On Einstein's Aether - The Way Forward For General Relativity

[exchemist]

I stand to be corrected, but I think the answer is in the zero point energy of the VACUUM, which not allow absolute zero to be reached. 

No, the ZPE of the vacuum does not need to be invoked here. It is the ZPE of the interatomic attraction between atoms - see above.

Edited September 4, 2019 by exchemist

[Vmedvil2]

I stand to be corrected, but I think the answer is in the zero point energy of the VACUUM, which not allow absolute zero to be reached. 

 

Yes, but isn't it because of microwave radiation that absolute zero cannot be reached, thus in effect is from a bosonic field being electromagnetic radiation causing the background energy of space, the Zero Point Energy of a Vacuum is not required for that.

Edited September 4, 2019 by VictorMedvil

[Dubbelosix]

What you wrote is the same as my own understanding of zero point energy.

 

I do, however, have a question along these lines that maybe you can answer.

 

As I understand it, the reason that liquid helium will not freeze, even if it were to reach the hypothetical value of 0 K, is because it has a very high zero point energy, higher than any other form of matter, or so I have read. (I can look for a link, if you need it)

 

So far so good, but if enough pressure is now applied to that liquid helium, still at 0 K, it will freeze. I think the pressure is on the order of 33 atms.

 

My question is, does this (the freezing) mean the zero point energy was extracted by applying pressure?

 

I expect this would not be an efficient way of extracting energy as much more energy would be needed to apply the pressure than any amount that can be extracted.

 

Still, I would be interested to know if now, with both 0 K T and high pressure, if all of the zero point energy of the helium is removed? Or would there still be some residual energy left even under these conditions?

 

I appreciate any thoughts about this.

 

What do you mean? I am not sure what you mean about this pressure thing, the liquid cannot freeze for a number of reasons, instead of thinking zero point energy, it is just another language to describe the uncertainty principle. There is always quantum movement and the fact this movement exists always means that a system cannot [ever] freeze over completely - that is, the system's constituents loses all momentum and motion. 

[Dubbelosix]

Yes, but isn't it because of microwave radiation that absolute zero cannot be reached, thus in effect is from a bosonic field being electromagnetic radiation causing the background energy of space, the Zero Point Energy of a Vacuum is not required for that.

 

No, because you can remove all the visible matter and energy from an area of space and still detect zero point fluctuations. 

Edited September 5, 2019 by Dubbelosix

[exchemist]

I don't think that completely agrees with SED. http://www.scienceforums.com/topic/36124-zero-point-energy/ 

Perhaps not, but SED is not so far accepted and we are not in the Alternative Theories section. What I have tried to give you is what the settled science has to say.  

[OceanBreeze]

What do you mean? I am not sure what you mean about this pressure thing, the liquid cannot freeze for a number of reasons, instead of thinking zero point energy, it is just another language to describe the uncertainty principle. There is always quantum movement and the fact this movement exists always means that a system cannot [ever] freeze over completely - that is, the system's constituents loses all momentum and motion. 

 

Liquid helium will freeze solid at 0 K and 2.5 Mpa pressure (that is ⁴He ³He is a bit different)

 

Edited September 5, 2019 by OceanBreeze

[OceanBreeze]

Nice one! 

 

This, as I understand (and I am rusty on this )  it refers to the zero point energy in a specific degree of freedom, namely that in the vibration of the interatomic "bond" between adjacent helium atoms.  

 

For anything to freeze, there has to be an interatomic or intermolecular attraction - in effect a sort of bond, however weak it may be  - which the atoms or molecules no longer have enough energy to break, when they are cooled. In the case of helium this is an interatomic attraction (because it does not form molecules) due to London (van der Waals) forces and it is very weak indeed.  

 

In the QM of the harmonic oscillator, which this very weak bond resembles, the confining potential is a parabola (restoring force proportional to displacement so potential is F x displacement, i.e. proportional to the square). If the parabola is narrow and deep the energy of the ground state - the zero point energy - is quite a long way above the minimum of the curve. If it is broad and shallow, the ground state is closer to the minimum.

 

In the case of helium the interatomic attraction is so feeble that by moving the atoms apart, even though this appears energetically unfavourable, the fall in the zero point energy overcomes this. And so the "bond" needed to make it solid never forms. If you pressurise it, the energy gain from this effect is no longer enough for it to be energetically worthwhile to move the atoms apart.

 

...I think.....

 

I am conscious I may not have explained this very well. But the zero point energy in question is the vibrational ZPE, associated with the solid atom-atom vibrational mode of motion. 

 

 

Thanks

[Dubbelosix]

Liquid helium will freeze solid at 0 K and 2.5 Mpa pressure (that is ⁴He ³He is a bit different)

 

 

A classical phase transition has nothing to do with the quantum mechanics of zero point energy - even this ''is not totally frozen'' as deducted from fundamental physics. 

[OceanBreeze]

A classical phase transition has nothing to do with the quantum mechanics of zero point energy - even this ''is not totally frozen'' as deducted from fundamental physics. 

 

I would listen to you if I thought you knew what you are talking about, but since you don't, I won't.

 

 

 

 

 

 

 

The reason for the different behavior of ⁴He and ³He is quantum mechanics. ⁴He is a boson. The appearance of the superfluid phase in ⁴He is related to Bose condensation, where a macroscopic fraction of the atoms is in the lowest-energy one-particle state. ³He is a fermion (like electron) and it is forbidden by the Pauli exclusion principle that more than one fermion is in the same one-particle state. The superfluidity arises from formation of weakly bound pairs of fermions, so called Cooper pairs. The pairs behave as bosons. In the superfluid state there is a macroscopic occupation of a single Cooper pair state.

