I heard in a documentary that : Irreversibility of time is linked to Heat Dissipation in thermodynamics. The theorist says "passage of time" is only a macro sensation,

the underlying effect is the heat dissipation (loss in energy) and 2nd law of thermodynamics.

I feel this theory looks very interesting and quite un-common, maybe some specialist may highlight their views and any papers on it.

1) Am referring to the relation between time and amount of heat dissipation. Time == Qty of Heat Dissipation

2) EDITED: Additional questions from the answer.

Arrow of time seems showing irreversibility....

1) Suppose we are at zero kelvin, no atoms move, nothing.

Is there any existence of time in this local environment ?

2) Is there any Equation linking Time variation and Entropy variation, like :

P[ dS/dt > 0 ] == 1

  • $\begingroup$ The thermodynamic "arrow of time" has been around for a long time and it's rather mainstream. On some level it has things the wrong way round. Time is not an abstract. Time is that which the clock shows. In order to have time one needs to have clocks. Clocks are physical systems that produce monotonously rising measurement that are ideally equidistant and that agree with other clocks that do the same thing using different principles. The only way we know how to build clocks is by using systems in "perfect disequilibrium", i.e. with an energy source and a heat sink. $\endgroup$ – CuriousOne Jul 28 '16 at 4:31
  • $\begingroup$ Am referring to the relation with amount of heat dissipation. Time == Qty of Heat Dissipation $\endgroup$ – quantCode Jul 28 '16 at 8:22
  • $\begingroup$ I know what you are referring to. It is pretty much acknowledge by physicists that one can't define time without an irreversible process. Systems in equilibrium don't have a dynamic, only systems far from equilibrium do. $\endgroup$ – CuriousOne Jul 28 '16 at 9:24
  • $\begingroup$ Question is how quantity of time is related by quantity of heat through equations ? $\endgroup$ – quantCode Jul 31 '16 at 18:16
  • $\begingroup$ It's related trough devices called clocks. $\endgroup$ – CuriousOne Jul 31 '16 at 21:44

You are talking about the thermodynamic arrow of time; I'll talk about this in reference to entropy. It's actually not an uncommon idea, especially in cosmology.

Let's think about a video of a teacup being dropped to the floor and shattering. We know (one might say instinctively) when the video is being played whether it is being played forwards or backwards. We know that teacups do not leap into our hands after lying shattered on the ground; instead, they fall from our hands and shatter - in other words, entropy increases. That's the second law: that most of the time, entropy will increase (as formulated by Boltzmann, $S = k*log W$, with k being Boltzmann's constant, S being the amount of entropy in a system and W being the number of indistinguishable arrangements of a system).

That's the thermodynamic arrow of time: we know the direction of time because entropy increases most of the time, so we know something is played backwards when entropy decreases, like a broken teacup becoming intact as it leaps into our hand. Now, the formulation of entropy is statistical. What this means is that for 99.99999999999999999999999999999% of the time (not an exact number) entropy increases, but occasionally, there'll be fluctuations. This could lead you into a whole description of the beginning of the universe and the question of why it started at low entropy (for more information on that, look at this question).

Dissipation is the result of an irreversible process. That is, it doesn't obey time symmetry; it looks different from the direction of past to future as opposed to the direction of future to past. Heat transfer is dissipative because it is the transfer of energy from a hotter body to a colder one. This dissipation obeys the second law of thermodynamics (aka, entropy). This is where the connection to the thermodynamic arrow of time comes in, of course. (If you'd like the equation, the entropy change of a system at temperature $T$ is $dS = δQ/T$, though that's simplifying it dramatically).

You can look at this website for more information.

Hope this helps!

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