# Doesn't sedimentation violate the second law of thermodynamics?

Entropy always increases or remains same.But what happens in sedimentation?The two mixtures separate out i.e disorder decreases.This violates the second law of thermodynamics. What all points am i missing?

• Sedimentation does not change the thermal energy in the system. So there is no change in entropy. – Aziraphale Sep 9 '17 at 6:06
• but doesnt the disorder decreases as particles are separated – spatialdelusion Sep 9 '17 at 6:25
• thermal energy also does not changewhen two things mix so does entropy not increase even then? – spatialdelusion Sep 9 '17 at 6:27
• You might receive more attention to this question by rephrasing the title to something like "Doesn't the process of sedimentation violate the the 2nd law of thermodynamics?" – docscience Sep 9 '17 at 13:56

You are conflating two different notions of 'entropy'. Essentially thermodynamic versus Shannon.

Firstly, what is entropy? Entropy is the logarithm of the number of microstates that correspond to a particular macrostate. But the crucial point you're missing is: what defines the macrostate?

In the context of the second law of thermodynamics, the macrostate is defined by the thermodynamic state variables of the system (such as energy, temperature, pressure, and so on). By this definition, the (thermodynamic) entropy will increase during sedimentation by the following mechanism: as the sediments collect on the floor they lose both kinetic and potential energy. This energy goes into heating the water (and thus increasing the entropy, $\delta Q/T$). If you do the maths you will find that the total entropy does indeed increase.

But what about the separation of sediments from water? Surely this is a more ordered system? It is only more ordered to the human eye, and the second law of thermodynamics does not apply to order as measured by the human eye!

To give a slightly more rigorous elaboration: Suppose you have a system in some microstate $x(t)$ which evolves with time $t$. You can define any function $f(x)$ that maps $x$ to a lower-dimensional space. So, for example, rather than consider the $\sim 10^{23}$ coordinates in $x$ that define every single atom in the water/sediment mixture, you may just project out the coordinates of a few hundred vaguely-defined 'sediments' in the water (this is essentially what our brains do when we look at things floating in water). Naturally, $f$ will provide a much simpler description of the system. And, as such, there will be multiple microstates $x$ that map to any given value for $f$. We may therefore define a (Shannon) entropy for $f$, call it $S(f_0)$, which counts the number of microstates $x$ such that $f(x)=f_0$.

In the case of sedimentation, this particular entropy, $S(f_0)$, that you've implicitly defined will indeed decrease. And that's okay because, as stated above, it is not the thermodynamic entropy (in which case $f$ would project out the thermodynamic state variables).

(As a side note, any projection $f$ will evolve in a way that minimises the free energy.)

• "By this definition, the (thermodynamic) entropy will increase during sedimentation by the following mechanism: as the sediments collect on the floor they lose both kinetic and potential energy. This energy goes into heating the water"how? – spatialdelusion Sep 9 '17 at 11:30
• @daboss Well where else is that energy going to go, other than the water (and seabed)? – lemon Sep 9 '17 at 11:49