Why would this star move outward?

[Buckethead]

I read this (relavant text made bold):

"In the study, published in the Monthly Notices of the Royal Astronomical Society, researchers analyzed Gaia space telescope observations of a large group of stars, the Hercules stream, which are in resonance with the bar -- that is, they revolve around the galaxy at the same rate as the bar's spin.

These stars are gravitationally trapped by the spinning bar. The same phenomenon occurs with Jupiter's Trojan and Greek asteroids, which orbit Jupiter's Lagrange points (ahead and behind Jupiter). If the bar's spin slows down, these stars would be expected to move further out in the galaxy, keeping their orbital period matched to that of the bar's spin."

Why would a star move outward if it is forced to slow its orbit? If a satellite around the Earth fires rockets to slow down, it falls toward the Earth.

 

[PeterDonis]

I read this (relavant text made bold)

Where? Please give a reference.

 

[Buckethead]

https://www.sciencedaily.com/releases/2021/06/210614153956.htm

 

[PeterDonis]

Why would a star move outward if it is forced to slow its orbit?

That's not what's being described. What's being described is stars continuing to match their orbital periods to the period of the bar's spin. So if the bar's orbital period increases, the stars, to continue to have the same orbital period as the bar, would have to move outward.

Without a reference it's impossible to know what mechanism is being claimed to enforce the matching of the stars' orbital periods to the bar's spin.

 

[Buckethead]

So if the bar's orbital period increases, the stars, to continue to have the same orbital period as the bar, would have to move outward.

Why would the star move outward? If it, and all the stars in the bar are slowing down, why wouldn't the whole bar start to collapse inward?

 

[PeterDonis]

If it, and all the stars in the bar are slowing down, why wouldn't the whole bar start to collapse inward?

Having now read the actual paper (a link to it is at the bottom of the Science Daily article you linked to), it's clearer to me what is being described.

The stars being referred to are not in the bar. The bar is at the center of our galaxy; the stars are much further out. But their orbits are in resonance with the bar; or, to put it another way, their orbits have just the right orbital period to be pulled by the bar's gravity the same way on each orbit. The simplest example would be a star that happens to have the same orbital period as the bar; the bar's gravity would tend to keep it in that orbit, i.e., the bar's gravity would counteract small perturbations from other sources of gravity (such as other nearby stars) that would otherwise change its orbit. Of course there are other more complicated resonances besides this one, and the paper is considering those as well.

The bar itself is not a single object; it is made up of closely packed stars near the galactic center, which have varying orbital periods. The way the orbital periods of the individual stars vary results in the bar itself having a net orbital period that is significantly slower than the orbital periods of the stars inside it. That is why the stars in resonance with the bar are not in the bar itself, but much further out.

As the bar's rotation slows, the effect on the stars in resonance, as far as I can tell, is similar to the effect on the Moon of the slowing rotation of the Earth and the consequent effects on the tidal bulge in the Earth that is caused by the Moon: the net effect is for the Moon's orbit to move further outward (and its orbital period to increase). Or, to put it another way, angular momentum is transferred from the Earth's spin to the Moon's orbital angular momentum. Similarly, as the bar in our galaxy slows its rotation (due to the drag from the dark matter halo), angular momentum is transferred from the bar's spin to the orbital angular momentum of the stars in resonance with the bar, causing them to move outward and their orbital periods to increase.

So to understand the general mechanism better (i.e., how the orbiting object ends up moving outward), I would recommend looking at more details about the Earth-Moon case.

 

[Buckethead]

Thank you so much for the detailed reply. I don't understand the transfer of momentum from the Earth to the Moon. Are you saying it has to do with the consequences of the Earth bulging, but how does that work? The Moon pulls on the Earth as it rotates, the Earth bulge moves relative to the ground so there is an energy loss, and this loss transfers to the Moon increasing it's momentum and that intrinsic increase in momentum forces the moon to move to a higher orbit?

 

[PeterDonis]

I don't understand the transfer of momentum from the Earth to the Moon. Are you saying it has to do with the consequences of the Earth bulging, but how does that work?

The Earth is rotating faster than the Moon revolves around it (rotation period 1 day vs. Moon orbital period 1 month). That means the Earth is rotating through the tidal bulge created by the Moon; the friction thus created causes the Earth's rotation to slow down. The angular momentum lost has to go somewhere since total angular momentum is conserved; if you work out the details, it goes into the Moon's orbital angular momentum, making the Moon's orbit move further away from the Earth (and increasing its orbital period).

The Moon pulls on the Earth as it rotates, the Earth bulge moves relative to the ground so there is an energy loss

There is an energy loss, yes (see below), but there is also a loss in the Earth's spin angular momentum; that is what I was talking about.

and this loss transfers to the Moon increasing it's momentum

This spin angular momentum loss in the Earth is transferred to the Moon, increasing its orbital angular momentum.

that intrinsic increase in momentum forces the moon to move to a higher orbit?

The increase in angular momentum is the Moon moving to a higher orbit. (The Moon's orbital energy also increases; again, this is balanced by a loss of spin energy in the Earth as its spin slows.)

 

[Buckethead]

The increase in angular momentum is the Moon moving to a higher orbit. (The Moon's orbital energy also increases; again, this is balanced by a loss of spin energy in the Earth as its spin slows.)

OK, got it. I take it the Moon's velocity is actually higher in its new orbit even though its angular velocity is less. Is this correct?

