# Tag Info

22

The question isn't silly. The speed of each molecule in the liquid is much higher than the speed of either the piston or the water shooting out from the nozzle. At room temperature, for water molecules the average is on the order of 500m/s. And yet, the speed of sound in water is three times higher than that, which implies that pressure can propagate in ...

21

Adjacent molecules in a liquid all repel each other because of the electron clouds that surround the nuclei that they contain. In that sense these molecules never even 'touch' each other (at least not in the intuitive sense of the word). When you apply pressure to the liquid you're squeezing them into a (very slightly) smaller volume, thereby increasing the ...

6

Let's simplify things down to the barest minimum: one dimension, one particle, and a wall. O | The particle moves to the right, hits the wall, and rebounds, perfectly elastically. If the wall is fixed in place, the particle will leave the collision with exactly the same kinetic energy as it came in with. But what if the wall is moving to ...

5

OP's question (v1) is essentially asking Does the operator identity $$e^{\frac{it}{\hbar}[\hat{H},~\cdot~]}\hat{A}~ =~ e^{i\hat{H}t/\hbar}\hat{A}e^{-i\hat{H}t/\hbar} \tag{1}$$ have an analog using functions/symbols $H$ and $A$ rather than operators $\hat{H}$ and $\hat{A}$, respectively? The answer is: Yes, in terms of the Groenewold-Moyal star ...

4

You start by solving the differential equation. It is a first order, linear differential equation with constant coefficients. So the solution of the homogenous system is quite simple: $U(t) = c\cdot e^{-t/RC}$. Now we solve the particular system with variation of the constant $c$, which means we try the Ansatz $U(t) = c(t)\cdot e^{-t/RC}$. This give the ...

3

Let us define: $$\hat L=i\sum ^N_{j=1}\bigg(\frac{{p}_j}{m_j}\frac{\partial }{\partial q^j}+\vec {F}_j(\boldsymbol q )\frac{\partial }{\partial p_j}\bigg)$$The Liouville operator can be expanded in terms of components and a basis like any vector field. Let $\xi =(q^1,\dots ,q^n,p^1,\dots p^n)$ be a phase space vector. ...

3

Then the number of microstates corresponding to this macrostate should be 1 (since all the particles are identical). The particles may be identical (have the same intrinsic physical properties), but that does not in any way mean that they are not distinct. It is most natural to assume the particles are distinct (they can be placed to different places, ...

2

In my opinion, the reason to be of the Keldysh formalism is that it is the way to write a path integral for non-equilibrium quantum systems. It provides an action which can be used to sample paths for out-of-equilibrium systems. This is great because it opens the door to the huge toolbox of equilibrium quantum field theory to non-equilibrium problems. I can ...

2

Entropy is not a force, no. It is a "chaos factor", if you will. The more entropy, the less structured a system is. A system will always move towards the state that is most probable. The most probable state (macro-state) will be the state with most micro-state configurations. Consider as an example four coins of heads H and tails T. Flip them and you can ...

2

Edit: Looks like Steeven beat me to it with a similar explanation, but I'll leave mine here for posterity. Entropy is just a property of a system (in a given state), in other words a state function. I hope that by elaborating on the statistical interpretation of entropy you will gain some intuitive notion of the meaning of entropy. Consider the following ...

2

The expansion of a gas into the vacuum is not a diffusive process. Depending on the initial conditions (density and density gradient) the expansion is either described by the Euler/Navier-Stokes equation of fluid dynamics, or the Boltzmann equation of kinetic theory. For the parameters that you mention the gas is sufficiently dense that most of it is in ...

2

Your misconception is that the water particles are moving very slowly or are stationary because they are not escaping. In fact, they are moving very quickly and are constantly bouncing off each other and the walls of the container. The pressure is basically how many collisions occur over a given time period. As you squeeze the piston, you are increasing the ...

1

No, most bosonic atoms (all except H and He) form solids. In fact, most of the BECs studied experimentally with ultracold atomic gases are metastable. The true ground state is a solid.

1

Try picking a tiny hole in a milk cardboard container. Squeeze the container. You only press is slightly, but milk is poured fast out of the tiny hole. This system is simply redistributing the total force that you provide onto a much smaller area. Same force on a smaller area equals larger pressure: $$p=\frac{F}{A}$$ The pressure on the particles at the ...

1

The energy $E$ of an oscillator is given by $$E=\frac{p^2}{2m}+\frac12m\omega_1^2x^2$$ This defines an ellipse in phase space! So now, when $E=E_1$ everything within the ellipse defined by $E_1$ will have energy less than $E_1$. To proceed with finding the limits of of integration, we consider the cases when the particles' have all kinetic or all ...

1

I think the answer is yes to some extent, but have you neglected potential energy? For example in the reaction $$H-H + F \rightarrow H + H-F$$ one of the electrons in the $H-H$ molecule ends up in a significantly lower energy state than it was in before the reaction so that the potential energy of the system is lower on the right hand side. Apologies ...

1

This formula is strictly applicable for gases. So what happens to this formula is it is no longer valid. Provided the gas density is low, this law holds for any single gas or for any mixture of different gases. However it is interesting that for osmotic pressure of a liquid we have a similar formula $$\pi=CRT$$ where $\pi$ is osmotic pressure and $C$ is ...

1

The ideal gas law applies in systems made of large number of particles (atoms or molecules) that are completely independent from each other as in a dilute gas. In that situation, the particles are able to move in all space directions without interfering with the others. That's an ideal situation, hence the name ideal gas. The law $$PV=nRT$$ is an immediate ...

1

The "Ideal Gas Law" (IGL) (pV = nRT) is an "equation of state" and any such equation is a relation between the macroscopical "state variables" of a matterial system in a given phase. Therefore, the form of a state equation depends on the system's phase. The basic assumption of the IGL is that the molecules do not interact by any other means besides point ...

1

You don't say whether your analysis includes rotational modes. I assume it does otherwise the disagreement between experiment and the ideal gas specific heat would be profound. A linear molecule will have two rotational modes, each adding $\tfrac{1}{2}R$, so the specific heat (excluding vibration) will be $C_p = \tfrac{7}{2}R$. Anyhow, hydrogen has quite a ...

1

It's a simple permutation problem. Suppose there are $N$ distinguishable particles. Let $n_L$ be the number of particles distributed in the left & $n_R$ distributed at the right compartment. First of all, you are to select a particle for the first place of $N$ places. How many ways can it be done? It can be done in $N$ ways; now the second place can ...

1

At the atomic/molecular level all chemical reactions are Quantum Mechanical phenomena where atomic and/or molecular electron orbitals of the reactants are being destroyed and new ones created in the reaction products. This is what you are correctly referring to as formation of bonds (although it has to noted that some bonds are also broken during chemical ...

1

Contrary to @Gert 's answer, creation of photons from reactions forming and breaking chemical bonds (typically reactions between molecules) is quite rare. When photons do occur, it can be (1) the result of a molecule, atom, or structure created or altered by the reaction being in an electronically excited state, which then decays, emitting a photon. There ...

1

This is because we want to avoid extensive mathematical calculations of pairwise interactions with all particle inside or outside the box. MIC is a way of providing a cutoff distance over which we are not calculating pairwise potential. The cutoff is usually half the box length. This means if distance between particle i and particle j is more than L/2, you ...

1

Yes, microstates possess individual information, as they are genuine physical objects and not just tools of statistics. Individual microstates may differ in positions and momenta of individual particles, or even in the total energy. The fact that you are treating microstates statistically is simply a tool for simplifying calculations.

1

It's the simplest useful case. There are more complicated ones, like when external magnetic/electric field needs to be taken into account, or when multiple chemical species are involved, so there is no single $n$, but several molar numbers $n_1,n_2,...$.

1

That information is not contained within the bb radiation - all that can be gleaned is an emitting area and a temperature. In practice the radiation can have arisen from any process where it is feasible for a photon at that frequency to be produced. Of course to actually be a blackbody emitter there must also be a 100% chance that a photon at that ...

1

I've only ever used the Ewald sum, I've never implemented it myself. However, you mention that you're not converging as $\kappa$ increases nor are you converging to the correct value. It would seem that regardless of the problem, if your implementation is correct it should converge at some point. If you do reach convergence wrt $\kappa$; as to the point ...

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