The average velocity of galaxies in a galaxy cluster [duplicate]

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.

• Spherical symmetry implies $\langle v_x^2 \rangle = \langle v_y^2 \rangle = \langle v_z^2 \rangle$. Also, taking the expectation value is a linear operation, so $\langle v^2 \rangle = \langle v_x^2 \rangle + \langle v_y^2 \rangle + \langle v_z^2 \rangle$ Feb 11 at 15:41
• @Shrey Can you please explain in more detail? To me it is still unclear. If someone explain this mathematically then that would be nice.
– SD11
Feb 13 at 13:17
• @ProfRob No my question includes "Does the symmetry imply that the galaxies moving randomly, and is the symmetry a valid assumption?"
– SD11
Feb 13 at 22:10
• The word "symmetry" is mentioned twice in your comment but not at all in your question. Feb 13 at 23:12
• @Shrey I think I had removed the part in my original post to simply the question, but now I added the question about the symmetry.
– SD11
Feb 14 at 7:42

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.

• Thanks for your answer, but let me ask some minor questions. (1) Are the single-brackets different from the double-brackets in the slides or in my post? (2) In a cluster, galaxies would be moving along orbits. Then is the symmetry still valid, or does it assume a non-realistic situation where galaxies really randomly move?
– SD11
Feb 13 at 22:12
• @SD11 No problem. 1. Yes, the double-brackets, defined on slide 23, define an average over the mass as well. However, the same argument applies. 2. I'm not sure how realistic this assumption is (i.e. to what extent it's a good approximation). Feb 13 at 22:43
• I'm afraid that I'm still having difficulties in understanding this. On the lecture notes, the author explains about the velocities of member galaxies. When that is spherically symmetric, then it would mean that $\langle\langle\vec{v}\rangle\rangle(\vec{r})=\langle\langle\vec{v}\rangle\rangle(r)$. Is this equivalent to what you wrote? Also is f(v) a quantity averaged over time and different galaxies?
– SD11
Feb 14 at 12:24