f/A phase space for a sin-
gle atom in one dimension, taking
momentum into account.g/Ludwig Boltzmann’s tomb,
inscribed with his equation for
entropy.5.4.3 Microscopic definitions of entropy and temperature
Two more issues need to be resolved in order to make a micro-
scopic definition of entropy.
First, if we defined entropy as the number of possible states,
it would be a multiplicative quantity, not an additive one: if an
ice cube in a glass of water hasM 1 states available to it, and the
number of states available to the water isM 2 , then the number of
possible states of the whole system is the productM 1 M 2. To get
around this problem, we take the natural logarithm of the number
of states, which makes the entropy additive because of the property
of the logarithm ln(M 1 M 2 ) = lnM 1 + lnM 2.
The second issue is a more trivial one. The concept of entropy
was originally invented as a purely macroscopic quantity, and the
macroscopic definition ∆S=Q/T, which has units of J/K, has a
different calibration than would result from definingS= lnM. The
calibration constant we need turns out to be simply the Boltzmann
constant,k.
Microscopic definition of entropy:The entropy of a system is
S=klnM, whereMis the number of available states.^4
This also leads to a more fundamental definition of temperature.
Two systems are in thermal equilibrium when they have maximized
their combined entropy through the exchange of energy. Here the
energy possessed by one part of the system,E 1 orE 2 , plays the
same role as the variableRin the examples of free expansion above.
A maximum of a function occurs when the derivative is zero, so the
maximum entropy occurs when
d(S 1 +S 2 )
dE 1= 0.
We assume the systems are only able to exchange heat energy with
each other, dE 1 =−dE 2 , so
dS 1
dE 1=
dS 2
dE 2,
and since the energy is being exchanged in the form of heat we can
make the equations look more familiar if we write dQfor an amount
of heat to be transferred into either system:dS 1
dQ 1=
dS 2
dQ 2.
In terms of our previous definition of entropy, this is equivalent to
1 /T 1 = 1/T 2 , which makes perfect sense since the systems are in
thermal equilibrium. According to our new approach, entropy has(^4) This is the same relation as the one on Boltzmann’s tomb, just in a slightly
different notation.
Section 5.4 Entropy as a microscopic quantity 329