## Science fleeting thoughts

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gmalivuk
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### Re: Science fleeting thoughts

Eebster the Great wrote:As you fall into a black hole, your acceleration does not approach infinity from any perspective, at least not until you reach the singularity. The gravitational redshift does. Hawking radiation is not a result of this infinite redshift being cancelled out.
Yes, it does. The coordinate acceleration adjusted for time dilation approaches to infinity. Otherwise, one would be able to escape the event horizon with a finite thrust. That's what the event horizon is, the point where not finite outward acceleration can cause you to escape.

The thrust necessary to hover (or turn around) at a given height increases to infinity as you get closer to the horizon. The proper acceleration in freefall does not, because the falling observer's time is further dilated by moving at great speed.

Why do you think the rigid cosmological horizon case experiences acceleration?
Recessional velocity is proportional to proper distance. Since the free falling object distance is increasing, it's recessional velocity must be increasing to maintain that constant relationship; a change in velocity is an acceleration. As the falling object is accelerating relative to the center, and the frame is not, therefore there is an acceleration between the two. This is also true if we start from looking at the change in proper or comoving distance between the shell and the falling object.
I think you should try to work through the actual math on this one.

If the frame has been accelerated up to velocity Hd toward the center, so that d is no longer changing, then no further acceleration is needed to maintain proper distance. The fact that a bit of dust (comoving with the center of the sphere) moves away with increasing recessional velocity due to the accelerating expansion of the universe means no more to the local apparent acceleration of the frame than it does to us, in the center. (After all, we're stationary relative to ourselves, and the recessional velocity of distant galaxies increases over time, and yet we don't feel any acceleration as a result.)

From the perspective of the sphere, after the initial boost, dust is just something with a particularly high peculiar velocity that then, as it recedes, gets augmented by metric expansion. Just like cosmic rays that fly past us here on Earth.
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### Re: Science fleeting thoughts

gmalivuk wrote:The thrust necessary to hover (or turn around) at a given height increases to infinity as you get closer to the horizon. The proper acceleration in freefall does not, because the falling observer's time is further dilated by moving at great speed..
That is absolutely true. An object in free-fall (by definition) experiences no acceleration, the acceleration I was discussing is the acceleration needed to entirely cancel out the effects of free fall: objects resisting free fall are experiencing proper acceleration.
...due to the accelerating expansion of the universe ...means
Just to check, you understand I haven't addressed the rate of the universe expanding changing at all?
I think you should try to work through the actual math on this one.
I absolutely have been.

1) By Hubble's law coordinate distance equals recessional velocity * Hubble's constant: v = H*s
2) take the derivative of velocity with respect to distance: dv = H ds
3) Coordinate acceleration, by definition : a = dv/dt
4) Replace dv : a = (H ds)/dt = H(ds/dt) = H*v = H2s (note that this is a coordinate acceleration, not a proper acceleration).
4) As this is a natural part of space-time, assume the proper acceleration of an object is the objects coordinate acceleration, minus the coordinate acceleration of a free-falling object at the same position, multiplied by local time dilation.

From the perspective of the sphere, after the initial boost, dust is just something with a particularly high peculiar velocity that then, as it recedes, gets augmented by metric expansion. Just like cosmic rays that fly past us here on Earth.
The salient difference is that the distance between the shell and the free failing object is zero. The difference in acceleration between the two doesn't start after they have gained some distance apart, but is exists when they are identical distance from the center.
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### Re: Science fleeting thoughts

gmalivuk wrote:The thrust necessary to hover (or turn around) at a given height increases to infinity as you get closer to the horizon. The proper acceleration in freefall does not, because the falling observer's time is further dilated by moving at great speed..
That is absolutely true. An object in free-fall (by definition) experiences no acceleration, the acceleration I was discussing is the acceleration needed to entirely cancel out the effects of free fall: objects resisting free fall are experiencing proper acceleration.

Sorry, I didn't mean "proper acceleration" in the sense of feeling a force (which of course it won't do), I meant in terms of an infalling observer measuring their height above the black hole in their own reference frame, the same way a skydiver could measure their height after jumping out of a plane.

The observer of course never feels any force, but also would only measure a finite acceleration (of 1/4M in units where c=G=1) when crossing the event horizon.

...due to the accelerating expansion of the universe ...means
Just to check, you understand I haven't addressed the rate of the universe expanding changing at all?

Of course you did. You said, "Since the free falling object distance is increasing, it's recessional velocity must be increasing to maintain that constant relationship; a change in velocity is an acceleration."

