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8

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, ...


6

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

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 ...


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 ...


4

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 ...


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

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 ...


3

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 ...


2

The intensive variabes $${T=\left(\frac{∂U}{∂S}\right)_V} \ \ \ \ \text{and}\ \ \ {-P=\left(\frac{∂U}{∂V}\right)_S}$$ give the amount of change of the internal energy $U(S,V)$ w.r.t the extensive variables $S$ and $V$. One is heat and the other is work. There are however other extensive variables $N_i$ with slopes ...


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 ...


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

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: ...


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

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

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 ...


1

I think you are confused between potential difference and energy. The energy used up can vary, if the current flowing through the resistor varies. First of all, a battery is 'not' a constant energy source. It's a constant potential source. Secondly, if more current flows, more energy is dissipated, even though the per capita energy is constant. Now, to ...


1

It feels like you are going to fast in your way to think. The difference between the chemical potential and the ground state energy is clear : 1.The ground-state energy of your system is here $\epsilon_\textbf{k}=0$ correponding to the ground-state $|\textbf{k=0}\rangle$, which is macroscopically occupied in a BEC (i.e. $N\sim N_0$). 2.The chemical ...


1

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 ...


1

I think you can consider the zero chemical potential to be an effect of popping up of photon. Both chemical potential and the temperature appears as a undetermined multiplier due to conservation of energy and number of particle respectively. For photon, there is no conservation of particle number and hence the corresponding undetermined multiplier, i.e. the ...


1

Just a hunch, but I bet Fick's law is really just a potential function gradient in disguise that simplifies down to the form you have for some situations. What you really want is something more like $$J = -D(p,T)\frac{\partial \psi(C,p,T)}{\partial z}$$ where $\psi(C,p,T)$ is probably related to the potential change from one point to another. How you ...


1

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 ...


1

For a bose gas where particle number is not conserved, e.g. blackbody photons, indeed $\mu=0$. How does that work? Well, as you approach zero frequency, the number of blackbody photons gets higher and higher -- yes, you bet, it approaches infinity. BUT nevertheless the total collective energy of those low-frequency photons gets lower and lower. (As you ...


1

rocket fuel is often characterised by its Specific Impulse. Loosely this is a measure of force applied to rocket per mass of fuel. I'm afriad I don't know what this would be for Bicarb...


1

If one goes to the wiki article on evaporation one sees that For molecules of a liquid to evaporate, they must be located near the surface, be moving in the proper direction, and have sufficient kinetic energy to overcome liquid-phase intermolecular forces. Only a small proportion of the molecules meet these criteria, so the rate of evaporation is ...


1

The use of the chemical potential $\mu$ as state variable is useful in situations where composition $N$ is variable and/or cannot be easily controlled. From an experimental point of view the chemical potential is fixed when the system is in contact with energy and particle reservoirs. At equilibrium, the chemical potential of the system equals that of the ...


1

One can interpret the Gibbs-Duhem relation as stating that there is necessarily one extensive degree of freedom in the system, so that the $k+2$ intensive parameters (with $k$ the number of components) only describe a $k+1$-dimensional "subspace". See the link below for a more detailed presentation: ...


1

When you say "reach space" do you mean a) get above most of the atmosphere, b) achieve orbit velocity, or c) achieve earth escape velocity? Either way, if you simplify the problem by ignoring air friction, it's only a matter of achieving a certain velocity. Robert H. Goddard showed that, if you built your rocket a certain way, you could achieve any desired ...


1

The pairs of electrons are more stable if you see the pair as a filled orbital (which can contain maximum of 2 different spin electrons via Pauli's Exclusion Principle). I think the simplest way of explaining the lower energy of two-nuclei orbital is that the size of the electron's 'playground' grows 2 times, and as the area over which the electron may ...


1

We want to find $$\frac{\partial\langle N\rangle}{\partial\mu}=\frac{1}{\beta} \frac{\partial^2}{\partial \mu^2} \ln \mathcal Z=(\frac{\mathcal Z'}{\beta \mathcal Z})'=(\frac{-\mathcal Z'^2}{\beta \mathcal Z^2}+\frac{\mathcal Z''}{\beta \mathcal Z})$$ where the prime denotes differentiation w.r.t. $\mu$. In order for this to be zero we ...


1

"Evaporation energy" is not the same as chemical potential, because you forgot about entropy. For example, at 100C, the chemical potential of liquid water and water vapor is exactly the same, because they are in "diffusive equilibrium" (the number of molecules going out of the vapor into the gas is, on average, equal to the number moving the opposite ...



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