Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Scaled Particle Theory

  1. Jun 8, 2015 #1
    Hi all,
    I'm reading an article titled "statistical mechanics of rigid spheres" by Reiss, Frisch and Lebowitz (1959). It tries to capture the equation of state of rigid spheres.
    In their article they write the probability that at a random point in the fluid the center of the nearest molecule is a distance ##\lambda ## away and in the range ##[\lambda,\lambda+d\lambda]##. They separate this probability to the product of the probability of finding a cavity of radius ##\lambda##, denoted ##p_0(\lambda)## times the conditional probability of finding at least one molecular center in the ##[\lambda,\lambda+d\lambda]## shell given that there is a cavity of radius ##\lambda## there. So far so good.
    My problem is that they assign things that doesn't look like probabilities to the two product terms. First, they claim that ##p_0(\lambda)=e^{-W(\lambda)/kT}## where ##W(\lambda)## is the reversible work needed to create the cavity. Since we don't know anything about this ##W(\lambda)## why do we assume that ##p_0(\lambda)## is normalized? Then they say that the conditional probability of finding at least one molecular center in the ##[\lambda,\lambda+d\lambda]## shell is ##4\pi \lambda^2 \rho G(\lambda) d\lambda## where ##\rho## is the average fluid density and ##G(\lambda)## is the radial distribution function at the cavity's surface. This too is not normalized since if we look for example at an ideal gas (where ##G(\lambda)=1##) and integrate over ##\lambda## between 0 and ##\infty## we get ##\rho V=N## where ##V## is the volume and ##N## the number of particles in the system. Can anybody comment on that please? Thanks!!
     
  2. jcsd
  3. Jun 8, 2015 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Can you find a possible ##W(\lambda)## where ##p_0(\lambda)## is not normalized?

    Isn't ##\rho## the particle number density?
     
  4. Jun 8, 2015 #3
    Regarding the first reply: let us take, for example, a work function ##W(\lambda)## that goes like the surface area of the sphere ##\lambda^2##. It is reasonable since the work needed to create a cavity of size 0 is 0 and it monotonically increases as ##\lambda## get's bigger, "exploding" at infinity. Then ##p_0(\lambda)## is monotonically decreasing, starting at 1 (when ##\lambda=0##) and reaching 0 at infinity. Integrating this over the range ##0<\lambda<\infty## gives some number, not necessarily 1.

    Regarding the second comment, ##\rho## is, indeed, the particle number density, as I have written. Is there anything I miss here?
     
  5. Jun 8, 2015 #4

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    OK - so what does the paper say about W? (You may have to read between the lines.)
    Is there an online copy of the paper I can see?

    Well you wrote:
    ... so what is wrong with that?

    It may be that you are expected to realize that the function should be chosen so that the result is normalized - they are building a model.
    Did you read the rest of the paper to see what they came up with?
     
  6. Jun 9, 2015 #5
    There it is: ftp://213.176.96.142/aip1c48329b-957e-20141114083409.pdf
    or: http://scitation.aip.org/content/aip/journal/jcp/31/2/10.1063/1.1730361

    Well, for [itex]p_0(\lambda)[/itex] to be a probability function, it has to be normalized so when you integrate over all possible values of [itex]\lambda[/itex] you get 1. Here we get some number which depends on your choice of [itex]W(\lambda)[/itex]. In their article they don't say anything special about [itex]W(\lambda)[/itex] except for the fact that it is the reversible work needed to create the cavity. It could equally be volume work, surface work... whatever.
    I did read the rest of the paper but still, no sign of any constraints on [itex]W(\lambda)[/itex].
     
  7. Jun 10, 2015 #6

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    It looks like it comes ut in the wash though - other papers running the same thing end up deriving functions for W and G.
    Do the authors not do anything like that here?
     
  8. Jun 10, 2015 #7
    The theory isn't about deriving the W function as far as I understand it. It uses a given W function to calculate all other things so it can't assume anything about it, besides the fact that it should go to [itex]\infty[/itex] when [itex]\lambda\rightarrow\infty[/itex] and to 0 when [itex]\lambda\rightarrow0[/itex].
    As I understand it, other papers do the same thing only with explicit W functions as opposed to this article.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook