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Pre Big Bang State


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#35 OceanBreeze

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Posted 17 January 2019 - 11:33 AM

I will save you some pain and just quote these sources for you:

 

 

Quantum mechanics carries the implication that no oscillatory system can lose all its energy; there must always remain at least a “zero-point energy” amounting to hν/2 for an oscillator with natural frequency ν (h is Planck’s constant). This also seemed to be required for the electromagnetic oscillations constituting radio waves, light, X-rays, and gamma rays.

 

and

 

Zero-point energy, vibrational energy that molecules retain even at the absolute zero of temperature. Temperature in physics has been found to be a measure of the intensity of random molecular motion, and it might be expected that, as temperature is reduced to absolute zero, all motion ceases and molecules come to rest. In fact, however, the motion corresponding to zero-point energy never vanishes.

 

 

You are wrong, as usual but I do not expect you to admit it.

 

You are a waste of time to discuss any physics with.


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#36 Dubbelosix

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Posted 17 January 2019 - 11:35 AM

Learn something you fool.

 

https://www.quora.co...-Why-or-why-not

 

https://chemistry.st...ro-unattainable

 

Just because a bit of bad terminology exists, will not prevent me from showing you why you are wrong.



#37 Dubbelosix

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Posted 17 January 2019 - 11:37 AM

Further misunderstanding. The radiation law predicts the radiation from matter, considered as a set of quantised oscillators. At absolute zero it predicts no radiation, which is what I have been saying.

 

However that does not mean there is no energy in the oscillators themselves (i.e. in the matter responsible for the radiation), because whatever energy they retain in the ground state does not allow them to emit radiation. 

 

Thus an object at 0K can have as much zero point energy as it likes and yet emit zero radiation, as would be expected for an object at absolute zero. 

Not entirely accurate, yes the power law describes radiation, but it also provides the lowest energy associated to the oscillator. Because temperature is approximately energy, and because that energy contains some circular motion from the term [math]\omega[/math] you will find that there has to be a lowest state of temperature, which is NOT zero like you keep claiming, I have provided two very useful discussions on this subject in my previous post you may want to look at.



#38 exchemist

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Posted 17 January 2019 - 11:45 AM

I will save you some pain and just quote these sources for you:

 

 

Quantum mechanics carries the implication that no oscillatory system can lose all its energy; there must always remain at least a “zero-point energy” amounting to hν/2 for an oscillator with natural frequency ν (h is Planck’s constant). This also seemed to be required for the electromagnetic oscillations constituting radio waves, light, X-rays, and gamma rays.

 

and

 

Zero-point energy, vibrational energy that molecules retain even at the absolute zero of temperature. Temperature in physics has been found to be a measure of the intensity of random molecular motion, and it might be expected that, as temperature is reduced to absolute zero, all motion ceases and molecules come to rest. In fact, however, the motion corresponding to zero-point energy never vanishes.

 

 

You are wrong, as usual but I do not expect you to admit it.

 

You are a waste of time to discuss any physics with.

Yes this is right of course. (Especially the last line  :winknudge: )

 

I think Dubbelsox has got in a muddle between matter and radiation, actually. He's really interested in the vacuum. However defining temperature at all in the absence of matter is highly problematic. It only makes sense if one has a black body radiation distribution, i.e. a radiation distribution equivalent to matter at a certain temperature.

 

P.S. I would, however, take issue with your earlier remark that absolute zero is attainable. It is normally said not to be attainable "in a finite number of steps", due simply to the non-existence of any heat bath below absolute zero, which is what one would need to extract the very last bit of extractable heat.  


Edited by exchemist, 17 January 2019 - 11:46 AM.

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#39 Dubbelosix

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Posted 17 January 2019 - 11:50 AM

Despite giving discussions which overwhelmingly prove my interpretation of physics, this is just amounting to trolling because you have no intentions discussing these things or following up any link I provide.

 

This is useless for me, so I will be ignoring you both from now on. Go stroke each others ego and keep to the wrong interpretations you live by, I really don't care any more.



