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Help:Displaying a formula

$dU=\underbrace {\delta Q} _{\delta Q}+\underbrace {\delta W_{irr}+\delta W_{rev}} _{\delta W}$

$dU=\underbrace {\delta Q+\delta W_{irr}} _{T\,dS}+\underbrace {\delta W_{rev}} _{-P\,dV}$

References {{note_label|Wood1992||}} {{ref_harvard|Wood1992|Wood, 1992|}}
References <ref name="???">reference</ref>,<ref name="???"/>,<references/>,{{rp|p.103}}
References (Harvard with pages)

<ref name="Rybicki 1979 22">{{harvnb|Rybicki|Lightman|1979|p=22}}</ref>
==References==
{{Reflist|# of columns}} 
=== Bibliography ===
{{ref begin}}
*{{Cite book|etc |ref=harv}}
{{ref end}}

equivlist

Template talk:probability distribution#Status of usage
Template talk:Prettytable and MediaWiki talk:Common.css
verify μ is mode of Levy distribution
Combine Heavy tail distribution and Long tail
Mutation-selection balance | Quasispecies model | http://www.biomedcentral.com/1471-2148/5/44

Thermodynamics

Chi-squared distributions


distribution	$\sigma ^{2}$	$\nu$
scale-inverse-chi-squared distribution	$\sigma ^{2}$	$\nu$
inverse-chi-squared distribution 1	$1/\nu$	$\nu$
inverse-chi-squared distribution 2	$1$	$\nu$
inverse gamma distribution	$\beta /\alpha$	$2\alpha$
Levy distribution	$c$	$1$

Heavy tail distributions

Distribution	character	$\alpha$
Levy skew alpha-stable distribution	continuous, stable	$0\leq \alpha <2$
Cauchy distribution	continuous, stable	$1$
Voigt distribution	continuous	$1$
Levy distribution	continuous, stable	$1/2$
scale-inverse-chi-squared distribution	continuous	$\nu /2>0$
inverse-chi-squared distribution	continuous	$\nu /2>0$
inverse gamma distribution	continuous	$\alpha >0$
Pareto distribution	continuous	$k>0$
Zipf's law	discrete	$s-1>-1???$
Zipf-Mandelbrot law	discrete	$s-1>-1???$
Zeta distribution	discrete	$s-1>-1???$
Student's t-distribution	continuous	$\nu >0$
Yule-Simon distribution	discrete	$\rho$
? distribution	continuous	$s-1>-1???$
Log-normal distribution???	continuous	$\rho$
Weibull distribution???	?	${\displaystyle$
Gamma-exponential distribution???	?	${\displaystyle$

Statistical Mechanics

More information Maxwell Boltzmann, Bose-Einstein ...

**A Table of Statistical Mechanics Articles**
	Maxwell Boltzmann	Bose-Einstein	Fermi-Dirac
Particle		Boson	Fermion
Statistics	Partition function Identical particles#Statistical properties Statistical ensemble\|Microcanonical ensemble \| Canonical ensemble \| Grand canonical ensemble
Statistics	Maxwell-Boltzmann statistics Maxwell-Boltzmann distribution Boltzmann distribution Derivation of the partition function Gibbs paradox	Bose-Einstein statistics	Fermi-Dirac statistics
Thomas-Fermi approximation	gas in a box gas in a harmonic trap
Gas	Ideal gas	Bose gas Bose-Einstein condensate Planck's law of black body radiation	Fermi gas Fermion condensate
Chemical Equilibrium	Classical Chemical equilibrium

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Others:

Continuum mechanics

http://online.physics.uiuc.edu/courses/phys598OS/fall04/lectures/

Work pages

To fix:

Degenerate distribution
Fermi-Dirac statistics - not continuous, necessarily
Bose gas (derive critical temperature)
Spectral density-SPD Cat:Physics = Power spectrum-Cat:signal processing

(subtract mean)	(no subtract mean)
Covariance	Correlation
Cross covariance	Cross correlation see ext
Autocovariance	Autocorrelation
Covariance matrix	Correlation matrix
Estimation of covariance matrices

Proof: Introduce an additional heat reservoir at an arbitrary temperature T₀, as well as N cycles with the following property: the j-th such cycle operates between the T₀ reservoir and the T_j reservoir, transferring energy dQ_j to the latter. From the above definition of temperature, the energy extracted from the T₀ reservoir by the j-th cycle is

dQ_{0,j}=T_{0}{\frac {dQ_{j}}{T_{j}}}\,\!