 

Link: here

 

But, don't let that stop you from carrying on  :good: 

[ralfcis]

Yeah I was kicked off a forum for arguing with idiots about  Cooper pairs. XC was there. Your explanation is good.

Edited September 5, 2019 by ralfcis

[Dubbelosix]

I would listen to you if I thought you knew what you are talking about, but since you don't, I won't.

 

 

 

 

 

 

 

The reason for the different behavior of ⁴He and ³He is quantum mechanics. ⁴He is a boson. The appearance of the superfluid phase in ⁴He is related to Bose condensation, where a macroscopic fraction of the atoms is in the lowest-energy one-particle state. ³He is a fermion (like electron) and it is forbidden by the Pauli exclusion principle that more than one fermion is in the same one-particle state. The superfluidity arises from formation of weakly bound pairs of fermions, so called Cooper pairs. The pairs behave as bosons. In the superfluid state there is a macroscopic occupation of a single Cooper pair state.

 

Link: here

 

But, don't let that stop you from carrying on  :good: 

 

Of course I know what I am talking about, I wouldn't parrot this stuff if I didn't.  zero point temperatures are unattainable, a classical phase transition from liquid to solid is not a true case of a frozen system. Only amateurs could get this wrong, like you. 

Edited September 5, 2019 by OceanBreeze

[OceanBreeze]

You might want to consider reading the site rules before posting.

 

 

Rude and offensive behavior will not tolerated. The staff reserves the right to decide what behavior is deemed offensive, and this rule applies to all areas of the site including private messages.

 

I consider your comment about puny little mind to be rude and offensive.

 

So, what I will do, this time only, is delete that part of your post, for your sake.

 

If you continue to be rude and offensive, I will delete your entire post.

 

If you still continue, I will delete YOU.

[Vmedvil2]

No, because you can remove all the visible matter and energy from an area of space and still detect zero point fluctuations. 

Ya, I am going to have to do some reading about Zero Point Energies before I comment about this anymore, I am not personally well versed about the dynamics of the Zero Point Energy itself, but my understanding is that you cannot have energy without a field giving rise to it.

Edited September 5, 2019 by VictorMedvil

[exchemist]

I would listen to you if I thought you knew what you are talking about, but since you don't, I won't.

 

 

 

 

 

 

 

The reason for the different behavior of ⁴He and ³He is quantum mechanics. ⁴He is a boson. The appearance of the superfluid phase in ⁴He is related to Bose condensation, where a macroscopic fraction of the atoms is in the lowest-energy one-particle state. ³He is a fermion (like electron) and it is forbidden by the Pauli exclusion principle that more than one fermion is in the same one-particle state. The superfluidity arises from formation of weakly bound pairs of fermions, so called Cooper pairs. The pairs behave as bosons. In the superfluid state there is a macroscopic occupation of a single Cooper pair state.

 

Link: here

 

But, don't let that stop you from carrying on  :good: 

Nice phase diagrams. 

 

For any readers who may not know, the abbreviations "hcp" and "bcc" refer to the packing arrangements of the atoms, "hexagonal close packed" and "body centred cube", respectively. 

[Dubbelosix]

Ya, I am going to have to do some reading about Zero Point Energies before I comment about this anymore, I am not personally well versed about the dynamics of the Zero Point Energy itself, but my understanding is that you cannot have energy without a field giving rise to it.

 

Zero point energy is a field, it is just the ground state of a field. 

[Vmedvil2]

Zero point energy is a field, it is just the ground state of a field. 

 

So you are saying there are amount of energy in a field even without the property present, that is true in gravitational and Electromagnetic fields as the range of them is infinite thus even matter or charge from one side of the universe will have some attraction to another object with a similar property thus it could be viewed as a static amount of energy gained, but at what point do you consider to be the ground state? The Magnitude of such a long distance field would still vary based on distance, Even Higg's like fields require a specific amount of energy dumped into them before they take effect and create something such as mass or to create a charge the mass of a electron or Up/Down Quark is required, so are you saying that these ground states act on smaller particles than Quarks and Leptons? What are the properties of a ground state particle, I guess I am asking? If they do exist then the only effect you may see is the effect of gravity as that effects all energy-mass in the universe.

Edited September 6, 2019 by VictorMedvil

[Dubbelosix]

So you are saying there are amount of energy in a field even without the property present, that is true in gravitational and Electromagnetic fields as the range of them is infinite thus even matter or charge from one side of the universe will have some attraction to another object with a similar property thus it could be viewed as a static amount of energy gained, but at what point do you consider to be the ground state? The Magnitude of such a long distance field would still vary based on distance, Even Higg's like fields require a specific amount of energy dumped into them before they take effect and create something such as mass or to create a charge the mass of a electron or Up/Down Quark is required, so are you saying that these ground states act on smaller particles than Quarks and Leptons? What are the properties of a ground state particle, I guess I am asking? If they do exist then the only effect you may see is the effect of gravity as that effects all energy-mass in the universe.

 

What I am saying, is that the fundamental forces, all fields of nature, but not necessarily gravity or even magnetism, all are described by a quantization of a field. These fields cannot ever be ''frozen'' in a classical sense, meaning it doesn't matter how much energy you pump into a refrigerator, it is not possible for a system to exist at zero point temperatures because there is always residual motion.