Annus MirabilisEinstein in 1905
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brownian motion

The equation explained

F=−RNTlg⁡∫e−ENRTdp1…dpl=−RTNlg⁡B.F = -\frac{R}{N} T \lg \int e^{-\frac{E N}{R T}} d p_1 \ldots d p_l = -\frac{R T}{N} \lg B.
In wordsThe free energy is minus the gas constant divided by the number of real molecules in a gram-molecule, times the absolute temperature, times the logarithm of the integral over all values of the state variables of e raised to minus the energy as a function of those variables times N over R T; and that integral is called B.OverviewSubtracting the temperature times the entropy from the energy leaves the free energy, and in it the system's own energy cancels: what remains is the logarithm of a single integral, which the paper names B. Everything the rest of the section does is done to that one integral.Full explanationThe free energy is the energy less the temperature times the entropy, and the paper writes down its value with a 'therefore', because the entropy of the display before is all it needs. Multiplying that entropy by TT gives Eˉ\bar{E} back in the first term, which then cancels against the Eˉ\bar{E} of the energy, and what is left is −2ϰT-2 \varkappa T times the logarithm of the integral. With 2ϰ=R/N2 \varkappa = R/N this is the first form printed. The second form gives the integral a name, BB: the paper will not evaluate it, and says on the next page that an exact calculation of FF could hardly be thought of even if the molecular picture were settled in every detail. It only needs to know how FF depends on the size of the volume V∗V^* in which the particles are held, and BB is where that dependence sits.Every stepF=Eˉ−TSF = \bar{E} - T SThe free energy in terms of the energy and the entropy. The paper uses it without printing it.F=Eˉ−T(EˉT+2ϰlg⁡∫e−E2ϰTdp1…dpl)F = \bar{E} - T \left( \frac{\bar{E}}{T} + 2 \varkappa \lg \int e^{-\frac{E}{2 \varkappa T}} d p_1 \ldots d p_l \right)The entropy of the display before, put in.F=−2ϰTlg⁡∫e−E2ϰTdp1…dplF = - 2 \varkappa T \lg \int e^{-\frac{E}{2 \varkappa T}} d p_1 \ldots d p_lThe temperature cancels in the first term inside the bracket, leaving Eˉ−Eˉ=0\bar{E} - \bar{E} = 0. The energy of the system has dropped out.F=−RNTlg⁡∫e−ENRTdp1…dplF = - \frac{R}{N} T \lg \int e^{-\frac{E N}{R T}} d p_1 \ldots d p_lWith 2ϰ=R/N2 \varkappa = R/N in the factor and in the exponent. This is the first form printed.B=∫e−ENRTdp1…dplB = \int e^{-\frac{E N}{R T}} d p_1 \ldots d p_lThe name the paper gives that integral. Page 552 calls it the integral B occurring in the expressions for SS and FF.F=−RTNlg⁡BF = - \frac{R T}{N} \lg BThe second form printed. The whole of the section's work is now a question about BB.Also

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