Now consider one cycle of the heat engine, accompanied by one cycle of each of the smaller cycles. At the end of this process, each of the N reservoirs have zero net energy loss (since the energy extracted by the engine is replaced by the smaller cycles), and the heat engine has done an amount of work equal to the energy extracted from the T₀ reservoir,

W=\sum _{j=1}^{N}dQ_{0,j}=T_{0}\sum _{j=1}^{N}{\frac {dQ_{j}}{T_{j}}}\,\!

If this quantity is positive, this process would be a perpetual motion machine of the second kind, which is impossible. Thus,

\sum _{i=1}^{N}{\frac {dQ_{i}}{T_{i}}}\leq 0\,\!

Now repeat the above argument for the reverse cycle. The result is

\sum _{i=1}^{N}{\frac {dQ_{i}}{T_{i}}}=0\,\!

(reversible cycles)

In mathematics, it is often desireable to express a functional relationship $f(x)\,$ as a different function, whose argument is the derivative of f , rather than x . If we let y=df/dx be the argument of this new function, then this new function is written $f^{\star }(y)\,$ and is called the Legendre transform of the original function.

References

[1]
Herrmann, F.; Würfel, P. (2005). "Light with nonzero chemical potential". Am. J. Phys. 78 (3). American Association of Physics Teachers: 717–721. doi:10.1119/1.1904623. Retrieved 2012-12-20. A necessary condition for Planck's law to hold is that the photon number is not conserved, implying that the chemical potential of the photons is zero. While this may be unavoidably true on very long timescales, there are many practical cases that are dealt with by assuming a nonzero chemical potential, which yields an equilibrium distribution which is not Planckian.

								Wall (Impermeable)
								Rigid	T-Ins	No IW	External
	δQh	δWx	dT	dS	dP	dV	dN	dV=0	δQh=0	δWx=0
Process
Isobaric process	-	-	-	-	0	-	0	no	-	-	constant pressure reservoir
Isochoric process	-	-	-	-	-	0	0	-	yes	-	-
Isothermal process	-	-	0	-	-	-	0	-	no	-	constant temperature reservoir
Isentropic process	0	0	-	0	-	-	0	-	yes	yes	-
Adiabatic process	0	-	-	-	-	-	0	-	yes	-	-
x-Isolated System
Mechanically Is. (Guha)	-	-	-	-	-	0	0	yes	-	-	-
Adiabatic Is	0	-	-	-	-	-	0	-	yes	-	-
Thermally Is.	0	0	-	0	-	-	0	-	yes	yes	-
Closed system	-	-	-	-	-	-	0	-	-	-	-
Isolated system	0	0	-	0	-	0	0	yes	yes	yes	-
Open system	-	-	-	-	-	-	-	no	no	no	-
Conserved Thermodynamic potential
U: Internal energy	0	0	-	0	-	0	0	yes	yes	yes	-
F: Helmholtz free energy	-	-	0	-	-	0	0	yes	no	-	constant temperature reservoir
H: Enthalpy	0	0	-	0	0	-	0	no	yes	yes	constant pressure reservoir
G: Gibbs free energy	-	-	0	-	0	-	0	no	no	-	constant pressure and temperature reservoir

δQh is heat, δWx is irreversible work, so TdS=δQh+δWx. If both are zero, then dS=0. For the 3 possible walls, "T ins" means thermally insulated, and "No IW" means no irreversible work. "External" specifies the region external to the system.

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