New Equivalence Principles?

[Shustaire]

Here try this methodology this article will cover all the essential equations including the redshift to entropy relations of a thermal heat state. See page 42 start with equation 4.1.1

http://www.wiese.itp.unibe.ch/lectures/universe.pdf

 

This chapter will provide the essential details including chemical potential, spin, Hamilton and the canonical partition functions involved. Which he proceeds to describe in the next equation.

 

 A useful application of this is one can calculate the particle number density for a given species via the statistics (Bose/Fermi) from any given blackbody temperature.

Edited August 1, 2018 by Shustaire

[Vmedvil2]

Here try this methodology this article will cover all the essential equations including the redshift to entropy relations of a thermal heat state. See page 42 start with equation 4.1.1

http://www.wiese.itp.unibe.ch/lectures/universe.pdf

 

This chapter will provide the essential details including chemical potential, spin, Hamilton and the canonical partition functions involved. Which he proceeds to describe in the next equation.

 

 A useful application of this is one can calculate the particle number density for a given species via the statistics (Bose/Fermi) from any given blackbody temperature.

 

See, like I said it has it relating back to curvature and the cosmological constant.

[Dubbelosix]

So the equation I suggested not long ago for the transformation of the volume lead to a change in the entropy:

 

 

[math]\Delta S = Nk_B\ \log_2(\frac{V_2}{V_1})[/math]

 

On my blog a while back I derived the following (with some adjustments for clarity here::

 

[math]\Delta S =  Nk_B \log_2(\frac{T_2}{T_1})[/math]

Edited August 1, 2018 by Dubbelosix

[Dubbelosix]

 

So the equation I suggested not long ago for the transformation of the volume lead to a change in the entropy:

 

 

[math]\Delta S = Nk_B\ \log_2(\frac{V_2}{V_1})[/math]

 

On my blog a while back I derived the following (with some adjustments for clarity here::

 

[math]\Delta S =  Nk_B \log_2(\frac{T_2}{T_1})[/math]

 

[math]\Delta S = Nk_B\ \log_2(\frac{V_2}{V_1})[/math]

 

 

 

 

 

So it stands to reason the identity may hold to satisfy a covariant temperature following the same transformation laws as temperature. 

 

[math]\Delta S = Nk_B(\log_2(\frac{V_2}{V_1}) = \log_2 (\frac{T_2}{T_1}))[/math]

Edited August 1, 2018 by Dubbelosix

[Vmedvil2]

Look at this dubbel, in sustaire's link.

 

[Dubbelosix]

Here try this methodology this article will cover all the essential equations including the redshift to entropy relations of a thermal heat state. See page 42 start with equation 4.1.1

http://www.wiese.itp.unibe.ch/lectures/universe.pdf

 

This chapter will provide the essential details including chemical potential, spin, Hamilton and the canonical partition functions involved. Which he proceeds to describe in the next equation.

 

 A useful application of this is one can calculate the particle number density for a given species via the statistics (Bose/Fermi) from any given blackbody temperature.

 

 

I have thought about application of chemical potential on black hole I have idea's towards this as well. Of course, I have idea's for the condensate. This is maybe to quick for me. Right now, I am trying to establish any flaws, because if there are no flaws, this whole idea in relativity that you cannot tell who is moving at relativistic speeds, is not something that holds now in my mind. That is not to say Einstein was wrong, there are very special cases, like the train and platform experiment really do show relativistic anomalies about who is actually in motion and who is at rest. They are interesting questions, but when it comes to a violation of the heat conduction of thermodynamic laws, we know heat can only transfer to cold, so bodies may appear warmer, when they are not. But red shift and blue shift corrections I expect will solve this ambiguity. It also seems that Einsteins transformation and Ott's transformation differs only by a factor of [math]\gamma^2[/math]. 

Edited August 1, 2018 by Dubbelosix

[Shustaire]

So can you get the entropy of a commoving volume

[math] S=\frac{\rho+p}{T}R^3[/math] equation 4.2.1 page 48 same article

[Dubbelosix]

Sure I will follow the link, but I take time reading papers, what I will say is that I can follow what Shustaire is talking about which is a bonus. I considered the chemical potential to explain possible phase changes in a black hole. 

 

 

Look at this dubbel, in sustaire's link.

 

[Dubbelosix]

So can you get the entropy of a commoving volume

[math] S=\frac{\rho+p}{T}R^3[/math] equation 4.2.1 page 48 same article

 

Perhaps.

 

I haven't looked strongly into this yet. 

[Shustaire]

my last post is a cross post with your last two replies.

[Shustaire]

Perhaps.

 

I haven't looked strongly into this yet. 

 

It should help address several questions for example the entropy under the above isn't changing due to relativistic corrections but the particle number density will increase on blueshift. This will also relate to the path integrals under QFT and QM treatments.

Edited August 1, 2018 by Shustaire

[Vmedvil2]

Well, to answer an earlier question with expansion N stays constant for entropy and stays in thermal equilibrium while expansion happens.

Edited August 1, 2018 by VictorMedvil

[Dubbelosix]

expansion N?

 

What do you mean, in my equations? N is just the particle number, which is required to measure a multiparticle state. 

[Vmedvil2]

expansion N?

 

What do you mean, in my equations? N is just the particle number, which is required to measure a multiparticle state. 

 

Yes in your equation with expansion your N stays constant.

[Dubbelosix]

It should help address several questions for example the entropy under the above isn't changing due to relativistic corrections but the particle number density will increase on blueshift. This will also relate to the path integrals under QFT and QM treatments.

 

 

Well, the only way to change the entropy is to see temperature vary with volume, there is no other way I can think of. You are right, it will involve relativistic corrections but as far as path integrals will go, this is way ahead of the simple approach I am taking just now. 

[Dubbelosix]

Yes in your equation with expansion your N stays constant.

 

 

yes, a change in particle density comes with a price, a derivative

 

[math]\frac{d}{dt}(\frac{N}{V}) = \dot{\mathbf{n}}[/math]

 

which is a particle production. There are entropy particle productions as well, but this is mentioned in the link, in a way.  

[Shustaire]

Not quite true if the volume expansion exceeds the particles interaction rate then the particle is no longer in thermal equilibrium and this results in an entropy change. As that particle now adds its effective degrees of freedom. lol another cross post this is in reply to previous post..

 

Part of the key is to understand the thermal equilibrium transition states in terms of expansion of a volume.

Edited August 1, 2018 by Shustaire