The average velocity of galaxies in a galaxy cluster On slide 24 of these lecture slides (Its title is 'worked example, what Zwicky did'), the author says that $\langle\langle v^{2} \rangle\rangle = 3 \langle\langle{v^{2}}_{s}\rangle\rangle$. Can someone please explain why this is the case? The line-of-sight velocity should be the speed of a galaxy relative to the observer. But What is the radial velocity? Also, why do they have such a relation?
It would be nice if you can also explain the meaning of spherical symmetry in the context (in the slide). To me it sounds like all member galaxies move in random directions, and this seems to contract the fact that they move along orbits.
 A: I will just answer why spherical symmetry implies $\langle v^2 \rangle = 3 \langle v_i^2 \rangle$ where $v_i$ is any Cartesian component since OP has requested this to be explained further.
$$\langle v_x^2 \rangle = \int dv'_x dv'_y dv'_z f(\mathbf{v}) v_x'^2 \tag{1}$$ where $f(\mathbf{v})$ is a probability density function for velocity. By this, we mean that $f(\mathbf{v}) dv_x dv_y dv_z$ is the probability to observe a velocity for a galaxy in the cluster within the range $[v_i, v_i + dv_i]$ for $i \in \{x,y,z\}$
Spherical symmetry means $f(\mathbf{v})$ can only depend on the magnitude of $\mathbf{v}$ because no particular direction of velocity should be privileged. In other words, $f(\mathbf{v}) = f(|\mathbf{v}|)$.
Therefore, $\langle v_y^2 \rangle$ and $\langle v_z^2 \rangle$ have the same values - the integration variables can be relabelled to make the associated integrals look equivalent to the RHS of equation 1.
Equation 1 also shows explicitly why $\langle v^2 \rangle = \langle v_x^2 \rangle + \langle v_y^2 \rangle + \langle v_z^2 \rangle$ after replacing ${v'_x}^2$ with ${v'_x}^2 + {v'_y}^2 + {v'_z}^2$ and separating this into 3 integrals.
