14
No you cannot.
we can show that (via conservation of kinetic energy and momentum) the speed of the particle before and after collision is the same.
Only in the center of mass of two elastically colliding particles the momentum remains the same. Each pairwise collision has a different center of mass. In the laboratory frame, which is the frame one is ...
9
Even if initial speed of all the particles was same, the molecular collisions will disrupt this uniformity.
I want to show this happens based on a diagram (on the left) from one of Maxwell's papers.
The centre and right diagrams are for two particles with the same mass travelling at the same speed $u_1 = u_2$ and for ease of presentation it is ...
9
"we can show that (via conservation of kinetic energy and momentum) the speed of the particle before and after collision is the same."
I think you'll find that for a head-on elastic collision between balls of equal mass, the particles swap velocities. But this is not the case for non-head-on (oblique) collisions. A naïve assumption would be that in this ...
5
The Maxwell-Jüttner distribution describes the distribution in a relativistic gas:
$$ f(\gamma) = \frac{\gamma^2\beta}{\theta K_2(1/\theta)}e^{-\gamma/\theta}$$
with
$$ \theta = \frac{kT}{mc^2}$$
The Boltzmann factor is defined in terms of energy, not velocity, and $E = \gamma m$, so the probability of any unbounded energy is not a problem.
Note that ...
4
No, you cannot write that particles are allowed to surpass the speed of light but with a vanishing probability. The fundamental reason is that statistical mechanics for classical particles is constructed in phase space $(\vec r,\vec p)$ Therefore, in special relativity, the partition function of an ideal gas reads
$${\cal Z}=z^N, \quad z=\int e^{-\beta\...
3
I think your perplexity originates from a too simple idea about what a photon is.
But, before discussing issues related to the full-flag concept of photon, let me address your question in a simpler way. For that you have to forget any idea about photons.
Let's go back to the blackbody radiation as a system made by electromagnetic radiation in thermal ...
3
Entropy isn't a property of a microstate (after there is only one of it so its entropy would be zero), but of the ensemble of states available under some set of conditions.
By the "some set of conditions" you consider in counting the states are generally defined in terms of macroscopic variables.
That is: the defining condition for the entropy is a ...
3
The function $p$, given by
$$p(t) = \frac{1}{\tau}e^{-t/\tau},$$
is the probability density for the time between collisions to be $t$. That is, the probability that the time between collisions is between $t$ and $t+\Delta t$ is given by
$$p(t)\Delta t = \frac{1}{\tau}e^{-t/\tau}\Delta t.$$
Once the probability distribution is established, we compute an ...
2
I suspect that what was meant by that statement (lacking saturation) is that no stable N-body state exists for the attractive case. For fermion systems the attractive case can be stabilized by the Pauli principle, but Bose systems lack this stabilization mechanism. Leib"s work has for many decades specialized in the study of stability in many body quantum ...
2
The product of the Boltzmann constant $k_B$ and the Avogadro constant $N_A\approx 6.022 \times 10^{23}\:\rm mol^{-1}$ (not the Avogadro number, which is the same but without the $\rm mol^{-1}$ qualifier) is equal to the molar gas constant:
$$
R = k_B N_A.
$$
The molar gas constant is the constant that shows up in the ideal gas law when it is expressed as $PV ...
1
You are missing a key point about averages.
In your first case you have assumed all energies are equally probable.
In your second case you have assumed energies have a probability distribution $P(E)$ of $e^{-\beta E}/Z$.
In general the average energy is $\sum E\cdot P(E)$.
1
The index $i$ in $n_i$ and $\epsilon_i$ refers to the label of the one-particle states. I.e., the Hamiltonian $\hat H_N$ of the non-interacting gas of $N$ particles is written as a sum of $N$ one-particle Hamiltonian $\hat h_{\alpha}$:
$$
\hat H_N= \sum_{\alpha=1}^N \hat h_{\alpha}
$$
and the eigenstates of $\hat h_{\alpha}$ are the states $\left| i\right>...
1
You are surprised that $S(T, 2V, 2N) = 2S(T, V, N)$ but this formula can be obtained with statistical physics considering a mix of two identical gas
When you remove the barriers and the two gas are identicals you do not increase the number of unknowns configurations of the system, therefore, entropy does not increase. (This is a really intuitive view of the ...
1
I think the answer is that it is a much more natural physical intuition to say that a hot body has more of something than to say a cold body has more of something. Put a cooking pot on a fire: it is very natural to say that the flames are providing something to the pot, rather than the pot delivering 'anti-heat' to the flames. When water boils, it becomes ...
1
To my understanding:
The heat bath has a fixed temperature but not a uniform energy distribution. The energy of the heat bath at any given point is constantly fluctuating, if during these fluctuations a large amount of energy happens to be concentrated around the system (which is inside the heat bath) the system is going to gain some of that energy.
1
Feynman derives an expression for the probability that the molecule survives for a time $t$ without a collision as $P(t)=\exp(-t/τ)$, ie the probability a particle, during a time interval of $t$, not colliding with anything.
At $t$ increases the probability of such an event (no collision) decreases which is what one would expect.
From the probability ...
1
For the Ising model, we have the power of exact solutions. In particular, the one-dimensional transverse-field quantum Ising model, defined by the Hamiltonian
$$
H = - J \sum_{i} \sigma^z_i \sigma^z_{i+1} - h \sum_i \sigma^x_i,
$$
turns out to be exactly solvable. One can use the Jordan-Wigner transformation to map it to free fermions, after which you can ...
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