The direction of entropy in the universe

[0xabc0]

I'm a little confused about entropy and it's increase along the arrow of time. My perception of entropy is that it is a measure of order in a system and that high entropy represents dissorder. The final result given sufficient time.

From what I have read, the universe has began with low and is moving toward high entropy. This seems contradictory to what I see.

Considering the arrow of time. Can the soup of subatomic particles than formed the primordal atoms be considered a move of low to high entropy? Can the gravitational collapse of primordal gas to a star be considered a similar move? Similarly, the synthesis of heavier nuclei within a star, the collapse of heavier molecular clouds to form solar systems, the evolution of life.

To me, it seems that the universe begins in a dense state of high entropy and through the fundamental forces evolves to a state of lower entropy, a minima, and slowly toward higher entropy given sufficent time.

This is at odds with what I have learn't. Any advice?

 

[torquil]

You must also include the entropy of the gravitational field itself (or entropy of spacetime if you will).

 

[JustinRyan]

http://tim-thompson.com/entropy1.html

I also had problems equating a large volume of high temperature homogeneous gas to a state of low entropy. But because of the the extremely high temperature at the beginning, and if you consider the vast majority of space is now around 2 degree K, it is not difficult to see entropy is increasing.
Entropy is energy per degree kelvin. Energy is constant. Temperature has fallen to near zero.

 

[ModusPwnd]

Might be time to consider the 'real' definition of entropy (from statistical mechanics), the logarithm of micro states associated with the macro state. Forget the logarithm, its there for convenience. Entropy is a measure of the number of micro states associated with a macro state. For the macro state of the early universe think about how many way can can arrange the constituents such that the early universe state is achieved. Now consider the current universe and consider how many ways you can arrange things to get a universe that looks like ours. The early universe is small and hot meaning energy states are filled up and there isn't much room to rearrange thigns. The current universe is large and cold. There is plenty of room in the energy states to rearrange things and get the same macroscopic picture.

Maybe some will disagree, but this is how I intuitively internalize it.

 

[MrMatt2532]

http://tim-thompson.com/entropy1.html

I also had problems equating a large volume of high temperature homogeneous gas to a state of low entropy. But because of the the extremely high temperature at the beginning, and if you consider the vast majority of space is now around 2 degree K, it is not difficult to see entropy is increasing.
Entropy is energy per degree kelvin. Energy is constant. Temperature has fallen to near zero.

You've got this backwards. See "heat death of the universe".

Entropy has units of energy per degree kelvin. That doesn't mean that to calculate entropy you find energy and divide by temperature.

Additionally, in viewing the average temperature of the universe, realize this is complicated by the fact that the universe is expanding.

Better to think of a simpler isolated system (besides the universe). Think of an isolated room of air with a spinning flywheel in it. Over time the friction between the flywheel and the air will cause the flywheel to slow down until it stops. All of that kinetic energy will have converted into internal energy and the average temperature of the system will have increased. This is an example of entropy increase as the energy in the system has dispersed as thermal energy and there is no more energy available to do work.

 

[JustinRyan]

OK, so in the heat death of the universe, when all available energy is converted to un-usable heat energy, and the universe has reached a thermodynamic equilibrium, the entropy is not calculated by energy divided by temperature?

In your isolated system, where did the energy of the flywheel originate?
If I were to put this room next to one without a flywheel, I could use a heat engine to couple the two rooms and extract that heat energy into work couldn't I?

I was not suggesting it is as a simple matter to estimate the entire amount of energy that exist, the proportion of that energy that is currently internal heat energy or the mean temperatue of the entire universe. Just a generalisation to demonstrate the direction of entropy with time is upward.

 

[chill_factor]

You seem to have units confused with actual concepts. You can't calculate the entropy by taking the energy and dividing it by temperature. Angular momentum, for example, has units of kg·m2/s but you don't get it by taking the mass, times the area of the thing, and dividing it by seconds.

Remember that dS = δQ/T. This is EQUILIBRIUM thermodynamics - it assumes temperature is constant, or that your system is in contact with a reservoir so big that the temperature of the reservoir might as well be constant.

But that isn't the case. What happens when there is a temperature change? Or a phase change that absorbs heat while not producing new entropy, such as in many solid-solid transitions?

Also, for the case of the universe, 99.999% of the entropy is contained in the cosmic event horizon and of the remaining 0.0001%, 99.999% is contained in supermassive black holes and of the remaining 0.0001% of 0.0001%, 99.999% is contained in the background radiation.

All matter and radiation derived thereof contributes a negligible footnote to universal entropy.

What I suggest is pick up a statistical mechanics book used in a standard chemistry or physics class and read the relevant parts on entropy.

 

[JustinRyan]

If there are two rooms with identical flywheels at identical speeds. Both expend all their energy and convert it into heat enery. Both rooms reach thermodynamic equilibrium, but one at a lower temperature due to a greater volume of air in the room. Which room represent a greater increase in entropy?

 

[MrMatt2532]

If there are two rooms with identical flywheels at identical speeds. Both expend all their energy and convert it into heat enery. Both rooms reach thermodynamic equilibrium, but one at a lower temperature due to a greater volume of air in the room. Which room represent a greater increase in entropy?

The larger room.

Another way of looking at it is you could just say you add heat to a constant volume closed system.

Then Δs=m*cv*ln(T_final/T_initial) for constant specific heat.

Now say T_initial is 300, and for the smaller room, Q causes the T_final to be 310. Now say the bigger room is twice the mass so its T_final would be 305.

So the bigger room must have a larger increase in entropy because 2*ln(305/300) is greater than ln(310/300). You could try this for any combination of masses, T_initial, and Q, and you will find that the entropy increases the most for the larger mass.

 

[JustinRyan]

OK, I had to consider that a while and now see the fault in my logic.
A lower temperature represents a GREATER dT.
Apologies for misleading the thread.