Acceleration of gravity from earth, on objects, traveling at near the SOL. - Page 1
Forum section: electronics / general

author: sourcecharge
author_rank: Regular Contributor
timestamp: December 18, 2020, 12:57:38 pm
content: So here's my first crack at stumping the forum:Time dilation and the conservation of energy law, A thought experiment to ponderhttps://www.eevblog.com/forum/chat/time-dilation-and-the-conservation-of-energy-law-a-thought-experiment-to-ponder/(Technically, I think it still hasn't been solved, as current can't be different in a 0 ohm transmission line.)But anyways, here the new mind bender:Time dilation, when it comes to traveling near the speed of light, hypothetically, time starts to slow down compared to an outside observer at rest. So if someone was traveling in a space ship at near the speed of light, people on earth would experience more time than those on the space ship. Let's call the amount of time that occurs on the space ship traveling at near the speed of light assec/10, and the amount of time that occurs on earth assec, for simplicity reasons.The question is, if the space ship traveled past earth at near the speed of light, how much acceleration would gravity act upon the space ship?Would it be 9.8 m/sec^2, or would it be 9.8 m/(sec/10)^2?

author: nctnico
author_rank: Super Contributor
timestamp: December 19, 2020, 01:55:33 am
content: AFAIK you can't apply special relativity theory this way. And then there is also a paradox involved but I forgot the details.

author: Ground_Loop
author_rank: Frequent Contributor
timestamp: December 19, 2020, 02:05:26 am
content: I'll have a guess.Since the relativistic mass of photons is acted upon by gravity without altering the gravitational constant I assume all mass is affected similarly.

author: Nominal Animal
author_rank: Super Contributor
timestamp: December 19, 2020, 10:03:34 am
content: Quote from: sourcecharge on December 18, 2020, 12:57:38 pmI'm tempted to ask, "How much color would a voltage of 2.34 V impart on the ship?"Your terminology is close enough to quibble about, since if one does not quibble a bit, serious misunderstanding is the most likely result.Newtonian (non-relativistic) gravitational force is \$F = G m_1 m_2 / r^2\$, where \$r\$ is the distance between the point-like masses \$m_1\$ and \$m_2\$, and \$G\$ isthe gravitational constant, \$G = 6.674 \cdot 10^11 \, m^3 \, kg^{-1} \, s^{-2}\$. The velocity of either point-like mass does not affect the force itself, only how the force varies as a function of time. The acceleration of a point like mass towards the center of mass of the pair is given by \$\vec{F} = m \vec{a}\$, so we can express the Newtonian acceleration of a point-like mass \$m\$ due to the gravitational effects of another point-like mass \$M\$ at \$\vec{r}\$ relative to the first point-like mass as$$\vec{a} = \frac{G \; M}{\lVert \vec{r} \rVert^2} \frac{\vec{r}}{\lVert\vec{r}\rVert} = \frac{G \; M \; \hat{\Delta}}{r^2}$$in barycentric coordinates (relative to the center of mass of the pair), with \$\vec{r} = r \hat{\Delta}\$, and \$\lVert\hat{\Delta}\rVert = 1\$ being the direction vector from the point-like mass towards the other point-like mass.Proper relativistic model uses differential calculus to express the relationships. If we approximate those solutions with polynomials, for small masses and small velocities we getEinstein-Infeld-Hoffman equations:$$\begin{aligned}\vec{a} & = \frac{G \; M \; \hat{\Delta}}{r^2} \\~ & + \frac{1}{c^2} \frac{G \; M \; \hat{\Delta}}{r^2} \left[ \vec{v} \cdot \vec{v} + 2 \vec{V} \cdot \vec{V} - 4 \vec{v} \cdot \vec{V} - \frac{3}{2}(\hat{\Delta}\cdot\vec{V})^2 + \frac{1}{2}\vec{r}\cdot\vec{A} \right] \\~ & + \frac{1}{c^2} \frac{G \; M}{r^2} \left[ \hat{\Delta} \cdot ( 4 \vec{v} - 3 \vec{V} ) \right] \left[ \vec{v} - \vec{V} \right] \\~ & + \frac{7}{c^2} \frac{G \; M \; \vec{A}}{r} \\~ & + O(c^{-4}) \\\end{aligned}$$The final term just explicitly tells what we omitted; the polynomial itself is infinite, but successive terms contribute less and less to the result. \$\vec{v}\$ is the velocity of the point-like mass itself, and \$\vec{V}\$ and \$\vec{A}\$ are the velocity and acceleration of the other point-like mass, all in barycentric coordinates (with respect to the center of mass of the system).Closer to the speed of light not only do we get Lorentz contraction, but relativistic effects of mass also. The effective mass of a point-like particle with velocity \$v\$ is$$m_{\text{rel}} = \frac{m_{\text{rest}}}{\sqrt{1 - \frac{v^2}{c^2}}}$$Relativistic time dilation is described by a similar relation,$$\Delta t_{\text{rel}} = \frac{\Delta t_{\text{rest}}}{\sqrt{1 - \frac{v^2}{c^2}}}$$and if we have time dilation ratio of about ten, we know that \$v/c \ge \sqrt{99/100} \approx 0.994987\$. Even with respect to the barycenter of the two-point system, their velocities are around half \$c\$ (because one is massive and almost stationary, and the other is light but moving at near lightspeed), causing a relativistic mass \$m_{\text{rel}} \approx \sqrt{4/3} m_{\text{rest}}\$, or \$m_{\text{rel}} \approx 1.1547 m_{\text{rest}}\$.So, the fact is, the gravitational interaction between the zipping spaceship and the planet is actuallyincreaseddue to the velocity, as it increases the effective mass of the spaceship. The higher velocity only means their relative distance \$r\$ is small (and interaction and therefore the acceleration large) for a short period of time.Thus, the answer is neither of the suggested ones. The force, and the accelerations in the barycentric coordinates, are somewhat larger in magnitude than they would be at very slow, non-relativistic velocities, because of relativistic effects of the effective masses of the spaceship and the planet in the barycentric coordinate system (coordinates with respect to the center of mass of the pair). The time dilation factor itself is not involved here at all.See how important that quibbling about terminology was?The repeating qualifier, "in the barycentric coordinate system", isn't very important in this system, because the center of the mass is where one expects it to be in this case. However, if we consider two spaceships traveling at near lightspeed passing by each other, and they have different masses, the picture becomes much more complicated; yet, using the above barycentric coordinates the definitions still hold.

