Why is the integrated velocity times change in velocity zero?

[sevbogae]

So consider a subject system, a galaxy, with stars in it. This galaxy is passed by another object, the perturber. When the interaction is fast, we call this interaction a high-speed encounter.

Now we consider that the interaction is really fast. This means that during the interaction, the positions of the stars in the subject galaxy don't change. As a consequence, the potential energy doesn't change. The change in energy during the encounter is thus the change in kinetic energy. We can write the change in kinetic energy as the kinetic energy after the encounter minus the kinetic energy before the encounter
$$ \Delta E = \Delta K = \frac{1}{2} \sum_\alpha m_\alpha \left[(\vec{v}_\alpha + \Delta \vec{v}_\alpha)^2 - \vec{v}_\alpha^2\right],$$
where the sum is going over each individual star $\alpha$. This can be rewritten to
$$ \Delta E = \Delta K = \frac{1}{2} \sum_\alpha m_\alpha \left[|\Delta\vec{v}_\alpha|^2 + 2 \vec{v}_\alpha \cdot \Delta\vec{v}_\alpha\right].$$
Now in a axisymmetric system, the last term, i.e.
$$ \sum_\alpha m_\alpha \vec{v}_\alpha \cdot \Delta\vec{v}_\alpha $$
equals zero. Why is this?

I tried to write the velocity $\vec{v}_\alpha=\omega R \hat{\vec{e}}_\phi$ and the change in velocity as a gradient of the gravitational potential but I can't seem to work it out. Any help? Thanks in advance

 

[PeterDonis]

Now we consider that the interaction is really fast. This means that during the interaction, the positions of the stars in the subject galaxy don't change.

Have you considered whether this assumption is actually valid for a galaxy-galaxy encounter?

As a consequence, the potential energy doesn't change.

Have you considered the potential energy due to the perturber?

Now in a axisymmetric system, the last term, i.e.
$$ \sum_\alpha m_\alpha \vec{v}_\alpha \cdot \Delta\vec{v}_\alpha $$
equals zero.

Where did you get this result from?

 

[sevbogae]

Have you considered whether this assumption is actually valid for a galaxy-galaxy encounter?Have you considered the potential energy due to the perturber?Where did you get this result from?

I got it from "James Binney and Scott Tremaine (2008). Galactic Dynamics: Second Edition" (section 8.2
High-speed encounters, page 655). My question is about page 657, between equation (8.29) and (8.30).

 

[PeterDonis]

I got it from "James Binney and Scott Tremaine (2008). Galactic Dynamics: Second Edition" (section 8.2
High-speed encounters, page 655). My question is about page 657, between equation (8.29) and (8.30).

Ah, ok. It looks like there is some more background explanation on the previous page, which addresses the two questions I posed.

Now in a axisymmetric system, the last term, i.e.
$$ \sum_\alpha m_\alpha \vec{v}_\alpha \cdot \Delta\vec{v}_\alpha $$
equals zero.

Not just an axisymmetric system, a static axisymmetric system. Big difference. Note that the discussion in this section considers the possibility of non static systems, for which $\Sigma_\alpha m_\alpha \vec{v}_\alpha \cdot \Delta \vec{v}_\alpha$ might not be zero; the "static" case is simply an approximation.

As for why $\Sigma_\alpha m_\alpha \vec{v}_\alpha \cdot \Delta \vec{v}_\alpha = 0$ in the static case, that is assigned as Problem 8.5 in the textbook. If you have trouble with that problem, you will need to post a separate thread in the homework forum (and abide by the rules of that forum, which require posting the applicable equations and your attempt at a solution). I would suggest reading that problem anyway since it gives a couple of hints.