- #1

sevbogae

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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