#40 OceanBreeze

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Posted 17 January 2019 - 11:56 AM

Yes this is right of course. (Especially the last line  :winknudge: )

 

I think Dubbelsox has got in a muddle between matter and radiation, actually. He's really interested in the vacuum. However defining temperature at all in the absence of matter is highly problematic. It only makes sense if one has a black body radiation distribution, i.e. a radiation distribution equivalent to matter at a certain temperature.

 

P.S. I would, however, take issue with your earlier remark that absolute zero is attainable. It is normally said not to be attainable "in a finite number of steps", due simply to the non-existence of any heat bath below absolute zero, which is what one would need to extract the very last bit of extractable heat.  

 

Yes, I meant to say it is theoretically possible, but I left out the "theoretically" part, so I stand corrected on that point.



#41 OceanBreeze

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Posted 17 January 2019 - 11:58 AM

Despite giving discussions which overwhelmingly prove my interpretation of physics, this is just amounting to trolling because you have no intentions discussing these things or following up any link I provide.

 

This is useless for me, so I will be ignoring you both from now on. Go stroke each others ego and keep to the wrong interpretations you live by, I really don't care any more.

 

Anyone can post to Quora, even you, so quoting those posts does not help your case. But even some of those posters disagree with you, if you bother to read them all and not just cherry pick a few.


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#42 Dubbelosix

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Posted 17 January 2019 - 01:03 PM

Physics stack is the best one, and you'll find the majority of posters are saying exactly what I have said.

 

If you knew anything about the third law of thermodynamics, no system, can be at absolute zero in any finite steps. It's a fact of physics which you and the other poster have never understood, and it's not my problem if you cannot understand basic physics.



#43 exchemist

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Posted 17 January 2019 - 02:32 PM

One special characteristic of Dubbelsox is his ability to quote someone as saying the exact opposite of what he actually said.

 

For example it was I who mentioned - in post 44 - that it is not possible to get to absolute zero in a finite number of steps. And yet, by post 48, Dubbelosox is saying I will never understand that it is impossible to get to absolute zero in a finite number of steps. WTF?

 

Another example: in post 23 he claims that for a while I was asserting that absolute zero could be reached. I defy him to produce evidence of any such statement on my part, as I have never said any such thing, here or on any forum. 

 

I've experienced this pattern of behaviour before with him, in previous sockpuppet incarnations (Simon's cat, Geon etc). This is another reason why actual discussion with him is useless. The best thing to do is simply to state the correct science when he makes incorrect statements. He does not do this very often, in fact, because most of the time he is engaged in his mathematical knitting, which is very unlikely to mislead anyone. 


Edited by exchemist, 17 January 2019 - 03:02 PM.


#44 Dubbelosix

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Posted 17 January 2019 - 03:01 PM

You have an extraordinary ability yourself to act like a total *******. How about that one?

 

Go ****ing learn science, you are not up to these debates. Who apart from you two, could argue with a ****ing link supporting my claims yet call it cherry picking... which was clear and concise concerning credible arguments which have been supporting everything I have said, yet can still be so disingenuous about what he has learned, or should I say haven't?

 

You are a total waste of my time.



#45 exchemist

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Posted 17 January 2019 - 03:19 PM

Anyone can post to Quora, even you, so quoting those posts does not help your case. But even some of those posters disagree with you, if you bother to read them all and not just cherry pick a few.

In fact even in the discussions Dubbelsox quotes, there are some nuggets. Someone quoted Richard Feynman:

Remember that when a crystal is cooled to absolute zero, the atoms do not stop moving, they still 'jiggle'. Why? If they stopped moving, we would know were they were and that they had they have zero motion, and that is against the Uncertainity Principle."

 

So he is stating quite plainly that zero point motion persists at absolute zero.

 

Now, who do we think is more likely to have got this right: Feynman or Dubbelsox? 


Edited by exchemist, 17 January 2019 - 03:20 PM.


