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20

First, strictly speaking a neutron star is not a nucleus since it is bound together by gravity rather than the strong force. Measuring a surface temperature for any star is deceptively simple. All that is needed is a spectrum, which gives the luminous flux (or similar quantity) as a function of photon wavelength. There will be a broad thermal peak somewhere ...

2

1. Hawking-Bekenstein entropy of a black hole is given by $S_{\text{BH}} = \frac{kAc^3}{4\hbar G}$ where $A$ is the area of the event horizon. Assuming a non-rotating black hole, there holds $r_s=\frac{2GM}{c^2}$ for the Schwartzschild radius, and therefore $A=4\pi r_s^2=\frac{16\pi G^2M^2}{c^4}$, which results in $$S_{BH}=\frac{4kGM^2}{\hbar c}$$ For ...

2

According to the first law of thermodynamics \begin{align}U=TS+YX+\sum_j\mu_jN_j.\end{align} Where $Y$ is the generalized force, $dX$ is the generalized displacement. Helmholtz Free Energy \begin{align}F=U-TS=YX+\sum_j\mu_jN_j. \end{align} Gibbs Free Energy \begin{align}G=U-TS-YX=\sum_j\mu_jN_j.\end{align} Therefore that ...

1

This is true because we have to use the grand canonical potential $\Omega$ which is directly related to the pressure: $\Omega /V=-P$ (This can be derived by Legendre transformations starting from $E=TS-pV+\mu N$) So if you maximize the pressure you actually minimize the free energy (density) (because of the minus sign) so you are looking for the state with ...

1

Well, first of all, it is important to realize that the integrals (1) and (3) are not merely ordinary double integrals over a single $x$- and a single $p$-variable. Instead they are (Wick-rotated) path integrals containing, heuristically speaking, infinitely many integrations. The path integral derivation of the free particle and the harmonic oscillator ...

1

(Not really an answer, but as one should not state such things in comments, I'm putting it here) You commented: "This seems to boil down to the relationship between the phase space and the Hilbert space." That's a deep question. I recommend reading Urs Schreiber's excellent post on how one gets from the phase space to the operators on a Hilbert space in a ...

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Yes, even though some answers to nearly equivalent cousins of this question claim that the answer is No. We really understand the right "theory of nearly everything" which means that using the microscopic theory – it's the "statistical mechanics" part that is most important here – we may calculate everything that happens in the world of everyday phenomena. ...

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