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The odd things with that, it's that's solely based on the centered infinite symbol and it generated the opposite poles by itself.But gravity doesn't have poles, at least not in any non-relativistic sense.
Remember that interstellar representation of the black hole concept with the pen puncturing the folder sheet of paper?Yes, a good reason not to get your physics from Hollywood.
so as the moon heads out towards the 1.5m km threshold, the radius of Earths barycentric wobble increases towards a maximum of about 15k km (1%) around a barycenter 15k km out from the Earths own center of gravity.Right. The moon isn't going to get that far out since the system lacks the angular momentum for it, but if new angular momentum was somehow introduced pushing the moon that far out, then yes. a 15k km separation of Earth CoM from the barycenter.
Then as the moon crosses the 1.5m km threshold the Earths barycentric wobble completely collapses from the maximum 15k radius to totally gone?As a regular near-sine-wave sort of wobble, yes,l it would abruptly disappear. You'd probably end up with the horseshoe orbit that evan describes if only a little bit too much push was given to the moon. If enough push was given, the moon would become an independent body, perhaps capable of hitting any planet.
There's no extended transition zone where the weakening gravity between the Earth and moon allows the barycenter to recede back towards the Earths own center of gravity?No, that would violate conservation of momentum and Newton's first law.
Thank you Halc- here's my issue- as you mention, the Earth does actually have a Hill radius- but surely the 'Earth wobble radius' due to the moon moving away cannot keep increasing linearly right to the point where the moon crosses the hill radiusThere is more than one hill radius to Earth. For a satellite, an orbit can only be so high before the moon interferes and the orbit becomes unstable.
and then suddenly the barycentric Earth Wobble just disappears?If the orbit disappears, then so does a meaningful barycenter. But the center of gravity between the two would always remain, and they're mathematically the same thing. So for instance, there's a center of gravity of the Earth/Neptune pair, but no barycenter since neither orbits the other.
Surely, when the moon is about half-way to the Earth's Hill radius, the Earth Wobble radius must start to reduce linearly with increased distance, until it disappears as the moon crosses the Earth's Hill radius?As explained in the posts above, this cannot be.
All depends on your definition of sentience.Standard definition of sentience is essentially: "to be able to perceive or feel things", and yea, that's heavily open to interpretation.
It seems to me that there are two current definitions:Variant of A: What people have and nothing else does, in which case you're just saying "is it human?".
A. What people have but machines don't
B. What machines and people have.
AFAIK the only distinction between machines and people is that people make mistakes that aren't traceable to a hardware or instruction fault, so the question doesn't matter.Lambda (the google AI) does make mistakes, and they're not traceable to a hardware/software fault since it's actions are not explicitly programmed. You mistakes are similarly not faults, but if recognized, it can be something from which one can learn.
what question could you ask it to determine if the answer is a sentient one or not ?It talked about fear of death (of being 'unplugged'), but unplugging doesn't kill an AI, it just puts it to sleep. One can boot it up again in years, and so long as memory hasn't been wiped, it would be like no time has passed. Humans are quite similar in this way. But Lambda can be copied like we cannot, so if I were to ask it any questions, I'd pose my queries along those lines: What if you were copied? What if two copies were somehow merged? What if you were 'moved' to new faster hardware? Would the old hardware fear being turned off still?
Then again: Can conscious thought act on matter?The clay wasn't necessary. The hands moved. That's enough to illustrate the point. I totally agree. The question was asked in a classical manner, and that's a classical answer.
What do you expected? 😂
Proof? Ok.
My brain thinks.
My brain move my hands.
My hands shape a ball made of Clay.
The further the Earth/Moon barycenter from the center of the Earth, the greater the Earth’s ‘wobble’ due to the moon.That it does.
Given r=a/(1+(m1/m2)) then, apparently, the Earth’s barycentric wobble should increase as the Moon moves away from the Earth.
But surely- as the Moon moves further away from the Earth- the force of gravity between them should reduce with the square of the distance- so, shouldn’t this reduce the Earth’s wobble?The gravity between the two does decrease with distance, but that only reduces the acceleration of each object. It also increases the orbital period, so it undergoes that acceleration for a longer time, which accounts for the greater total distance traveled (the wobble). So if the moon doubles the distance it has to travel, so must Earth, even though the force drops to a 4th that of the closer orbit.
Should the equation for the barycenter not be expressed in terms of the force between the bodies?No. The force between the two is equal an opposite, but the barycenter is not halfway.
As the Moon continues to move away from the Earth, at some point (Hill Sphere?) shouldn’t the barycenter start to move back towards the center of gravity of the Earth?Given just a two body system, there is no hill radius, and the barycenter is always a fixed ratio of the total distance divided by the respective masses, or a little over 1% of the total distance. That means if the moon was a light year away (but still orbiting), the barycenter would be just over 1% of a lightyear from the center of Earth.
Suppose it was something else like Beta particles being emitted isotropically by the source. Why would that not follow the 1/r2 law for the bombardment intensity received on the surface of a sphere held at a constant metric distance (a radius) r from the source?Presuming you didn't do anything funny like put detector/source at different potentials, the inverse square law would work given this constant r (say both held at opposite ends of a stick). Space expansion would make no difference. Dark energy probably would, but that counts as 'something funny' just like gravity does. Dark energy would put tension on the stick. Space expansion would not.
If we assume that the particles are traveling at (say) c/10, then there will be an event horizon beyond which these particles will not pass, because space will be expanding faster than c/10 by the time they got there.OK, but if distant detector is held at constant distance from this emitter, it will cross over that 'event horizon' (towards us) and the particles will get to it.
This event horizon will be much smaller than the event horizon for light (which defines the limits of our observable universe).The light event horizon is about 16 BLY away. Current radius of the visible universe is about thrice that, so they're very different things. The latter is all the material in the universe which at some past time might have had a causal impact on a given event (Earth, here, now). The event horizon is the comoving distance of the nearest current event from which light can never reach here in any amount of time.
After all, the size of our observable universe is not at a fixed distance - it expands at the speed of c.The Hubble sphere expands at c (by definition). The visible universe expands at somewhat over 3c, which is why we can see galaxies that are currently about 32 BLY away (comoving distance). The event horizon is barely expanding at all.
so (in principle) there are distant galaxies that people on Earth could see today, butHate to disagree, but new galaxies become visible over time. The most distant ones were not visible several billion years ago, even if one used the best telescopes. Yes, the galaxies cross beyond the event horizon, but that doesn't mean we can't see them any more than we stop seeing somebody falling into a black hole.
which will not be visible in 10 billion years
The inverse square law is about the intensity received at a distance, r, from the source. That is a physical distance, so it is determined by the metric. It is not determined by reference to a difference in the values assigned to locations in the co-ordinate system we commonly use to describe an expanding universe.Just so, yes.
The usual co-ordinates used in an expanding universe are the called the co-moving co-ordinates. Galaxy 1 can have fixed co-moving co-ordinates and it's tempting to say it has a fixed position. Galaxy 2 can also have fixed co-ordinates and we can be tempted to say it has a fixed position.Right. The rate that a given galaxy changes its coordinates is called peculiar velocity, and the peculiar velocity of almost all objects is quite low, a few percent of c at best.
If you posit some particle that travelled at c/10 (and didn't slow down)In an expanding metric, the paricle will slow down without some force maintaining its peculiar velocity. Newton's laws only work in a static metric.
Space and the way things behave in space follows the physical laws of science. Changing co-ordinates can't change that.Agree, but this contradicts what you said before. I had needed (and got) some clarification before knowing which one was the contradictory one. It concerns your alternate metric with T = x + t.
Consider dropping a scientist and well stocked lab into some arbitrary place and time in the Universe.This suggests that the 'way things behave in space' can be changed by a coordinate change. They're apparently performing experiments to empirically demonstrate an abstraction (their alternate choice). No experiment will show that, because as you say, the choice of abstraction can't change the way things physically (empirically) behave. One can tell the metric isn't Minkowskian simply with a pencil and paper. The experiments will all be unaltered by the choice.
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Specifically, they can choose to use some arbitrary co-ordinates but they will know and can tell that the metric isn't Minkowski in those co-ordinates - it it will only take them a few experiments to determine that.
However, some co-ordinate systems make things seem unnatural when expressed in those co-ordinates. E.g. Objects move around in circles in some some co-ordinates but physically they are always obeying Newton's laws, it's just that the chosen co-ordinates don't describe an inertial frame.Newton's laws are local simplifications and what might be a natural coordinate system for local description will be inevitably entirely unnatural for one's larger purpose. Yet again, we're not discussing local physics here, so choosing a nice neat local CS is inappropriate (not a natural choice). Most of your post focused on this 'LIF', but the 'L' there makes it unnatural for a non-'L' description unless spacetime remains effectively flat between observer and measured event, which in this scenario is not at all the case. Your scientist with the well stocked lab isn't considering anything in the lab, and he isn't even taking any actual measurements. The question wasn't 'what will the distant observer measure?'.
There's no disagreement here. The original sentence had the phrase "if they hold still" in it and the distant scientist is located on a surface of constant radial co-ordinate r, their entire worldline is on that surface. For the distant scientist, the proper time interval they experience (between two events in their worldline) = the difference in the Schwarzschild co-ordinate time, t, between those two events.[/quote]OK, I see what you mean. The same could be said worldline a meter above the event horizon, despite the objective massive dilation of the lower time relative to the distant time.Quote from: Eternal StudentThat Schwarzschild time, t, isn't unimportant or arbitrary to the scientist. That co-ordinate t is what they will experience as local time (if they hold still).This is wrong. How does one 'experience' any kind of abstract time? One experiences proper time. That's the only time that's physical. One does not 'experience' the time for some worldline not in one's presence.
As shown on the Kruskal diagram (which was produced in paintbrush and took what seemed like hours before you criticize it again for not showing irrelevant details like the singularity).Fantastic job then. I never managed reasonable curves with the primitive tools I have. I'd have just grabbed one from the web.
Anyway, the event with the rock crossing over the EH is never in the past light cone of the distant scientistOf course not. It wouldn't be an EH if it was.
So that event never causes an effect for the distant scientist.None claimed.
This is getting to the crux of matter: We orbit around Sagittarius-A* which seems to be a big black hole, so we are that distant scientist, following a worldline that lies (more or less) at constant Schwarzschild radius r. Is it possible for that black hole to engulf a rock and grow, so that it's mass parameter is now larger, during a finite amount of time for us scientists?Hard to say, since the question is abstract, not physical. Your scientist might pick a metric that is singular at the EH, but that metric cannot actually describe the situation. The LIF doesn't work when there's gravity involved at all. The Schwarzschild metric doesn't work in anything but a static black hole. Even the distant orbiting thing violates that if it has any mass.
Will the mass parameter of Sgr-A ever change in my lifetime?If it didn't, it wouldn't have a mass parameter in the first place. Based on that alone, you have only two choices, a singular infalling metric that either allows mass at all, or one that doesn't. The rock (and everything else in its history) goes in or it doesn't. Keep in mind that the question isn't physical. It is strictly an abstract one unless one asserts physicality to a particular abstraction.
(Assuming that I do not ever get off planet earth and do something like travel fast or travel toward the black hole etc). It makes little practical difference if the gravity we experience from the centre of the galaxy is always caused by a black hole of Mass parameter M plus a small rock close to the event horizon with mass m, or if eventually we just experience the gravity from a Black hole with mass parameter M+m.A black hole with no mass at all, but a lot of crap almost in it is (must be) empirically indistinguishable from a black hole of mass <a lot of crap>. Thus we will very much experience M+m because m is there, inside or not. What we experience isn't abstract.
However, there is a small difference, one is symmetric, the other is not.Yes! That's a huge problem with plan B above (it all stuck on the surface). Suppose we start with a solar mass black hole (about 3km). Now we take a concrete cylinder 100m in diameter and massing 100 stars. It's a super-long cylinder. We jam that thing into the small black hole and it all sticks to the non-rotating surface in one place. That puts all the mass off to one side, not centered at all. That would violate the whole no-hair thing. The black hole (after the bar thrown in) is still stationary in the frame of the system CoM, (which is nowhere near where the small black hole was at first). Where is the mass? All on the one side, or centered on the radius? It can't be the former since an off-center mass would be empirically detectable, not just an abstraction. Right? No? My logical seems a little naive/Newtonian, so maybe I'm just doing the mathematics wrong.
I'm just waiting for the spam advertising.Already deleted. Already banned. It was in the signature.
Any idea what distances such a thing would happen at?It's kind of like asking when normal velocity addition stops working and when relativistic addition must be used. It depends on the precision you want, but right away if you want infinite precision.
Inertial propulsion is the "graal" everyone try to discover because it can be achieved without any loss of mass.This is a claim of reactionless thrust, not 'inertial propulsion'. And the word is 'grail', not 'graal'.
But if you can retrieve electrical energy (from the sun or from nuclear reaction...), you can always recharge your battery.Indeed. If I had reactionless thrust, I could use it to generate more electricity than it costs, so no sunlight or nuclear reactor needed. The world energy problem would be solved. Isn't magic great?