#46 Dubbelosix

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Posted 17 January 2019 - 03:39 PM

There is no [at] zero temperature, and you are an absolute moron for thinking so.



#47 Dubbelosix

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Posted 17 January 2019 - 03:42 PM

What happens when T goes to 0 Kelvin, in your equations. ?

 

 

 

A theory I tend to agree with, is zero energy universe. This also includes a zero charge, not equal amounts of matter and antimatter. You will be aware looking at the fundamental particles quarks electrons etc the summed charges are zero. You will also be aware in beta decay a proton made of quarks can emit a positron. so what is this antimatter matter problem. I dont think one exists, except in some theory that is clearly not backed up by observation. A charge balance is all that is retained.

 

I suspect you are correct and a cold pre big bang stage must have occurred, to maintain the zero energy universe theory. I also suspect you are correct in that the pre big bang stage was populated by matter, (due to some variation on HUP which allows at near 0 kelvin permanent particles to appear) which would have been in some kind of condensate of quarks electrons etc. In order to have zero energy, gravity would also have to have existed. A problem you will note from this is that things would tend to contract rather than expand, unless a negative mass effect occurs as can happen in a condensate, which can only form at very low temperatures ie less than 2.7Kelvin.

 

Could a negative mass effect, as found in a condensate, drive the expansion(inflationary stage) of the universe, and keep space cold until it eventually starts collapsing and compressing then getting hot in nebulae and eventually exploding in super novae etc?. See the Boomerang nebulae 1Kelvin! Could we even detect  a nebulae in space at 0 kelvin except perhaps via gravitational lensing, the supposed dark matter effect. Everything unexplained in the universe is not dark matter and could easily be explained differently.

 

To repeat my question what happens in your theory when T = 0K ? Pre Bang 

 

 

So I've come back to think, and to show how in principle that temperature can only approach zero, is best fitted when we model the fluctuations into an expanding model, and they can even generate cosmic seeds. To ''put in'' the laws of thermodynamics is relatively easy. So I did some calibrating of the modified equation to show Plancks law will hold.

 

[math]\frac{\dot{R}}{R} (\dot{H} + H^2 + \frac{kc^2}{a}) = \frac{8 \pi G}{3}\ \sum_k\ ( nQ_k + nT S_k +n T S_{ik})\frac{\dot{T}}{T}[/math]

 

Right hand side is concerned now with the thermodynamics: This as it turns out is my preferred form of the Friedmann equation. The energy of each oscillator needs to be attributed to the Planck law as (using dimensionless form of entropy this time):

 

[math]\frac{\dot{R}}{R} (\dot{H} + H^2 + \frac{kc^2}{a}) = \frac{8 \pi G}{3}\ \sum_k\ ( [n_k + n S_k + n S_{ik}]\ \frac{\hbar \omega}{\frac{\hbar \omega}{k_BT} - 1})\frac{\dot{T}}{T}[/math]

 

Absorbing the derivative into the entropy (entropy production) and allowing the correction of zero point energy we get:

 

[math]\frac{\dot{R}}{R} (\dot{H} + H^2 + \frac{kc^2}{a}) = \frac{8 \pi G}{3}\ \sum_k\ ( [\dot{n}_k + n \dot{S}_k + n \dot{S}_{ik}]\ \frac{\hbar \omega}{\frac{\hbar \omega}{k_BT} - 1} + \dot{N}\frac{\hbar \omega}{2})[/math]

 

This shows that temperature cannot go to zero. Even in a number of infinite steps, it is unclear how a vacuum can truly lose the quantum property of fluctuations that do owe its existence to the uncertainty principle. The number density should also be properly understood as the wave number density: [math]\int\ k\ dk^3[/math] and in such a case we should obtain ~

 

[math]\frac{\dot{R}}{R} (\dot{H} + H^2 + \frac{kc^2}{a}) = \frac{8 \pi G}{3}\ \int\ ([\frac{d}{dt} + \dot{S}_k + \dot{S}_{ik} ] \frac{\hbar c}{\frac{\hbar \omega}{k_BT} - 1} + \frac{\hbar c}{2})\ \frac{dk}{dt}\ dk^3[/math]


Edited by Dubbelosix, 18 January 2019 - 08:40 AM.


