Logarithmic Cosmic Evolution and Temperature Dependence

[MrOrlock]

I'll start with a set of simple relationships

$$\dot{R} = \frac{k_B}{m}T$$

$$K_B = \frac{mR^2}{t^2T}$$

Plugging the second into the first gives

$$(\frac{\dot{R}}{R})^2 = \frac{\ddot{T}}{T}$$

We find as a statement of the algebra that the temperature has a first order in relation to the fluid expansion of the universe were we take the temperature as an additive component as

$$(\frac{\dot{R}}{R})^2 + \frac{\ddot{T}}{T} =\frac{8\pi G}{3} \rho$$

This is the simplest pressureless and shape-free model of a temperature dependent Friedmann equation. Rearranging

$$T\dot{R}^2 + R^2\ddot{T} =\frac{8\pi GR^2T}{3} \rho$$

Using the dimensions of G ie.

$$G = \frac{Rc}{m}$$

We get

$$T\dot{R}^2 + R^2\ddot{T} =\frac{8\pi R^3c^2T}{3m} \rho$$

Applying variational calculus defining R^3 as the unit volume we get

$$\delta T \dot{R}^2 + \delta R R \ddot{T} =\frac{8\pi Tc^2}{3m}(\delta \rho V +\rho \delta V)$$

Dividing through by R^2T we get a logarithmic form for Cosmic Evolution

$$\frac{\delta T}{T} (\frac{\dot{R}}{R})^2 + \frac{\delta R}{R} \frac{\ddot{T}}{T} =\frac{8\pi c^2}{3mR^2}(\delta \rho V +\rho \delta V)$$

I can identify (c/R)^2 as the invariant time operator, and by using the logarithmic law, obtain this dimensionless case

$$[\frac{\delta T}{T} (\frac{\dot{R}}{R})^2 + \frac{\delta R}{R} \frac{\ddot{T}}{T}] (\frac{R}{c})^2$$

$$= \frac{8\pi}{3m}(\delta \rho V +\rho \delta V)$$

$$= [\delta \log T (\frac{\dot{R}}{R})^2 + \delta \log R \frac{\ddot{T}}{T}](\frac{R}{c})^2$$

$$ = \frac{8\pi}{3m}(\delta \rho V +\rho \delta V)$$

$$= \delta \log (T,R)[(\frac{\dot{R}}{R})^2 + \frac{\ddot{T}}{T}] (\frac{R}{c})^2$$

$$= \frac{8\pi}{3m}(\delta \rho V +\rho \delta V)$$

There's also a possible logarithmic entropy encoded in the equation

 

[Michael 345]

I'll start with a set of simple relationships

$$\dot{R} = \frac{k_B}{m}T$$

$$K_B = \frac{mR^2}{t^2T}$$

Plugging the second into the first gives

$$(\frac{\dot{R}}{R})^2 = \frac{\ddot{T}}{T}$$

We find as a statement of the algebra that the temperature has a first order in relation to the fluid expansion of the universe were we take the temperature as an additive component as

$$(\frac{\dot{R}}{R})^2 + \frac{\ddot{T}}{T} =\frac{8\pi G}{3} \rho$$

This is the simplest pressureless and shape-free model of a temperature dependent Friedmann equation. Rearranging

$$T\dot{R}^2 + R^2\ddot{T} =\frac{8\pi GR^2T}{3} \rho$$

Using the dimensions of G ie.

$$G = \frac{Rc}{m}$$

We get

$$T\dot{R}^2 + R^2\ddot{T} =\frac{8\pi R^3c^2T}{3m} \rho$$

Applying variational calculus defining R^3 as the unit volume we get

$$\delta T \dot{R}^2 + \delta R R \ddot{T} =\frac{8\pi Tc^2}{3m}(\delta \rho V +\rho \delta V)$$

Dividing through by R^2T we get a logarithmic form for Cosmic Evolution

$$\frac{\delta T}{T} (\frac{\dot{R}}{R})^2 + \frac{\delta R}{R} \frac{\ddot{T}}{T} =\frac{8\pi c^2}{3mR^2}(\delta \rho V +\rho \delta V)$$

I can identify (c/R)^2 as the invariant time operator, and by using the logarithmic law, obtain this dimensionless case

$$[\frac{\delta T}{T} (\frac{\dot{R}}{R})^2 + \frac{\delta R}{R} \frac{\ddot{T}}{T}] (\frac{R}{c})^2$$

$$= \frac{8\pi}{3m}(\delta \rho V +\rho \delta V)$$

$$= [\delta \log T (\frac{\dot{R}}{R})^2 + \delta \log R \frac{\ddot{T}}{T}](\frac{R}{c})^2$$

$$ = \frac{8\pi}{3m}(\delta \rho V +\rho \delta V)$$

$$= \delta \log (T,R)[(\frac{\dot{R}}{R})^2 + \frac{\ddot{T}}{T}] (\frac{R}{c})^2$$

$$= \frac{8\pi}{3m}(\delta \rho V +\rho \delta V)$$

There's also a possible logarithmic entropy encoded in the equation

No shallow end at your location?

