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So this is something I was thinking about recently. Given that quantum harmonic oscillation indicates that quanta are actually defined as harmonics of a fundame

So this is something I was thinking about recently. Given that quantum harmonic oscillation indicates that quanta are actually defined as harmonics of a fundamental frequency, what if dark matter is simply all energy at a frequency that isn’t a harmonic? Likewise, what if dark energy is simply the quantum harmonic oscillator that is our universe undergoing damping? That is to say, it’s not so much that space is expanding as it is that the frequency of our universe is diminishing: lower frequency = higher wavelength, which appears to us like space is growing. The graph of an oscillator undergoing damping looks an awful lot like models of dark energy expansion of the universe. If this were the case, we could think of black holes as nodes on this wave.

Justcreatednow 7y ago
Wow! What about posting this on /r/asksciencediscussion or Physics.SE and get some review? I would be interested in reading discussions about this
fallingfrog 7y ago
The s orbital electrons around an atom is the fundamental, the p and other orbitals are harmonics. I think you’ve stretched your analogy too far.
- aalleavitch 7y ago
  Couldn’t the s orbital itself be a harmonic of an even lower fundamental?
  - fallingfrog 7y ago
    No, because the energy of the electron is known and is always quantized.
    - tedsanders 7y ago
      When you say this what do you mean? Here are some statements of my own, typed quickly on a phone before I head to work: -Electrons do not in general have quantized energy; only bound electrons have a discrete energy spectrum -Electrons do not necessarily have a single energy; although energy eigenstates form a basis that can describe all states, very often the electron won't be in a pure eigenstate -Single electron models are simplifications; in general, atoms will have multielectron states that are very difficult to write down or reason about; it's deep and non-obvious why single electron models do so well at describing behavior of multielectron wavefunctions
  - ninkendo 7y ago
    I think if that were true, we’d see other orbitals that are harmonics of the lower fundamental, but not a harmonic of the s orbital... which we I don’t think we see in practice? (If we did, we wouldn’t consider the s orbital to be fundamental.)
    - aalleavitch 7y ago
      In this case, the lower fundamental would be something that permeates space (zero-point energy?), so the orbitals that are harmonics of the lower fundamental but not an atom's s orbital would be on other atoms entirely.
    - ta1234567890 7y ago
      This is very interesting. So, if we interpret all matter to be different expressions of a fundamental frequency, it means us, the observers are also made up of the same, and to observe you would be basically sampling reality. Even if you consider measurement instruments to be an observer, they also sample other things that are made up of the same. Then maybe there's a fundamental limit to what we can sample, hence observe. And that could explain why we only observe harmonics of the s orbital.
  - fallingfrog 7y ago
    By the way you seem really smart and I love the enthusiasm, I don’t want to seem like a downer.. if you haven’t already taken some classes in physics, you should
axilmar 7y ago
I think you've hit the nail in the head. I recently had a similar thought that the 'expansion' of the cosmos may simply be "the same distances divided into more chunks". I.e. initially, at the big bang, the analogy was 1/1, i.e. the whole universe was of distance '1' divided by '1', and as the universe 'expanded', it didn't really expand, it was actually the divident being incremented, which created more chunks in the universe. It is as if we have a computer display of 1x1 meter, and we gradually increase the resolution; the actual size of the display wouldn't change, but its resolution would. Of course your explanation is much more scientific than mine, which is just a hunch. Such 'resolution' increase of the universe would explain nicely its expansion. I wonder if the creation of wormwholes that would allow travelling to distant systems in human lifetimes would simply be the opposite effect of damping of this frequence of our universe. By reversing the damping, the "resolution" of the universe may decrease and then moving from one place to the other would require a lot less time. Also, could gravity be a change of this frequency by mass?
PopsiclePete 7y ago
Does this change the whole heat death/big freeze question about the End of the Universe?
ur-whale 7y ago
>what if dark energy is simply the quantum harmonic oscillator Is there anything at all in the whole of physics that isn't, at some point, shoehorned in the "mass on a spring" framework </sarcasm>
- semi-extrinsic 7y ago
  The reason everything gets "shoehorned" into "mass on a spring" is that this is the first term in the Taylor expansion. No matter how ugly an expression you have for force as a function of distance, if you do the Taylor expansion you inevitably get a constant force term (often zero) and a "mass on a spring" term.
Steuard 7y ago
