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By the time you posted, I already had a first WIP done -- but was too tired and went to bed. It is here. Some of the first few levels aren't optimal -- notably, the first level and the eighth (the one with the swing sets) -- but I didn't want to go back and edit them once I was done.
Here is an encode:
Link to video
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@Patashu: you are confusing what we know with what there is; we don't know (yet) how to describe gravity and quantum mechanics correctly at the same time; we have only knowledge of gravity on the large scale, and quantum knowledge on the small scale. That we can't integrate them both does not mean it is impossible, or that it doesn't exist, only that our knowledge of the universe is still incomplete.
@DarkKobold: GR is totally and completely deterministic; spacetime geometry is determined by the contents (energy/matter/momentum/stress/pressure) of the universe, which in turn determines how matter moves. Spacetime is an integral part of dynamics, and you can't separate the motion of matter from it. Finally, GR is ultra-deterministic: all of the history of spacetime exists and is completely defined from the outset. In fact, GR is even more deterministic than either Newtonian physics or special relativity -- since it allows (in theory, at least) time travel ("closed timelike curves") under some special circumstances, it results in a history consistent with time travel in any spacetime that allows it.
In QM, the position and momentum of matter/energy are "undefined" until you measure (the most "orthodox" interpretations are probabilistic theories); spacetime is the "backdrop" against which QM happens. QM is inherently probabilistic, and you can only predict the statistics of any system -- you are powerless to predict the outcome of a single run of an experiment.
Since you can't define the locations and motion of matter in QM, you can't compute the spacetime geometry, and in turn you can't define how matter moves.
There is more to it, but this is one of the most important points.
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You managed to mangle the quotes and tags quite horribly on your post. You also managed to mangle the notion of entanglement quite horribly. The important points about entanglement are:
(1) The spin is a vector quantity, which can be measured along different axis and which is either -1 or +1 (photon) or -1/2 or +1/2 (electrons, protons, neutrons) in any axis you measure;
(2) you can measure the spins of the entangled particles in different, uncorrelated axis (possibly "randomly" chosen);
(3) when you do that, there will be a correlation of the values you measure;
(4) this correlation (as predicted by QM) is higher than what is possible in any local (nothing moves faster than c), realistic (spins exist and are defined all the way, before measurement), hidden variables (uses information not in QM, such as the spin of the particle along the way) theory (Bell's theorem, Einstein-Podolsky-Rosen paradox).
Experiments aren't good enough to show (4) conclusively because of many things; further, there are other limitations to these no-go theorems in the form of seemingly reasonable assumptions which ultimately can't be justified -- such as the "random" selection of the axis you want to measure the spin.
Moreover, you can't even say that for any given pair of entangled particles, you will measure spins in opposite directions; only when you do statistics with many of them you actually get the correlation. So, to refine your layman explanation: after putting two balls, one blue and one red, in boxes, you would look at the balls and see them both blue, both red or one blue and one red. Only after doing statistics you would see that the preponderance of cases are one blue and one red.
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Warp wrote:
I think that the relevant question is not so much whether quantum effects are random, but whether they are predictable.
QM has no say in this either, being a theory that predicts outcomes as probabilities. We simply don't know at present if quantum effects truly are unpredictable or if they would be predictable with information we can't access at present. The strictly positivist reply would be "yes" to unpredictability; and so would be the reply of anyone basing their opinion on Bell's theorem or other no-go theorems. But either case suffers from assumptions that can't be justified with current evidence (although this evidence is very tempting).
Warp wrote:
Am I correct in assuming that if the so-called quantum entanglement phenomenon does indeed exist, and that measuring for example the spin of one of the particles determines the spin of its entangled pair (regardless of their physical separation), and that if one could somehow predict the spin of the particle in advance, this could be used to transmit information (by choosing between a group of entangled particles) with no delay whatsoever regardless of distance?