#48 Dubbelosix

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Posted 17 January 2019 - 03:47 PM

In fact, I would go as far to extend the third law which states it as a matter of finite steps. In addition, zero temperature cannot be attainable in an infinite amount of steps because of the scale invariance of fluctuations, which has a lot of interest in theoretical physics.



#49 Dubbelosix

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Posted 17 January 2019 - 04:24 PM


In the previous derivation, the gauge that measured changes in temperature, also known as anistropies, takes the form of [math]\frac{\dot{T}}{T}[/math], and not only did we absorb that into the entropy, it is also true that it can be interpreted the following way by considering the short wavelength ~


[math]\frac{\partial S}{\partial U_{\lambda}} = \frac{1}{T}[/math]


Rayleigh's 1900 heuristic formula approximately meant that energy was proportional to temperature (see reference) below,

http://canhttps://en...ki/Planck's_law

https://en.wikipedia...ro-point_energy

And so, we can also state the ratio in the following way

[math]\frac{\partial S\dot{T}}{\partial U_{\lambda}} = \frac{\dot{T}}{T}[/math]

For long wavelengths the expression is used

[math]\frac{\partial^2 S\dot{T}}{\partial U^2_{\lambda}}[/math]

The greater the energy, the larger the frequency and the shorter (smaller) the wavelength. When we speak about wavlengths in these cases, we tend to think of deBroglie thermal wavelengths. But as shown in the link, Wien showed that his formula lead to second derivatives for short wavelengths and it is believed Planck being aware of this, constructed a formula for both large and small wavelengths as:

[math]\frac{\partial^2 S}{\partial U^2_{\lambda}} = \frac{\alpha}{U_{\lambda}(\beta + U_{\lambda})}[/math]
 



#50 Dubbelosix

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Posted 17 January 2019 - 07:02 PM

We can define

 

 

[math]\frac{\partial S}{\partial U_{\lambda}} = \frac{S}{k_BT}[/math]

 

multiplying through a factor of temperature, with entropy becoming dimensionless and taking a derivative I get:

 

[math]\frac{\partial^2 S}{\partial U^2_{\lambda}}\ k_B\dot{T} = \frac{\partial S}{\partial U_{\lambda}} \frac{\dot{T}}{T}[/math]

 

 

I wanted to see how that  derivative might enter the main equation. The equation we formed was:

 

[math]\frac{\dot{R}}{R} (\dot{H} + H^2 + \frac{kc^2}{a}) = \frac{8 \pi G}{3}\ \int\ ([\frac{d}{dt} + \dot{S}_k + \dot{S}_{ik} ] \frac{\hbar c}{\frac{\hbar \omega}{k_BT} - 1} + \frac{\hbar c}{2})\ \frac{dk}{dt}\ dk^3[/math]

 

Letting the entropy take on the dimensions of the Boltzmann constant allows we to write in the following form, after distributing a partial derivative:

 

[math]\frac{\partial}{\partial U_{\lambda}}\frac{\dot{T}}{T}\ (\dot{H} + H^2 + \frac{\mathbf{k} c^2}{a}) = \frac{8 \pi G}{3}\  ([n\frac{\partial S}{\partial U_{\lambda}} + \frac{\partial \mathbf{S}_k}{\partial U_{\lambda}} + \frac{\partial \mathbf{S}_{ik}}{\partial U_{\lambda}} ] \frac{\dot{T}}{\frac{k_BT}{\hbar \omega} - 1} + \int\ \frac{\partial S}{\partial U_{\lambda}}\frac{\dot{T}}{2}\ dk^3)[/math]

 

Even though the derivative appears to be of first power on the left hand side, we have also shown that the second power can also be considered

 

 

