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Why is thermal noise Gaussian distributed in voltage, but Rayleigh distributed in amplitude?

Maybe this will help a bit. Consider two normally distributed variables $X_1,X_2\sim N(\mu=0,\sigma=1) $. Then $X_1,X_1^2,X_1^2+X_2^2$ and $\sqrt{X_1^2+X_2^2}$ will produce the following plots. Notice ...
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3 votes

Why is thermal noise Gaussian distributed in voltage, but Rayleigh distributed in amplitude?

The amplitude is $\rho^2=x^2+y^2$, which is by definition positive, and which will obey Reyleigh distribution, if $x,y$ are a bivariate normal. Note also that detector is always a quadratic detector - ...
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1 vote

Calculating average of a function of molecule's orientation (Euler angles)

Maybe this is just a silly comment, but it is too long to be a comment itself. @GiorgioP 's answer explains why the "$\sin$" has to be in your expression, but is still interesting to notice ...
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5 votes

Calculating average of a function of molecule's orientation (Euler angles)

The reason is the same why, in the case of the spherical coordinates, the surface area element contains the $\sin(\theta)$ factor: we would like to have a uniform measure on the unit sphere surface ...
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2 votes

Properties of random-walk in infinite and finite two-dimensional space: probability of two particles being in the same location at time t

In general, for this kind of stochastic processes, and $t_f > t > t_i$, we have : $$G(x_f,t_f|x_i,t_i) = \int G(x_f,t_f|x,t)G(x,t|x_i,t_i)\text dx$$ Representing the fact that the particle must ...
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2 votes

How does an exothermic reaction release energy?

The energy release din chemical reaction comes from the difference in binding energies of the reactants and the reaction products. Thus, if we consider two atoms $A$ and $B$ that can join in a ...
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1 vote

On-shell Poisson brackets and time derivative

According to Liouville's theorem $$\frac{d\rho}{dt}(S^t(p,q),t)=0\iff\frac{\partial\rho}{\partial t}(S^t(p,q),t)+\{H,\rho\}=0\tag{A}$$ where the Poisson Brackets are evaluated on the Hamiltonian flow ...
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3 votes

How does an exothermic reaction release energy?

Here is more information related to the basic question "How does an exothermic reaction release energy?" Classical thermodynamics accounts for the release of energy using such concepts as &...
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Thermal noise phasor amplitudes are Rayleigh distributed. How are voltages at the antenna distributed?

Let $\mathbf z = \mathbf x + \mathfrak j \mathbf y$ where the random variates $\mathbf {x,y}$ are independent and normal distributed with zero mean, that is: $\mathbf {x,y} := \mathcal N (0, \sigma)$. ...
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Do Stochastic Differential Equation models conserve energy?

My question is this: Does this model conserve energy at all points in time? Alternatively, how do stochastic models like this deal with energy conservation? You actually answer this question in the ...
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8 votes

How does an exothermic reaction release energy?

For example, one (or more) of the reaction partners is after the chemical reaction not in the ground state, but in an excited state. The electron will sooner or later fall back into the ground state ...
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4 votes
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Derive Canonical Ensemble from Maximum Entropy Principle

A good way to start the proof is to first select some basis such that the density matrix $$\rho=\sum_{j}{p_j|\psi_j\rangle \langle\psi_j|}$$ where $p_j>0$ and $\{|\psi_j\rangle\}$ is orthonormal. ...
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1 vote

The probability of a circular region being "invaded" by moving spheres as a function of time

I assume classical mechanics applies to this problem. If by the spheres "do not interact with each other" you mean no collisions, then only those spheres initially headed in a direction that ...
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0 votes

Why does adiabatic expansion occur in the carnot process?

In the second step, the system is separated from the reservoir and is now thermally insulated, resulting in another small adiabatic expansion. I ask myself, why is a little internal energy converted ...
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Why does adiabatic expansion occur in the carnot process?

You are in control of the piston, the piston does not move spontaneously, the external pressure is set up (controlled) by you at every time so that the process is quasistatic. So the system is at ...
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Why does adiabatic expansion occur in the carnot process?