author: sourcecharge
author_rank: Regular Contributor
timestamp: December 19, 2020, 12:46:20 pm
content: Quote from: nctnico on December 19, 2020, 01:55:33 amPlease expand your explination, I'm not sure exactly what you are thinking. What way can't it be applied, and what is the paradox?@ Nominal Animal & Ground_LoopRelativistic mass does not create gravitational force like real mass.

author: tszaboo
author_rank: Super Contributor
timestamp: December 19, 2020, 01:03:02 pm
content: As I understand, if you accelerate an object, which is going at relativistic speeds, not all the energy is increasing the speed. Most of the energy is going to increase the mass of the object, not the speed, the faster it goes, the higher percentage of the energy goes into the mass. Slowly reaching 100%. So none of the answers.

author: vad
author_rank: Frequent Contributor
timestamp: December 19, 2020, 01:33:08 pm
content: Quote from: sourcecharge on December 18, 2020, 12:57:38 pmFor simplicity, assume the spacecraft is moving along Earth radial axis, so both the acceleration and motion vectors are parallel.Lorentz transformation for time dilation:t’ = gamma * tFrom you example, gamma = 10.Lorentz transformation for acceleration along axis of motion is:a’ = gamma^3 * aIn Earth reference frame, the spaceship accelerates at 9.8 m/s^2 towards Earth. In other words a’ = 9.8. Therefore:a = a’ / gamma^3 = 0.0098 m/s^2.

author: Nominal Animal
author_rank: Super Contributor
timestamp: December 19, 2020, 01:36:44 pm
content: Quote from: sourcecharge on December 19, 2020, 12:46:20 pmOf course it does. Energy and momentum of whatever matter and radiation are present, affect the curvature of spacetime.Even light, which has as close to zero rest mass as we can ascertain – photon rest mass is less than 10-18eV/c2, or less than 1:100,000,000,000,000,000,000,000th of the rest mass of a single electron –, bends space; exactly because they havemomentum. In Einstein's field equations, it is the energy-momentum tensor that determines the curvature of space.In fact,massis not directly related to spacetime curvature (and therefore gravitational acceleration) at all; only energy and momentum is. Mass is just one way to affect the energy and momentum present; and those determine the curvature of space.

