The entropy of systems

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15 years 3 months ago #23426 by GD
Replied by GD on topic Reply from
F=ma is no longer valid since it follows Galilean transformations. Einstein modified it to work with Lorentz transformations and Maxwell's energy equations. Unfortunately the theory of special relativity still deals only with particle motion which is referred to 'relativistic' <i>mechanics</i>.

What we need is 'relativistic' <i>thermodynamics</i>.

So I think the equation would be a mix between equations from Maxwell, Einstein, Boltzmann, Planck, and Lorentz.

This equation has to show that energy transfer causes a force (gravity).

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15 years 3 months ago #22958 by PhilJ
Replied by PhilJ on topic Reply from Philip Janes
Jim: 24 Jul 2009 : 14:10:07 <blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Momentum of a photon assumes zero mass and rather than putting zero into the math calculation you eliminate the unit of mass and then do the math-not a good thing. If a photon has momentum is also has acceleration and mass and no matter how you hide F=ma it is still there. Isn't F=p/t or dp/dt as good as F=ma?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

I assume zero rest mass because a photon cannot rest; it must go at the speed of light. I do not assume zero mass (1) because energy is equivalent to mass moving at the speed of light, (2) because acceleration is equivalent to gravity, (3) because inertial mass is equivalent to gravitational mass (at least withing our universe), (4) because mass reacts to gravity and (5) because even a photons mass is surrounded by a gravity well to which all other masses must react. Assuming zero mass for a photon means you assume that a photon is not surrounded by its own gravity well. If a photon changes direction, the direction of its momentum changes; according to Newtons third law, an equal and opposite change must occur to the momentum of something else.

A photon does accelerate laterally when bending around a gravity well. F = ma can be applied to that lateral acceleration, but only at the instant when the gradient of gravity is perpendicular to the path of the photon. At all other times, f is not equal to ma because some of the momentum goes into the changing relativistic mass. We can avoid all that complexity by saying "f = ma" for slow particles and by applying "f = dp/dt" to photons and to particles with relativistic speeds.

Note: I exclude the ether from equivalence of inertial mass and gravitational mass. I believe the ether is incredibly inert, but has no gravity of its own.
Fractal Foam Model of Universes: Creator

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15 years 3 months ago #22960 by GD
Replied by GD on topic Reply from
PhilJ: I would have the following comments:

... I do not assume zero mass (1) because energy is equivalent to mass moving at the speed of light,

GD: I can argue the opposite: I assume light has zero mass because mass moving at the speed of light is equivalent to energy.

(2) because acceleration is equivalent to gravity,

GD: The acceleration of radiating bodies in a non-equilibrium dissipative system is equivalent to saying "gravity".

(3) because inertial mass is equivalent to gravitational mass (at least withing our universe),

GD: same as (2)

(4) because mass reacts to gravity and...

GD: mass reacts to the change in momentum as the internal energy of matter tends to zero exergy.

(5) because even a photons mass is surrounded by a gravity well to which all other masses must react. Assuming zero mass for a photon means you assume that a photon is not surrounded by its own gravity well. If a photon changes direction, the direction of its momentum changes; according to Newtons third law, an equal and opposite change must occur to the momentum of something else.

GD: Newton's laws do not apply to thermodynamics: acceleration due to the exchange of energy from a mass to the environment.
Also direction of momentum can change according to the general theory of relativity (curved space-time). This is why our galaxy has curved paths towards the main attractor (black hole).

A photon does accelerate laterally when bending around a gravity well. F = ma can be applied to that lateral acceleration, but only at the instant when the gradient of gravity is perpendicular to the path of the photon. At all other times, f is not equal to ma because some of the momentum goes into the changing relativistic mass. We can avoid all that complexity by saying "f = ma" for slow particles and by applying "f = dp/dt" to photons and to particles with relativistic speeds.

GD: Gravity wells are the same as saying: entropy change.
Entropy: A concept unknown in Newton's days. If he would have known, he would have called Gravity: Entropy change.

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15 years 3 months ago #22961 by Jim
Replied by Jim on topic Reply from
Have you guys ever looked at photochemical events such as photosynthesis? Photons come to rest in these reactions and mass must be gained as a result of that acceleration. The photon is accelerated from light speed to what ever speed the photochemical reaction requires just as stated in F=ma. All photons are eventually transformed into mass and entrophy has nothing to do with it.

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15 years 3 months ago #22962 by GD
Replied by GD on topic Reply from
Jim,

here is a paper which indicates that energy transfer in a process involving photosyntisis is 80 to 90% efficient:

arxiv.org/PS_cache/arxiv/pdf/0806/0806.4725v2.pdf

"...Exciton transfer among chlorophyll molecules is the energy transport mechanism of the initial step of the photosynthetic process

The role of quantum coherence and the environment in the dynamics of excitation energy transfer is not fully understood. In this work, we introduce the concept of dynamical contributions of various physical processes to the energy transfer efficiency

In this work, we have addressed the role of quantum coherence and the environment in excitonic energy transfer.

To this end, we have characterized the underlying processes constituting the open quantum walk of the excitation in terms of their contribution to the transfer efficiency.

The methods presented here are general and can be applied to a large class of transport systems in the presence of Markovian environments.

Within both the Greens function and the energy transfer susceptibility formalisms we conclude that the major part of the high efficiency of the Fenna-Matthews-Olson protein complex of about 80% or 87.5% is due to environment-induced relaxation down to the lowest energy site.

The role of quantum coherence induced by the Hamiltonian dynamics can be quantified at around 10% or 7.5% respectively. Furthermore, we used the susceptibility measure to assign percentage-wise contributions to exciton relaxation pathways in the molecular basis. The detailed analysis of the open quantum dynamics presented in this work could be harnessed for engineering artificial materials such as quantum dots [25] to achieve optimal energy transport in realistic environments..."


Conclusion: energy transfer is only 80 to 90% efficient due to energy losses caused by gravity (the flow of energy which produces work).

For example energy transfer in the sun is alot less efficient to produce work between complex systems as it is here on Earth. The mass of the sun is closest to zero exergy than it is for the mass of the Earth.

Therefore the efficiency of energy transfer of a system is related to the size and motion (acceleration) of its mass.

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15 years 3 months ago #23704 by Jim
Replied by Jim on topic Reply from
GD, Thankyou for the link. The key idea of this paper is:"energy transfer is not fully understood". All that follows is bs in my opinion and all such projects lead no where. The transfer of energy has to begin with the unit we call the electron volt being what is transfered. The way these units are counted matters little compared to understanding the transfer is always one unit at a time.

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