The Relativity of Temperature and the Weinberg Formula and Hubble Expansion

Discussion in 'Physics & Math' started by curvature, Aug 11, 2018.

  1. curvature Registered Member

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    To effects or order no larger than third after a Taylor series on Einstein's suggested kinetic equation yielded:


    \(K_0 - K_1 = \frac{E}{2}\frac{v^2}{c^2}\)


    So for a black hole the formula should still hold and was seen the emission reducing the mass of the system and so its energy content had to be a measure of its mass.


    It is an equation for discrete processes and since the emission of radiation from black holes also reduces its total mass, then maybe the equation has a more broadened spectrum to theorize with.


    So for a black hole the formula should still hold and was seen as the emission reducing the mass of the system and so its energy content had to be a measure of its mass - additionally, I showed a way to talk about those discrete processes in terms of the energy levels for a black hole ~


    \(\mathbf{S} = \frac{\hbar c}{k_BT}\frac{p}{4 \pi \hbar} = \frac{\hbar c}{k_BT}\frac{\sqrt{2mE}}{4 \pi \hbar} = \frac{1}{k_BT} \frac{\hbar c}{\lambda_0} = \mathbf{R}(\frac{Gm^2}{n^2_1k_BT} - \frac{Gm^2}{n^2_2k_BT})\)


    It is an equation for discrete processes and since the emission of radiation from black holes also reduces its total mass, then maybe the equation has a more broad spectrum to theorize with.


    \(K_0 - K_1 = \frac{E}{2}\frac{v^2}{c^2}\)


    Let’s say a few things about the ratio \(\frac{v^2}{c^2}\). It is often given beta notation \beta for shorthand but may be related also to the beta matrix found in Dirac’s theory - but I am working from memory here so I could be wrong. Anyway, it’s been related to a fine structure interpretation. But most importantly \(\frac{v^2}{c^2}\) is the inverse of the square of the Lorentz gamma function. Such an inverse would appear like this:


    \(K_0 - K_1 = \frac{E}{2}\sqrt{1 - \frac{1}{\gamma^2}}\)


    This ''asymptotic form'' for the Lorentz factor is usually used to quickly calculate velocities from large values in \gamma. By ‘’asymptotic form,’’ it is usually meant in the case of a series:


    \(\beta = 1 - \frac{1}{2} - \frac{1}{2}\gamma^{-2} - \frac{1}{8}\gamma^{-} - \frac{1}{16}\gamma^{-6} + ... higher\ terms\)


    Any formula like the ones we have featured that measure the change in the energy of the system can be written in other dynamics explaining different but related physics. Maybe even analogue cases? Who knows right now. We shown that temperature too would follow relativistic laws, but which law was disputed. I claim to have found the answer: They were in fact both correct.


    \(\Delta \mathbf{S} = 2k_BT \sqrt{1 - \frac{v^2}{c^2}} \frac{1}{1 - \frac{v^2}{c^2}} = \frac{2k_BT}{\sqrt{1 - \frac{v^2}{c^2}}} = \gamma^{-1}k_BT\)


    how do you transform between Ott's law and Einstein's? Well, you can see the above as already satisfying Einstein's version. In which case, to get Ott's law you just multiply through by \(\gamma^{-2}\)


    \(\Delta \mathbf{S} = 2k_BT \sqrt{1 - \frac{v^2}{c^2}} = \gamma^{-2} \frac{2k_BT}{\sqrt{1 - \frac{v^2}{c^2}}} = \gamma k_BT\)


    Moving relative to a thermodynamic body was only one part of the problem and the problem is the two solutions make sense that both laws hold when you take into account the relativistic effects of redshift and blueshift. Before we speak a bit more on that, I want to make very clear the equations of Ott and Einstein:


    The Ott covariant transformation is


    \(T = T_0 \sqrt{1 - \frac{v^2}{c^2}} = \gamma T\)


    And the suggestion by Einstein and Planck was


    \(T = \frac{2T_0}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{2T}{\gamma}\)


    par a factor of two that we have picked up through this derivation. The equation will be universal in understanding the relationship between the two transformation laws:


    \(T’ = \gamma T\)


    \(T’ = \frac{T}{\gamma}\)


    There was not much dispute for many years until Ott showed that a relativistically moving body can appear hotter. I came to [eventually] realize from my own investigation, that both Einstein and Ott where right and in fact temperature is not alone subjected to the relativistic motion of systems and observers at rest, or the converse, but is also subject to redshift and blueshift which means it [also] depends on the direction of motion, so I concluded the two laws for cooler and hotter descriptions of relativistic bodies depend on additional relativistic premises and so both laws should hold in different circumstances. Perhaps the more cooler equation was Hubble expansion related to a transverse and longitudinal time direction.