author: JohnnyMalaria
author_rank: Super Contributor
timestamp: December 19, 2020, 01:46:02 pm
content: Quote from: sourcecharge on December 19, 2020, 12:46:20 pmIt's the twin paradox. One twin blasts off from earth and travels at very close to the speed of light for what that twin experiences as two years. Upon return to earth, more than 100 years have passed and the other twin is dead.How so? If everything is relative, why not the other way around?It's because *acceleration* is absolute. If it wasn't, every time an object accelerated away from you, you would feel the force. Of course, we don't.

author: sourcecharge
author_rank: Regular Contributor
timestamp: December 19, 2020, 07:40:13 pm
content: Quote from: Nominal Animal on December 19, 2020, 01:36:44 pmThis explains why relativistic mass does not have gravitational force.Can you go fast enough to get enough mass to become a black hole?https://wtamu.edu/~cbaird/sq/2013/06/18/can-you-go-fast-enough-to-get-enough-mass-to-become-a-black-hole/Relativistic mass is really only kenetic energy.Quote from: NANDBlog on December 19, 2020, 01:03:02 pmIt was a hypothetical question...how the space ship got to it's speed is improbable, this is true, but that part was supposed to be hypothetical.Quote from: vad on December 19, 2020, 01:33:08 pmI don't think I've seen this before, do you have a reference link to this?

author: tszaboo
author_rank: Super Contributor
timestamp: December 19, 2020, 07:54:56 pm
content: Quote from: sourcecharge on December 19, 2020, 07:44:02 pmBut it is still an answer. Your answer(s) assume that the gravitational pull will increase the speed. What actually happens, that the gravitational pull will accelerate the object with some newtons. For most falling objects (in vacuum) this will increase the speed. 9.8 m/s2is an oversimplification, because the mass of the object increases the force (newtons). For a relativistic speed object, the force (newtons) will be spent on increasing the mass, not the speed. So even 0.98m/s2could be a reasonable answer. Probably not though.

author: sourcecharge
author_rank: Regular Contributor
timestamp: December 19, 2020, 08:03:30 pm
content: Quote from: NANDBlog on December 19, 2020, 07:54:56 pmI think it might be possible that the acceleration of gravity on the ship is lower, that is why I was interested in vad's answer and wanted more information.

author: Non-Abelian
author_rank: Contributor
timestamp: December 19, 2020, 08:27:53 pm
content: Quote from: sourcecharge on December 18, 2020, 12:57:38 pmNot enough to be noticed. In the relativistic limit, you can neglect the mass of the rocket, so E~pc, in which case you're asking how much will a light ray be bent by the earth's gravitational field.QuoteNone of the above. First of all, the acceleration due to gravity depends on how far from the earth you are. Second, for the relativistic case, gravity is involved, so you would need to use the schwarzschild metric to calculate something, which in this case isn't enough of an effect to calculate. But, the solution is that the trajectory will be deflected by approximately theta ~ 4m/r, where m is the mass of the earth in kilometers (the mass of the earth in these units is 0.8 cm = 0.000008 km) and the radius at the grazing angle of the earth is its radius (about 6400km).

author: Nominal Animal
author_rank: Super Contributor
timestamp: December 19, 2020, 08:29:35 pm
content: Quote from: sourcecharge on December 19, 2020, 07:40:13 pmUnfortunately, Dr. Baird is wrong. It is obvious general relativity is way outside his area of expertise. I'm sure that a PhD specializing in optics finds the fact that even light bends space frightening.Like I said, in Einstein field equations energy and momentum cause the curvature of spacetime. Mass affects curvature only through energy and momentum. Dr. Bairds fundamental error is assuming that mass itself is what curves spacetime. That is simply not true.Relativistic mass is actually just a notional measure. It is equal to the energy that causes the same spacetime curvature as the momentum of the point-like mass. It is only needed when you need something that has an analog in Newtonian physics. To correctly model spacetime, you need to use Einstein field equations.

author: vad
author_rank: Frequent Contributor
timestamp: December 19, 2020, 09:05:03 pm
content: Quote from: sourcecharge on December 19, 2020, 07:40:13 pmhttps://en.m.wikipedia.org/wiki/Acceleration_(special_relativity)Equation (1c).If we choose axis x to be the Earth’s radial along which spacecraft is moving, then:a’x = a’, a’y = a’z = 0, ux = v, uy = uz = 0.Therefore:a = a’x = a / (gamma^3 * (1 - v^2/c^2)^3)Where:gamma = (1-v^2/c^2)^-0.5Hence:a’ = a’x = a / (gamma^3 * gamma^-6) = gamma^3 * a