If H is constant, expansion is exponentially accelerating.

I think you should try to work through the actual math on this one.
I absolutely have been.

1) By Hubble's law coordinate distance equals recessional velocity * Hubble's constant: v = H*s
2) take the derivative of velocity with respect to distance: dv = H ds
3) Coordinate acceleration, by definition : a = dv/dt
4) Replace dv : a = (H ds)/dt = H(ds/dt) = H*v = H2s (note that this is a coordinate acceleration, not a proper acceleration).
5) As this is a natural part of space-time, assume the proper acceleration of an object is the objects coordinate acceleration, minus the coordinate acceleration of a free-falling object at the same position, multiplied by local time dilation.

Pick a distance s (say, the distance to your favorite quasar). Accelerate yourself to H*s in some direction (say, the direction toward said quasar). Then stop accelerating.

You're now maintaining constant proper distance to that quasar. This will continue for as long as H remains constant and you don't undergo any further proper acceleration.

You just did the same thing to that quasar as we'd have to do with the sphere frame in the power generation scenario. After boosting it out of the center's comoving rest frame (and into the center's coordinate rest frame), no further change in velocity is required. Ergo no further acceleration.

From the perspective of the sphere, after the initial boost, dust is just something with a particularly high peculiar velocity that then, as it recedes, gets augmented by metric expansion. Just like cosmic rays that fly past us here on Earth.
The salient difference is that the distance between the shell and the free failing object is zero. The difference in acceleration between the two doesn't start after they have gained some distance apart, but is exists when they are identical distance from the center.

How do you figure that? Metric expansion is the same everywhere. The shell and the piece of dust both measure the same H that we do, and both observe that things distance d away recede at H*d, plus or minus any relative peculiar velocity.

For the situation of the sphere and the dust particle to be different from the situation of us and a nearby cosmic ray, there'd have to be not just some universal rest frame but also a universal origin point. You seem to be arguing that physics works differently out near the cosmological event horizon, but that's a pretty damn extraordinary claim to be justifying with zero evidence.

(Though on the upside if physics does change from place to place it means conservation of momentum isn't a thing, so maybe you could get the EM-drive to work!)
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### Re: Science fleeting thoughts

gmalivuk wrote:The observer of course never feels any force, but also would only measure a finite acceleration (of 1/4M in units where c=G=1) when crossing the event horizon.
That's true, but that's also a different number. The question is how much proper acceleration would they feel if they completely resisted moving into the black hole, as the rigid frame does. The proper acceleration required to resist/ escape the black hole's gravity increases hyperbolically toward infinity as the in-faller approaches the horizon. Otherwise, either a finite acceleration could escape the horizon, or the required acceleration would have a discontinuity, jumping from a finite to an infinite amount.

If you must also look at this from the in-faller's perspective. The proper acceleration require to get back to the distant observer approaches infinity as they approach the the coordinate location of the event horizon. Crossing the coordinate location of the horizon doesn't have a dramatic consequence to the in-faller, but they will see the distant observer disappear.
You're now maintaining constant proper distance to that quasar. This will continue for as long as H remains constant and you don't undergo any further proper acceleration.
This is not possible. You can achieve an instantaneous velocity equal to the distant quasar, but absent something to balance it, you and your quasar will be accelerated away from each other.

In an expanding universe, it is not the default case that two objects will remain the same distance apart, or maintain the same relative velocity, or relative acceleration, or jerk, et cetra with all the derivatives. It is the very nature of expanding space that those things are (by default) changing. If you want to hold any of those constant, you need a force (counting local gravity as a force).
For the situation of the sphere and the dust particle to be different from the situation of us and a nearby cosmic ray, there'd have to be not just some universal rest frame but also a universal origin point
The reason the rigid frame is different is because it is attached to hugely distant parts of itself.

And a preferred frame is not the same thing as a having consequences of matching or not matching the rest frame of the observable universe. A a preferred frame is is something that can be measured in a closed box, and exists beyond the scope of the box. A preferred frame would have to have consequences beyond the observable universe.