Back to post #6, In this case we have a hypothetical dark matter halo that has significantly more mass than the galactic center. The bars in the center slow down due to the pull of the DM. If this is the case, I am guessing that the transfer of energy is not strictly between the DM and the bars but includes the star in question. Is this correct? I would think that the slowing of the bars would simply force the star to slow due to simple gravitational pull of the bar on the star whereby the star can't move too far ahead without the bar pulling it backward and that this is not related to transfer of angular momentum as is the case with the Moon. If that is true then this looks more like the case of an Earth satellite just slowing down due to the firing of retro rockets and so should instead fall instead of moving to a higher orbit.

Also, the DM halo should be more or less symmetrical, so the galactic center shouldn't have a bulge and therefore should not have a loss of angular momentum as would the Earth.

 

[Bandersnatch]

OK, got it. I take it the Moon's velocity is actually higher in its new orbit even though its angular velocity is less. Is this correct?

No, the tangential velocity is lower in the higher orbit. The torque from the tidal bulge continuously acts prograde, much like a low-powered rocket thruster would. This momentarily increases orbital velocity above what is required for a stable orbit at the current radius, which forces the orbit to raise. Raising the orbit reduces the orbital velocity to below the initial value, but still a bit above what is required for the new orbit (due to the tidal torque still pulling prograde) - continuing the process. Since this is a continuous process, the slope of the velocity change is always negative, i.e. the Moon decelerates as it recedes.

Also, the DM halo should be more or less symmetrical, so the galactic center shouldn't have a bulge and therefore should not have a loss of angular momentum as would the Earth.

The mechanism here is not the same as with tidal bulges - this is dynamical friction against the surrounding DM. Check Wikipedia for intuitive descriptions.

I won't comment on the rest of the post, as I find the nuts and bolts of resonances confusing. Maybe I'll just say that these involve interactions over multiple orbits with different periods, with forces between bodies (or structures) varying in magnitude and direction. So it's generally a mistake to think of those as one-off or static interactions.

 

[PeterDonis]

The mechanism here is not the same as with tidal bulges - this is dynamical friction against the surrounding DM.

Dynamical friction against the surrounding DM is how the bar's rotation slows, yes. I didn't mean to imply that the slowing of the bar was due to tidal forces from the stars in the rest of the galaxy.

Given the slowing of the bar, as far as I can gather from the paper, the mechanism that affects the orbits of the stars in resonance with the bar is similar in effect to the effect on the Moon's orbit of the Earth's rotation slowing. However, I admit the correspondence is heuristic and the details of the mechanism might not be the same. My only purpose was to at least give some idea of why we would expect the stars in resonance to move outward and increase their orbital period as the bar slows.

 

[PeterDonis]

The bars in the center

There is only one bar in the center of our galaxy.

slow down due to the pull of the DM

Due to dynamical friction with the DM. The DM here is best thought of as a viscous fluid through which the stars in the bar are moving; friction with the fluid slows them down. It is true that the underlying interaction that causes the viscosity is gravitational, but it's not a gravitational pull of a single "DM" object, so it doesn't work the way your intuitive grasp of gravity between ordinary planets or stars works. The fluid model is a much better approximation to what is going on.

I am guessing that the transfer of energy is not strictly between the DM and the bars but includes the star in question. Is this correct?

Not as I understand the model in the paper. The only interaction that involves the DM is the viscous fluid interaction I described above, which only involves the bar. The interaction between the bar and the stars in resonance, which are well outside the bar, is more like a normal gravitational interaction, at least if my understanding of the paper is correct.

I would think that the slowing of the bars would simply force the star to slow due to simple gravitational pull of the bar on the star

It depends on the specific resonance that the star is in. I only mentioned the 1:1 resonance (where the star's orbital period is the same as the bar's) as a simple example, not as a claim about the actual resonances that the stars in our galaxy are in. The paper does not describe the specific orbital characteristics of the resonances, but I suspect that the normal case is for the star's orbital period to be slower than the bar's by some ratio of small integers (like 2/3, for example), which would mean that the bar's gravitational pull on the star is prograde, like the pull of the Earth's tidal bulge on the Moon. In that case, the effect would be as @Bandersnatch described.

 

[sophiecentaur]

No, the tangential velocity is lower in the higher orbit.

Bit of a problem here. To go to a higher orbit you need to speed up. (Also, a rocket slows itself down to reach a lower orbit.) Of course, the orbital frequency reduces as its radius increases.
To move away, the Moon's total orbital energy must increase.

 

[Bandersnatch]

Bit of a problem here. To go to a higher orbit you need to speed up. (Also, a rocket slows itself down to reach a lower orbit.) Of course, the orbital frequency reduces as its radius increases.
To move away, the Moon's total orbital energy must increase.

Yes, you do need to speed up for higher orbits. That's pretty much what the couple sentences after what you quoted describe.
You will still end up decelerating as your orbit spirals outwards, trading your kinetic energy for potential energy. The total orbital energy increases all the same.

If you were to plot a graph of velocity required for a circular orbit vs orbital radius, it will be sloping downwards with higher R ( $V=\sqrt{GM/R}$ ). With continuously applied thrust (as is the case with tidal acceleration) in the prograde direction, you displace the curve upwards towards higher velocities. So that at any radius the velocity is too high for a circular orbit. But the curve still slopes downwards.

Or, think about the Hohmann transfer to a higher orbit. Here you make two one-off burns at two points, in the prograde direction, each burn adding to orbital energy. The final velocity at the new orbit is lower than the initial velocity, even though all you ever did was speed up.

 

[sophiecentaur]

You will still end up decelerating as your orbit spirals outwards, trading your kinetic energy for potential energy. The total orbital energy increases all the same.

Of course there is interaction with the other masses as well as the central attractor so everything is in a form of orbit around a significant amount of mass at all distances from the centre. The resonances are not intuitive!