brownian motion

The equation explained

S = \frac{\bar{E}}{T} + 2 \varkappa \lg \int e^{-\frac{E}{2 \varkappa T}} d p_1 \ldots d p_l.

In wordsThe entropy of the system is the energy of the system divided by the absolute temperature, plus twice the constant kappa times the logarithm of an integral, taken over every combination of values of the state variables the problem allows, of e raised to minus the energy as a function of those variables divided by twice kappa times the temperature.OverviewFor such a system the entropy is its mean energy over the temperature, plus a term that counts the states available to it: the logarithm of an integral in which a state of high energy counts for little and one of low energy for much.Full explanationThe letters are named in the sentence under the display.

T

is the absolute temperature,

\bar{E}

the energy of the system,

E

the energy as a function of the

p_\nu

, and the integral is to be taken over all combinations of values of the

p_\nu

compatible with the conditions of the problem. The bar matters:

\bar{E}

is one number, the energy the system has, while

E

under the integral is a function that takes a different value at each combination of the state variables. The constant

\varkappa

is not left free. The text ties it to the

N

of section 1 by

2 \varkappa N = R

, so

2\varkappa

is the gas constant

R

divided by the number of molecules in a gram-molecule: the gas constant for a single molecule. Put that in and the exponent becomes

-EN/RT

, which is the form the free energy of the next display is printed in.Every step

S = \frac{\bar{E}}{T} + 2 \varkappa \lg \int e^{-\frac{E}{2 \varkappa T}} d p_1 \ldots d p_l

As printed. The integral runs over the whole space of the state variables, restricted only by the conditions of the problem, and

d p_1 \ldots d p_l

is its element of volume.

2 \varkappa N = R

The relation the same paragraph states between

\varkappa

and the

N

of section 1, the number of real molecules in a gram-molecule.

2 \varkappa = \frac{R}{N}

Divide by

N

. The gas constant belongs to a gram-molecule, so

R/N

is what belongs to one molecule.

\frac{E}{2 \varkappa T} = \frac{E N}{R T}

The exponent with

2\varkappa

replaced. Dividing by

R/N

is multiplying by

N/R

S = \frac{\bar{E}}{T} + \frac{R}{N} \lg \int e^{-\frac{E N}{R T}} d p_1 \ldots d p_l

The same entropy in the letters of section 1, with

R

and

N

in place of

\varkappa

in both the factor and the exponent.Historian's marginThe paper fixes

\varkappa

only through

2 \varkappa N = R

, so

2\varkappa

is the quantity later written

k_B

and

\varkappa

is half of it. The notation concordance records the halving, because reading

\varkappa

as Boltzmann's constant would double every energy this section counts.AlsoOverviewFull explanationEvery stepHistorian's margin

Read it in the paper