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Entropy and sliding block with friction

  1. Jun 12, 2009 #1
    Something strange showed up while I was thinking about the following problem: I'd like to prove, by the only mean of the principle of maximum entropy, that the thermodynamic destiny of a block sliding on an infinite horizontal plane is to stop and come to thermal equilibrium with the plane itself.

    Let [tex]T_a, T_b, v, F[/tex] be respectively the temperature of the block, of the plane, the speed of the block and the friction force.

    Now the first principle of thermodynamics tells us that

    [tex]\displaystyle dQ = T_a dS_a = dU_a + dL = C_v dT_a + m v dv + F dx[/tex]. Now, solving for [tex]dS_a[/tex], we find that [tex]\frac{\partial S_a}{\partial v}=\frac{mv}{T_a}[/tex] and [tex]\frac{\partial S_a}{\partial T_a}=C_v[/tex]

    Oddly enough, I then noticed that [tex]\frac{\partial^2 S_a}{\partial T_a \partial v}=-\frac{mv}{T_a^2}[/tex] and [tex]\frac{\partial^2 S_a}{\partial v \partial T_a}=0[/tex], which is something we wouldn't expect because of Schwarz's theorem.

    If one goes on mindlessly, integration (which is something we couldn't actually do, because of the problem with mixed partial derivatives) gives

    [tex]\Delta S=C_v \log \left( \frac{T_a}{T_0} \right)+C_b \log \left( \frac{T_B}{T_{0B}} \right)+\frac{m}{2} \left(\frac{v^2}{T_a}-\frac{v_0^2}{T_0}\right)[/tex]
    which is maximum for [tex]v=0, T_a=T_b[/tex], as I'd like to prove...

    So, what am I missing here? I'm sure it's something stupid, but at the moment I can't really see it.

    Thank you all!
  2. jcsd
  3. Jun 12, 2009 #2


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    Just to get me up to speed, could you tell me

    1. How are you writing the differential energy of your system (which looks like it's the block)? For example, if a closed system experiencing [itex]P-V[/itex] work has [itex]dU=T\,dS-P\,dV[/itex], what energy terms are you adding / ignoring?

    2. By adding the [itex]F\,dx[/itex], are you saying that the frictional energy goes entirely into the block? Can you give a reference for your equation [itex]dQ = T_a dS_a = dU_a + dL = C_v dT_a + m v dv + F dx[/itex]? I haven't seen friction treated this way before, and would like to read more.

    3. How are you defining your heat capacity [itex]C_v[/itex]? Holding the velocity constant only?
  4. Jun 13, 2009 #3
    First things first: thanks for your quick reply!
    Now, my system is actually composed of both the block and the plane (indeed, the term F dx vanishes in the total entropy formula because there is an equal term -F dx in the expression for the differential variation of entropy of the plane). For the whole system, I wrote the differential energy as

    [tex]dE= C_v dT_a + C_{plane} dT_b + m v dv[/tex]

    For your second point, please keep in mind I'm only an humble student, and I was simply trying to get something "mechanically" interesting out of thermodynamics principles... I've never seen anything like this on my books, so I'm afraid I can't point you any decent source for such a treatment. Sorry if I gave you the idea to be knowing more precisely what I was doing.

    Coming now to your third point, I think you've centered the problem. I kept thinking to the standard heat capacity at constant Volume ( :blushing: ), and I didn't think that it can be function of velocity, too. Anyways, keeping track of this (and calling the heat capacity simply C) one gets

    [tex] - \frac{mv}{T^2}= \frac{1}{T} \frac{\partial C}{\partial v}[/tex]

    which gives an interesting relation between C and v... that unfortunately I don't think is correct, since it seems to imply a quite fast change in C even for small changes in v, and where I don't like the minus sign, either (i can find an explanation for changes of C in terms of cinetic energies, since slowing down of a fixed amount a fast-moving particle it's harder than doing so for one which is not, but 1) the speed of the block is really small, compared to that of thermal agitation, and 2) it would seem that C has to rise, not to decrease, with speed, if this was the case...)

    I thank you once again, and I hope to have answered, at least partially, to your questions.
  5. Jun 13, 2009 #4


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    OK, I'm on board with

    [tex]dE=C_{V,box}\,dT_{box} + C_{V,plane}\,dT_{plane} + mv\,dv=T_{box}\,dS_{box} + T_{plane}\,dS_{plane} + mv\,dv=0[/tex]

    where we're assuming constant volume for each of the components and overall conserved energy. At equilibrium we also have

    [tex]dS_{box} + dS_{plane}=0[/tex]

    But you have to be careful in incorporating dissipative, irreversible effects like friction into the thermodynamics equations of equilibrium systems. I don't have any experience here; perhaps our resident expert Andys or Count Iblis could weigh in.

    Verbally, at least, we can make the connection between (1) the tendency of entropy to increase, (2) the fact that dissipative processes like friction create entropy, and (3) the fact that friction will slow the box relative to the plane, to conclude that friction will spontaneously slow the box to a stop. At that point it's easy to show from the above equations that at equilibrium [itex]T_{box}=T_{plane}[/itex].
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