[math]\frac{\partial^2 }{\partial U^2_{\lambda}}\ k_B\dot{T} = \frac{\partial }{\partial U_{\lambda}} \frac{\dot{T}}{T}[/math]

 

Best not to write it in explicitly but at least we know it is there. The left hand side of the equation

 

[math]\frac{\partial}{\partial U_{\lambda}}\frac{\dot{T}}{T}\ (\dot{H} + H^2 + \frac{\mathbf{k} c^2}{a}) = \frac{8 \pi G}{3}\  ([n\frac{\partial S}{\partial U_{\lambda}} + \frac{\partial \mathbf{S}_k}{\partial U_{\lambda}} + \frac{\partial \mathbf{S}_{ik}}{\partial U_{\lambda}} ] \frac{\dot{T}}{\frac{k_BT}{\hbar \omega} - 1} + \int\ \frac{\partial S}{\partial U_{\lambda}}\frac{\dot{T}}{2}\ dk^3)[/math]

 

Looks a bit messy, but keep in mind we can just write the whole thing as:

 

[math]\frac{\partial }{\partial U_{\lambda}}\frac{\dddot{R}}{R} = \frac{8 \pi G}{3}\  ([n\frac{\partial S}{\partial U_{\lambda}} + \frac{\partial \mathbf{S}_k}{\partial U_{\lambda}} + \frac{\partial \mathbf{S}_{ik}}{\partial U_{\lambda}} ] \frac{\dot{T}}{\frac{k_BT}{\hbar \omega} - 1} + \int\ \frac{\partial S}{\partial U_{\lambda}}\frac{\dot{T}}{2}\ dk^3)[/math]

 

Another form of the equation I looked at a while back considered the quantum thermal application of the system to form a condensate.

I suggest an approach to how we might come to describe the conditions required for the phase transition. A simple equation of state is

 

[math]\frac{d}{dt}(NV) = \dot{N}V + N \dot{V} = (\dot{N} + 3 \frac{\dot{R}}{R}N)V = NV\Gamma[/math]

 

Where [math]N[/math] is the particle number and [math]R[/math] is the radius of a universe and [math]\Gamma[/math] is the particle production which is also related dynamically to the fluid expansion [math]\Theta = 3\frac{\dot{a}}{a}[/math]. To find the required object to describe a condensate, we divide through by [math]N^2\lambda^3[/math] where [math]\lambda[/math] is the thermal wavelength and [math]\lambda^3[/math] will replace the role of the density term,

 

[math]\frac{d}{dt}(\frac{NV}{N^2 \lambda^3}) = \frac{\dot{N}V}{N^2\lambda^3} + \frac{N \dot{V}}{N^2\lambda^3} = (\frac{V}{N\lambda^3})\Gamma[/math]

 

In which we can measure the statistics from the interparticle distance where

 

[math]\frac{V}{N \lambda^3} \leq 1[/math]

 

In which the interparticle distance is smaller than its thermal wavelength, in which case, the system is then said to follow Bose statistics or Fermi statistics. On the other hand, when it is much larger ie.

 

[math]\frac{V}{N \lambda^3} >> 1[/math]

 

Then it will obey the Maxwell Boltzmann statistics. The latter here is classical but the former, the Bose and Fermi statistics describes a situation where classical physics are smeared out by the quantum. In this picture, it may make describe the ability to construct a condensate universe from a supercool region that existed before the big bang (the stage in which the universe began to heat up).

 

Using the previous information, you can construct a Friedmann equation with the necessary terms, which are attached to the effective density. The equation, in a similar non-conserved form as given in the opening post we have:

 

[math]\frac{\dot{R}}{R}(\frac{\ddot{R}}{R} + \frac{kc^2}{a}) = \frac{8 \pi G}{3}(\dot{\mathbf{q}}_{rev} + [(\frac{\rho V}{n \lambda^3}) + 3P_{irr}(\frac{V}{n \lambda^3})]n\Gamma)][/math]

 

The pre big bang phase, if it existed, would imply a supercool region obeying non-classical statistics. As the universe underwent a collapse, the liquid state of the universe changed into the radiation vapor we see today. As the universe got larger the interparticle distances clearly changes and this picture allows us to look at the universe in a completely different light. How is temperature related to motion?
 