author: Non-Abelian
author_rank: Contributor
timestamp: December 19, 2020, 09:40:51 pm
content: Quote from: Nominal Animal on December 19, 2020, 08:29:35 pmNo, he is not wrong. His point is that the useful concept in relativity is the invariant mass, which is not a frame dependent quantity (which seems to be what you are trying to say, but don't realize it). He isn't trying to explain general relativity to anyone. Second, the curvature is in spacetime, not space. Third, you can always pick a frame in which spacetime is flat at a point.QuoteIn the rest frame of a mass, E=m. (The c^2 is irrelevant since c is just a way to convert meters to seconds. It doesn't mean anything physically.)QuoteSo, what's the problem? That is what the web page you are railing against was pointing out.QuoteAs noted above, the energy in the rest frame of a mass is just its mass. T_00 is its mass density.

author: Nominal Animal
author_rank: Super Contributor
timestamp: December 19, 2020, 10:06:09 pm
content: Quote from: Non-Abelian on December 19, 2020, 09:40:51 pmYou managed to pick the one point where I miswrote spacetime, and made that a point.Perhaps you should read my first post (fourth in this thread), and point out why you believe Dr. Bairds post invalidates my point.

author: MIS42N
author_rank: Frequent Contributor
timestamp: December 20, 2020, 12:33:51 am
content: I have always been suspicious of so called time dilation of fast moving objects. There isn't a way to measure time, we rely on observing repetitive phenomena (like atomic vibration), and counting how many repetitions of the phenomenon occur between some other events (like the sun appearing at the same angle above the horizon). It is determined that repetitive phenomena in an object being accelerated occur slower (or faster, I don't know which) than one not being accelerated. This explains (to me) why clocks on satellites run at a different repetition rate than those on earth's surface. We live in a higher acceleration field (a different spacetime?) than the satellite. The satellite has to move around the earth at speed to stay in space, but is the different rate of its clock affected by the speed, or just by being in a different spacetime?Consider an object moving away from earth near SOL. If we are in continuous contact with the object, it's clock will slow down while it is being accelerated, then it will appear to be running slower because of redshift in the received signal. But is it actually running slower when the object reaches its desired speed and is not being accelerated? From the point of view of the object, it will see the earth recede at the same speed, and signals from earth will be redshifted.If a spaceship were able to go to Alpha xxx our nearest stars (say 4 light years away) at 1/2 the speed of light, then return at 1/2 the speed of light, I think what we would see is the spaceship taking 12 years to get there and 4 years to get back. Any people in the spaceship would see it take 8 years to get there and 8 years to get back. Apart from the slowdown due to acceleration, the people on earth and on the spaceship will age about the same amount. It seems to me any other outcome results in a paradox. Using the spaceship as a reference or using the earth as a reference, shouldn't the result be the same? The spaceship and earth start in the same spacetime and end in the same spacetime. But during the journey observations are distorted due to the speed of light. Does relativity deal well with correlating observers in different reference frames?

author: Non-Abelian
author_rank: Contributor
timestamp: December 20, 2020, 12:38:25 am
content: Quote from: Nominal Animal on December 19, 2020, 10:06:09 pmHis point was the same as the one you seem to be trying to make - that the invariant mass is the physically relevant quantity, which is correct. You can only disagree with that by trying to say physical quantities depend on what coordinates you choose, which would be ridiculous.

author: nctnico
author_rank: Super Contributor
timestamp: December 20, 2020, 01:30:03 am
content: Quote from: MIS42N on December 20, 2020, 12:33:51 amUnfortunately it doesn't work that way. You can only apply special relativity theory on objects which have a constant velocity compared to the reference object. As soon as there is accelleration / de accelleration involved you need general relativity theory. Examples of objects to which the special relativity applies to are satellites orbiting the earth.