We are (in fact) quite capable of saying the observable universe is (on average) following some specific rest frame, or that a distant observer will see their observable universe (on average) following some other specific rest frame.
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### Re: Science fleeting thoughts

gmalivuk wrote:The observer of course never feels any force, but also would only measure a finite acceleration (of 1/4M in units where c=G=1) when crossing the event horizon.
That's true, but that's also a different number. The question is how much proper acceleration would they feel if they completely resisted moving into the black hole, as the rigid frame does. The proper acceleration required to resist/ escape the black hole's gravity increases hyperbolically toward infinity as the in-faller approaches the horizon. Otherwise, either a finite acceleration could escape the horizon, or the required acceleration would have a discontinuity, jumping from a finite to an infinite amount.
Okay, then we're not disagreeing, you're just talking about something different from what Eebster was saying.

To review, Eebster said: "As you fall into a black hole, your acceleration does not approach infinity from any perspective"
This is absolutely 100% true and now it seems like you're acknowledging that it's true. But if that's the case, why did you initially disagree with that statement?

If you must also look at this from the in-faller's perspective. The proper acceleration require to get back to the distant observer approaches infinity as they approach the the coordinate location of the event horizon. Crossing the coordinate location of the horizon doesn't have a dramatic consequence to the in-faller, but they will see the distant observer disappear.

Why would the distant observer disappear? Light from the distant observer is still able to cross into the black hole, even though nothing can get back out.

You're now maintaining constant proper distance to that quasar. This will continue for as long as H remains constant and you don't undergo any further proper acceleration.
This is not possible. You can achieve an instantaneous velocity equal to the distant quasar, but absent something to balance it, you and your quasar will be accelerated away from each other.
What will accelerate us away from each other? Hubble's law says stuff accelerates away from other stuff if the distance changes (and if H doesn't decrease fast enough), but you can't suppose the distance changes in order to prove the distance changes.

(I'm open to being wrong about the conclusion, but before I changed my mind I'd need to see non-circular reasoning for it.)

For the situation of the sphere and the dust particle to be different from the situation of us and a nearby cosmic ray, there'd have to be not just some universal rest frame but also a universal origin point
The reason the rigid frame is different is because it is attached to hugely distant parts of itself.
That can't be why it's different, because the rest of your argument doesn't say anything about what is attached to what. The example where you accelerate toward the quasar doesn't require that you're attached to anything even farther away from the quasar, so why does the attachment make a difference now?

We are (in fact) quite capable of saying the observable universe is (on average) following some specific rest frame, or that a distant observer will see their observable universe (on average) following some other specific rest frame.
Yes, we are also capable of slapping coordinates on the universe that put us at the origin.

The problem is that you're saying physics behaves differently as you get far from that origin.
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### Re: Science fleeting thoughts

Suppose we have a universe with two observers, A and B, and a uniform distribution of everything else (matter, radiation, dark energy). Further suppose that all that matter is comoving (i.e. at every position, all the nearby dust appears to be mutually stationary).

At time t0, A and B are comoving at distance r0 from each other, and H = H0. In other words, dr/dt = v0 = H0r0. A and B are both stationary relative to the nearby matter.

1) Now, let A accelerate toward B up to speed H0r0. This is A's velocity relative to the surrounding dust. Surely you agree that, at least for the moment, dr/dt = 0, right? (If you don't agree, please explain why.)

So, A is moving at v0 relative to its surrounding dust, B is moving at 0 relative to its surrounding dust, and dr/dt = 0.

2) Without any force acting on A, it will remain at v0 relative to its surrounding dust, because inertia exists regardless of cosmological expansion, right? (If you don't agree, please explain why.)

3) If H remains constant, the dust at distance r0 from B will always increase its distance from B at recessional velocity H0r0, right? (If you don't agree, please explain why.)

4) If dust at a particular location moves away from B at v0, and A moves through the dust at relative velocity v0 toward B, then A does not change its distance to B, right? (If you don't agree, please explain why.)

5) If you agree with all of the above, but you still think A will start accelerating away from B, please explain why.
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### Re: Science fleeting thoughts

gmalivuk wrote: But if that's the case, why did you initially disagree with that statement?
Because I interpreted his statement as a response to my statement that immediately preceded it. The situation involving both a black hole and acceleration that I listed was the rigid case.

Why would the distant observer disappear? Light from the distant observer is still able to cross into the black hole, even though nothing can get back out.
You are correct, I made a mistake there. The distant observer would continue to be visible. The distant observer would appear blue shifted and aged, but wouldn't disappear.

What will accelerate us away from each other? Hubble's law says stuff accelerates away from other stuff if the distance changes (and if H doesn't decrease fast enough), but you can't suppose the distance changes in order to prove the distance changes.

