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Well I have the answer it is $\frac{N(N+1)}{2}$ but what the procedure to derive it .

I tried this. 1).I have $N$ number of translation freedom.

To calculate the number of rotational freedom I tried this.

I have $N^2$ number of constraints. From expression.

$L_iL_j=\delta_{ij}$

$i,j$ can be chosen in $N$ different ways so $N \times N$. So what next?

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There are n degrees of freedom which come from its position (say of its center).

Furthermore, there are n(n-1)/2 rotational degrees of freedom. This is since they correspond to SO(n), the special orthogonal group of n-dimensional matrices which has dimension n(n-1)/2.

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  • $\begingroup$ Ohh that's the reason . thanks $\endgroup$ Mar 31, 2022 at 11:48

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