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


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#18 Flummoxed

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Posted 04 September 2019 - 05:02 AM

Sure. The zero point energy of any quantum system is the energy remaining in the ground state.

 

To take an example, a hydrogen atom at absolute zero has no translational motion (obviously) and its one electron will be in the 1s orbital, which is the ground state. But that electron still has energy. So that is a form of zero point energy. It is unextractable energy, because there is no lower state available to the system, but it is still energy. 

 

Similarly, a diatomic molecule has a bond between the two atoms and and this bond can vibrate. But the vibration is quantised, so the vibrational energy of the bond can only be present in certain fixed amounts (this is how the lines in the IR spectrum arise, from transitions between these vibrational energy levels.) In the ground state, there is still some residual vibration present. So at absolute zero, the bond still vibrates a bit. That is zero point energy.

 

So, when matter is at absolute zero, it is in the ground state in all its degrees of freedom, but there is still energy present in some of these modes. This is zero point energy.

 

The vacuum is a special case, arising from the QED idea that fields too cannot be said to have definitively zero energy - hence the vacuum itself has a zero point energy, just as matter does.

 

So in the context you have in mind, to be accurate you should speak of the zero point energy of the vacuum or of the various fields, because you are not talking about the zero point energy of matter.

 

People get a bit lazy sometimes and forget this distinction, because physicists are all excited about hip and trendy virtual photons and things, but to a chemist, zero point energy is just a humdrum thing: the energy of the ground state of whatever atom or molecule one is talking about. :)

Thanks for the clarification. 



#19 Dubbelosix

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Posted 04 September 2019 - 06:03 AM

Zero Point energy is made up of virtual particles, it is electromagnetic in nature. 

 

Not just electromagnetic in nature, all quantum fields possess a ground state. 



#20 Dubbelosix

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Posted 04 September 2019 - 06:05 AM

Sure. The zero point energy of any quantum system is the energy remaining in the ground state.

 

To take an example, a hydrogen atom at absolute zero has no translational motion (obviously) and its one electron will be in the 1s orbital, which is the ground state. But that electron still has energy. So that is a form of zero point energy. It is unextractable energy, because there is no lower state available to the system, but it is still energy. 

 

Similarly, a diatomic molecule has a bond between the two atoms and and this bond can vibrate. But the vibration is quantised, so the vibrational energy of the bond can only be present in certain fixed amounts (this is how the lines in the IR spectrum arise, from transitions between these vibrational energy levels.) In the ground state, there is still some residual vibration present. So at absolute zero, the bond still vibrates a bit. That is zero point energy.

 

So, when matter is at absolute zero, it is in the ground state in all its degrees of freedom, but there is still energy present in some of these modes. This is zero point energy.

 

The vacuum is a special case, arising from the QED idea that fields too cannot be said to have definitively zero energy - hence the vacuum itself has a zero point energy, just as matter does.

 

So in the context you have in mind, to be accurate you should speak of the zero point energy of the vacuum or of the various fields, because you are not talking about the zero point energy of matter.

 

People get a bit lazy sometimes and forget this distinction, because physicists are all excited about hip and trendy virtual photons and things, but to a chemist, zero point energy is just a humdrum thing: the energy of the ground state of whatever atom or molecule one is talking about. :)

 

While you have made some good points, please be more careful in your language. An object is never at absolute zero. 



#21 OceanBreeze

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Posted 04 September 2019 - 12:01 PM

Sure. The zero point energy of any quantum system is the energy remaining in the ground state.

 

To take an example, a hydrogen atom at absolute zero has no translational motion (obviously) and its one electron will be in the 1s orbital, which is the ground state. But that electron still has energy. So that is a form of zero point energy. It is unextractable energy, because there is no lower state available to the system, but it is still energy. 

 

Similarly, a diatomic molecule has a bond between the two atoms and and this bond can vibrate. But the vibration is quantised, so the vibrational energy of the bond can only be present in certain fixed amounts (this is how the lines in the IR spectrum arise, from transitions between these vibrational energy levels.) In the ground state, there is still some residual vibration present. So at absolute zero, the bond still vibrates a bit. That is zero point energy.

 

So, when matter is at absolute zero, it is in the ground state in all its degrees of freedom, but there is still energy present in some of these modes. This is zero point energy.

 

The vacuum is a special case, arising from the QED idea that fields too cannot be said to have definitively zero energy - hence the vacuum itself has a zero point energy, just as matter does.

 

So in the context you have in mind, to be accurate you should speak of the zero point energy of the vacuum or of the various fields, because you are not talking about the zero point energy of matter.

 

People get a bit lazy sometimes and forget this distinction, because physicists are all excited about hip and trendy virtual photons and things, but to a chemist, zero point energy is just a humdrum thing: the energy of the ground state of whatever atom or molecule one is talking about. :)

 

 

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.



#22 Flummoxed

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Posted 04 September 2019 - 03:06 PM

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. 



#23 exchemist

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Posted 04 September 2019 - 03:20 PM

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.

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. 


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#24 exchemist

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Posted 04 September 2019 - 03:22 PM

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 by exchemist, 04 September 2019 - 03:22 PM.


#25 VictorMedvil

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Posted 04 September 2019 - 05:35 PM

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 by VictorMedvil, 04 September 2019 - 05:37 PM.


#26 Flummoxed

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Posted 05 September 2019 - 03:29 AM

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.

 

I don't think that completely agrees with SED. http://www.sciencefo...-point-energy/ 

 

 

 

I've skimmed the intro and conclusions and it is certainly very interesting. They turn QM on its head, by proposing that it is interaction between matter and the ZPF that gives rise to the wavelike behaviour of matter, including quantisation and even Heisenberg's principle of indeterminacy. I notice they also come down in favour of viewing Schrödinger's equation as applying only to ensembles and not individual particles. 

 

This seems to be part of a development called Stochastic Electrodynamics: https://en.wikipedia...electrodynamics . I note that Anna Maria Cetto, who has played a prominent role in SED, is a co-author of the paper.

 

It is a type of hidden variable theory, it seems.  


Edited by Flummoxed, 05 September 2019 - 03:31 AM.


#27 Dubbelosix

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Posted 05 September 2019 - 04:40 AM

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. 



#28 Dubbelosix

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Posted 05 September 2019 - 04:42 AM

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 by Dubbelosix, 05 September 2019 - 04:42 AM.


#29 exchemist

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Posted 05 September 2019 - 05:09 AM

I don't think that completely agrees with SED. http://www.sciencefo...-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.  


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#30 OceanBreeze

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Posted 05 September 2019 - 07:51 AM

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)

 

He4PD.gif


Edited by OceanBreeze, 05 September 2019 - 07:55 AM.


#31 OceanBreeze

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Posted 05 September 2019 - 07:53 AM

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



#32 Dubbelosix

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Posted 05 September 2019 - 10:07 AM

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

 

He4PD.gif

 

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. 



#33 OceanBreeze

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Posted 05 September 2019 - 10:37 AM

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.

 

He4PD.gif

 

 

 

Phasehe3log.gif

 

 

 

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: 


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#34 ralfcis

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Posted 05 September 2019 - 10:42 AM

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


Edited by ralfcis, 05 September 2019 - 10:46 AM.