What if dark energy is just orphan energy floating around in the universe?

[Micheth]

I believe this is the proper forum for this question...
More properly perhaps it should be posted under the forum "Crazy ideas that are a waste of time" :-)

So, basically when I first learned of the mysterious phenomenon of dark energy, this sort of occurred to me as being an "obvious" explanation. To me :-)
And of course I wanted to know why it's not true.
But, at least in the popular literature, I've never heard any reference even remotely resembling it, even simply in order to dismiss it as ridiculous, so evidently it's nuttier than even I imagine, or more likely based on my fundamental misunderstanding.
If someone has the time perhaps they could quickly point out why it's not even worth considering.

Ok, enough intro.
Quantum physics has of course confirmed that energy exists in discrete packets. Emitted as such and absorbed as such. All matter, all of our detectors, our eyes, etc. - any way that matter can interact with energy - requires the energy to be at least a quantum level, else it cannot knock an electron to a higher orbital, for example, and affect it. And matter can only emit energy in quanta.

But it seems that, simply because we find energy traveling in packets (which just happens to be the only energy we can possibly detect), would not exclude the possibility that there could conceivably be a great deal of energy waves distributed throughout the universe that are *not* present as discrete units. (Most of it, in fact?)
Amounts smaller than the minimum quantum value, I mean.
Certainly if such energy abounded it would not be absorbed and thus not detectable in any way?
In other words, not some different "kind" of exotic energy but simply, perhaps more boringly, just "orphan" energy oscillating around in the universe in arbitrary levels less than the minimum "required" in quantum physics.
(That is, not detectable except, I suppose, for its bouncing off matter and thus producing radiation pressure similar to photons, which would allow for the universe expansion acceleration, which as I understand is the only evidence we have for dark energy in the first place...)

Maybe some physicist long ago considered this for half a second and enjoyed a nice laugh :-)
But then again, maybe not...? :-)

Cheers,

 

[bhobba]

Quantum physics has of course confirmed that energy exists in discrete packets

First error - not true.

Thanks
Bill

 

[Micheth]

First error - not true.

Thanks
Bill

Mmmmm, not very helpful :-) but OK.
So, I thought that was most definitely true?

 

[bhobba]

Mmmmm, not very helpful :-) but OK.
So, I thought that was most definitely true?

It isn't.

Here is what's really going on:
http://www.physics.ox.ac.uk/Users/smithb/website/coursenotes/qi/QILectureNotes3.pdf

Have a look at the free particle.

Thanks
Bill

 

[Micheth]

Here is what's really going on:
http://www.physics.ox.ac.uk/Users/smithb/website/coursenotes/qi/QILectureNotes3.pdf

Have a look at the free particle.

Thanks
Bill

Hm? That's quite a bit beyond my comprehension level...
I did search for "quant" in this article to try glean something from it, but only found that because the Schrodinger and other equations include Planck's constant h (which I think Planck had discovered gives the right explanation for how we observe energy to behave), solutions to the equations can only be in quanta.
But I didn't see something saying something like "energy we don't observe or interact with can't be less than a quantum because...".

I'm puzzled because a general definition for "quantum" that you find is: "In physics, discrete bundles in which radiation and other forms of energy occur."

 

[bhobba]

but only found that because the Schrodinger and other equations include Planck's constant h (which I think Planck had discovered gives the right explanation for how we observe energy to behave), solutions to the equations can only be in quanta.

I don't know where you got that from but its incorrect. The free particle solution can have any value of energy. Just as an aside the deep reason for Schroedinger's equation, and what h really is, is symmetry - but that is advanced. This shows h is simply a constant used to write equations in our usual units. One can easily use units where it's one - it doesn't really mean that much.

Here is a modern take on what QM is REALLY about:
http://www.scottaaronson.com/democritus/lec9.html

As you can see its different from the usual pop-sci accounts - and much less sensational.

I'm puzzled because a general definition for "quantum" that you find is: "In physics, discrete bundles in which radiation and other forms of energy occur."

Its a misnomer.

What they may be alluding to is what's called Quantum Field Theory (QFT) where a continuous field and particles are integrated into the one formalism. But that is even more advanced than ordinary QM - anything you have read outside a Quantum Field textbook is likely wrong.