(I'm open to being wrong about the conclusion, but before I changed my mind I'd need to see non-circular reasoning for it.)
"distance changes" is the entirety of what Hubble's law says; that and how it changes. Distance changes proportionally to how much distance there already is, which is the very definition of an exponential change.

The derivative of a exponential change (with a non zero rate) is non-zero, is is the derivative of that, ad infinitum; this is a very proven characteristic of exponential functions. I specially, mathematically proved in an earlier post that acceleration equals distance * the Hubble parameter squared.

That can't be why it's different, because the rest of your argument doesn't say anything about what is attached to what. The example where you accelerate toward the quasar doesn't require that you're attached to anything even farther away from the quasar, so why does the attachment make a difference now?
Rigidity and solid objects imply something is attached to something, so I've been discussing attached objects the entire time I've been discussing rigid things.

This shell can maintain a constant proper distance because there is a force involved. The dust particle can't. The ship can change it's velocity between that of the dust particle and ship, but it can't maintain a constant proper distance once it cuts it's engines off.

2) Without any force acting on A, it will remain at v0 relative to its surrounding dust, because inertia exists regardless of cosmological expansion, right? (If you don't agree, please explain why.)
The surrounding dust is changing. While dust and A approach each other, they have a non-zero distance and accelerate away from each other. A would asymptotically approach the same velocity as the dust (but very slowly, since H2 is so small).
4) If dust at a particular location moves away from B at v0, and A moves through the dust at relative velocity v0 toward B, then A does not change its distance to B, right? (If you don't agree, please explain why.)
The dust's instantaneous velocity is v0, but both the dust and A are being accelerated away from B at an identical rate.
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### Re: Science fleeting thoughts

Quizatzhaderac wrote:The derivative of a exponential change (with a non zero rate) is non-zero, is is the derivative of that, ad infinitum; this is a very proven characteristic of exponential functions. I specially, mathematically proved in an earlier post that acceleration equals distance * the Hubble parameter squared.

That's the acceleration for comoving objects, but by hypothesis A and B are no longer comoving, because A has accelerated out of the comoving frame.

Comoving objects start out with an instantaneous recessional velocity of H*s, but I'm talking about something that starts out with an instantaneous recessional velocity of 0. Your reasoning in that earlier post doesn't apply to non-comoving objects. (You can't start with v = Hs and reach a conclusion about when v != Hs.)

2) Without any force acting on A, it will remain at v0 relative to its surrounding dust, because inertia exists regardless of cosmological expansion, right? (If you don't agree, please explain why.)
The surrounding dust is changing. While dust and A approach each other, they have a non-zero distance and accelerate away from each other. A would asymptotically approach the same velocity as the dust (but very slowly, since H2 is so small).

By what mechanism? How, without any force acting on you, would you not remain in motion relative to your surroundings? The surrounding dust will become less dense over time as the universe expands, but what will change your velocity relative to the dust that's passing you?

Like I said before, your hypothesis amounts to denying conservation of momentum.

4) If dust at a particular location moves away from B at v0, and A moves through the dust at relative velocity v0 toward B, then A does not change its distance to B, right? (If you don't agree, please explain why.)
The dust's instantaneous velocity is v0, but both the dust and A are being accelerated away from B at an identical rate.
And if they're accelerating at the same rate, how do their velocities change relative to each other?
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### Re: Science fleeting thoughts

The more I think about it, the more confident I am that the "dustverse" thought experiment is the way to conceptualize local versus cosmological motion. Everything behaves as we expect relative to the local dust (from Newton or, if speed or gravity gets too high, from basic Einstein), while the dust itself is (by assumption) all comoving and can therefore be modeled with cosmological equations.

An object at velocity v0 relative to the dust will remain at v0 relative to the dust unless acted upon by an external force. If H decreases over time, A would start moving toward B, because v0 would be greater than the decreasing value of H*r0. If H increases, A would move away as H grows and v0 ends up being smaller than H*r0.
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### Re: Science fleeting thoughts

(Assume we're working in one dimension.)

vc = coordinate velocity
vh = Hubble flow velocity
vp = peculiar velocity (defined as velocity that differs from Hubble flow, and equal to velocity relative to local dust under the assumption that all dust is mutually comoving)

vc = vh + vp = ds/dt
vh = H*s
ds/dt = H*s + vp

If we assume H is constant and vp = 0, then d2s/dt2 = H2s, as calculated in the earlier post.