 

[MrOrlock]

No shallow end at your location?

I guess not, the first equation should be R'^2 instead of R' just to make a note

 

[exchemist]

I'll start with a set of simple relationships

$$\dot{R} = \frac{k_B}{m}T$$

$$K_B = \frac{mR^2}{t^2T}$$

Plugging the second into the first gives

$$(\frac{\dot{R}}{R})^2 = \frac{\ddot{T}}{T}$$

We find as a statement of the algebra that the temperature has a first order in relation to the fluid expansion of the universe were we take the temperature as an additive component as

$$(\frac{\dot{R}}{R})^2 + \frac{\ddot{T}}{T} =\frac{8\pi G}{3} \rho$$

This is the simplest pressureless and shape-free model of a temperature dependent Friedmann equation. Rearranging

$$T\dot{R}^2 + R^2\ddot{T} =\frac{8\pi GR^2T}{3} \rho$$

Using the dimensions of G ie.

$$G = \frac{Rc}{m}$$

We get

$$T\dot{R}^2 + R^2\ddot{T} =\frac{8\pi R^3c^2T}{3m} \rho$$

Applying variational calculus defining R^3 as the unit volume we get

$$\delta T \dot{R}^2 + \delta R R \ddot{T} =\frac{8\pi Tc^2}{3m}(\delta \rho V +\rho \delta V)$$

Dividing through by R^2T we get a logarithmic form for Cosmic Evolution

$$\frac{\delta T}{T} (\frac{\dot{R}}{R})^2 + \frac{\delta R}{R} \frac{\ddot{T}}{T} =\frac{8\pi c^2}{3mR^2}(\delta \rho V +\rho \delta V)$$

I can identify (c/R)^2 as the invariant time operator, and by using the logarithmic law, obtain this dimensionless case

$$[\frac{\delta T}{T} (\frac{\dot{R}}{R})^2 + \frac{\delta R}{R} \frac{\ddot{T}}{T}] (\frac{R}{c})^2$$

$$= \frac{8\pi}{3m}(\delta \rho V +\rho \delta V)$$

$$= [\delta \log T (\frac{\dot{R}}{R})^2 + \delta \log R \frac{\ddot{T}}{T}](\frac{R}{c})^2$$

$$ = \frac{8\pi}{3m}(\delta \rho V +\rho \delta V)$$

$$= \delta \log (T,R)[(\frac{\dot{R}}{R})^2 + \frac{\ddot{T}}{T}] (\frac{R}{c})^2$$

$$= \frac{8\pi}{3m}(\delta \rho V +\rho \delta V)$$

There's also a possible logarithmic entropy encoded in the equation

This farcical attempt at algebra is horribly familiar.

If R'² = KT/m and K= mR²/t²T, then R'² =R²/t²,

so (R'/R)² = 1/t².

T doesn't appear anywhere, as it cancels.

 

[exchemist]

No shallow end at your location?

It's Reiku again

 

[MrOrlock]

Let's do it

$$\dot{R}^2 = \frac{k_B}{m}T$$

$$K_B = \frac{mR^2}{t^2T}$$

Plug eq 2 into 1

$$\dot{R} = \frac{mR^2}{t^2T}\frac{1}{m}T$$

We choose not to cancel Just because T is balanced by T does not mean they need cancel. Say you have some extensive knowledge of Newtonian or Kepler derivations, factors of a variable often shows up in the numerator and denominator without cancelling towards a certain goal.

Dividing through by R^2

$$(\frac{\dot{R}}{R})^2 = \frac{m}{t^2T}\frac{1}{m}T$$

And by stipulation we allow only a mass to simplify so that our temperature satisfies the symmetry of the fluid expansion yielding


$$(\frac{\dot{R}}{R})^2 = \frac{\ddot{T}}{T}$$

Both sides are dimensionless and are appropriate to any algebra.

 

[exchemist]

Let's do it

$$\dot{R}^2 = \frac{k_B}{m}T$$

$$K_B = \frac{mR^2}{t^2T}$$

Plug eq 2 into 1

$$\dot{R} = \frac{mR^2}{t^2T}\frac{1}{m}T$$

We choose not to cancel Just because T is balanced by T does not mean they need cancel. Say you have some extensive knowledge of Newtonian or Kepler derivations, factors of a variable often shows up in the numerator and denominator without cancelling towards a certain goal.

