## 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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gmalivuk
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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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gmalivuk
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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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