Because both heat transfer steps must be isothermal to be reversible and generate no entropy. The Carnot cycle works by exploiting the temperature difference between two reservoirs. It uses reversible ...
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1 vote

Why does adiabatic expansion occur in the carnot process?

At the end of the reversible isothermal expansion the external pressure is deliberately further slowly reduced, but now according to $Pv^{\gamma}$ = constant, instead of $Pv$=constant, to allow the ...
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How do solids transfer heat?

As the highly influential 20th century physicist, Rudolph Peirls said in his article "Quantum Theory of Solids" in 1951: "It seems there is no problem in modern physics for which there ...
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2 votes

Do Stochastic Differential Equation models conserve energy?

Physically motivated Langevin equations usually contain dissipative terms alongside the noise terms, whose parameters are related via the fluctuation-dissipation theorem (of which the Einstein ...
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2 votes

Do we need atomic theory to do thermodynamics?

Thermodynamics and statistical physics are two different ways of looking at the same thing. The former indeed does not require atomic theory, see, e.g., here and here. Also relevant: Timeline of ...
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1 vote
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Explaining internal energy from a macroscopic perspective

What you need really is a proper starter-course in thermodynamics, with a good textbook (and reader can guess which one I recommend!) The logic goes as follows. Various experiments (especially those ...
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1 vote

Explaining internal energy from a macroscopic perspective

The concept of internal energy is introduced in the first law of thermodynamics, based on Joule's work. Based on his experiments with heating and mixing water with a paddle wheel, he realized the ...
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1 vote

What is the theoretical value of this phase space invariant?

After pondering for a while, I thought it might be worth sharing a different angle to this problem. So let's start with the stress energy tensor $T^{\mu \nu}$ of a perfect fluid: $$T^{\mu \nu} = \left(...
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2 votes

Explaining internal energy from a macroscopic perspective

The Joule historic experience measuring carefully how mechanical work can raise the temperature of water is the basic idea of internal energy in my opinion. We know from mechanics that the work of the ...
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1 vote

Explaining internal energy from a macroscopic perspective

What came up in comments to this answer to the question quoted in the OP, is that one has to distinguish, on the one hand, phenomenological vs. microscopic models/theories and on the other hand, ...
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2 votes

What is the theoretical value of this phase space invariant?

I'm going to start by using some relations that appear explicitly in the text you linked. Specifically: \begin{equation} \mathcal{N} = \frac{g_{s}}{h^{3}} \eta \end{equation} where, for the era of ...
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3 votes
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Why is electronic specific heat divided by $V$ rather than $M$?

Start with the definition of (extensive) specific heat: $$C_v=\frac{dE}{dT}$$ from there, you can define three different intenstive quantities: massic : $\displaystyle c_v=\frac{C_v}{m}$ in J/K/kg ...
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1 vote

What actually are microscopic and macroscopic viewpoints in thermodynamics?

I am not a physicist, and I won't give a technically precise answer, but I think a simple question deserves a simple answer, even if it leaves a lot more to be said. I think the essence of your ...
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Energy dissipated by friction and entropy

It would be correct to assume that the entropy increase of the universe after the block has stopped is: $$\Delta S = \frac{K}{T}$$ In response to my query, you indicated that the "ambient ...
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1 vote
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Integration by Parts in Liouville's Theorem

The first thing to note, which I leave to you to verify, is that $D_H$ is a derivation on smooth functions, meaning it is linear and satisfies a product rule: \begin{align} D_H(FG)=(D_HF)\cdot G+F\...
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-3 votes

What actually are microscopic and macroscopic viewpoints in thermodynamics?

There is no meaningful concept of micro vs. macro in thermodynamics. There is no natural scale to the universe. Consider this in relation to Einstain's claim that there is no absolute reference ...
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Unitary evolution and von Neumann entropy

Another neat method to prove this: write the von Neumann entropy as a limit of the Renyi entropies: $$ S[\rho] = \lim_{n \to 1} S^{(n)}[\rho] = \lim_{n \to 1} \frac{1}{1-n} \log \text{Tr} \rho^n $$ ...
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1 vote

What quantity can a microstate have?

