Why is the integrated velocity times change in velocity zero?

In summary, the assumption that the positions of stars in a galaxy don't change during an interaction is valid for a galaxy-galaxy encounter, but there is a possibility that the potential energy due to the perturber might not be zero.
  • #1
sevbogae
2
0
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
 
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  • #2
sevbogae said:
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?

sevbogae said:
As a consequence, the potential energy doesn't change.
Have you considered the potential energy due to the perturber?

sevbogae said:
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?
 
  • #3
PeterDonis said:
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).
 
  • #4
sevbogae said:
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.

sevbogae said:
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.
 

FAQ: Why is the integrated velocity times change in velocity zero?

1. Why is the integrated velocity times change in velocity zero?

The integrated velocity times change in velocity is zero because it represents the area under the velocity-time graph. When the velocity is constant, the area under the graph is a rectangle with a base of change in time and a height of the constant velocity. Since the rectangle has a width of zero, the area and therefore the integrated velocity times change in velocity is also zero.

2. Can the integrated velocity times change in velocity ever be non-zero?

Yes, the integrated velocity times change in velocity can be non-zero if the velocity is not constant. In this case, the area under the velocity-time graph will not be a rectangle, but rather a trapezoid or some other shape. This will result in a non-zero value for the integrated velocity times change in velocity.

3. How does the concept of acceleration relate to the integrated velocity times change in velocity?

The concept of acceleration is closely related to the integrated velocity times change in velocity. Acceleration is the rate of change of velocity over time, and the integrated velocity times change in velocity represents the total change in velocity over a given time interval. Therefore, acceleration can be calculated by dividing the integrated velocity times change in velocity by the change in time.

4. Why is the integrated velocity times change in velocity important in physics?

The integrated velocity times change in velocity is important in physics because it represents the total displacement of an object over a given time interval. This is a fundamental concept in understanding the motion of objects and is used in many equations and principles in physics, such as the equations of motion and the principle of conservation of energy.

5. How can the integrated velocity times change in velocity be used to calculate the displacement of an object?

The integrated velocity times change in velocity can be used to calculate the displacement of an object by dividing it by the change in velocity. This will give the average velocity of the object over the given time interval. Then, multiplying the average velocity by the change in time will give the displacement of the object. This is known as the average velocity formula: displacement = average velocity x time.

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