Re: 2nd law of thermodynamics


"spintronic" <spintronic@xxxxxxxxxxx> wrote in message
news:1184594235.729409.148550@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Jul 16, 11:59 am, "Ross Langerak" <rlange...@xxxxxxxxxxxxx> wrote:

"Disorder" is not a scientifically defined term, at least not in
thermodynamics.

Yes it is. Its the sum of combinations of possible states. Or degrees
of freedom.

No, it isn't. On my desk, I have a classical thermodynamics textbook, a
statistical thermodynamics textbook, and a quantum mechanics textbook. None
of them provides a definition of order, disorder, or complexity.

Your definition of disorder looks suspiciously similar to the definition of
statistical entropy. Pure (unitless) entropy is the natural logarithm of
the number of equivalent states of a system. Degrees of freedom are not
relevant to this definition of entropy.

Imagine a large square box called (Box A).

Firstly we put a smaller box, called (Box B) into (Box A).

(Box B) is 1/3 the length of any of the sides of (Box A).

(Box B) has 27 possible states in (Boc A).

Now take (Box B) out of (Box A)

We wil now put a larger box, (Box C) into (Box A).

(Box C) is 1/2 the length of any of the sides of (Box A)

(Box B) Had 27 possible states. Yet
(Box C) Only has 8 Possible states.

Therefore (Box C) has less freedom, and is thus more organised, due to
its larger volume! ;-) Kapische??

Being more ordered it has LESS entropy!

Assuming the sizes of the boxes are supposed to be analogous to the energy
levels of an atom, your use of order is unnecessary since entropy is already
defined independently of any definition of order (or disorder) that you may
provide.

Also, for each position of either of the smaller boxes in Box A, there would
only be one equivalent state (the existing state, not another state), so the
entropy in all cases would be the same (zero).

Take your most basic system of entropy.
2 atoms. (a & b)
Atom (a) is in energy level 1, and Atom (b) is in energy level 2.
Atom (a) in more highly ordered. (smaller volume)
Again, you are using an undefined, and therefore meaningless, term.
Atom (b) now transferres a photon to Atom (a).
According to thermodynamics, because of the transfer of thermal
energy, the entropy of this system *MUST* increase. But it CANT!


Classical thermodynamics deals with the transfer of heat in macroscopic
systems, not microscopic systems consisting of one or a few atoms. For
your
system, you need to use statistical thermodynamics.

So;


I assume you mean that the transfer of a photon from atom (b) to atom (a)
results in a change in the energy levels of the electrons in the atoms,
rather than the energy levels of the atoms.

Actually its both (but anyhows) Well concentrate on the electron.
After the photon is emitted, it has a shorter wavelength due to the
loss of EPE.

What is EPE? What is "it"? Is "it" the photon or the electron?

EPE is twice the KE of the now faster orbital speed, the photon is the
diff between the 2.

The energy of the photon is equal to the difference between the energy
levels of the electron. Is this what you were getting at?

Anyhows!

Our Atom was in a more highly organised state, due to its Larger
volume (as proved by the boxes) before emission.

Again, organization is not a scientifically defined term. For one atom, for
any state, there is only one equivalent state, so the entropy will always be
the same. Volume is irrelevant.

After emission, it is less organised, yet the other atom absorbes the
photon, and becomes more ordered.

Again, order is not a scientifically defined term. For a two atom system,
with one atom excited and the other not, there are two equivalent states,
and the entropy is the natural logarithm of 2. Order or disorder, even if
they were defined, would be irrelevant.

In other words, the entropy of the system hasnt changed!

Because the system has moved from one state to another equivalent state.

Assuming both atoms are of the
same element,

Derrr!!!!

You didn't specify.

according to statistical thermodynamics, the entropy remains
the same.

Good, you agree with what ive been saying all along!

But not for the same reasons. Getting lucky doesn't validate your argument.

Atom (a) receives the photon from (b). (Transfer of thermal energy
therefore increased entropy), but at the COST of rearranged
organisation between the two atoms.

Again, "organisation" is not a scientifically defined term.

Ive just defined it for you!

Your definition doesn't work.

According to statistical thermodynamics, there is no change in entropy.

Objection, your honour, ther witness is repetetive!

Sometimes it's necessary to be repetitive.

Atom (a) is now in energy level 2, and is thus more *DISORGANISED* as
you would expect!

Again, "disorganised" is not a scientifically defined term. What
equation
are you using to calculate "disorganisation"?

The equations for the box's! For a fusion event (say 2 hydrogen
atoms) we use degrees of freedom. Same diff!

What equations? You've provided no equations. What equation are you using
to calculate "disorganization"?

In your box example, each of the smaller boxes had three degrees of freedom
(x, y, and z) in the larger box. (You do know what a degree of freedom is,
don't you?) The number of degrees of freedom was the same for each of the
smaller boxes. The number of possible (non-equivalent) states was different
(assuming the possible positions along each axis were quantized based upon
the relative sizes of the constrained and constraining boxes) for each of
the smaller boxes, but the number of degrees of freedom was the same (3).

As for the fusion reaction, how do you propose to use degrees of freedom to
determine the change in entropy? Be specific.

HOWEVER & MORE IMPORTANTLY!!!!!

Atom (b) is now in energy level 1, and has become more organised. Due
to its (NOW) smaller volume.

Volume is only relevant to the macroscopic, classical thermodynamics.

No, It isnt!

Yes, it is.

And for particles with no volume you use their wavelength to show its
amount of freedom, in its 4 vectors!

You'll have to show how that is done. Your statement makes no sense to me.
Is the particle a photon? What are the four vectors? By "amount of
freedom", do you mean the number of degrees of freedom? What exactly are
those degrees of freedom?

On a whole. Entropy, like energy, momentum, mass is conserved! :)
Enjoy!

No, entropy is not conserved. Entropy, like temperature, is not a thing
that can be moved from place to place.

Temperature can be moved from place to place!

No, it can't. If it's 60 degrees outside and 70 degrees inside, I can't
grab two degrees outside, making it 58 degrees, and put them inside, making
it 72 degrees. How would I even grab a degree?

If I want to increase the
temperature of a room by two degrees, I can't just go outside and grab
two
degrees and bring them inside. Instead, I have to add sufficient energy
to
the room to raise the temperature by two degrees. The change in
temperature
is the result of the movement of energy.

And the movement of energy, is the result of the
"rearrangement" (movement) of matter.

That would be convection, which is just one of three forms of heat transfer.
The other two are conduction and radiation.

This rearrangement in order
compensates for the transfer of thermal energy.

Why would we need to compensate for the transfer of thermal energy (I assume
you mean heat)? Just add energy (heat) to the room and the temperature
increases.

And the entropy stays constant!

Actually, no, the entropy of the room does not stay constant when you add
energy.

Likewise, if I want to decrease the entropy of a system by two joules per
degree Kelvin, I can't steal two joules per degree Kelvin from another
system and add them to the first. Entropy is a state function. Whatever
state a system may be in, it has a particular entropy. If I want to
change
the entropy of the system, I must change the state of the system by
rearranging the energy of the system. As with temperature, the change in
entropy is the result of the rearrangement of energy in the system.

You cant rearrange the temperature, (without) rearranging the matter!

Yes, you can. (Remember conduction and radiation?) Take a hot brick and a
cold brick, and hold them together. What will happen to the relative
temperatures of the bricks? Will there be a rearrangement of matter between
the bricks?

But that last paragraph was a statement about entropy, not temperature, so
what was your point? Rearrangement of matter that results in a
rearrangemnet of energy may still result in a change in entropy.

.