85

The anthropomorphic formulation "tries to" is misleading. Under the effect of ambient noise, matter explores the possible configurations around its current state: e.g., two single hydrogen atoms wiggle around and meet. If they happen to bind, this releases energy which goes away, and we say that the energetic state of this new $H_2$ molecule is lower than ...


44

Consolidating some of the points made in the answers to the question you linked, and comments: When constructing a chair, 4 legs is easy when you use traditional (wooden) construction - 90 degree angles, and easy to make stackable. A little bit harder than three legs because you have to make sure they are all the same length (or the chair will wobble). ...


41

Arnold Neumaier's comment about statistical mechanics is correct, but here's how you can prove it using just thermodynamics. Let's imagine two bodies at different temperatures in contact with one another. Let's say that body 1 transfers a small amount of heat $Q$ to body 2. Body 1's entropy changes by $-Q/T_1$, and body 2's entropy changes by $Q/T_2$, so ...


31

Nature has no preferences, and therefore entropy tends to increase. Sounds paradoxical? The point is that each microscopic state (describing the exact position and velocity of each atom in the system) is equally likely. However, what we typically observe is not a micro state, but a course-grained description corresponding to incredibly many micro states. ...


27

From a fundamental (i.e., statistical mechanics) point of view, the physically relevant parameter is coldness = inverse temperature $\beta=1/k_BT$. This changes continuously. If it passes from a positive value through zero to a negative value, the temperature changes from very large positive to infinite (with indefinite sign) to very large negative. ...


25

The question doesn't make sense, as not every system has to be in thermal equilibrium. The argument you're making works for classical mechanics too. "Consider a ball lying on the ground." "That's impossible, since by statistical mechanics, the position of the ball must obey the Boltzmann distribution."


25

In the centre of a bowl there is equilibrium. Put a ping pong ball in it. If this ball is ever shaken slightly away from equilibrium, it will immediately roll back. A "shake-proof" equilibrium is called stable. Now, turn the bowl around and put the ball on the top. This is an equilibrium. But at the slightest shake, the ball rolls down. This "non-shake-...


25

It entirely depends on what you think "thermodynamics" is. The traditional idea of thermodynamics dealing with systems whose macrostate can be fully described by e.g. temperature, pressure and volume indeed only applies to systems in equilibrium. Of course, as an approximation it also applies to systems "not far" from equilibrium, for some suitable notion ...


24

"A state of rest" is a relative term. Relative means - measured in comparison to the things around it. When you sit in a train and sip from a cup of coffee, you can do so because the cup is still relative to you even though both of you might be hurtling through the countryside at 200 km/h. For most experiments, objects can be considered "at rest" if they ...


22

This is a consequence of the second law of thermodynamics, which states that In a closed system with fixed internal energy (i.e. an isolated system), entropy is maximized at equilibrium. It can be shown that this statement is equivalent to the following: In a closed system with fixed entropy, the energy is minimized at equilibrium. Callen in his ...


22

Indeed, you need friction. Otherwise this object will slide to equator just as you expect. In fact, all the objects will slide there to form equatorial bulge and eventually the sphere will be turned into new shape with surface orthogonal to $\vec{g}+\omega^2 \vec{r}$ at each point. That is exactly what happens to planets in reality while they are still ...


20

It is wrong to think potential energy is stored in the object. The earth pulls the object down, but the object pulls the earth up. They share the potential energy. The object fails to fall down because the tabletop pushes it up. The earth fails to fall up because the bottom of the table legs push the earth down. The table pushes up and down because it is ...


17

This is really a statistical effect, as pretty much all of thermodynamics. You have two free hydrogen atoms. They tend to move around the space they have, and when conditions are favourable (there's enough energy, the atoms come "close enough" together), they might interact - chemically or otherwise. Now, "enough energy" is the important bit here. When a ...


15

First, if ${\rm d}S\neq 0$, then the entropy will change, and because something is changing, it's obviously not an equilibrium. If the physical system doesn't maximize the entropy and it's composed of many parts that may interact with each other, directly or indirectly (it's not disconnected), then any path that allows the entropy to be increased (given ...


