Brian Bi

Are there discrete (quantised) units of space (and time)? Does the expression for entropy S = k.lnW suggests this must be so? If not, why not?

We have no evidence of space and time being discrete. As Mark Barton points out, GR and QFT both require continuous space-time, but it's possible that some future theory will describe space-time as discrete.

However, I see that the second part of the question has not been addressed yet. The argument to the logarithm must be dimensionless, and it is supposed to represent the phase space of possible microstates of the system consistent with a given macrostate. The phase space is obtained by integrating over all positions and momenta of a given particle. When you do this integral, you end up with [math]3N[/math] factors of length and [math]3N[/math] factors of momentum (one for each particle and one for each dimension). It is highly tempting to say that phase space is discrete at some sufficiently small scale, so that the entropy should really be defined as

[math]S = k_B \log \frac{\Omega}{N! C^{3N}}[/math]

where [math]C[/math] is some constant with the dimensions of action (length × momentum, or energy × time). In fact, it is really, extremely tempting to take the constant [math]C[/math] to be equal to Planck's constant, which has the correct dimension, and comes along with a hand-wavy justification based on the uncertainty principle: we can say that if two microstates have [math]\Delta x \Delta p \lesssim h[/math], then the uncertainty principle prevents us from being able to distinguish them---the two classical states belong to the same cell.

But I would caution you against taking this argument too seriously. Otherwise you quickly run into absurd conclusions. What's more, it turns out that the value of the constant [math]C[/math] does not affect anything experimentally observable, so it cannot be used to argue for a particular scale at which space-time must be discrete. For example, notice that the ideal gas law contains no Planck’s constant. It is not hard to see why this must be so: the choice of [math]C[/math] only shifts the entropy by a constant, and shifting the entropy by a constant doesn't affect where it attains its maximum. The entropy itself is not observable, so it is not possible to measure the constant directly.