New answers tagged

0

What you write down is not the pressure of a BEC, but that of a free Bose gas at zero chemical potential (you also missed a factor of m). The pressure of a non-interacting BEC is equal to this result only at $T=T_c$, where the condensate fraction is zero. Below $T_c$, the BEC is a mixture of a superfluid component (with zero pressure), and a normal component ...


0

The pressure of a Bose-Einstein condensate $p = kT\frac{g}{\lambda^3}\zeta(5/2)$ that composes a star, which is an interesting conjecture in a way, is countered by gravity. The hydrostatic condition for a star is $$ \frac{dp}{dr} = \frac{GM(r)\rho(r)}{r^2} $$ The temperature is set by the number of particles, which for a star is considerable $n \sim ...


2

From the links I provided in the comments below your question it should become clear that entropy "meters" do not exist, you calculate it from other measured variables. If this does not satisfy you requisites for an experimental measurement, then your conclusion that the claims are only theoretical is justified. However, having said that, with your ...


1

More of an extended comment, but here are two thoughts: 1) I'm not sure I completely agree with the statement Textbook discussions of the Second Law of Thermodynamics (SLT) often stress that this law applies only to "closed systems". Or, differently stated: if the system is not closed, its entropy can go down. Okay, this is correct, of course, ...


0

Use the definition $<v_x> = \int d^3 v v_x (exp(\frac{E}{k_B T})+1)^{-1}/ Z$ with sum of all states $Z$ and $E = \frac{m}{2}(v_x^2+v_y^2+v_z^2)$, introduce spherical coordinates and make the substitution $z = |\vec{v}|/(\sqrt{2 k_B T/m})$. What is if $T=0$?


1

A thermodynamic process is called reversible if an infinitesimal change of the external conditions is capable of reversing the process. A thermodynamic process is called quasi-static if it is a dense succession of equilibrium states. Roughly speaking, a quasi-static process is one that can be represented by a continuous curve in the thermodynamic ...


3

This question is a little too big, entire text books have been written to answer it. A standard reference is van Kampen, Stochastic processes in physics and chemistry. Roughly speaking, for a Markov process Master equation -> Kramers-Moyal expansion -> Fokker-Planck equation where the master equation gives the microscopic probabilistic rule for ...


0

The mechanism for increasing the thermal conductivity is phonon assisted hopping. For a disordered system, one which do not preserve the long range order, the electronic wave function becomes localized. The wave function extent is typically much smaller than the system size and is characterized by the localization length $\xi$, a parameter in theory. In this ...


0

Kinetic energy in 1D, method 1. Free electrons. Assume no potential energy at the moment. Zero temperature. \begin{equation} n_e=\int_0^{E_F}g(\epsilon)d\epsilon=\int_0^{E_F}\frac{1}{\pi\hbar}\sqrt{\frac{m}{2\epsilon}}d\epsilon=\frac{\sqrt{2mE_F}}{\pi\hbar} \end{equation} \begin{equation} E_F=\frac{\pi^2\hbar^2n_e^2}{2m} \end{equation} Kinetic energy, ...


0

Normalization Factor Let us define a generalized Gaussian probability density function (PDF) as: $$ f_{s}\left( x \right) = A_{o} \ e^{^{\displaystyle - \frac{ (x - x_{o})^{2} }{ 2 \sigma^{2} } }} \tag{0} $$ where $A_{o}$ is the normalization constant, $x$ is the argument, and $s$ denotes the set of distributions (e.g., particle species), $x_{o}$ is the ...


95

A "perfectly efficient" computer can mean many things, but, for the purposes of this answer, let's take it to mean a reversible computer (explained further as we go). The theoretical lower limit to energy needs in computing is the Landauer Limit, which states that the forgetting of one bit of information requires the input of work amounting $k\,T\,\log 2$ ...


4

As stated in the comment by Peter Diehr, the question is in principle no different whether you ask it for electromagnetic, gravitational or any other kind of wave. The wave's entropy is simply the conditional Shannon entropy of the specification needed to define the wave's full state given knowledge of its macroscopically measured variables. A theoretical ...


2

The answer is a definite yes and no. Gravitational waves have entropy in that we can think of them travelling from their source to our detector as a channel that sends units of information in the sense of the Shannon formula. The ringing of our detector is then the reception of that information. The Shannon formula $S=-k\sum_n p_n log(p_n)$ would give a ...


2

The noise spectrum is the Fourier transform of the correlation function, as long as the correlation function is invariant under global time translation. This is nice, because the Fourier transform of the correlation function you have there produces a delta function in frequency, which then makes the frequency integral totally trivial. \begin{align} ...