The virial theorem, the potential to be clear, is also related to the thermodynamics of the system.

[math]2<k_BT> =\ - \sum^N_{k=1}\ <\mathbf{F}_k \cdot r_k>\ = n\ <V_{tot}>[/math]

And as stated before, the temperature is roughly related to the kinetic energy of the system

[math]k_B T = \frac{1}{2}\sum^N_{k=1} m_kv^2_k[/math]

and can be related to a mass tensor, with Dirac notation:

[math]= \frac{1}{2}\sum^N_{k=1} <\frac{d\mathbf{q}_k}{dt}|\mathbf{M}| \frac{d\mathbf{q}}{dt}>[/math]

using generalized coordinates. Above we saw that the temperature was twice the magntitude of the potential, it is also true the temperature related to generalized coordinates as:

[math]2k_BT = \mathbf{p} \cdot \dot{\mathbf{q}} = m(\frac{ds}{dt})^2[/math]

with [math]ds[/math] as a metric term and this relationship is well-known under the Maupertuis principle.


Edited by Dubbelosix, 17 January 2019 - 08:24 PM.


#51 Dubbelosix

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Posted 17 January 2019 - 08:15 PM


 

[math]\frac{\dot{R}}{R} (\dot{H} + H^2 + \frac{kc^2}{a}) = \frac{8 \pi G}{3}\ \int\ ([\frac{d}{dt} + \dot{S}_k + \dot{S}_{ik} ] \frac{\hbar c}{\frac{\hbar \omega}{k_BT} - 1} + \frac{\hbar c}{2})\ k\ dk^3[/math]

 

Now here comes the coolest part. The correction of the presence of fluctuations in cosmology is, in a much more simple form:

 

[math]m\dot{R}^2  = \frac{8 \pi GmR^2}{3}(\rho + \hbar c \mathbf{R} \int k\ dk)[/math]

When [math]R \approx 0[/math] (but not pointlike) then the fluctuations are in their ground state.

Though inflation is not required to explain the cosmic seeding, there are alternatives themselves to cosmic inflation such as one particular subject I have investigated with a passion; rotation can mimic dark energy perfectly which is thought to explain the expansion and perhaps even acceleration (if such a thing exists). It is possible to expand the Langrangian density of the zero point modes on the background spacetime curvatuture in a power series, a result of Sakharov concerning corrections to gravity:

[math]\mathcal{L} = \hbar c\ \mathbf{R}\ \int\ k\ dk... + \hbar c\ \mathbf{R}^n\ \int\ \frac{dk}{k^{n-1}}[/math]

 

Where [math]\mathbf{R}[/math] is the Ricci curvature.

 

The cool thing is that we can do the same thing for the expanding model, in which the Sakharov term becomes similar to a gravitational pressure term which acts like an antigravity. This gravitational pressure is also responsible for the particle creation process. For a universe in with fluctuations in the ground state, then only one term in our modified equation will describe such a system and the equation reduces to the zero point energy term:

 

[math]\frac{\ddot{R}}{R}  + \frac{kc^2}{a} = \frac{8 \pi G}{3}\ \int\ \frac{\hbar c}{2} k dk^3[/math]

 

And likewise, the contribution of curvature to the fluctuations is with [math]8 \pi G = c = 1[/math]:

 

[math]\frac{\ddot{R}}{R} + \frac{\mathbf{k}c^2}{a} = \int\ \frac{\hbar c}{2}\ k\ dk^3... +\ \int\ \hbar c\ \mathbf{R}^n\ \int\ \frac{dk}{k^{n-1}}\ k\ dk^3[/math]


Edited by Dubbelosix, 18 January 2019 - 09:24 AM.