15

I'm going to take a slightly different approach and say it's because we defined energy to make it so. In other words, systems "try" to find the lowest energy state because energy is a concept humans invented in order to describe what we observe. This is the reason that for any given set of constraints, you might need a different "energy" to describe the ...


13

You need to read this paper by Jaynes. I can't explain it as well as him, but I will try to summarise the main points below. The first thing is to realise that the entropy is observer-dependent: it depends on what information you have access to about the system. A finite temperature means that you don't have access to all the information about the state of ...


13

You have to be careful to distinguish between microstates and macrostates. Thermodynamic equilibrium is a macrostate which consists of a mixture of all possible microstates of energy $E$ weighted by a Boltzmann weight $e^{- \beta E} / Z$. A state in macroscopic thermal equilibrium can be thought of as "moving through phase space" ergodically (i.e. the ...


13

An $\text{O}^{2+}$ ion, with 6 electrons, is stable in a vacuum. If you put one in the depths of interstellar space, far from any other matter, then it will sit in that state until the end of time. Similarly, you can have a high-elevation lake, and it can be absolutely stable if there is no path for water to drain out. Also, is an atom being stable ...


11

The reason is, quite simply, that the walls of the container exert a force on the fluid given by the pressure at each point, and orthogonal to the container's surface. In your second container, the wall is slanted and therefore provides an upward force which helps stabilize the fluid elements next to it. If you draw a free-body diagram of the bit of fluid ...


11

does it also leads to ... No, it doesn't. A simple Counterexample: Consider the figure below (the bar $\textrm {AB}$ and forces $F$ are on a plane parallel to $\textrm {xy}$ plane) We have $\Sigma \vec F=\vec 0$, but, if we calculate vector sum of torques about point $\textrm A$ we will obtain $\Sigma \vec M_A=F (\overline{AB})\vec k\neq \vec 0$ ($\vec k$...


10

If you made the most perfect cone possible, so that its tip was a single atom, and stood it on the most perfect surface possible (a perfectly smooth, perfectly hard sheet of atoms), and completely removed all forces other than gravity, it would still topple. This is because those atoms are all jiggling around due to thermal motion. This effect fundamentally ...


10

"Equilibrium" means thermal equilibrium. The solid has one well defined temperature, and a constant Fermi energy. The Fermi energy is an energy value against which energy levels are compared to determine how fully occupied (or not) an energy level is. Generally when the Fermi level is constant throughout a solid electrons diffuse equally in all directions....


10

Let us look at a one-dimensional example: Recall that $f(x) = \dot{x}$, so $f$ encodes the time evolution of $x$. If $f < 0$, then $x$ will move to the left. If $f > 0$, then $x$ will move to the right. If $f = 0$, $x$ will not move at all, this is why $f(x_0) = 0$ is the equilibrium condition. Now, look what happens if you perturb the equilibria $...


10

A system in perfect equilibrium is not perfectly stationary in space or time. But, any change in the system is offset by another change such that there is no net change. Let's look at a simple example. Take a perfectly insulated box isolated from the environment and fill it with a gas. The gas molecules are all bouncing around, hitting each other and the ...


10

Your example does not contradict Earnshaw's theorem for electrostatics, because it rules out stable equilibrium in a region without charge, possibly containing fields made by charges outside that region. Here you're doing the exact opposite, looking at the only point in your situation with charge.


10

You exchange heat with the objects around you. If the objects around you are hotter than you, you'll heat up. If the objects around you are cooler than you (neglecting the heat you're generating due to metabolic processes), you'll cool off. In space, the objects around you (mostly interstellar medium) is cooler than you so you radiate more heat away from ...


9

To answer your question, you should first understand when is a system most stable. Firstly it shouldn't have a tendency to move or change state, thus it should be under equilibrium conditions, i.e. the net Force should be zero. We know that $$F = - \frac{dU}{dx}$$ Putting $F=0$, we get $$\frac{dU}{dx}=0 \tag{1}$$ Secondly, it should be able to maintain ...


9

This is a very interesting, important and at the same time subtle matter which is useful not only in thermodynamics, but other areas as well, and not everyone quite understands it. First off, two statements that you'd better just memorize for now: In a reversible process, equilibrium is never left. Yes, that is a paradox. That's why all real processes are ...


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