Dubbelosix Posted August 1, 2018 Author Report Share Posted August 1, 2018 (edited) Not quite true if the volume 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.. Ah right you mean other cases, no you are absolutely right. The simple approach I suggest is just keep thermodynamics to an equipartition description of the average kinetic energy of the particles, which make a good rough estimate. Clearly an advanced model will need to explain the temperature in a non-macroscopic way, when we know it arises from individual constituents. It's also possible what I just suggested is too difficult and arduous task. Edited August 1, 2018 by Dubbelosix Quote Link to comment Share on other sites More sharing options...
Shustaire Posted August 1, 2018 Report Share Posted August 1, 2018 (edited) correct the methodology above applies the ret mass aand momentum terms via equation 4.1.1 so you have your kinetic energy term. [math]E=\sqrt{k^2+m^2}[/math] note this is is a derivitave of the energy momentum equation. The next equations will sum over momenta and show the particle number density relation via equation 4.1.7. The elegance of this is it will also relate to the energy density terms under the FRW metric for commoving volumes in so far as the statistical mechanic applications for the equations of states Edited August 1, 2018 by Shustaire Quote Link to comment Share on other sites More sharing options...
Vmedvil2 Posted August 1, 2018 Report Share Posted August 1, 2018 correct the methodology above applies the ret mass aand momentum terms via equation 4.1.1 so you have your kinetic energy term. [math]E=\sqrt{k^2+m^2}[/math] note this is is a derivitave of the energy momentum equation. The next equations will sum over momenta and show the particle number density relation via equation 4.1.7. The elegance of this is it will also relate to the energy density terms under the FRW metric for commoving volumes in so far as the statistical mechanic applications for the equations of states You know that is actually the same one I used for my equation. Quote Link to comment Share on other sites More sharing options...
Dubbelosix Posted August 1, 2018 Author Report Share Posted August 1, 2018 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] To suggest the above though, would mean the ideal gas law would require a modification leading to an entropy of ideal law is: [math]Nk_BT = PV[/math] and so from this you may be able to form the entropy as: [math]\Delta \mathbf{S} = N\log_2(\frac{T_2}{T_1}) = \log_2(\frac{V_2}{V_1})[/math] In which the last term would have been the coefficient with the pressure associated to the gas law. It falls out of this though because the entropy is dimensionless. I haven't derived this though so its more speculation at this moment in time. Quote Link to comment Share on other sites More sharing options...
Dubbelosix Posted August 1, 2018 Author Report Share Posted August 1, 2018 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. Ah! With one exception of course, black holes, in which N does increase and the volume increases. You could actually argue that for any body really, but black holes are such a nice choice. So yeah, N can vary with the expansion of something. If you mean expansion of the universe, then sure, there is not enough energy today to allow virtual particles to become ''real'' or on-shell. But I still stand to the assertion that this may not have been the case for a short period only in the earliest phase of the big bang. Quote Link to comment Share on other sites More sharing options...
Vmedvil2 Posted August 1, 2018 Report Share Posted August 1, 2018 (edited) Ah! With one exception of course, black holes, in which N does increase and the volume increases. You could actually argue that for any body really, but black holes are such a nice choice. So yeah, N can vary with the expansion of something. If you mean expansion of the universe, then sure, there is not enough energy today to allow virtual particles to become ''real'' or on-shell. But I still stand to the assertion that this may not have been the case for a short period only in the earliest phase of the big bang. When the BB was still a singularity it would have not been constant but since that time, it was constant, but I think the writer of that paper meant with the expansion of the universe via DE. You will have to take in account for both states. Edited August 1, 2018 by VictorMedvil Dubbelosix 1 Quote Link to comment Share on other sites More sharing options...
Shustaire Posted August 1, 2018 Report Share Posted August 1, 2018 (edited) Yes but recall your dealing with blackbody temperature not a true temperature. The blackbody temperature is the amount of absorbtion of the EM spectrum. Hence (black). So one has to be careful to remember this when applying the ideal gas laws. Edited August 1, 2018 by Shustaire Dubbelosix 1 Quote Link to comment Share on other sites More sharing options...
Dubbelosix Posted August 1, 2018 Author Report Share Posted August 1, 2018 Anyway this has been a nice discussion guys. Thank you. hopefully Poly will stop playing his mind games and just get on with it. Quote Link to comment Share on other sites More sharing options...
Dubbelosix Posted August 1, 2018 Author Report Share Posted August 1, 2018 Yes but recall your dealing with blackbody temperature not a true temperature. The blackbody temperature is the amount of absorbtion of the EM spectrum. Hence (black) For a black hole, the black body temperature is its Hawking emission which is a measure of the bolometric flux from near the surface. So in a sense, the are all related. I only work in this special case. Quote Link to comment Share on other sites More sharing options...
Shustaire Posted August 1, 2018 Report Share Posted August 1, 2018 yes but you need to be careful in thinking of blackbody in terms of heat different devils. Dubbelosix 1 Quote Link to comment Share on other sites More sharing options...
Vmedvil2 Posted August 1, 2018 Report Share Posted August 1, 2018 (edited) For a black hole, the black body temperature is its Hawking emission which is a measure of the bolometric flux from near the surface. So in a sense, the are all related. I only work in this special case. Ya, the Hawking radiation emission of the BB while in singularity form will be non-constant while the DE emission of the universe is constant this says after the BB did its transformation into the universe it had a constant N value for particles, which also says the accelerated expansion of the universe has a constant acceleration value based on the Hawking radiation or DE value of the universe since the BB, which is interesting is DE actually hawking radiation from the universe. Furthermore it says that even DE has been conserved since the BB's initial energy. Edited August 1, 2018 by VictorMedvil Quote Link to comment Share on other sites More sharing options...