What I would suggest is the following lay book may be of value:
https://www.amazon.com/dp/0473179768/?tag=pfamazon01-20

The Kindle version is cheap and would give you a bit of background to continue this discussion. Be warned though - its a lay book about QFT - they are rare and will give you a different view than you may have read elsewhere. Also, as I said before, QFT outside an actual textbook will not really be correct - but we all have to start somewhere, and that book is reasonably good.

Thanks
Bill

 

[my2cts]

The original question is not "educate me on QFT".
As I read it, it is "why can't we detect the DE density ?". There is no need to delve into QFT.
At first sight, the answer is that a) the actual energy density is very low and b) no one knows what to look for.
Obviously, we have no clue, but is the DE density small ? From wikipedia:
"on a mass–energy equivalence basis, the density of dark energy (6.91 × 10−27 kg/m3) is very low, much less than the density of ordinary matter or dark matter within galaxies."
Dividing by c^2 I get 7 × 10−10 J/m3. This is a whopping 17.500 times the total cosmic microwave background energy density of 4×10-14 J/m3. Since we can detect the CMB then we would see DE if it would interact with ordinary matter with the same coupling strength as electromagnetism. The conclusion must be that the interaction constant is much weaker.
How does DE compare to the energy density of sunlight? The formula is P/d^2/c. The sun emits P=4 × 10+26 W, so beyond a distance of d=4 × 10^10 km, 10 times the distance to Neptune, the solar radiation energy density is overtaken by the dark energy density. DE actually is not a small energy density at all.
https://en.wikipedia.org/wiki/Cosmic_microwave_background#Features

 

[ddd123]

Just as an aside the deep reason for Schroedinger's equation, and what h really is, is symmetry - but that is advanced.

Can you remind me what that would be specifically? Never thought of h as related to symmetry, I probably should know though.

 

[bhobba]

Can you remind me what that would be specifically? Never thought of h as related to symmetry, I probably should know though.

See page 81 Ballentine.

Thanks
Bill

 

[Micheth]

Re: "a general definition for "quantum" that you find is: "In physics, discrete bundles in which radiation and other forms of energy occur."

Its a misnomer.
What they may be alluding to is what's called Quantum Field Theory (QFT)
Bill

Hm? How can a definition be a misnomer?
Even wiki teaches:
"In physics, a quantum (plural: quanta) is the minimum amount of any physical entity involved in an interaction. ... the magnitude can take on only certain discrete values. A photon is a single quantum of light... the energy of an electron bound to an atom is quantized... As incorporated into the theory of quantum mechanics, this is regarded by physicists as part of the fundamental framework for understanding and describing nature at the smallest length-scales."

I don't see how that's different from my statement that "Quantum physics has of course confirmed that energy exists in discrete packets"
(which you said is wrong).

 

[Micheth]

The original question is not "educate me on QFT".
As I read it, it is "why can't we detect the DE density ?". There is no need to delve into QFT.

Hi, while I also don't get the connection with bringing up QFT, unless it somehow answers the question, my original question was actually somewhat different.

As i understand it (= layman) there is a minimum quantized amount in which we find energy to exist, i.e. E = 1hv, although it can be 2hv, 3hv, etc.
Notwithstanding Bill's comments, every physics description I read says exactly that.
We only "find" energy to exist in discrete packets (photons), and apparently that's the only way it can interact with matter and thus be detected.
The OP question was, are we justified in assuming that merely because discrete packets are all that are detectable, that there couldn't be lots of energy milling about that is not in such discrete units, and thus not detectable?
Many thousands of smart theoretical physicists can't have ignored that possibility, so I'm just wondering what are the reasons for its dismissal. Certainly good reasons, but I don't know what they are.

 

[bhobba]

Hm? How can a definition be a misnomer?

Easy - its not correct as the free particle solution proves.

Wiki articles about QM are generally good - but there is some of it that isn't. You have found some.

The reason I raised about QFT is it may have been where this misconception came from.

Thanks
Bill

 

[meBigGuy]

e=h*wavelength

The wavelength can be any value, and not just integers.

https://en.wikipedia.org/wiki/Planck_constant

 

[bhobba]

"In physics, a quantum (plural: quanta) is the minimum amount of any physical entity involved in an interaction.

That's not correct. There are quantum interactions in which no minimum amount of energy is involved.

As i understand it (= layman) there is a minimum quantized amount in which we find energy to exist, i.e. E = 1hv, although it can be 2hv, 3hv, etc.