However, if we set vp to a constant v0 other than 0 (and H remains constant), then the differential equation becomes
s'(t) = H*s(t) + v0
Take the derivative to get
s''(t) = H*s'(t) = H(H*s(t) + v0) = H2s(t) + H*v0

If v0 = -H*s(0), then s''(t) = H2(s(t) - s(0)), which is 0 at t = 0, meaning the thing doesn't accelerate.

---

The exact solution to s'(t) = H*s(t) + v0 is
s(t) = (eHt(H*s(0) + v0) - v0)/H

If v0 = -H*s(0), that simplifies to
(eHt(H*s(0) - H*s(0)) + H*s(0))/H = H*s(0)/H = s(0)

---

If you want to argue that the distance changes, you have to argue that peculiar velocity changes without any additional force.

Then you'll have to explain how whatever argument you just made doesn't violate conservation of momentum.
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### Re: Science fleeting thoughts

A lot of this is going over my head, but if conservation of energy is already being violated by the metric expansion of space, is it really such a problem if conservation of momentum is being violated as well? Can you even violate one without violating the other?
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### Re: Science fleeting thoughts

Even if that were true globally, the problem is that Quizatzhaderac's account would seem to violate conservation of momentum even locally.

Also, yes, conservation of momentum and conservation of energy could conceivably be violated independently of each other.
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### Re: Science fleeting thoughts

A tangentially related thought I've been trying to sort out for a while:

If you have two distant objects moving toward each other at high velocity, emitting identical (same energy) photons at each other, each of them will receive the other's photons at a higher energy than the photons they're emitting themselves, because of blueshift, right? So they would each be gaining energy from each other. But the difference in radiation pressure caused by that would mean that they would be pushing each other away and so slowing down their relative velocity to each other, so in effect that energy is coming from converting each other's relative velocity into another form of energy, and when that velocity reaches zero then there's no more energy to be gained.

If they were falling toward each other due to mutual gravitational attraction, and exchanging photons like this, they could reach an equilibrium state, where they are hovering a static distance away from each other because radiation pressure is exactly countering the attraction of gravity. Any inefficiency in the return of like photon for like, however, such as if the energy received from the other object were used to do anything other than send more photons right back, would come at the expense of the objects' gravitational potential energy relative to each other: they would still fall toward each other.

If space were expanding, however, surely that equilibrium point would be at a different distance than if it were not? Or for another illustration: two identical objects in a perfect circular orbit in a non-expanding spacetime would start to drift apart from each other if that spacetime were to then start expanding, no? So these two objects hovering at a fixed distance due to exchanging photons, in a non-expanding space, would start to drift apart if space were to start expanding, no? Or if they were rather still falling toward each other because they were not perfectly re-transmitting photons back to each other, but instead using them for other things or just losing energy to inefficiency, would not they be pushed back to an equilibrium state (at a static distance from each other) if space were to start expanding at an appropriate rate?
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### Re: Science fleeting thoughts

If space suddenly went from static to expanding, I suspect there are all sorts of dynamic systems that would change behavior.
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### Re: Science fleeting thoughts

I'm only talking about "changing from static to expanding" as a way of comparing the behavior of a system in a non-expanding space to the behavior of a system in an expanding space.

In this case, it seems like you could set up a system (as described above) where two objects would be exchanging gravitational potential energy for usable work in a non-expanding space, but in an expanding space would remain the same distance from each other (because new space is created between them) and so be able to keep doing that indefinitely.
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### Re: Science fleeting thoughts

Well if space is expanding then those photons are redshifted even if the distance isn't changing.
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### Re: Science fleeting thoughts

Also, shouldn’t the photons be emitted isotropically? Only a small fraction of the photons from each object will reach the other unless all emissions, including thermal blackbody emissions, are highly directional.

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### Re: Science fleeting thoughts

Instead of black bodies, they can be tiny bots communicating with each other. Each occasionally sends a single photon to the other, and since they are falling together, those photons are blue-shifted. Emitting these photons comes at the cost of some battery, but whenever they receive a photon, that charges the battery, and by Doppler shifting, they seem to be getting a free lunch, emitting as many photons as they receive yet charging the battery.

As you say, they will eventually fall to their common center of mass and nothing else will happen. But what if they are held a constant coordinate distance apart by Hubble expansion? This seems like the ultimate loophole. Unfortunately, that expansion itself causes a redshift, a cosmological redshift, and I believe that should exactly cancel the blueshift you get due to the proper motion of the ships. So it should all exactly balance, and htere is no free lunch.

EDIT: gmalivuk said exactly this an hour and a half ago, but I'm leaving it here BECAUSE.