Dubbelosix Posted August 1, 2018 Author Report Share Posted August 1, 2018 https://entropytemperature.quora.com/A-Universal-Equation-for-the-Relativistic-Transformation-of-Temperature Quote Link to comment Share on other sites More sharing options...
Dubbelosix Posted August 1, 2018 Author Report Share Posted August 1, 2018 (edited) edited: Edited August 2, 2018 by Dubbelosix Quote Link to comment Share on other sites More sharing options...
Dubbelosix Posted August 1, 2018 Author Report Share Posted August 1, 2018 (edited) 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 Right now back to this good question i have not properly covered yet. I suspected I could, let's just some idea's that pop out my head so far... I expect to write the entropy in the following form: Ideal gas law is [math]PV = n\mathbf{R}T[/math] The dimensionless entropy: [math]dS = \frac{1}{k_BT}(d \rho + PdV)[/math] I assume your volume term normalized this from the equation i just suggested. And into this, I will need to look for relationships with the comoving volume. The comoving object is a volume of [math]dV = D(\frac{(1 + z)^2 D^2}{E(z)})\ d\Omega\ dz[/math] And Shustaire has led me to an equation in which deals with the square of the red shift. Very nice. Will definitely have implication to the square of the redshift power formula I derived. I will certainly need to leave a passage in my blog to dedicate thanks to helpful discussions here. Oh Shustaire will know this, perhaps, but the term [math]\frac{1}{(1 + z)}[/math] is in fact equivalent to the scale factor [math]a(t)[/math] in the Freidmann equation! Nice hack to remember. https://ned.ipac.caltech.edu/level5/Hogg/Hogg9.html Edited August 1, 2018 by Dubbelosix Quote Link to comment Share on other sites More sharing options...
Shustaire Posted August 1, 2018 Report Share Posted August 1, 2018 (edited) Another nice hack to remember is that temperature is approximately the inverse of the scale factor.[math]T=\frac{1}{a}[/math] Freidmann was a genius when he developed his equations using the fluid equations of the FRW metric. Edited August 1, 2018 by Shustaire Dubbelosix 1 Quote Link to comment Share on other sites More sharing options...
Dubbelosix Posted August 1, 2018 Author Report Share Posted August 1, 2018 Hey shustaire, the latex sign for approximation is \approx just for future reference will save you having to mention it to save you time. And as promised: https://entropytemperature.quora.com/A-Few-Notable-Mentions-of-Contributors I only do this because I am a great believer in archiving a true history, especially on a blog intended to do this kind of stuff, which is why I like to include references to everything that requires it. Quote Link to comment Share on other sites More sharing options...
Dubbelosix Posted August 1, 2018 Author Report Share Posted August 1, 2018 (edited) The next idea, if anyone is interesting, the topic I decided to discuss had to do with the classical coupling of gravity and magnetism. I did a treatise on the couplings suggested by Sciama and later by Motz into a post that covered in an extensive way, why those ideas held in physics. Gravitomagnetism is not some ad hoc hypothesis, but there is strong mathematical evidence suggesting the Coriolis effect is not just gravitational - it is itself an analogue of magnetism. Gravity is line dependent, like electric forces, so gravity and electric force in this sense can play interchangeable roles in different scales of energy - I have ever been since convinced the concept of gravitimagnetism can be understood to unify a quantum force with a pseudo-one, a unification I call semi-quantization method of gravitoelectromagnetism. https://spinorbitcoupling.quora.com/ I also suspect strong formal analogies to Maxwell's equations. (Excerpt) '' For the ideas contained in the treatise I provide, and will follow with an excerpt, these couplings where the first couplings of the gravitational charge to its own gravitimagnetic field I studied. His cross product terms can be understood from the following derivations: Motz has explained, that a moving gravitational charge [math]\sqrt{G}m[/math] with a velocity v will couple not only to the Newtonian gravitational and gravielectric fields but also the Coriolis field in which the gravimagnetic field is defined by him as [math]\frac{2(\omega \times v)}{\sqrt{G}}[/math] He also states that the coupling of the charge to the gravimagnetic field is [math]\sqrt{G}m \cdot \frac{v}{c} \times \frac{2 \omega \times v}{\sqrt{G}}[/math] He didn’t go into much explanation why, but we can show why for the purpose of this work. The Lorentz force is of course [math]F = e(v \times \mathbf{B})[/math] and the coriolis force is [math]-2m (\omega \times v)[/math] You can set the two quantities equal and this gives [math]e(v \times \mathbf{B}) = -2m (\omega \times v)[/math] If we from here, we divide through by the gravitational charge we get, [math]v \times B = -\frac{2(\omega \times v)}{\sqrt{G}}[/math] Here you may notice we have retrieved Motz’ and Sciama’s coupling equation: It also leads to the gravitational analogue of the Lorentz force with G = 1: [math]F = m(v \times \mathbf{B})[/math] '' Concerning the gravimagnetic force (ie. Coriolis force) we simply square the equation [math]\sqrt{G}m \cdot \frac{v}{c} \times \frac{2 \omega \times v}{\sqrt{G}}[/math] Doing so will lead to the inverse of the Gamma function for relativistic velocities a subject I covered in an algebra blog post recently. [math]Gm^2 \cdot \beta^2 \times (\frac{2 \omega \times v}{\sqrt{G}})^2[/math] Edited August 1, 2018 by Dubbelosix Quote Link to comment Share on other sites More sharing options...
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