0

Some good texts about it: Giuseppe Massardo - Statistical Field Theory Mehran Kardar - Statistical Physics of Fields


2

A classic on the subject is Giorgio Parisi's Statistical Field Theory. It is a complete book written by one of the most influential physicists in the field. The book starts with a brief recap on statistical mechanics and then introduces the Ising Model, were the basic techniques of statistical field theory are introduced. It then moves on to (in this ...


0

I'll give an alternate view of how second order phase transitions may look like. Let us study a parameter $F$. If there is a second order phase transition in $F$, then the second derivative would be discontinuous but won't diverge, as for the heat capacity below: and first derivative should look like: Interestingly, the plot of $F$ itself may be ...


0

The definition of $v_{avg}$ for 3 dimension is $$v_{avg}=\int_0^\infty |v|f(v)dv$$ However, it needs to be clear that $$v^2=v_x^2+v_y^2+v_z^2$$ For 3D, the distribution can be derived from 1D distribution but a little different. $$f_{3D}(v) = \left(\frac{m}{2 \pi k T}\right)^{3/2} \exp\left(-\frac{m v^2}{2kT} \right)$$ After similar integration as above, ...


1

This picture explains better for creating a rabbit. The picture you showed is good for understanding Gibbs energy, but not good for describing creation of a rabbit.


0

A Boltzmann distribution is system dependent--it depends on the energy eigenstates. Moreover, if the system lacks symmetry, then Vx, Vy, and Vz may have very different distributions. However, if you're talking about an ideal gas--then it's the standard Maxwell-Boltzmann distribution. To get just Vx, you can use the equal-parition property, and rewrite a 1-d ...


0

The reason is an n doped material has additional atoms (donor atoms) added near the conduction band. They provide electrons which can easily go to the conduction band, as the energy gap between the donor level and the conduction band is low. There are now more levels above the Fermi level, thus skewing the distribution of electrons higher. This is seen by ...


0

The one dimensional Maxwell distribution for the $i-$component of the velocity vector is $$f_{1D }(v_i) = \left(\frac{m}{2 \pi k T}\right)^{1/2} \exp\left(-\frac{m v_i^2}{2kT} \right)$$ Let's drop the $i$ and $1D$ subscripts for simplicity. You are looking for the average of the absolute value of $v$, $|v|$. To find $\langle |v| \rangle $, we have to ...


2

My main concern is: does this blurring/loss of knowledge come from any well-known physical law/principle? For instance, should we link the quantization of the phase space to ΔxΔp≥ℏ2ΔxΔp≥ℏ2, or to some kind of observer effect? The description in the Wikipedia article is misinformed and misguiding. The blurring, or coarse-graining, is merely one possible ...


-1

A canonical ensemble is a collection of weakly interacting systems, in thermal equilibrium with each other. The individual systems can be a single particle/molecule if the energy of interaction between the particles is negligible in comparison with their own (kinetic) energy. The energy of ensemble is then just the sum of energies of individual particles, ...


-1

One should distinguish here de Broglie-Bohm theory for the general situation outside the equilibrium, and that for quantum equilibrium. Entropy is defined as usual by $H=-\int \rho \ln \rho dq$. Outside the quantum equilibrium it is useful to split it into the entropy relative to the quantum equilibrium $H=-\int \rho \ln (\rho/|\psi|^2) dq$. This relative ...


2

The physical principle being invoked is the finite resolution of any experiment, independent of the value of $\hbar$, together with coupling between observable and microscopic degrees of freedom, i.e. it applies to both classical and quantum systems. Technically energy conservation and Liouville's theorem, or unitarity in QM, are also needed to prevent ...


2

Their average speed would be non zero but their average velocity would be zero as long as they are not moving preferentially in one direction.


0

A carnot engine is essentially any device that produces work from temperature differences in a reversible way. Ideal gas expansion is an obvious mechanism to accomplish this. A similar example are fluids that undergo a phase change from liquid to gas in the engine, like refrigerants. The principle is the same. For something involving electrical energy, ...


0

They are the same expression, just with different accounting of states. (Your energy relation at the end is incorrect.) However, it is sometimes useful to carry the $g_j$ coefficients around, because certain quantities are most easily found by differentiating with respect to the $g_j$. This can even be done when there is no degeneracy; you just set ...


2

I don't believe there is a mathematical reason, especially if there is latitude in reverse-engineering the field theory or stat mech system to evince such a behavior. Indeed, if Lorentz-nonivariant systems are examined, things like limit cycles , e.g. this one are not hard to concoct. As for physical reasons, they might well be easy to bypass/moot if one ...


2

Two systems belonging to the same universality class will have the same critical exponents. There are many things that determine the universality class of a system, one being its dimension. The 2D Ising model is one of the most studied system in statistical mechanics because it admits an exact soultion, found by Lars Onsager in 1944. Its critical exponents ...


1

Energy is a secondary concept, at least in Newtownian mechanics, so let's start with the fundamentals. There exists a force field. A gravitational one, say. An object in this field feels a force. This force "wants" to make the object accelerate. From forces, define work as $force \times distance$ and now we can give precise mathematical meaning to the ...


11

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 ...


9

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 ...


4

I think such a function may only exist in the Maxwell-Boltzmann limit. Here's why: For simplicity let us parametrize everything in terms of $\beta = 1/T$ and denote $Z(\beta) = \int{d^3p\; f_{eq}(p, \beta)}$. Rewrite the latter as $$ Z(\beta) = 4\pi \int_0^\infty{dp\;\frac{p^2}{e^{\beta E_p}\pm 1}} = 4\pi \int_m^\infty{dE\;\frac{E\sqrt{E^2-m^2}}{e^{\beta ...


17

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 ...


-6

It is caused by the fundamental forces, and Tendency of energy to escape Number 2 itself is caused by fundamental forces, so it is the forces. Most questions end up at why/how there are fundamental forces.


74

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 ...


0

You can allso call it a force free system, meaning that the net force in the system is 0. Lets asume you have simple quasi-closed homogen system filled with helium gas. Then we asume you introduce a force on small volume of the gas in the system. This increases the energy with in this volume. When the excited particles, that we asume have the same energy, ...


0

If you can define a perfectly closed system which eventually arrives back at its starting state, that system is reversible. A reversible system is not irreversible, and can never "become" irreversable without changing the system. However, in the real world, there are limits which come into play. The first is that the systems are never actually isolated ...


0

So I think I have managed to figure it out in the end. So we have Bose-Einstein Distribution: $$ n_i = \frac{1}{e^{\beta(e_i-\mu)}-1}$$ Now, going off an assumption that the gas is degenerate and that chemical potential is zero, this becomes" $$ n_i = \frac{1}{e^{\beta e}-1}$$ Now $\beta = \frac{1}{k_bT}$ I will make another assumption that $T= T_{crtitical} ...


1

Poincaré theorem holds for Hamiltonian systems of finite phase space. We do not know whether Universe is finite or infinite and thus whether it is better to describe it with mechanical model of finite or infinite phase space.


1

Every deterministic system with chaotic trajectories is losing information on account of infinitely small deviations leading to completely different results. In your case, if the ball eventually settles into a stable orbit after some time, no matter where it started, then computing backwards would amplify any uncertainty so that after a few cycles, you know ...


3

Let's take the 1D equivalent of your problem for simplicity: a particle bouncing back and forth along a segment, reversing its velocity every time it hits the boundaries of the segment. If we know perfectly the initial state of the particle, i.e. its position and velocity at time $0$, $(x(0),v(0))$, we will know exactly what the motion of the particle will ...


0

If you assume that the ball is in the state you describe, and will continue to run through the same path over and over again, then you assume, that your system is deterministic, that means, from one position and momentum of the ball, you will be able to calculate its path for all the times that will come after that. In your special case: If the time goes ...


2

There is a really neat way to prove this. The equilibrium probability distribution $P_{e}$ for the canonical ensemble is that which minimizes the free energy functional $F[P]$, i.e. $$F[P] \geq F[P_e]$$ using the method of lagrangian multipliers, we impose that the functional derivative of the free energy with the constraint of normalization $$\int P(C) ...


3

The quantity $\left(\frac{\partial S}{\partial T} \right)_{\{\alpha\}} = TC_v$ is essentially proportional to the heat capacity of the thermodynamic system under study. As far as I know, there is no principle of thermodynamics that forbids such a quantity to be negative. Considerations such as "yes otherwise matter would not be stable" lie outside the ...


1

It is because the only force acting on the molecule is acting perpendicular to the wall, i.e. in the $x$ direction. The force is acting perpendicular to the wall because we assume that there is no friction between particle and wall (they are both perfectly smooth and rigid). In the absence of friction, there can be no force not perpendicular to the wall. ...


0

No. Under negative thermal temperatures the entropy of a system increases as T is lowered. For example, if the magnetic field is reversed quickly enough around a ferromagnetic material, then the system is in negative thermal temperature.


0

Why only x component changes is : The definition of elastic collision (https://en.wikipedia.org/wiki/Elastic_collision) states exactly that. You can see it as ping pong ball that hits the table. All the speed in the direction of the table will make it bounce away from it : if you hit the ball down, it will come up after the bounce on the table. But the other ...



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