# Tag Info

17

If the particles are not point-like, they will take up some volume. As the gas is compressed, the collision frequency will rise more quickly, which will make the pressure-volume curve change. The corrections in the Van der Waals model of a real gas accounts for the volume of the particles. Also if they have internal structure, that structure can have ...

4

Refs. 1 and 2 define a canonical transformation (CT) $$\tag{1} (q^i,p_i)~\longrightarrow~ (Q^i,P_i)$$ [together with choices of Hamiltonian $H(q,p,t)$ and Kamiltonian $K(Q,P,t)$] as satisfying $$\tag{2} (p_i\mathrm{d}q^i-H\mathrm{d}t)-(P_i\mathrm{d}Q^i -K\mathrm{d}t) ~=~\mathrm{d}F$$ for some generating function $F$. On the other hand, Wikipedia (March ...

4

I think the following was the argument that the authors used to arrive at their expression: The first particle can choose to be in any one of the g states; the second again. So N particles have a total of g*g*g*... choices - which is the equation given. When they are indistinguishable you divide by N factorial since that is how many times each ...

4

It seems to me that you're looking for the Boltzmann transport equation: $$\frac{\partial f}{\partial t}+\frac{\mathbf p}{m}\cdot\nabla f+\mathbf F\cdot\frac{\partial f}{\partial\mathbf p}=Q+\left(\frac{df}{dt}\right)_{\rm coll}$$ with $f$ the distribution in phase-space, $\mathbf p$ the particle momentum, $Q$ some source term, and the RHS an interaction ...

3

Because if those particles aren't point objects, you must also take into consideration that they take some space in the system and have properties like density and size which have to be taken into consideration when formulating the laws. This makes everything extremely complicated. A huge part of classical mechanics is only true for point objects for the ...

3

You're running into a tricky property of statistical variables: what is true for an individual particle is not necessarily true when averaged across a distribution. In particular, you can say that the distance one particle travels between collisions, its free path length $\ell$, is equal to that one particle's speed times its free path time $t$: $$\ell = ... 2 This is a classic conundrum and it is called the "problem of the adiabatic piston". You can find it discussed in books on thermodynamics by Landau & Lifhsitz and by Callen. Another very thorough analysis is by Gruber "Thermodynamics of systems with internal adiabatic constraints: time evolution of the adiabatic piston". (You can find Gruber's article ... 2 The argument goes something like this: Suppose we have a system in contact with a large reservoir at temperature T with which it can exchange energy. The Boltzmann distribution tells us that the probability of finding the system in some small region of phase space \mathrm{d}V is given by$$ \rho(q_i,p_i)\,\mathrm{d}{V}= \frac{1}{Z} ...

2

Yes, it is possible. The simplest qualitative answer to this is that, at the microscopic level, the electrons in a conductor are dictated by quantum mechanics, which is inherently probabilistic. Velocities and positions are rarely ever totally excluded from a given value; it's just insanely unlikely for a single electron to attain that given value. ...

2

TylerHG: Yes it is easy to calculate the density of states. But what I'm really asking here is "why." Note that a thin circular ring in $\mathbf{k}$-space of thickness $dk$ has area $dA=2\pi k\,dk$ (by elementary geometry). In $E$-space, since $E\propto k^2$, that ring corresponds to a patch of width $dE=2k\,dk$. Thus $$\frac{dA}{dE}=\pi.$$ But ...

2

It is easy to show that the total number of electrons in a 3D fermi sphere is : $$N(e)=\frac{V}{3\pi^2}*k_F^3$$ Where $k_F$ is the Fermi wave vector and $V$ is the real space volume of your sphere. Now if you rearrange for $k_F$ in terms of the total number of electrons you'll get a particular equation. It is know that ...

2

I will not directly answer your question, rather I'll try to make plausible the connexion between QFT and statistical physics. To my mind the mathematical details are somehow obscure and confusing, whereas using the theory is worth a deal, and give interesting results, especially in condensed matter and nuclear matter problems. For more details you can have ...

1

The term you are looking for is premelting or "surface melting." It is an observed phenomenon (which could explain how ice skating works) with some thermodynamic descriptions. Basically what happens is the system is separated into two distinct phases, a solid (ice) and a vapor (air). There is a surface energy associated with this interface. If it happens ...

1

A (3d) gas of particles with a gravitational interaction is an example of a system with long range interactions, where the energy is not additive and thus many basic results of classical statistical mechanics are not valid, including the equivalence of the microcanonical, canonical and grand-canonical ensembles. For a general introduction to the subject see ...

1

I am not quite sure I understand your question. In your second equation, $E$ is treated as a variable and not as function anymore, so why do you want to find $E$ function? The density of states is basically a function counting the number of state in $x\in\mathbb{X}$ that give the same energy: $\Omega(E_0)=|\{x:E(x)=E_0\mbox{ and } x\in\mathbb{X}\}|$ ...

1

An ideal gas in equilibrium cannot be supposed to have reached its equilibrium from a non-equilibrium state by interaction of its particles, because by definition the particles of an ideal gas do not interact: Hard spheres of radius $a$ that collide elastically do interact in the time average - even when they interact "almost never". (Among other things ...

1

Ideal gas does only mean that there are no forces between the particles. They do not have to be point-like. For example 2-atomic gasses could have 3 translatory and 2 rotational degrees of freedom in kinetic gas theory, while still no forces act between the molecules. So ideal but not point-like. For one-atomic gasses the atoms are often taken to be hard ...

1

The short answer is that $\bar E = -\frac{1}{Z}\frac{\partial Z}{\partial \beta}$ ($\dagger$). $\bar E$ is the average energy of a quadratic degree of freedom in this case and $Z$ is the partition function which sums over all possible values of $E$ associated with your degree of freedom $q$. So it's just how the math works out. There is a short derivation ...

1

There are multiple ways of looking at the ideal gas model. One is to say we have point particles colliding elastically (as an ideal scenario) and proceed to obtain the exact equation of state, i.e $pV=nRT$. The other approach, is to state a priori the relevant length scales and time scales at which the system is going to be studied are sufficiently ...

1

You are absolutely right that the limit in which this approximation holds is $$\beta(\epsilon - \mu) \gg 1 \,,$$ which is not trivially the 'high-temperature limit', and indeed looks rather like the low temperature limit. However, it also looks like the limit of large negative $\mu$. If we want to know how temperature will affect the exponent, we need to ...

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