt a quantum theory of radiation (e.g., quantum electrodynamics for EM radiation).

For a particular atom, emission due to electrons changing energy levels in that atom can only produce radiation with particular values for frequency/wavelength (though even there spectral lines have finite width so you will not get single discrete values). But there are lots of different kinds of atoms (and molecules, etc.), and lots of other ways for matter to emit radiation besides electrons changing energy levels in single atoms. So given any desired frequency/wavelength for a transmitter, in principle there will be some way to find a piece of matter that can emit it.

 

[Silly Questions]

The idea I was aiming at was to have a detector that can distinguish between electrons changing energy levels in atoms versus other sources of EM, like broadcasts from antennae.

Is there an apparatus that can definitively determine one of these possibilities:
"That signal source is definitely electrons changing energy levels."
"That signal source is definitely not."

If a broadcast from an antenna throws a splash of wavelengths while electrons changing energy levels emit only on very specific wavelengths that could be a means of detecting the difference.

I am confused however by "black-body radiation" being different from "electrons changing energy levels". If the two are not interchangeable then I have been badly misusing the former term. How is black-body radiation different from electrons changing energy levels?

 

[Nugatory]

How is black-body radiation different from electrons changing energy levels?

That’s in the last paragraph of #19 above. And take the first sentence of #20 more seriously.

 

[PeroK]

I am confused however by "black-body radiation" being different from "electrons changing energy levels". If the two are not interchangeable then I have been badly misusing the former term. How is black-body radiation different from electrons changing energy levels?

What's confusing you is that you want a universal quantum that applies across all systems. Some specific minimum energy value that all energy values are whole number multiplies of.

What we have is a different quantum for each system. Each system has an energy spectrum consisting of integer multiples of a fundamental energy value. But the fundamental value itself varies by system.

The most common system is the quantum harmonic oscillator, whose energies are:$$E_n = \hbar \omega(n + \frac 1 2)$$where $\omega$ is the characteristic angular frequency of the particular system. Each harmonic oscillator has a minimum energy: $E_0 = \frac{\hbar \omega}{2}$, but there is no universal minimum for $\omega$ that applies across all physical systems.

 

[vanhees71]

Thank you for a very precise answer.

I noticed you put zero in the domain of j. What exactly happens when j = 0? What does that look like?

$j=0$ is "the vacuum". In the context here it's the state that no photon of the frequency under consideration is present.

 

[vanhees71]

What's confusing you is that you want a universal quantum that applies across all systems. Some specific minimum energy value that all energy values are whole number multiplies of.

What we have is a different quantum for each system. Each system has an energy spectrum consisting of integer multiples of a fundamental energy value. But the fundamental value itself varies by system.

The most common system is the quantum harmonic oscillator, whose energies are:$$E_n = \hbar \omega(n + \frac 1 2)$$where $\omega$ is the characteristic angular frequency of the particular system. Each harmonic oscillator has a minimum energy: $E_0 = \frac{\hbar \omega}{2}$, but there is no universal minimum for $\omega$ that applies across all physical systems.

Yes, and these zero-point energies add to a total energy of $\infty$ for the em. field. On the other hand all that's observable are energy differences. That's why at this point already you have to do the first renormalization and define the vacuum state to be an energy eigenstate with eigen value 0. Formally that's achieved with "normal ordering".

 

[PeterDonis]

Each system has an energy spectrum consisting of integer multiples of a fundamental energy value.

This is not true as you state it, because some systems (typically free systems rather than bound systems) have a continuous energy spectrum.

 

[vanhees71]

Only the harmonic oscillator has an energy spectrum of integer multiples of $\hbar \omega$ (if you shift the ground-state energy to 0).

 

[valenumr]

You are not reading other people's responses very carefully. You have already been told that frequency and wavelength of radiation are not "quantized" in the sense you mean: the entire continuum of values for frequency/wavelength is possible. This is true even when you adopt a quantum theory of radiation (e.g., quantum electrodynamics for EM radiation).

For a particular atom, emission due to electrons changing energy levels in that atom can only produce radiation with particular values for frequency/wavelength (though even there spectral lines have finite width so you will not get single discrete values). But there are lots of different kinds of atoms (and molecules, etc.), and lots of other ways for matter to emit radiation besides electrons changing energy levels in single atoms. So given any desired frequency/wavelength for a transmitter, in principle there will be some way to find a piece of matter that can emit it.

Or you could just make the transmitter relativistic.

 

[Silly Questions]

I think I get what you're all saying about EM quantization. Let me see if I've got it straight:

1. The EM spectrum itself has no quantization rules.
2. Every EM emitter has is own rules, which may or may not include quantization.
3. Even with those rules applied relativity shifts the frequency, so even a "quantized" wavelength could be anything once relative motion is factored in.
4. Planck Length quantization applies only to electrons changing energy levels.

On the topic of black-body radiation I've re-read the relevant answers and only wound up with a third question: What is the difference between black-body radiation, thermal radiation, and electrons changing energy levels? I get that black-body radiation has parameters imposed on the EM waves by interactions with the walls, but what finally comes out the hole? Is that not thermal radiation? If I think all these things are the same thing and they're not what exactly am I missing? It's probably something so big nobody even understands what I'm asking -- indeed, my original question suffered from this -- so try to imagine a misconception so huge it makes it impossible for me to tell the difference between black body radiation, thermal radiation, and atoms changing energy levels. I'm missing something big here.

And thanks to everyone for great answers. Everything I thought I knew was wrong in a way that was not likely to ever be corrected. I can't pretend I can drink from the fire hose in one gulp, but I've washed off a lot of mud.

 

[vanhees71]

I think "black body radiation" is indeed synonymous with electromagnetic radiation in thermal equilibrium at a given temperature $T$. It can be realized by a cavity with the walls kept at a well-defined constant temperature for a very long time. The radiation emitted from a little hole then is with high accuracy indeed "thermal radiation".

Concerning your general questions, you should read some introductory book on quantum field theory (or more specifically QED). It cannot be explained in a forum posting.

 

[Silly Questions]

If those critera are not met does black body radiation then cease to be the same as thermal?

 

[vanhees71]

Sure, if the radiation is not in thermal equilibrium it's described by another state and thus you'll not measure a Planck spectrum.

 

[PeterDonis]

does black body radiation then cease to be the same as thermal?

No, the radiation just ceases to be either thermal or black body (since "black body" and "thermal" mean the same thing for radiation).

 

[Silly Questions]

So is there anything wrong with thinking that black body radiation, thermal radiation, and electrons changing energy levels are the same thing?

For the Planck length quantization to apply does the material radiating need to be in pure elemental form? I think I've been told in this thread that molecules have their own rules, but I wanted to make sure: quantization by Planck lengths only applies to the emissions from electrons changing energy levels within the energy shells of atoms, not the various hybrid orbitals of molecules?

Does this mean the inner walls of the black body apparatus need to be made out of elemental carbon? Black to absorb light and in element form to get the proper electron shells?

 

[PeterDonis]

So is there anything wrong with thinking that black body radiation, thermal radiation, and electrons changing energy levels are the same thing?

Yes. Only the first two are the same thing. Electrons are not radiation. Also, individual electrons changing energy levels do not emit black body/thermal radiation.