Notwithstanding Bill's comments, every physics description I read says exactly that.

The free particle solution to Schroedinger's equation I linked to didn't say that. For a free particle energy can exist at any value.

Your issue is you are reading lay explanations which often take liberties with the truth - you need to read the more serious literature that includes the math.

Thanks
Bill

 

[Micheth]

e=h*wavelength

The wavelength can be any value, and not just integers.

https://en.wikipedia.org/wiki/Planck_constant

Thank you.
Yes I understand that the wavelength/frequency can be any value, but that the energy cannot be.
That is, the energy of a photon of frequency x must be hx or 2hx, etc.
But I assume one way we know this is because every photon of frequency x that we've ever encountered has had energy (K)hx.
I'm simply wondering, is there a way that we are able to know that there cannot exist energy, of frequency x, that is in an amount of less than hx, such as 3/4hx?
In other words, less than the "discrete packet" of hx.
(Because It seems to me that, if it hypothetically could exist in such an amount, it would be undetectable as normal energy, being unable to interact with matter.)

 

[bhobba]

That is, the energy of a photon of frequency x must be hx or 2hx, etc.

The energy of a photon is E = hν. Both v and E can be anything. As I said before h is basically a conversion factor between units.

Thanks
Bill

 

[Micheth]

The energy of a photon is E = hν. Both v and E can be anything. As I said before h is basically a conversion factor between units.

Thanks
Bill

Sure, v and E may each independently be anything, but for a given v, E can't be 1/2hv can it?

 

[bhobba]

Sure, v and E may each independently be anything, but for a given v, E can't be 1/2hv can it?

It can't be any value other than hv - it expresses an equality. And by varying v you can change E and conversely. Its easier to see in units h is 1. In those units E = v.

It applies not just to photons but to free electrons. We can put an electron in an electric field and give it any energy - or change it to any energy - that's the example that came to mind as energy is not quantised in interactions - there may be others as well. For free particles energy is not quantised.

Thanks
Bill

 

[Micheth]

It can't be any value other than hv - it expresses an equality. And by varying v you can change E and conversely. Its easier to see in units h is 1. In those units E = v.
It applies not just to photons but to free electrons. We can put an electron in an electric field and give it any energy - or change it to any energy - that's the example that came to mind as energy is not quantised in interactions - there may be others as well. For free particles energy is not quantised.
Thanks
Bill

I understand about h being needed to make the equality, but how did we determine h? Was it not by dividing the energy of photons we observe by their frequency?
Perhaps that's not right but I think certainly some empirical method is used?
Which is why I ask, if there were energy of frequency v mulling about in an amount of less than 1hv, having no observable effects, it would not have affected our calculation of h, right?

Stated differently and hopefully not too awkwardly, if there had existed some energy of frequency v in an amount of say, 1/2hv, that did affect matter in an observable and measurable way, then would we not have defined that 1/2h simply as h, as that would have expressed the minimal equality of what we observed between E and v.
But, and this gets to my question, if "1/2hv amounts of energy" somehow exist but don't do anything observable, then the minimal possible amount of energy as we can observe it would still be hv as we know it (h being the constant that empirically satisfies the equality).

(Incidentally, if energy is not quantized for free particles like electrons then I guess that's another issue, I'm only asking about where energy is quantized.)

 

[bhobba]

I understand about h being needed to make the equality, but how did we determine h?

It pops up in all sorts of equations such as, for example, the Ryberg formula. That many seemingly disparate phenomena have it allows it to be determined in multiple ways.

if there had existed some energy of frequency v in an amount of say, 1/2hv

Then QM would be proven incorrect.

(Incidentally, if energy is not quantized for free particles like electrons then I guess that's another issue, I'm only asking about where energy is quantized.)

Ok then. can you state your query in those terms.

Thanks
Bill

 

[Micheth]

It pops up in all sorts of equations such as, for example, the Ryberg formula. That many seemingly disparate phenomena have it allows it to be determined in multiple ways.
Bill

Hm? wiki says the h in the Rydberg formula is something different, not Planck's constant.
Ah well.

 

[bhobba]

Hm? wiki says the h in the Rydberg formula is something different, not Planck's constant..

http://chemwiki.ucdavis.edu/Physica...ion_of_the_Rydberg_Equation_from_Bohr's_Model

Thanks
Bill