It would be cool to see exactly what sort of model would produce behavior like this. I'm a theoretical physicist, and at first glance I have absolutely no idea how one could write down a model that had "energy at a frequency that isn't a harmonic": it doesn't really fit with anything that I know about quantum systems or quantum field theory, which pretty solidly force us into discrete energy spectra for the types of systems that you're talking about. But I'd love to see a mathematically consistent model that does what you're talking about: it sounds fresh and different, if it turns out to be possible at all. If you come up with some math to share, please do! (Hmm... in the back of my mind, there was some interest years ago in "unparticle" models that might maybe possibly be related, or maybe I'm just completely fuzzy on what that idea was about.)
- aalleavitch 7y ago
  I am not a physicist or a mathematician, but wouldn't a virtual particle be more or less what I'm talking about? They exist for a limited time because they are producing interference with our universe's 'wave' but they don't stick around because they are not at a harmonic frequency. Likewise they cannot have a precisely defined energy or momentum because they break our frame of reference by not being at a harmonic frequency.
ssivark 7y ago
Some comments, since this post is generating some discussion: 1. If you squint, the inflation model is something in this spirit. The "inflaton" field corresponds to the harmonic oscillator like variable that ismovingunder a potential, and causing space to expand. 2. The quantization of frequencies for a harmonic oscillator (or a particle in a box) comes from spatial bounds on it's profile -- through some potential function. The quanta are separated by inverse of the region size (like aliasing effects a la Fourier). If we're imagining a field spread all throughout space then the "harmonics" are very closely spaced enough to be considered a continuous spectrum, sort of. 3. All that was just quantum mechanics. A more appropriate framework to have this conversation is quantum field theory (and not just in flat spacetime, but curved spacetime). In that context, quanta refer to number of inflaton particles at a specific frequency, and that goes in a direction different from what we're interested in. So that was just a paranthetical aside. 4. As far as we can tell from observations space hasn't been globally oscillating. So it's more like an overdamped oscillator (with a lot of "friction") than an underdamped one -- see "slow roll inflation". (For examples of where space was locally affected by underdamped oscillations, see baryonic acoustic oscillations). Beyond just a few technical nuggets, what I hope to have conveyed is that the way to research these things is to take every word/facet seriously and try to operationalize it in an appropriate framework (which is very hard) and verify whether it matches our experience of reality.
gus_massa 7y ago
No. The solution to the problem you are proposing is well known and is part of any (advanced?) QM course, and it doesn't involve dark matter or dark energy. In an harmonic oscillator (an ideal perfect spring with an ideal punctual mass) the energies are (2n+1)ħ/2*ω. They are all multiples of the minimal value for this oscillator. Note that it is a different set of values for each oscillator. Each of these energies is associated with a wavefunction with a well defined energy. Combining all these wavefunctions you can represent any wavefunction. (The technical term is "orthonormal base", or simply "base".) This is easy to prove. More details in https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator So if you try to put the oscillator in a state that is not a wavefunction with a well defined energy, the state can be represented as a linear combination of the wavefunction with a well defined energy, and it will evolve following the laws of QM. It will not get transformed to dark energy or dark matter or get lost by damping. If you later make measure the energy of the harmonic oscillator, the arbitrary wavefunction will colapse to a wavefunction with a well defined energy, and the probabilities are easy to calculate knowing how the initial state was decomposed in wavefunction with a well defined energy. If you measure the position or speed, the calculation is similar but there are a few more technical details. In a more realistic system like an Hydrogen atoms, the energies are not multiples of a minimal value, the energies are like -E0/n^2. The corresponding wavefunctions are a base anyway, so any initial state can be decomposed in wavefunction with a well defined energy (that sometimes are calles "orbitals", but sometimes more general states are called "orbitals"). So there is no a fundamental frequency, there is no a minimal energy that is the divisor of all the energies, there is no dark energy. In atoms with more electrons the calculations are more complicated, there is no nice formula for the energies, but is anyway possible to decompose any initial state without using dark energy.
- alfiedotwtf 7y ago
  Interesting question, but also interesting answer. Thanks for the detailed response!
comnetxr 7y ago
The quantum states only have discrete frequencies because they are bound states contained within a trap. If you consider H = p^2/2m, a free particle, the frequencies are continuous. The universe isn't a trap.
Bombthecat 7y ago
What about rotation of the universe?

