I want to find the maximum (or a relatively tight upper bound on the maximum) of $$\sum_{i,j\in [n]} \left(\arccos\left(\left\langle \vec{v}_i, \vec{v}_j\right\rangle\right) - \arccos\left(\left\langle \vec{w}_i, \vec{w}_j\right\rangle\right)\right)$$ subject to the constraints that all of the vectors are unit vectors in $\mathbb{R}^n$ and $$\sum_{i,j\in [n]} \left\langle \vec{v}_i, \vec{v}_j\right\rangle = \sum_{i,j\in [n]}\left\langle \vec{w}_i, \vec{w}_j\right\rangle.$$
I asked a similar question but without the relation to unit vectors (i.e., maximize $\sum_{i = 1}^m (\arccos(x_i) - \arccos(y_i))$ subject to $x_1 + \dots + x_m = y_1 + \dots + y_m$). I am wondering if this restriction will lead to a slower rate of growth. I think this might be the case because we cannot make $\sum_{i,j\in [n]} \arccos\left(\left\langle \vec{v}_i, \vec{v}_j\right\rangle\right)$ arbitrary large. For example, if we choose $\vec{v}_1, \dots, \vec{v}_n$ to maximize the minimum $\arccos\left(\left\langle \vec{v}_i, \vec{v}_j\right\rangle\right)$, we must choose $\vec{v}_1, \dots, \vec{v}_n$ to be equally "spread out" according to this post, yielding $$\sum_{i,j\in [n]} \arccos\left(\left\langle \vec{v}_i, \vec{v}_j\right\rangle\right) = n(n-1)\arccos\left(-\frac{1}{n-1}\right).$$