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Cohesive energy calculation

  1. Jan 28, 2016 #1
    When the calculation for the cohesive energy in a molecular solid is carried on, there appears a summation for the interaction of all the molecules or atoms in the solid, (see for example Kittel Introduction to solid state Physics 3rd edition, page 87). For noble gases, this interaction could be a Van der Waals interaction, then for any two atoms the interaction potential is given by

    ##U=4 \epsilon \left [ \left(\frac{\sigma}{R} \right )^{12}-\left(\frac{\sigma}{R} \right )^{6} \right ]##.

    So the total cohesive energy in the solid will be given by

    ##U_T=\frac{1}{2} \sum_{i,j} 4 \epsilon \left [ \left(\frac{\sigma}{R_{i,j}} \right )^{12}-\left(\frac{\sigma}{R_{i,j}} \right )^{6} \right ]##


    Now, this double sum can be converted into a single sum, and here is my doubt, which is how the reasoning is made to get that the sum over all i's gives a term of N, you can see this result in the book by Kittel, where besides it is taken a geometrical factor for the distance bewtween atoms, where appears the distance to first neighboors ##R##, so that:


    But basically my doubt is on this step:

    ##U_T=\frac{1}{2} \sum_{i,j} 4 \epsilon \left [ \left(\frac{\sigma}{R_{i,j}} \right )^{12}-\left(\frac{\sigma}{R_{i,j}} \right )^{6} \right ]=\frac{N}{2} 4 \epsilon \sum_{j \neq i} \left [ \left(\frac{\sigma}{R_{i,j}} \right )^{12}-\left(\frac{\sigma}{R_{i,j}} \right )^{6} \right ]##

    Which I interpret that states, for an arbitrary potential between atoms ##u_{i,j}## that:

    ##\sum_{i,j} u_{i,j}=N \sum_{j\neq i} u_{i,j}##.

    That result is the one which I can't understand, I don't know how to get it. I think that it is a fundamental fact to demonstrate this result that ##u_{i,j}=u_{j,i}=u(|\vec{r}_i-\vec{r}_j|)##, but I still don't know how to show that the sum over one index gives N times the same thing.

    Thanks in advance.
  2. jcsd
  3. Jan 29, 2016 #2
    Best to express the term in brackets (where the sigmas and R's are in) as something like U_ij and then manually perform a double sum and collect like terms. You should see the N pop out. Note that for this to work U_ij = U_ji as you stated.
  4. Jan 29, 2016 #3
    Thanks. I got it. Theres still something. When it is computed the case for ionic solids, there appears a factor of 2N, that I don't know where it comes from. Instead of having the sum:

    ##U_T=\frac{1}{2}\sum_i \sum_j u_{i,j}=\frac{N}{2} \sum_{j\neq i} u_{i,j}##

    It comes a factor of 2:

    ##U_T=\frac{1}{2}\sum_i \sum_j u_{i,j}=\frac{2N}{2} \sum_{j\neq i} u_{i,j}##

    And I don't know why that is.
  5. Jan 29, 2016 #4
    I would find the factor of 2 to be normal as it is cancelled out by the 1/2. The purpose of 1/2 is to eliminate the double counting. This is typical for pair interaction potentials.
    If you're comparing with the book, there is a possibility of a typo.
  6. Jan 29, 2016 #5
    No, its ok. The book just considers N molecules, where each molecule is a pair of ions, thats where it comes from. I was a little bit annoyed because for molecular solids (noble gases) that factor of 2 wasn't taken in account, and for ionic solids it was. But now I get what it means. Thanks.
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