Justcreatednow 7y ago

Wow! What about posting this on /r/asksciencediscussion or Physics.SE and get some review? I would be interested in reading discussions about this

fallingfrog 7y ago

The s orbital electrons around an atom is the fundamental, the p and other orbitals are harmonics. I think you’ve stretched your analogy too far.

aalleavitch 7y ago
Couldn’t the s orbital itself be a harmonic of an even lower fundamental?
- fallingfrog 7y ago
  No, because the energy of the electron is known and is always quantized.
  - tedsanders 7y ago
    When you say this what do you mean? Here are some statements of my own, typed quickly on a phone before I head to work: -Electrons do not in general have quantized energy; only bound electrons have a discrete energy spectrum -Electrons do not necessarily have a single energy; although energy eigenstates form a basis that can describe all states, very often the electron won't be in a pure eigenstate -Single electron models are simplifications; in general, atoms will have multielectron states that are very difficult to write down or reason about; it's deep and non-obvious why single electron models do so well at describing behavior of multielectron wavefunctions
- ninkendo 7y ago
  I think if that were true, we’d see other orbitals that are harmonics of the lower fundamental, but not a harmonic of the s orbital... which we I don’t think we see in practice? (If we did, we wouldn’t consider the s orbital to be fundamental.)
  - aalleavitch 7y ago
    In this case, the lower fundamental would be something that permeates space (zero-point energy?), so the orbitals that are harmonics of the lower fundamental but not an atom's s orbital would be on other atoms entirely.
  - ta1234567890 7y ago
    This is very interesting. So, if we interpret all matter to be different expressions of a fundamental frequency, it means us, the observers are also made up of the same, and to observe you would be basically sampling reality. Even if you consider measurement instruments to be an observer, they also sample other things that are made up of the same. Then maybe there's a fundamental limit to what we can sample, hence observe. And that could explain why we only observe harmonics of the s orbital.
- fallingfrog 7y ago
  By the way you seem really smart and I love the enthusiasm, I don’t want to seem like a downer.. if you haven’t already taken some classes in physics, you should

axilmar 7y ago

I think you've hit the nail in the head. I recently had a similar thought that the 'expansion' of the cosmos may simply be "the same distances divided into more chunks". I.e. initially, at the big bang, the analogy was 1/1, i.e. the whole universe was of distance '1' divided by '1', and as the universe 'expanded', it didn't really expand, it was actually the divident being incremented, which created more chunks in the universe. It is as if we have a computer display of 1x1 meter, and we gradually increase the resolution; the actual size of the display wouldn't change, but its resolution would. Of course your explanation is much more scientific than mine, which is just a hunch. Such 'resolution' increase of the universe would explain nicely its expansion. I wonder if the creation of wormwholes that would allow travelling to distant systems in human lifetimes would simply be the opposite effect of damping of this frequence of our universe. By reversing the damping, the "resolution" of the universe may decrease and then moving from one place to the other would require a lot less time. Also, could gravity be a change of this frequency by mass?

PopsiclePete 7y ago

Does this change the whole heat death/big freeze question about the End of the Universe?

ur-whale 7y ago

>what if dark energy is simply the quantum harmonic oscillator Is there anything at all in the whole of physics that isn't, at some point, shoehorned in the "mass on a spring" framework </sarcasm>

semi-extrinsic 7y ago
The reason everything gets "shoehorned" into "mass on a spring" is that this is the first term in the Taylor expansion. No matter how ugly an expression you have for force as a function of distance, if you do the Taylor expansion you inevitably get a constant force term (often zero) and a "mass on a spring" term.