    The two formulas that were presented by Weinberg were:


    \(m = (\frac{H_0\hbar^2}{Gc})^{\frac{1}{3}}\)


    In which H_0 is the Hubble parameter. The alternative form was


    \(m^3 = \frac{H_0 \hbar^2}{Gc}\)


    The equation has been verbally described as ‘’mysterious’’ as no one knows truly what Weinberg was suggesting - does the formula present three particle masses, or does it reduce to a quantization of some kind of just the one mass, as supported from the first equation?


    It’s intriguing for me because I think maybe both equations may hold, so long as you consider the third power of the mass to describe three particle mass states: Which as we know now, the Neutrino does possess a three-particle mass through its phase transitions. In terms of the Hubble parameter, I have shown that it can obey an equality for the transverse and longitudinal time:


    \(H_l = \frac{c \sqrt{1 - \frac{v^2}{c^2}}}{2r_{HS}} \frac{1}{1 - \frac{v^2}{c^2}} = \frac{c}{2r_{HS}}\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = H_t\)


    *Note, to see if there are any consequences from unitin the theories like this, Weinbergs Hubble constant can be seen in longitudinal and transverse directions.


    And so I began looking into mass or a mass formula, the longitudinal and tranverse masses are


    \(m_L = \frac{m_0}{(\sqrt{1 - \frac{v^2}{c^2}})^3}\)


    \(m_T = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}\)


    So that the cube of the longitudinal mass becomes:


    \(m^3_L = \frac{m_0^3}{(\sqrt{1 - \frac{v^2}{c^2}})^6} = \frac{m_0^3}{(1 - \frac{v^2}{c^2})^3}\)


    You can do the same for the transverse mass:


    \(m^2_T = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}\)
     
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  3. curvature Registered Member

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    More on Weinbergs' formula:



    In natural units,the gravitational fine structure constant is equal to the square of the mass of a particle


    \(\alpha_G = m^2\)


    and its quantization is


    \(n\hbar = m^2\)


    The hope or immediate realization of this are attempts to find quantization of mass depending on factors of \(n\hbar\).


    The \(Gm^2\) can be thought of as the squared gravitational charge of the system analogous to the electric charge \(e^2\) - both mass and charge are conserved, treating them as coefficients to the Lie Generators, for a purey Noetherian view. Some mass formula have been suggested in literature (see What is special about the Planck mass? Sivaram). Before we look at the more advanced mass formula that will be suggested here, let’s take a look at what some have called ‘’the mysterious’’ Weinberg formula:


    \(m = (\frac{H_0\hbar^2}{Gc})^{\frac{1}{3}}\)


    In which H_0 is the Hubble parameter. This will correspond to either one particle mass, or Weinberg had in mind a spectral property leading to different masses. In any case, this equation cannot predict all particle masses with this description alone. Alternatively, the Weinberg formula has been written as


    \(m^3 = \frac{H_0 \hbar^2}{Gc}\)


    In this formulation, it has been suggested by Arun and Sivaram that it is ''unclear'' whether now he associates the equation to three particle masses [1] or as an attempt to find a fundamental unit of mass. Either way, we can see his ultimate aim would have been to describe fundamental particle masses from cosmological features, like the Hubble constant. Schwinger, Motz (et al). have demonstrated what it means to quantize a charge: Charge and mass are in fact so similar, you can indeed describe both in similar physics. Nature seems to posit some kind of universality with the presence of an electric charge being synonymous with mass. The only exception it appears in nature is a neutrino, which is expected to have no charge and a very small mass, but if it has a very small mass (you can also argue) it may possess a vanishing, but non-zero charge with a proportionality of \(e^2 \propto Gm^2\). To try and articulate how small this charge would be, it would have a mass and charge 1 millionth of that of an electron. Either way, the ultimate idea is that large numbers can determine the small in dynamic ways.


    We can in fact restate Weinberg’s formula for one that satisfies the gravitational interpretation of the charge. I provide this as a simple manipulation of his equation:


    \(Gm^2 = \frac{H_0 \hbar^2}{mc}\)


    and for a single charge


    \(\sqrt{G}m = \sqrt{\frac{H_0 \hbar^2}{mc}}\)


    Let’s have a quick look at a more advanced mass formula candidate


    \(M = nk[\frac{mc}{e}\frac{1}{2 \sqrt{T}}]^n m_e = nk [\frac{\sqrt{G}m}{2 e}]^n m_e\)


    Immediately we can notice the use of the gravitational charge in the last term \sqrt{G}m - the only difference is that it has focused on the Planck mass definition of the charge. The Planck mass should not necessarily be considered fundamental, it seems like too much a basic unit of matter for any particles we have observed in the standard model. Though the middle term is good for string dynamics and superstring tension, the last term appears to be made of more fundamental assumptions which included the gravitational charge of the system. The adjustable parameters is what allows us to predict particle masses, for example n = 2, k = 3 gives the muon mass, n = 2, k = 4 gives the pion mass, n = 3, k = 4 the \(\Delta\)-resonance, n = 3, k = 6 the D meson n = 3, and many many more particles can be predicted from it.


    Anyway, the main point may have became a bit clearer: The more advanced suggested mass formula does indeed predict a wide range of particle masses, but more importantly, the Weinberg formula can be described as a gravitational charge which can fit into the more advanced formula. Let’s see how that would look ~


    \(\frac{m}{m_e} = nk [\frac{\sqrt{G}m}{2 e}]^n = nk [\sqrt{\frac{H_0 \hbar^2}{mc}}\frac{1}{2 e}]^n = nk [\sqrt{\frac{H_0}{mc}}\frac{\hbar}{2 e}]^n\)


    So in this sense, we get to keep Weinberg’s attractive idea about determining fundamental parameters from cosmic parameters. And hopefully a more robust understanding into the Hubble transformation laws could maybe lead to particles being dynamically determined by a number of factors, shaped when the universe was young. This term \(\frac{\hbar}{2 e}\) can be seen using a more complicated object using the magnetic moment ~


    \(\frac{m \mu_B}{e^2} = \frac{\hbar}{2e}\)


    in which


    \(\mu_B = \frac{e \hbar}{2m}\)


    Which is the Bohr magneton. Since this term arises \(\frac{\hbar}{2e}\) then we can even speculate it in the following form as:


    \(\frac{m}{m_e} = nk [\frac{\sqrt{G}m}{2 e}]^n = nk [\sqrt{\frac{H_0 \hbar^2}{mc}}\frac{1}{2 e}]^n= nk [\sqrt{\frac{H_0}{mc}}\frac{m \mu_B}{e^2} ]^n\)


    It's only theoretical looking for alternative ways to write something, but sometimes messing around like this can be intuitive.
     
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  5. origin In a democracy you deserve the leaders you elect. Valued Senior Member

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    So you like to copy other peoples work and add a little bit of inane bull shit of your own to make you feel intelligent. Got it.
    Do you have to do it here, reiku?
     
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  7. NotEinstein Valued Senior Member

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    Oh my gosh, are you trolling!!!111one!

    Please Register or Log in to view the hidden image!



    It's kinda sad seeing somebody with that much interest in physics have (even after all this time) so little competency in it.
     
  8. origin In a democracy you deserve the leaders you elect. Valued Senior Member

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    I agree. Farsight also comes to mind. He has spent close to a decade spouting bull shit when he could have taken enough courses at just a community college to be able to actually have a fair understanding of physics. I guess bull shit is just alot easier.
     
  9. curvature Registered Member

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    One of the most stopid things I have ever heard, only intended to get a reaction out of me.
     
  10. origin In a democracy you deserve the leaders you elect. Valued Senior Member

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    I don't want any reaction from you, just stating facts.
     
  11. curvature Registered Member

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    Well look, it is common practice to work from to add things to physics, its called progression.
     
  12. origin In a democracy you deserve the leaders you elect. Valued Senior Member

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    I am not going to have a conversation with a person who dishonestly sneaks back on the forum to post nonsense. Bye reiku hopefully you won't be here long.
     
  13. curvature Registered Member

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    I don't really care what else you assert - I will only call out the nonsense that you seem to be under the delusion that science does not progress from already known established equations. I can only imagine what made you think otherwise. Bit silly wasn't it?
     
  14. origin In a democracy you deserve the leaders you elect. Valued Senior Member

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    10,579
    But you don't understand the equations and you are only capable of doing low level high school algebra. You are doing a caricature of science. You know it we all know it.
     
  15. curvature Registered Member

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    You just keep saying that to yourself. Quite a few people of reasonable credentials have had chances to object - it's funny the only people that tend to do that are groups of internet trolls.
     
  16. origin In a democracy you deserve the leaders you elect. Valued Senior Member

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    I would have liked to have responded to you but you were banned because you're an internet troll. How ironic....
     
  17. NotEinstein Valued Senior Member

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    I remember the last time I objected to the contents of your posts: it turned out you couldn't even spot a simple minus sign mistake in your own math. No, instead, you threw insults at me. That's why "quite a few people of reasonable credentials" haven't responded to you: because you're incapable of intellectually handling such a response. Have fun being banned again.
     

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