- #1
Telemachus
- 835
- 30
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 ]##
##R_{i,j}=|\vec{r}_i-\vec{r}_j|##.
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:
##R_{i,j}=p_{i,j}R##.
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.
##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 ]##
##R_{i,j}=|\vec{r}_i-\vec{r}_j|##.
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:
##R_{i,j}=p_{i,j}R##.
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.