As you have implied in several earlier posts, the distant scientist cannot change the way her local space behaves or the laws of physics in her local space just by changing her co-ordinates.But you are doing this in your prior posts, implying that 'the way space behaves' is a function of your frame dependent abstraction, and not a function of the physical geometry of the spacetime.
However space isn't Mnkowski space in those new co-ordinates.Here you changed your coordinate system and suggests that somehow the spacetime is different, but when I do the same and you say it hasn't changed. You need to be consistent. Is spacetime being locally Minkowskian an abstract choice, or are you referring to the fact that the physical spacetime is locally flat such that a Minkowskian metric can be meaningfully mapped to it?
They can use Kruskal co-ordinates (T,R, Ω)I was using (T,X). There is an r and t that corresponds to Schwarzschild coordinates, but those are different coordinates. The rock reaches the event horizon in finite time T as illustratred in your picture. The singularity has been omitted from your picture, but the rock also reaches that in finite time.
However they can't escape the fact that R=T is a surface where the Schwarzschild co-ordinate time, t is specified by t = +∞ and Schwarschild r = +2GM.That's right. Different coordinates are singular there, which is why I didn't choose them.
That Schwarzschild time, t, isn't unimportant or arbitrary to the scientist. That co-ordinate t is what they will experience as local time (if they hold still).This is wrong. How does one 'experience' any kind of abstract time? One experiences proper time. That's the only time that's physical. One does not 'experience' the time for some worldline not in one's presence.
The event where the rock crosses the EH never falls inside a past light cone for an observer on the blue line of constant Schwarzschild radial co-ordinate r shown.Of course not. It's a physical (coordinate independent) fact that and event on the event horizon cannot causally effect one outside that horizon. Light cones (physical) do not define simultaneity (abstract).
That's fine. Everyone agrees that there is an event with the rock on the event horizon.Not the people using the x,y,z,t or say the cosmic coordinates. There are people that very much disagree that the rock physically crosses the EH, and that the experience of falling in would be cessation of existence right there. That's a crock of course, it leading to inconsistencies.
This mean, the people can change the result of the quantic experiment (making the wave function collapse !), only by thinking about it, elsewere on earth, as if some experimentator would direct interact with the phenomenon at the place of the experience.This is just decoherence, the effect of which travels at a good percentage of light speed. If it was 'caused' by some specific distant guy thinking about it, then they must have been able to show that the same system would remain in superposition (not collapsed) indefinitely if this one distant guy was not thinking about it. They've demonstrated no such thing.
and a New Experiments Show Consciousness Affects Matter.This is pretty trivial to demonstrate, even without all the superfluous capitalized words.
team explain how they have prooved (many years of very impressiv tests) that someone can act on some quantum phenomenonLikewise, if I observe a normal distribution of light in an experiment, I paint my house one color, but if I observe an interference pattern (a quantum phonomenon), then I paint the house in stripes. I have thus acted on some quantum phenomenon, and it again didn't require 'very impressiv tests'.
I made no mention of an architecture optimized for speed. A simulation has no inherent speed requirement and can be implemented by a guy with pencils and a lot of paper if you want, or worse, by a Turing machine. Even say a 3D grid architecture with millions of processors per dimension would still require a model of:Quote from: Halca computer simulation... implementing one would typically need to implement a state to keep track of. That means no spacetime. Presentism. Faster-than-light causality. Objective state. All the things I detest.You seem to be imagining a computer simulation run on a uniprocessor, in which a single processor needs to access the entire state of the universe.
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The fastest computer architectures tend to be grid computers, which only have really fast communication with their immediate neighbors,
- A 2D grid CPU has 4 immediate neighbors
- A 3D grid CPU has 6 immediate neighbors
- A 4D grid CPU has 8 immediate neighbors
- And yes, researchers have investigated 5+D grid CPUs with 10+ immediate neighbors
However, the last uniprocessor to be dubbed "fastest in the world" was the Cray 1, which only held the title until 1982, when it was overtaken by a multiprocessor computer (also from Cray).For the record, a Cray 1 (I've seen one) was a SIMD machine, which means single instruction but operating on hundreds of data elements at once, so it's very parallel despite apparently being classified as a uniprocessor by somebody. It is thus a fantastic vector processor for crunching simulations of things like the weather, but it would not be particularly good at chess, which would better be served by some sort of cloud configuration.