A microstate is defined as a specific microscopic configuration that a system can have. One of the primary goals of statistical physics is to see the relation between microscopic proprieties and ...
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12 votes
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What actually are microscopic and macroscopic viewpoints in thermodynamics?

The question is about the meaning of terms such as 'microscopic' and 'macroscopic' in thermal physics generally. To define these terms it is best to put to one side for a moment the idea that large ...
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4 votes

What actually are microscopic and macroscopic viewpoints in thermodynamics?

Temperature is the measure of the average KE of the molecules of a system. Clearly, we're talking about molecules when we talk about temperature then why it is a macroscopic concept? One should ...
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6 votes

What actually are microscopic and macroscopic viewpoints in thermodynamics?

A full answer to your question is simply reading a book on statistical physics. The thermodynamic limit There's no microscopic thermodynamics. At the microscopic level, you simply use ordinary ...
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5 votes
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Unitary evolution and von Neumann entropy

Hint: Use the spectral decomposition to write $$\rho(0) := \sum\limits_k \lambda_k \,|k\rangle \langle k| \tag{1} ,$$ and then find an expression for $\rho(t)$ in terms of $\lambda_k$. Especially ...
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0 votes

What's the significance of a quasi-static process?

If a system initially at equilibrium in state 1 is moved to state 2, then it will not be necessarily in equilibrium in this latter state, so one will have to wait for some time that it relaxes to ...
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What's the significance of a quasi-static process?

A quasi static process is an (ideal) process where the system remains in internal equilibrium at all time, but not in equilibrium with the external environment. Internal equilibrium implies that ...
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1 vote

Is the free energy of the 3 and 4 state Potts model in a positive magnetic field analytic?

To my knowledge this is not know, but expected. In the paper https://arxiv.org/abs/2206.07033 we prove that for sufficiently large strength of the magnetic field $h>0$ the free energy is analytic.
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Proof of pressure of ideal gas from first principles

Before getting into the proof, I should note that an ideal gas can interact only via elastic collisions. Essentially, the particles' individual velocities can change but their energy remains constant. ...
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2 votes

Proof of pressure of ideal gas from first principles

Statistical physics does make assumptions about interactions: these are essential for establishing the thermodynamic equilibrium, but do not determine the actual form of this equilibrium. So the non-...
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Issue with path integrals for the partition function

I'm thinking of the below reason. Please let me know if someone sees an issue with it. There are two Jacobians to consider a) $\sqrt{\frac{1}{\Delta \tau a^3}}$ and b)The Jacobian $\sqrt{\beta^3 V}$, ...
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Partition function for $H=-\vec \mu \cdot \vec B$

I don’t feel that the preceding answers address the question fully. Granted, the canonical ensemble dictates that your microstates have a Boltzman weight. However, you need to apply these weights to a ...
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1 vote
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Intuition behind entropy and its differentiation

This is a development of my comment below the original question. I think that the calculation is wrong, even though the final result is correct. For a monoatomic ideal gaz: $$dU=\frac{3}{2}\,nR\,dT$$ ...
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1 vote

Partition function for $H=-\vec \mu \cdot \vec B$

The calculation of the microcanonical ensemble involves integrating over all possible energetic microstates, which in this case translate to all possible spin (or magnetic dipole) orientations. You ...
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Partition function for $H=-\vec \mu \cdot \vec B$

Magnetic field here is a preferred axis. So, if we simply integrate over $\theta$ our result will be dependent on the direction of this axis. One way to do it consistently is to start with Cartesian ...
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0 votes

Energy dissipated by friction and entropy

It would be correct to assume that the entropy increase of the universe after the block has stopped is: $$\Delta S = \frac{K}{T}$$ This would be correct only provided the temperature of the "...
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1 vote
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Probability of collision between two particles (Statistical Mechanics)

$\iiint P(\vec{r}) dV = {1 \over N} \iiint n(\vec{r}) dV = {1 \over N} \iiint dN(\vec{r}) = 1$ Where $N$ is the total number of particle in $dV$. Now these functions can be generalized to the ...
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1 vote

Energy dissipated by friction and entropy

If the initial temperature and the final temperature of the block are both T, then applying the complete version of the first law of thermodynamics to the block gives $Q=-K$, where Q is the amount of ...
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