Ok, so I found that (a^2+a)x_4 = ab+b (ignore #3, I made a miscalculation).
a^2+a = 0 => a=0 or a=-1
No solution:
a^2+a=0 and ab+b != 0
a=0, b \in R except 0
One solution:
ab+b != 0 <=> ab != -b
a \in R except -1 and b \in R or the opposite.
No unique solution:
a^2+a=0 and ab+b = 0
a=0 and...
No, the picture just shows the system of equations.
This is it: http://peecee.dk/uploads/042009/Screenshot-1.png
Should I reduce it even further to the "reduced row echelon form", if I want to use your suggestion?
http://peecee.dk/uploads/042009/Screenshot4.png
I have to find (a,b) \in R when the system of linear equations
a) have no solution
b) have one solution
c) have infinitely mane solutions.
I've converted the equations so they're on a row echelon form. My question is, what do I do from here?
There are a lot of different formulations of the 2LT (2nd law of thermodynamics). I think the arrow of time is derived from the macroscopic, phenomenological 2LT:
"In an isolated system, the entropy has a tendency to increase."
The "has a tendency"-part accounts for fluctuations. I think this...
Time, or rather the arrow of time, is defined within the 2nd law of thermodynamics: The future is in a higher entropy state than the present, which in turn is in a higher entopy state than the past.
But when the universe, our isolated system, reaches the maximum entropy state - heat death -...
I'm reading an old, maybe outdated, paper by Karl Popper about the 2. law of thermodynamics, brownian motion and perpetual motion.
Popper writes:
Before that, Popper has described Planck's law as:
So, my question is: Is brownian motion considered to be a violation to the 2. law of...
I agree.
Really? - How? AFAIK "normal" computers don't come anywhere near that lower limit.
More likely 40 years, Landauer published his paper in 1961, Bennett published his solution to MS in the same decade.
But also in "classical" entropy?
So what is preventing us from being such a demon? Do you imply that the act of information acquisition increases the entropy?
The entropy of a system is
S = k*ln W
If we obtain some information about the microstates, e.g. know the location and velocities of some of the molecules, then W is decreased by a amount w, so
S1 = k*ln(W-w)
That is an decrease in entropy, i.e. S1 < S.
Do physicists still "believe" so?
That is way over my knowledge of the subject, but I may have to look into it. Thanks anyway.
Thanks for the search term, I was not aware that I had already been adressed by Gibbs.
Yes, the two side are at the same T. They are perfectly symmetrical.
The real problem is when the two gases in each side is of two different kinds. In that case, there is an entropy increase.
Do I need elaborate #3 a bit more?