Dividing through by R^2

$$(\frac{\dot{R}}{R})^2 = \frac{m}{t^2T}\frac{1}{m}T$$

And by stipulation we allow only a mass to simplify so that our temperature satisfies the symmetry of the fluid expansion yielding


$$(\frac{\dot{R}}{R})^2 = \frac{\ddot{T}}{T}$$

Both sides are dimensionless and are appropriate to any algebra.

You have T in both numerator and denominator so they cancel one another out, meaning the expression is independent of T.

T'' appears nowhere.

Give it up Gareth, you don't know what you're doing.

 

[MrOrlock]

You have T in both numerator and denominator so they cancel one another out, meaning the expression is independent of T.

Based on dimensional anslysis, we make the assumption that T/t^2 holds for the interpretation of a double derivative. It's a very minor stipulation you seem to have missed. We're not proving a statement outside that the length parameter must be squared $$\dot{R}^2 \propto \ddot{T}.$$

 

[MrOrlock]

And no, T/T does not need to cancel. Do the ratio's of masses in a fine structure cancel? No they don't.

R/R in the Friedmann equation, has the denominator fixed while the fluid expansion varies in the numerator. A similar property is found here. By Simplifying T with T is a case of oversimplified crunching. For someone who claims to know algebra, your arguments seem to be lost on the deeper meaning of what is posted here.

 

[exchemist]

Based on dimensionsl anslysis, we make the assumption that T/t^2 holds for the interpretation of a double derivative. It's s very minor stipulation you seem to have missed. We're not proving a statement outside that the length parameter must be squared $$\dot{R}^2 \propto \ddot{T}.$$

Makes no difference. If you have T in both numerator and denominator the expression is independent of T. There is no way of getting away from that.

"T/t^2 holds for the interpretation of a double derivative" is gibberish.

 

[exchemist]

And no, T/T does not need to cancel. Do the ratio's of masses in a fine structure cancel? No they don't.

R/R in the Friedmann equation, has the denominator fixed while the fluid expansion varies in the numerator. A similar property is found here. By Simplifying T with T is a case of oversimplified crunching. For someone who claims to know algebra, your arguments seem to be lost on the deeper meaning of what is posted here.

x/x = 1.

And T/T = 1. That remains true however you try to dress it up.

 

[MrOrlock]

Makes no difference. If you have T in both numerator and denominator the expression is independent of T. There is no way of getting away from that.

"T/t^2 holds for the interpretation of a double derivative" is gibberish.

No it's not gibberish, I assume you know dimensional algebra? What are the dimensions of 1/R^2 d^2R/dt^2 and what are the dimensions of (R/Rt)^2? The math doesn't need to be carbon copies under dimensional algebra, that's the whole point.

 

[MrOrlock]

x/x = 1.

And T/T = 1. That remains true however you try to dress it up.

Not always, this is a very simplified view of an argument that isn't been applied here. Is R'/R or R/r equal to 1? No it's not, any more than (R''/R)^2 or T''/T and you haven't addressed the issues that quanties need not be exact to cancel, a good case I explained was the masses that calculate a fine structure.

 

[exchemist]

Not always, this is a very simplified view of an argument that isn't been applied here. Is R'/R or R/r equal to 1? No it's not, any more than (R''/R)^2 or T''/T and you haven't addressed the issues that quanties need not be exact to cancel, a good case I explained was the masses that calculate a fine structure.

Nonsensical gibberish.

But, presuming you to be Reiku, you'll soon be gone, anyway.

Until next time, then.................

 

[MrOrlock]

So, you don't realise the difference between a ratio of T/T where T=T to T/t where T not equal to t? And you don't understand the subtle issues of mathematical objects like t/t(0)?

Some people think they know it all until they demonstrate they know little.

 

[MrOrlock]

You've not addressed anything I've pointed out to you, so I'm not playing with you. I don't like one sided arguments.

 

[Michael 345]

I guess not, the first equation should be R'^2 instead of R' just to make a note

Could be R2D2 for all the sense it makes to me

Came to SciForum for few reasons. One was to swim in science

Thanks to yourself I'll be coughing up undigested equations for a few days

 

[MrOrlock]

Is there a time limit on editing? The gravitational constant should be
G = Rc^2/m
As defined by its parameter.

 

[MrOrlock]

It follows correctly though so no harm done

 

[origin]

It's Reiku again

Bingo! I got 2 lines in, and it was pretty clear. The guy never gives up. If he spent half the effort to learning physics that he does making up this crap he'd be pretty learned.