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10

The chemical potential can be thought of as how accepting the system is of new particles -- how much work you have to do to stick a new particle in the system. Since you can stick as many bosons in a given state as you want, the system is always accepting of new particles. At worst, you have to do zero work to add a boson (corresponding to $\mu=0$), and ...


9

No, it's not a problem. The reason is that, in order for expressions like $$\mu=-T\left(\tfrac{\partial S}{\partial N}\right)_{E,V}.$$ to be meaningful, you have to be using the grand canonical ensemble (or a generalisation thereof), in which particles are able to enter and leave the system. Consequently, $N$ stands not for an integer number of particles, ...


8

There simply doesn't exist any container with $\mu\gt \epsilon_0$; that's what the quoted sentence says. What you could try is to try to increase the chemical potential. But the Bose-Einstein distribution says $$ \langle n_i\rangle \sim\frac{g_i}{e^{(\epsilon_i-\mu)/kT}-1} $$ and if you chose values $\mu\gt \epsilon_i$, then the exponent in the denominator ...


7

Throughout, let's assume that the ground state energy of the system under consideration is zero. Chay Paterson has addressed your question in the case of a gas of bosons in which the number of particles is not conserved, but from the wording of your question, it seems that you're concerned about the case in which the total number of particles is fixed. ...


6

Usually the driving force for diffusion is a concentration gradient, however that is not strictly true. It is actually a gradient in chemical potential which is the driving force for diffusional processes. For some simplifying assumptions this reduces again to concentration gradient as driving force. For osmosis, the chemical potential is the driving force ...


5

Consider for a moment, a cell that is not connected to a circuit, i.e., there is no path for current external to the cell. The chemical reactions inside the cell remove electrons from the cathode and add electrons to the anode. Thus, as the chemical reactions proceed, an electric field builds between the anode and cathode due to the differing charge ...


4

Gibbs' thought on this was (Elementary principles, page 204 footnote) "Strictly speaking, $\psi_{\rm gen}$ is not determined as function of $\nu_1,\ldots\nu_h$, except for integral values of these variables. Yet we may suppose it to be determined as a continuous function by any suitable process of interpolation." Here $\psi_{\rm gen}$ is the free energy of ...


4

The problem here is that the thermodynamic potentials are functions of three thermodynamic variables each. Now, each thermodynamic potential has a set of natural variables. For the internal energy $U$, these are S, V and N. Now, your partial derivatives should explicitly state which other variables are held constant. For example, $$ \mu = (\frac{\partial U}...


4

The formula you write down is one of thermodynamics. In the statistical mechanics version it is valid in the grand canonical ensemble only if you interpret the extensive variables as expectation values. (See, e.g., Chapter 9 of my online book Classical and Quantum Mechanics via Lie algebras, arXiv:0810.1019.) But expectation values are continuous even when ...


4

When the chemical potential is 0 the extra free energy needed to add or remove a particle to the system is 0(i.e $\mu=\frac{dA}{dN}=0$. So particles can leave and enter the system without changing the (free) energy. In A BEC all particles have condensed to the ground state of the system. Particles entering or leaving the system will be added to the ground ...


4

To determine the upper limit on chemical potential for a gas of $\mathcal N$ bosons, look at the form of the Bose distribution in the grand canonical ensemble with $\langle N \rangle = \mathcal N$. When using the GCE, it's easiest to work at chemical potential $\mu$ and to then choose $\mu(\mathcal N)$ so that $\langle N\rangle(\mu)=\mathcal N$. Each state $...


4

Ok, lets look at how we determine $\mu$ in a cosmological setting. In order to determine $\mu_i$, we can use the fact that, in equilibrium, $\mu$ is conserved in all reactions. This means that if we have a scattering process $i + j \rightarrow a+b$, then we know that $\mu_i + \mu_j = \mu_a + \mu_b$. Fermions in equilibrium, like electrons and neutrinos ...


3

By passing to the grandcanonical ensemble, the particle number is allowed to fluctuate. You can imagine coupling your system to a 'bath' with which particles can be exchanged. Nevertheless, in thermal equilibrium the relative fluctuation of the particle number behaves as $$\Delta N/N \propto 1/\sqrt{\langle N \rangle} $$ thus vanishing in the ...


3

Suppose we take some object, for example a conducting sphere, and start with it at the same electrical potential as its surroundings. Now we add one electron the sphere, and because the sphere now has a higher negative charge that its surroundings the potential of the sphere will be slightly lower (i.e. more negative) than its surroundings. Adding the ...


3

The chemical kinetics of air depend on both how fast you are flying and your altitude. Fortunately, NASA has studied these issues. The figures below are from NASA Report NACA-TN-4359. The predominant chemistry in the stagnation region of an airfoil as a function of flight speed and altitude are shown below: You say $M=7$. If your vehicle is near sea ...