To be annoyingly strict, I have to say "maybe" to the question of entanglement; most experiments about entanglement assume QM to be true, hence explain everything under those assumptions. Those that don't are usually those testing no-go theorems, such as the aforementioned Bell's theorem -- and they aren't good enough (at present) to rule out theories that explain "entanglement" by other means.
p4wn3r wrote:
May I ask for further clarifications on this? I mean psi(t) = exp(Ĥ i t / h) psi(0) clearly depends on time and the initial state
You are assuming that t means time -- in most interpretations, it does, and it is even associated with what we call time at the macroscopic (classic) scale. In one particular interpretation (I can't for the life of me remember the name; and yes, it is one of the more esoteric ones), all possibilities (and all "instants") exist "simultaneously", and t is a measure of how "far", and thus how likely, each possibility is from another possibility. Classical time is a product of the mind (I said it was an esoteric one...), which stitches together the "instants" in order to make a consistent narrative. Most physicists don't subscribe to this interpretation, but a few (very few) do.
p4wn3r wrote:
and thus, there would be at least one notion of time assimmetry by looking at quantum states, no matter how someone may interpret this. Further, wouldn't an interpretation that has no causality assimmetry mean that every quantum event state(A)->state(B) could also happen as state(B)->state(A) and wouldn't this only hold together with the second law of thermodynamics if all of them didn't change entropy and so, are all reversible?
Unless there is a clear temporal asymmetry of the system (magnetic fields go here, unless you swap charge and parity too*), every quantum event state(A)->state(B) does happen as state(B)->state(A). Thermodynamics are for thermodynamic systems -- quantum systems generally don't apply -- hence talk of entropy is meaningless.
* See CPT symmetry.
In a more modern description, the emergence of the classical world comes about as a result of decoherence -- a shifting of wavefunction phases that causes the interference (quantum) effects to disappear; this, along with statistical mechanics, would lead to the second law of thermodynamics.
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There have been many people making models of QM based on classical stochastic mechanics (with a classical stochastic force). There is also Bohm's interpretation of QM which treats QM as normal particle subjected to a (random) quantum potential. There is a theorem that says no realistic, local hidden-variable theory can reproduce all the results of QM. There is even work on simulating the results of QM in a perfectly classical computer using stochastic classical mechanics in just such a way despite that theorem, because the theorem makes assumptions that are incorrect.
What is more, there are many different interpretations of QM, some of them not even having a defined notion of history or past -- hence, no true way to talk about "randomness" (as "randomness" means that an event is causally disconnected from its past -- no past, no randomness).
The result is that any claims about QM being stochastic or random are nonsense -- nevertheless, many respectable physicists do just that. Any such talk is based on unsupported assumptions about the nature of the universe at the quantum level. The bottom line is that QM is a logical positivist theory -- it describes what can be measured -- and a statistical theory -- the results it gives are statistical. You can't reverse it to infer what "goes on" on the quantum level because the information isn't relevant to, or contained in, QM.
Earlier I mentioned that interpretations of QM are pseudo-science -- this is the reason why: as long as they give the same results as QM (and as long as there is no experiment that can tell it apart from QM), any interpretation will make unsupported and untestable assumptions about how the world works "down there".
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Regarding the CNZ video:
(1) In my video, there weren't 2 frames waiting for the spike to go down, and I told you so. I charge a full spindash and release it as soon as it is done charging. Releasing it sooner (or charging it closer to the spikes) results in Sonic bumping against the spikes. You can watch it in frame advance.
(2) The bottom route is faster, yes -- I had already tried it, and managed to knock off 40 frames faster than my previous video. I still didn't need to go all the way down, as I managed to get the spikes to get out of the way.
(3) Neither of the videos have been released yet, and I will wait until I have finished the run up until at least CNZ before posting a WIP.
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Edit: This was supposed to have been just below Dada's post, not in another page; so here is the relevant quote:
Dada wrote:
Quantum physics is a great example. One relatively popular theory right now is called the many-worlds interpretation,
Just a quibble, the "many worlds interpretation" is not a theory -- it is an interpretation, while Quantum theory is a theory. The difference is this: a theory is supported by evidence, while there is no way to differentiate between the many interpretations of Quantum theory based on the evidence -- they are all equal in terms of predictive power. This effectively makes all interpretations of Quantum theory pseudo-science at the present time.
Back to the topic of free will: there is no such thing. Your brain processes information and decides how to act. It then sends a signal that creates the conscious experience of making the decision and delays any signals to the limbs/mouth/eyes/whatever in order to reinforce the illusion that the conscience decided what to do. Since you aren't aware of all of this processing, the notion of free will seem plausible ideas -- you just "decided" what to do, after all. [1]
One example that uses this is a button pushing experiment that was first performed the 1960s: people were told that they would be watching a series of slides, and that pushing a button would to advance to the next slide. There were several sensors attached that measured neural activity in areas related to hand motion; the participants weren't told that it was the output of these sensors, and not their button presses, which was what actually advancing the slides. The result: the participants were amazed to discover that the slide-show was able to predict their decisions. [2]
[1] D. Wegner (2002). The Illusion of Conscious Will. The MIT Press, Cambridge, MA.
[2] For a more thorough description of this experiment, see "Time and the Observer" by D. C. Dennett and M. Kinsbourne, in The Nature of Consciousness: Philosophical debates, (Ned Block, Owen Flanigan, et al., eds., 1997), The MIT Press, Cambridge, MA, page 168.
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Some happened in the underwater palette ($FFF080-$FFF0FF); some in the plane data buffer ($FFF100-$FFF57F); others in the VRAM buffer ($FFF580-$FFF5FF). I got errors at various addresses in these ranges depending on how I moved and where I turned off debug mode; I don't know if all addresses on these ranges are equally affected, and I can't claim that the address errors are restricted to these regions of RAM.
But in any event, your description makes it clear that it is unique to debug mode; hence, I don't expect it to appear on many TASes or speedruns. Unless another way of filling up the object tables like that, which I doubt.
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After doing some testing with Regen (the debug version), I found out that the crash in real hardware is due to a word- or long-sized access at an odd address -- an address error. Moreover, there are multiple such crashes, depending on where exactly you turn off the debug placement mode, all at different addresses -- some involving reads, others involving writes. It doesn't appear to be doable without debug mode, and it seems that the boss only appears due to an emulation error -- the fact that Gens does not generate an address error exception.
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The crash could be due to an address error, which Gens doesn't emulate; if you try on Regen and the same crash happens, it is definitely it. I tried replicating it without debug by using Hyper Sonic (as he can get there without debug mode), but no dice -- nothing special happened. It may be tied to debug mode.
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Look at the video: "there is a better way to do computer graphics, which is used in medicine and the sciences". Those are voxels, alright. And yeah, TASeditor, you only watched bad examples :-)
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So... voxels with another name and a smaller size? While impressive, it is neither new nor innovative -- I have seen voxel renderes written in VB a decade and a half ago, many games use it, it is used in medical gear... and not to mention that they aren't forthcoming with the limitations. Notch of Minecraft has called it a "scam": see here.
My ears! They bleed! What exactly did you do that made the video sound that badly?
Anyways, here is the Youtube video from DMTM's channel: http://www.youtube.com/watch?v=0L0KQKtxk6c.
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Well, whenever you reach IceCap, you can base it on this WIP. I used DMTM's and Qwerty's trick, but going right instead of left. It is probably possible to improve, I am not that good at TASing Knuckles.
Here is an encode of the WIP, skipping the stage selection code:
Link to video
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DarkKobold wrote:
Aqfaq wrote:
Well, the situation here could be interpreted so that the idling teamplayer controller is actually delivering input....
The other issue is that this may very well just be an error in the emulation of the team-player attachment. Without testing this on a console, requiring a genesis, team player adapter, and a copy of bible adventures, I'd be hard pressed to believe in the legitimacy of it.
It is not an emulation bug. The teamplayer attachments have specific communication protocols to work around the fact that there is a limited number of input pins and ports for the controllers in the Genesis (Sega's teamplayer and EA's teamplayer are different, but the Sega teamplayer MKII can work in compatibility mode with EA's protocol). Without going into details about the way the teamplayers work (this is not the thread for it anyway), it suffices to say that there is a lot of chatter going on which the game misinterprets as input, since it is not designed to handle the teamplayer attachment.
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p4wn3r wrote:
I understand that many users won't have the time nor the interest to do all the necessary algebra though. I'll consider the problem solved if someone gives a precise description of the method and laws/theorems used to achieve the solution.
In either case, I suppose I am disqualified because I not only already knew the problem, I already saw different solutions in the form of articles that were published about it.
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p4wn3r wrote:
Also, I believe this is a typo, because it should have a square root, but didn't influence the posterior steps:
marzojr wrote:
Ut(τ) = dt(τ)/dτ = 1 / {1 - v(τ)2 / c2}
You are correct, there should have been a square root there.
p4wn3r wrote:
I like classical mechanics more. I'll post a problem I could solve with some considerable amount of work. I think people here will find it interesting.
[snip]
Hrm. That problem is more annoying than difficult.
I will give a hint to those going on the wrong direction: the friction is completely irrelevant except that it prevents sliding. Note that at each stage of the motion, there is one edge of the pencil touching the inclined plane; this edge will act as the axis of rotation for the pencil for about 60 degrees, when one face of the pencil will hit the incline and the axis of rotation will switch to the next edge of the pencil. Hence, at each such stage, the friction acts at the axis of rotation, hence it does not generate torque and does not alter rotational kinetic energy.
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Bobo the King wrote:
Nevertheless, I have to ask: what is wrong with interpreting a as dv/dt?
The word "interpreting": you don't interpret something to be what it is defined to be. A definition in science has to be unambiguous and as free of interpretation as possible or it is useless. This is why the meaning of "v" and "t" have to be specified also when defining acceleration (whether in classical or relativistic physics) -- for example, 't' being 'time' is unambiguous in classical mechanics (universal time), but ambiguous in relativity (because time is relative); the most straightforward measure to remove ambiguity is to use the proper time of the thing whose velocity is being measured (because this isn't contingent on the observer).
Here is a perfect way of explaining the difference between definition and interpretation: quantum physics. You have several definitions (what a wave function is, mathematically speaking, and how you extract observable values from it) and several interpretations (Copenhagen, Bohm, consistent histories, many worlds). In practical terms, the interpretations are irrelevant -- they all give the same experimental results and are indistinguishable (so far), while the definitions are all-important.
Bobo the King wrote:
Okay, if a is commonly understood to be the "constant proper acceleration", I suppose I can't argue with that, though physicists are generally more precise than most with their nomenclature. If we cannot discard the possibility of a=dv/dt (which seems quite sensible to me), then we must at least hint to our interpretation of acceleration by choice of units and words.
Bobo the King wrote:
I still maintain that velocity cannot increase monotonically indefinitely (and if it does not increase monotonically indefinitely, it technically does not increase monotonically at all).
For t >= 0:
v(t) = c(1 - e-t)
dv(t)/dt = ce-t > 0 for all t
This asymptotically approaches c and is monotonically increasing throughout the domain of t (from 0 to infinity).
If you mean physically, the answer is 'duh': it would need an infinite power supply to do it, which not only does not exist but would collapse the entire universe if it did (infinite Schwarzschild radius and all).
Warp wrote:
As you might remember, mass increases as velocity increases, which AFAIK is related to the explanation of why the astronaut can feel a constant acceleration forever.
That explanation has long fallen in disfavor and is no longer used. The more correct explanation takes into account the contraction of space and dilation of time (or rather, the changing hyperplanes of simultaneity) -- the proper acceleration remains constant from the rocket's point of view, but because of the contraction of space and dilation of time, it is always getting smaller from the point of view of an external observer.
Warp wrote:
I think that the point was that m/s2 is a unit of acceleration (it's what eg. an accelerometer could report) and hence it's completely valid to use it.
And it is, indeed, used by professional physicists working with relativity. Although for expediency when doing math by hand, it is usually measured in m-1 using c to convert between time and distance units (measuring time in m, with speed being adimensional and c then being set to 1), with dimensional analysis to recover the missing lost factors of c.
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p4wn3r wrote:
marzojr wrote:
We must calculate the final value of τ we are interested in; this corresponds to the half-way point x(τm) = (4.365/2) ly. To do this, lets invert the equation for x(τ) -- we can do that in this interval because x(τ) increases monotonically with τ in this interval -- to obtain
τ = (c / a) * acosh{(a / c2) * x(τ) + 1}
Plugging in x(τm) = (4.365/2) ly = 2.1825 ly, a = 1.0315512 ly/y2 and c = 1 ly/y, we find
τm = 1.791172306 y
For the Earth reference frame, this corresponds to
t(τm) = 2.999132811 y
Multiplying these by 4 for the four segments of the journey, we see that 7.164689224 y transpire at the rocket, while 11.996531244 transpire on Earth. The difference to p4wn3r's result is probably a result of numerical errors in his part, as his math seems correct.
Thanks for looking over my "solution". I didn't do the results very precisely, but still the time for the rocket is differing by a large value. Now I have a question, why can you you plug half the distance from the star directly into that formula? Shouldn't the distance for the ship's frame shrink because of the Lorenz factor?
The distance for the ship's frame indeed gets reduced; but the formulas I obtained at the end, for t(τ) and for x(τ), give (respectively) the time and position of the rocket on Earth's reference frame as a function of the rocket's proper time τ. Thus, when the rocket is halfway to Alpha Centauri (according to Earth), x(τ) will be half the distance to Alpha Centauri (as seen by Earth).
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Warp wrote:
Let's assume that only special relativity is in effect (general relativity probably would make a slight difference because the solar system is a gravity well, and Alpha Centauri is another, but just for the sake of simplification let's consider it negligible).
An astronaut starts traveling from Earth to Alpha Centauri, a distance of 4.365 light years. The spaceship maintains a comfortable constant acceleration of 9.8 m/s2 until half of this distance has been traveled, after which it makes a 180-degree turn and starts decelerating at the same 9.8 m/s2 so that when it reaches Alpha Centauri, it will stop. The trip back to Earth is started immediately in the same way.
1) How long did the overall trip take from the point of view of the astronaut? (In other words, how much older is he after he's back to Earth?)
2) How long did the overall trip take from the point of view of the Earth? (In other words, how much older is someone on Earth?)
I'm actually curious to see how this calculation is done.
For an extra difficult additional (but optional) assignment: How much of a difference would it make if we took general relativity into account?
I will give the more modern version of the solution using techniques from differential geometry -- for this case, only knowledge of 4-vectors are enough for it.
I will use light-years (ly) for distance and years (y) for time. The only relevant piece of information needed to convert is the acceleration; the conversion factor is 1 m/s2 = 0.10526033 * 1 ly/y2, making the acceleration a = 1.0315512 ly/y2.
For simplicity, we can assume the movement to be done in one direction. We will work on the reference frame of the Earth/Sun, assuming it is inertial; the coordinates we need to track the rocket in this frame are the time t and the position at this time, x(t). We also need to take care of the rocket's proper time, τ. Now, both t and τ are monotonically increasing -- they mark the passage after all. Moreover, given any time t, we will have a unique corresponding value for τ (the converse also being true) -- hence they are related. We are yet to find this relation, but we can write it in implied form: we can write t and x as functions of the proper time τ: t(τ), x(τ).
This is a mere reparametrization of the rocket's path, by the way: instead of using the current reference frame's time to determine its position and using another equation to determine the relation between frame time and proper time, we mix them both and use proper time to determine frame time and position. This has the advantage that the proper time is an "universal" parameter -- every reference frame (even non-inertial ones) will agree how much proper time elapses between two points in the rocket's trajectory. The proper time has the great advantage, thus, of being a Lorentz-invariant parameter: you can perform Poincaré transformations before or after you apply the operator(s) dn/dτn.
Thus we can define the 4-velocity and the 4-acceleration vectors by
U(τ) = {Ut(τ), Ux(τ)} = {dt(τ)/dτ, dx(τ)/dτ}
A(τ) = {At(τ), Ax(τ)} = {dUt(τ)/dτ, dUx(τ)/dτ} = {d2t(τ)/dτ2, d2x(τ)/dτ2}
Using the chain-rule for the derivative of composite functions, it is easy to see that
(c * Ut(τ))2 - Ux(τ)2 = (c * Ut(τ))2 * {1 - Ux(τ)2 / (c * Ut(τ))2} = (c * Ut(τ))2 * {1 - v(τ)2 / c2}
Moreover, using the time-dilation formula, it is easy to see that
Ut(τ) = dt(τ)/dτ = 1 / sqrt{1 - v(τ)2 / c2}
hence, the Minkowski norm of the 4-velocity is
(c * Ut(τ))2 - Ux(τ)2 = c2
Doing similar steps for A(τ), we reach the Minkowski norm of the 4-acceleration:
(c * At(τ))2 - Ax(τ)2 = -a2
where a is the (constant, in this case) proper acceleration, that is, the acceleration measured by an accelerometer inside the accelerating rocket. These relations also show that the 4-acceleration A and the 4-velocity U are orthogonal in the Minkowski norm:
c2 * Ut(τ) * At(τ) - Ux(τ) * Ax(τ) = 0
(this follows from the constancy of the Minkowski norm of the 4-velocity by doing a proper-time derivative).
Taking the Minkowski norms above and writing them in terms of derivatives of t(τ) and x(τ), we have
c2 * (dt(τ)/dτ)2 - (dx(τ)/dτ)2 = c2c2 *(d2t(τ)/dτ2)2 - (d2x(τ)/dτ2)2 = -a2
This set of equations is all we need to answer the special relativistic version of the problem. To that, I claim that the following is the solution:
t(τ) = c * sinh{a * (τ - τ0) / c} / a + t0x(τ) = c2 * cosh{a * (τ - τ0) / c} / a + x0
Direct substitution is the best way to check this.
As p4wn3r observed, the special relativistic version has several symmetries: it can be divided in 4 essentially identical subproblems, whose sole difference is the value of the constants t0, x0 and τ0. It is enough to solve one of these sub-problems, so we will take the first -- the constantly accelerating portion for the first half of the trip to Alpha Centauri. Our values for t0, x0 and τ0 are then defined by our boundary conditions that the rocket is at rest at the Earth when t = τ = 0, so that t0 = τ0 = 0 and x0 = -c2 / a; thus, we will work with
t(τ) = (c / a) * sinh(a * τ / c)
x(τ) = (c2 / a) * {cosh(a * τ / c) - 1}
We must calculate the final value of τ we are interested in; this corresponds to the half-way point x(τm) = (4.365/2) ly. To do this, lets invert the equation for x(τ) -- we can do that in this interval because x(τ) increases monotonically with τ in this interval -- to obtain
τ = (c / a) * acosh{(a / c2) * x(τ) + 1}
Plugging in x(τm) = (4.365/2) ly = 2.1825 ly, a = 1.0315512 ly/y2 and c = 1 ly/y, we find
τm = 1.791172306 y
For the Earth reference frame, this corresponds to
t(τm) = 2.999132811 y
Multiplying these by 4 for the four segments of the journey, we see that 7.164689224 y transpire at the rocket, while 11.996531244 transpire on Earth. The difference to p4wn3r's result is probably a result of numerical errors in his part, as his math seems correct.
For the GR version: I started doing some work on it; but it is ugly and messy. FYI, here a few of the confounding factors:
The relative motion of the star systems will greatly complicate matters because of the changing gravitational field;
Alpha Centauri is a binary star system, where each stars have a mass near to the Sun's own mass; not only that causes nonlinearity problems (see next item), but there is also the additional issue of dealing with gravitational radiation;
Due to the non-linear nature of Einstein's field equations, the metric tensor for the 3 stars is not the sum of the metrics of the individual stars;
The exact starting position near the Sun and ending position near the binary Alpha Centauri star is crucial to avoid the Schwarschild radius (unless you model the star interior as well);
The difference in mass between the star systems will remove some of the symmetries that allowed us to calculate 1 leg of the trip and multiply by 4 in the special relativistic version.
There are ways around these issues, using several simplifications; but the math still comes out ugly. I will leave this for the future...
Now for replies to posts made while I was writing the above:
Bobo the King wrote:
Still, I will take the time to point out that to properly define the acceleration, you should refer to the force per unit mass. In particular, the problem is well-stated if you demand that the spaceship's momentum, not its velocity, increasesa monotonically. Your units imply that the velocity increases monotonically, which is obviously not possible.
You haven't yet taken (non-introductory) courses in relativity, have you? If you had, you would already know that "constant acceleration" in relativity is shorthand for "constant proper acceleration", which is perfectly well defined and leads to hyperbolic motion. His choice of units is totally irrelevant for that; he can chose to measure acceleration in m/s2, ly/y2, feet/min2, m-1 (using c as a conversion factor between s and m), s (the reverse conversion), kg-1 (using G/c2 as a conversion factor between Kg and m and c as a conversion factor between m and s) or any other exotic units of his choice and his question would still be well formed. Moreover, his velocity can increase monotonically whether or not this is relativity; in relativity, it also has to grow asymptotically to the speed of light, but that does not prevent monotonic growth.
Warp wrote:
The acceleration felt by the astronaut (which can be unambiguously measured by the astronaut using an accelerometer) can be constant forever, even under SR/GR, as far as I understand.
Absolutely correct.
Bobo the King wrote:
Said this way, yes, you are absolutely correct, though I think that speaking of the force felt by the astronaut is even more accurate.
So far so good. Irrelevant, but correct nevertheless. Sadly, the rest of the paragraph, does not have any single correct sentence:
Bobo the King wrote:
In fact, I think the key word there is "felt". The earlier problem came from ambiguity in the word "acceleration" and your choice of units, m/s^2, which, taken in a perfectly valid way, implies a constant dv/dt.
See above.
Bobo the King wrote:
In this interpretation, the astronaut would not feel a constant acceleration; he would instead feel an asymptotically increasing force as he approaches the speed of light, with death and annihilation of the spaceship surely occurring sometime before he reaches v=c (in a finite time).
This "interpretation" is not a result of any real understanding of relativity. The force he feels is the force measured in his reference frame, period (same for acceleration). If the force is constant in this reference frame, so is the acceleration (and the converse also holds). If you were to demand constant acceleration as seen by any other reference frame, the ship would break apart quickly due to a matter of simultaneity (that is, enormous stresses due to parts of the ship being accelerated before others in the ship's reference frame); but no one but you is talking about that.
Bobo the King wrote:
Hope that clears things up. I didn't have time to answer the problem, so I instead settled for a nitpick.
You can clear things up with a confused explanation about an invalid nitpick. Try harder.
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rhebus wrote:
Integrating distance feels strange because you don't normally have situations where distance is a function of time. Velocity and acceleration are often a function of time, but distance?
Let me rewrite what you just wrote:
Integrating distance feels strange because you don't normally have situations distance is a function ofwhere things move over time. Velocity and acceleration are often a function ofoften change in time, but distance?
See how absurd that sounds now? Yeah, I thought so.
Even for straight motion with constant speed, you have that x(t) = v*t.
With that out of the way, lets see DarkKobold's question:
DarkKobold wrote:
If you were to integrate distance... you'd get (m * s). Perhaps I'm just too far out of physics... but that doesn't seem like it would have any meaning.
As rhebus correctly answered after the above snippet, it is related to position averages: int_a^b{x(t) dt}/(b - a) is the average position over the time interval [a, b].
Derakon wrote:
Oh, here's a question I've been wanting to hurt people with for awhile.[snip]
That is an interesting one. I will be working on it, but it may take some time.
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Another new WIP, this time up to AIZ2. AIZ2 is finished in 1:04::37, 92 frames faster than the previous version. The new zip was finally pulled off by MrSweed, after I had long given it up; the rest of the improvement is better optimization. I used the cutscene skip, but using a flame shield this time; this made the skip 13 frames faster than the version in the any% run (although I was worried for a time that it was slower, because the fire shield version has a smoother start).
Before anyone asks: yes, I loiter a few frames after entering the ground on the first zip of AIZ2. If I didn't, I wouldn't grab the bubble shield monitor and I wouldn't be bounced back to the ground, thus avoiding quite a speed loss. Also, by the end of the level, flying off from the tube is 6 frames faster than jumping out of it because the setup for it makes me traverse the tube slower, and Hyper Sonic's acceleration is high enough that I can keep some of the difference after the tube.
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Sonikkustar wrote:
...marzojr, You amaze me at your Sonic TASing skills. I feel ashamed now.
Seriously, Its amazing you managed to improve AI1. I never thought of using Tails like that!
Good luck with the timed platform.
Thanks for the kind words, but you have nothing to be ashamed of -- you are very good too, as the S2 run proves.
All in all, I did manage to have good luck in that timed platform. Better use of Tails right after it also gave me 5 more frames of improvement, for a total time of 0:37::56. I don't think that a better time is possible in this zone (except for maybe 1 frame or 2 at the final ascent before the boss) without Hyper Sonic.
