Homework SoursePlease take time to show all workSolutionHow many ways are t
Please take time to show all work.
Solution
How many ways are there to get total moment zero for N = 4 and N = 6?
N = 4
(0)1(0)2(0)3(0)4, (+1)1(-1)2(0)3(0)4, (+1)1(0)2(-1)3(0)4, (+1)1(0)2(0)3(-1)4, (-1)1(+1)2(0)3(0)4, (0)1(+1)2(-1)3(0)4, (0)1(+1)2(0)3(-1)4, (-1)1(0)2(+1)3(0)4, (0)1(-1)2(+1)3(0)4, (0)1(0)2(+1)3(-1)4, (-1)1(0)2(0)3(+1)4, (0)1(-1)2(0)3(+1)4, (0)1(0)2(-1)3(+1)4, (+1)1(+1)2(-1)3(-1)4, (+1)1(-1)2(+1)3(-1)4, (+1)1(-1)2(-1)3(+1)4, (-1)1(+1)2(+1)3(-1)4, (-1)1(+1)2(-1)3(+1)4, (-1)1(-1)2(+1)3(+1)4
So, admittedly, the most probable configuration is where two moments take the value 0 and two others take the value +1 and -1.
N = 6
[To keep things more readable, I\'m just going to represent, say five states, (+1)1(-1)2(0)3(0)4(0)5(0)6, (+1)1(0)2(-1)3(0)4(0)5(0)6, (+1)1(0)2(0)3(-1)4(0)5(0)6, (+1)1(0)2(0)3(0)4(-1)5(0)6, (+1)1(0)2(0)3(0)4(0)5(-1)6, by 5 * [(+1)1(-1)*2(0)3(0)4(0)5(0)6] where (+1) is constant in all the set and (-1)* is taken by different moments in different sets so that the total moment is always zero and the combinations do not overlap.]
(0)1(0)2(0)3(0)4(0)5(0)6, 5 * [(+1)1(-1)*2(0)3(0)4(0)5(0)6], 5 * [(0)1(+1)2(-1)*3(0)4(0)5(0)6], 5 * [(0)1(0)2(+1)3(-1)*4(0)5(0)6], 5 * [(0)1(0)2(0)3(+1)4(-1)*5(0)6], 5 * [(0)1(0)2(0)3(0)4(+1)5(-1)*6], 5 * [(-1)*1(0)2(0)3(0)4(0)5(+1)6], 6 * [(+1)1(+1)2(-1)*3(-1)*4(0)5(0)6], 6 * [(0)1(+1)2(+1)3(-1)*4(-1)*5(0)6], 6 * [(0)1(0)2(+1)3(+1)4(-1)*5(-1)*6], 6 * [(-1)*1(0)2(0)3(+1)4(+1)5(-1)*6], 6 * [(-1)*1(-1)*2(0)3(0)4(+1)5(+1)6], 6 * [(+1)1(-1)*2(-1)*3(0)4(0)5(+1)6], 6 * [(+1)1(-1)*2(-1)*3(0)4(+1)5(0)6], 6 * [(+1)1(-1)*2(-1)*3(+1)4(0)5(0)6], 6 * [(+1)1(0)2(+1)3(-1)*4(-1)*5(0)6], 6 * [(0)1(+1)2(-1)*3(-1)*4(0)5(+1)6], 6 * [(0)1(+1)2(-1)*3(-1)*4(+1)5(0)6], 6 * [(0)1(+1)2(0)3(+1)4(-1)*5(-1)*6], 6 * [(-1)*1(-1)*2(+1)3(0)4(0)5(+1)6], 6 * [(-1)*1(-1)*2(+1)3(0)4(+1)5(0)6], 6 * [(-1)*1(-1)*2(0)3(+1)4(0)5(+1)6], 10 * [(+1)1(+1)*2(+1)*3(-1)*4(-1)*5(-1)*6], 6 * [(-1)1(+1)2(+1)*3(+1)*4(-1)*5(-1)*6], 3 * [(-1)1(-1)2(+1)3(+1)*4(+1)*5(-1)*6], (-1)1(-1)2(-1)3(+1)4(+1)5(+1)6
The calculation gets complex with the increasing N value. However, it is evident from the above numbers that the most probable state is the one where two of the moments take +1 value, 2 of the other moments take -1 value and the two other moments take 0 value.
N = 4 and total moment NT = 2
(+1)1(+1)2(0)3(0)4, (0)1(+1)2(+1)3(0)4, (0)1(0)2(+1)3(+1)4, (+1)1(0)2(+1)3(0)4, (+1)1(0)2(0)3(+1)4, (0)1(+1)2(0)3(+1)4
N = 6 and total moment NT = 3
{(+1)1(+1)2(+1)3(0)4(0)5(0)6, (+1)1(+1)2(0)3(+1)4(0)5(0)6, (+1)1(+1)2(0)3(0)4(+1)5(0)6, (+1)1(+1)2(0)3(0)4(0)5(+1)6, (0)1(+1)2(+1)3(+1)4(0)5(0)6, (0)1(+1)2(+1)3(0)4(+1)5(0)6, (0)1(+1)2(+1)3(0)4(0)5(+1)6, (0)1(0)2(+1)3(+1)4(+1)5(0)6, (0)1(0)2(+1)3(+1)4(0)5(+1)6, (+1)1(0)2(+1)3(+1)4(0)5(0)6, (0)1(0)2(0)3(+1)4(+1)5(+1)6, (+1)1(0)2(0)3(+1)4(+1)5(0)6, (0)1(+1)2(0)3(+1)4(+1)5(0)6, (+1)1(0)2(0)3(0)4(+1)5(+1)6, (0)1(+1)2(0)3(0)4(+1)5(+1)6, (0)1(0)2(+1)3(0)4(+1)5(+1)6, (+1)1(+1)2(0)3(0)4(0)5(+1)6, (+1)1(0)2(+1)3(0)4(0)5(+1)6, (+1)1(0)2(0)3(+1)4(0)5(+1)6, (+1)1(0)2(+1)3(0)4(+1)5(0)6, (0)1(+1)2(0)3(+1)4(0)5(+1)6}
[(+1)1(+1)2(+1)3(-1)4(+1)5(0)6, (+1)1(+1)2(+1)3(-1)4(0)5(+1)6, (+1)1(+1)2(+1)3(+1)4(-1)5(0)6, (+1)1(+1)2(+1)3(+1)4(0)5(-1)6, (+1)1(+1)2(+1)3(0)4(+1)5(-1)6, (+1)1(+1)2(+1)3(0)4(-1)5(+1)6]--- this set is a variable of [(+1)1(+1)2(+1)3(0)4(0)5(0)6] where in two other moments have taken +1 and -1 values. This way, each set in the {} will have six more configurations when two other moments in the set +1 and -1 values, where the total moment remains N/2 = 3
But as can be seen, in this case, the most probable configuration is the one where 4 of the moments take +1 , one of the moments takes -1 and the other takes 0 values.
Thus, we can confidently say that, in the limit of large N, fluctuations away from the most probable configuration become rare.