3

You can think of the chemical potential as the amount of free energy needed to add one additional particle to the system. Because the ground state of a BEC is degenerate and can hold an infinite number of particles, there's no energy cost to add another particle to that state. So, $\mu = 0$.


3

No, this is not possible at least in the way you are implying. Chemical potential as temperature are abstractions that apply to large systems as a whole, because they depend on equilibrium conditions, or in other words, they rely on fluctuations being much smaller than average values. You don't measure temperature directly either, once your thermometer is ...


2

Your last sentence is exactly right: the energy cost for fusion is almost all in those last few femtometers, at which electronic effects are negligible. Although there is in principle a difference between colliding neutral atoms and nuclei, at the energies required for fusion the effect is tiny. The energy differences associated with the presence, absence, ...


2

It essentially is the energy required to start breaking up the molecules involved. This happens through collisions. So, the activation energy is a measure of the amount of heat (via temperature, which is a measure of translational energy of molecules) required for the collisions to start breaking things up when they hit. For instance, CH4 and O2 at 200K won'...


2

Your quotation of the equation is actually wrong. The correct version is given by $\rho^{ind}(r)=-e[n_0(\mu+e\phi(r))-n_0(\mu)].$ Now you have to expand this expression around $\phi=0$ $\rho^{ind}(r)=\rho^{ind}(r)\mid_{\phi=0}+\frac{\partial}{\partial \phi}\rho^{ind}\mid_{\phi=0}\phi+\ldots$ Now we have to evaluate the zero and first order terms: $\rho^...


2

If in your system the number of of photons is non conserved, the Gibbs free energy cannot depend on the number of photons. So you will have $$\mu_{\gamma}=\left(\frac{\partial G}{\partial N_{\gamma}}\right)_{T,P}=0$$ This also implies that $\mu_{matter}+\mu_{\gamma}=\mu_{matter}$. However, in systems that conserve the number of photons, you return to the ...


2

It seems to me that what you need is much more an introduction to superconductivity than to Majorana modes physics. I suggest you to open any book called superconductivity to have more details than the ones I give below. A standard description of a superconductor consists in saying it is a perfect metal with an electron-electron attractive potential. A ...


2

I recommend this paper: http://www.elp.uji.es/masterNNM/docencia/refs/1995%20AJP%20Cook%20chem%20potential.pdf The authors explain in a simple and clear way the physical meaning of the chemical potential in various systems, including ideal Bose gas.


2

The concentration equation is only valid for an ideal solution that obeys Raoult's law. The assumption is that the solute-solute, solvent-solute and solvent-solvent molecular interactions are all identical. "Physical Chemistry, 3rd ed." by Levine, pages 238-241 gives further explanation (although not specific to crystallization) and says "For a more ...


2

Happy birthday. Once upon a time, there was a king and he had a sister who also liked coffee... Or let me omit this portion of the answer. A battery is motivating the electrons to buy a round trip ticket around the circuit by the voltage $V$. The voltage is nothing else than the energy $E$ per unit charge $Q$, i.e. $V=E/Q$; the unit 1 volt is nothing else ...


2

This question [ Chemical potential in Thermodynamics ] turns out have most of the answer. It wasn't tagged with the 'chemical-potential' label so I didn't see it when I was asking. Anyway, I'll re-state it as it applies to this question: chemical potential is the energy added per particle if entropy and volume are kept constant. The real Fundamental ...


2

Let's start from the second TD law: $$ \tag 1 dU(S,V) = TdS - pdV, $$ In case of variable number of particles it is modified by the adding $\mu dn$ term to the right side: $$ dU(S, V, n) = TdS - pdV + \mu d n. $$ The other TD potentials are related with $U$ by Legendre transformation with variables $T, S, p, V$, but not $\mu$ or $n$. So it's obviously that ...


2

The chemical potential is sort of the potential energy needed to add another particle from the surrounding reservoir to the system. Thus to add another particle to a particular single particle level requires energy if the chemical potential is larger than the energy of single-particle level. If the chemical potential was smaller than the energy level, then ...


2

Let's set up an example. Imagine a $10kg$ platform, a spring, and a $1kg$ mass. Suppose the spring is compressed to hold sufficient energy to push the items apart with a relative velocity of $11m/s$. Beginning from rest, this will give the object a forward speed of $10m/s$ and the platform a rearward speed of $1m/s$. The total kinetic energy after this ...


2

John Grant Watterson claims in his paper "What drives osmosis?" that a fall in free energy, which is the rigorous thermodynamic criterion for a spontaneous change, cannot be the drive in osmotic processes: Our models and theories require the introduction of a parameter that explicitly represents structure in liquids, which until now has had no place ...



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