Steuard 7y ago

It would be cool to see exactly what sort of model would produce behavior like this. I'm a theoretical physicist, and at first glance I have absolutely no idea how one could write down a model that had "energy at a frequency that isn't a harmonic": it doesn't really fit with anything that I know about quantum systems or quantum field theory, which pretty solidly force us into discrete energy spectra for the types of systems that you're talking about. But I'd love to see a mathematically consistent model that does what you're talking about: it sounds fresh and different, if it turns out to be possible at all. If you come up with some math to share, please do! (Hmm... in the back of my mind, there was some interest years ago in "unparticle" models that might maybe possibly be related, or maybe I'm just completely fuzzy on what that idea was about.)

aalleavitch 7y ago
I am not a physicist or a mathematician, but wouldn't a virtual particle be more or less what I'm talking about? They exist for a limited time because they are producing interference with our universe's 'wave' but they don't stick around because they are not at a harmonic frequency. Likewise they cannot have a precisely defined energy or momentum because they break our frame of reference by not being at a harmonic frequency.

ssivark 7y ago

Some comments, since this post is generating some discussion: 1. If you squint, the inflation model is something in this spirit. The "inflaton" field corresponds to the harmonic oscillator like variable that ismovingunder a potential, and causing space to expand. 2. The quantization of frequencies for a harmonic oscillator (or a particle in a box) comes from spatial bounds on it's profile -- through some potential function. The quanta are separated by inverse of the region size (like aliasing effects a la Fourier). If we're imagining a field spread all throughout space then the "harmonics" are very closely spaced enough to be considered a continuous spectrum, sort of. 3. All that was just quantum mechanics. A more appropriate framework to have this conversation is quantum field theory (and not just in flat spacetime, but curved spacetime). In that context, quanta refer to number of inflaton particles at a specific frequency, and that goes in a direction different from what we're interested in. So that was just a paranthetical aside. 4. As far as we can tell from observations space hasn't been globally oscillating. So it's more like an overdamped oscillator (with a lot of "friction") than an underdamped one -- see "slow roll inflation". (For examples of where space was locally affected by underdamped oscillations, see baryonic acoustic oscillations). Beyond just a few technical nuggets, what I hope to have conveyed is that the way to research these things is to take every word/facet seriously and try to operationalize it in an appropriate framework (which is very hard) and verify whether it matches our experience of reality.

gus_massa 7y ago

No. The solution to the problem you are proposing is well known and is part of any (advanced?) QM course, and it doesn't involve dark matter or dark energy. In an harmonic oscillator (an ideal perfect spring with an ideal punctual mass) the energies are (2n+1)ħ/2*ω. They are all multiples of the minimal value for this oscillator. Note that it is a different set of values for each oscillator. Each of these energies is associated with a wavefunction with a well defined energy. Combining all these wavefunctions you can represent any wavefunction. (The technical term is "orthonormal base", or simply "base".) This is easy to prove. More details in https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator So if you try to put the oscillator in a state that is not a wavefunction with a well defined energy, the state can be represented as a linear combination of the wavefunction with a well defined energy, and it will evolve following the laws of QM. It will not get transformed to dark energy or dark matter or get lost by damping. If you later make measure the energy of the harmonic oscillator, the arbitrary wavefunction will colapse to a wavefunction with a well defined energy, and the probabilities are easy to calculate knowing how the initial state was decomposed in wavefunction with a well defined energy. If you measure the position or speed, the calculation is similar but there are a few more technical details. In a more realistic system like an Hydrogen atoms, the energies are not multiples of a minimal value, the energies are like -E0/n^2. The corresponding wavefunctions are a base anyway, so any initial state can be decomposed in wavefunction with a well defined energy (that sometimes are calles "orbitals", but sometimes more general states are called "orbitals"). So there is no a fundamental frequency, there is no a minimal energy that is the divisor of all the energies, there is no dark energy. In atoms with more electrons the calculations are more complicated, there is no nice formula for the energies, but is anyway possible to decompose any initial state without using dark energy.

alfiedotwtf 7y ago
Interesting question, but also interesting answer. Thanks for the detailed response!

comnetxr 7y ago

The quantum states only have discrete frequencies because they are bound states contained within a trap. If you consider H = p^2/2m, a free particle, the frequencies are continuous. The universe isn't a trap.

Bombthecat 7y ago

What about rotation of the universe?