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Consider a system of $N = \sum_i n_i$ identical fermions, with the discrete hydrogen spectrum. In this case the energy levels are, in Coulomb units, well known to be $E_n = - 1/2n^2$ or, since I used "i" to refer to the $i = 1,2,...$'th energy level: $$E_{i} = - \frac{1}{2i^2},$$ so that the total energy is $E = \sum_i n_i E_i$ where there are $n_i$...


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They didn't associate entropy with a black hole at all. They had no intuition about it at all, basically because they saw no internal structure to a black hole and evaluating entropy requires counting the number of micro-states. It was the intervention by Bekenstein that raised this question and in fact he was able to write out Hawkings entropy formula ...


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Assumption 4 isn't needed to talk about the Boltzmann distribution. Each state has a probability which is proportional to $\exp(-E/k_B T)$, and if there are $g(E)$ different states with the same energy (degeneracy), then the probability to be at any state of energy E is just proportional to $g(E) \cdot \exp(-E/k_B T)$.


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The question asks for intuition. The equation shown in the question can be derived from the fact that $\sigma_E^2$ and $C$ can both be expressed in terms of derivatives of the partition function with respect to $\beta=1/kT$, as shown here. We can extract some intuition from that derivation. In the canonical ensemble, the probability of a state with energy $E$...


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Regarding your first question about rms error: Say the true value of $X$ is $\bar{X}$, and you measured $X_i$ (which on average should be $\bar{X}$). The measurement error would be: $X_i - \bar{X}$. The mean of the square of the errors would be $\langle (X_i - \bar{X})^2 \rangle $ which is exactly the variance. The root of the mean of the squares is the ...


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Let's say you do $n$ measurements and add up the results of the measurements. The reason the rms error in the sum is not $n$ times the rms of a single error is that the errors are assumed to be independent ("each time with an independent error"), so they cancel out each other to some degree (i.e., one experiment might give a positive error, the ...


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Wellcome, @rs_physics. I found it a bit messy how you proceed after the definition of $v$. I would introduce a Taylor expansion in the Chapman-Kolmogorov equation (your first equation). $P(x,t)=(r+p+q)P(x,t) -\Delta t (r+p+q) \partial_tP(x,t)+(q-p )\Delta x\partial_x P(x,t)+(q+p )\Delta x^2\partial_x^2 P(x,t)$ Using p+q+r=1 and rearranging terms you'll ...


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The multiplicative measure that you seek is perhaps most sensibly understood to simply be $W$ itself - the number of microstates that are compatible with the macrostate. You cannot do the exponentiation $$W^{k_B}$$ because $k_B$ has units. Unitful quantities cannot appear as exponents - just ask yourself, "what is $2^\mathrm{m}$, i.e. '2 to the power of ...


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