author: Non-Abelian
author_rank: Contributor
timestamp: December 20, 2020, 02:10:16 am
content: Quote from: MIS42N on December 20, 2020, 12:33:51 amSure, but that's how you measure time. And you are not counting repititions in the sense of a deterministic repetitive process so that the precision of the measurement depends on the statistics, not the time for the process itself to occur (which is probabalistic). You end up with a better clock by having large statistics on probabilistic events than trying to create a clock from a fixed repetition rate.QuoteUsing the speed of light in relativity to understand relativity is an archaic way of doing relativity. The reason Einstein did it that way is that he was trying to explain electrodynamics by modifying Newtonian echanics rather than the other way around, which is what everyone else was trying to do at the time. Einstein had no concept of geometry at that time and it wasn't until Minkowski pointed out in 1908 that reltivity could be understood as a 4-dimensional spacetime that it began to be understood as geometry. I don't have any idea why it is still taught this way. when a physicist says "speed of light," he/she doesn't really mean that (depending on context). It's just that other physicists know what it means in context and it's easier than saying something else.The speed of light really has nothing to do with it except for the fact that massless particles propagate at the same velocity in every frame and the photon is known to be massless to the best limits of experiment. A theory of electromagnetism in which the photon is NOT massless and which would require Maxwell's equations to be changed was known as early as 1914 (the Proca Lagrangian). So, really, the speed of light thing properly belongs in a theory of electromagnetism.Relativity itself simply treats space and time on equal footing. You can posit two different theories of space and time which have physical plausibility. One is Galilean relativity in which time is parameter and relativity (special or general) in which time is a coordinate and "proper time" is an invariant quantity. In regular euclidian space you measure the lengths of lines (loosely speaking) via the "metric" ds^2 = dx^2 + dy^2 + dz^2, i.e. the pythagorean theorem. The equivalent in relativity is ds^2 = dx^2 + dy^2 + dz^2 - dt^2 (where time here is measured in meters since the constant c just converts seconds to meters and really doesn't mean anything physically). Experiments show that galilean relativity is not correct, which leaves general/special relativity.QuoteAs noted, the speed of light has nothing to do with it. Relativity is almost the same as changing the origin point on a circle from (x,y) to (x',y').x' = x cos(a) + y sin(a)y' = y cos(a) - x sin(a)Instead of using sines and cosines, you use sinh and cosh:x' = x cosh (a) - t sinh (a)t' = t cosh (a) - x sinh (a)In this case a velocity is just a hyperbolic rotation v = tanh(a). Feel free to change the t and t' to ct and ct' if you're uncomfortable with time in meters in which case you also insert a c before the tanh for the velocity. As an exercise, show that x^2 - t^2 = x'^2 - t'^2, so that the result x^2 - t^2 is not frame dependent.QuoteSure. Since you have invariants that are frame independent, different observers relate what they observe by changing coordinates, just like you would if you and a friend were sitting in two different places and each of you wanted to use where you were sitting as the origin of your coordinates.As an aside, this also ends up meaning that the physically meaningful quantities for capacitance and inductance are geometric. (1pf ~ 1cm if I recall). Not so useful for engineering.

author: Non-Abelian
author_rank: Contributor
timestamp: December 20, 2020, 02:16:07 am
content: Quote from: nctnico on December 20, 2020, 01:30:03 amActually, you don't need general relativity to deal with accelerations. In the case of satellites, you do need general relativity for gps satellites. The general relativistic effects are small, but enough that gps would not work without those corrections. I'm not sure if anyone bothers with it for satellites that don't need to be syncronized.

author: MIS42N
author_rank: Frequent Contributor
timestamp: December 20, 2020, 02:22:19 am
content: Quote from: nctnico on December 20, 2020, 01:30:03 amI would say satellites orbiting earth need general relativity because they are in a gravitational field and therefore being accelerated toward earth. They negate that by travelling in a straight line in a curved spacetime. And once the spaceship gets up to speed, it DOES have constant velocity relative to earth. The observed velocity should be constant both from the spaceship and earth, just they are different. The measured elapsed time on the spaceship will be slightly different to that on earth because we are in both the earth's and the sun's gravitational well and the spaceship isn't. But I don't think it makes the sort of difference people call time dilation.

author: Nominal Animal
author_rank: Super Contributor
timestamp: December 20, 2020, 02:37:20 am
content: Quote from: Non-Abelian on December 20, 2020, 12:38:25 amI'm trying to explain how spacetime curvature depends only on energy (which, if we areprecise, is not just invariant mass, but also e.g. the electric fields) and momentum.Consider his title:"Can you go fast enough to get enough mass to become a black hole". Yes, you can interpret his text to make a point about invariant mass; or you can read his text at the face of it, and see that he completely ignores the effect of momentum on the curvature of space. (I believe the title is technically misleading: it's like posing a question like"Will I live forever, if I eat a young, healthy, unrestrained, Siberian Tiger every day?", then completely ignoring the context of "tiger blood" and traditional remedies, and self-answering the question by stating"No, because your stomach isn't big enough."See? You can argue it is technically correct, but in the context of claiming to explain things and answering the stated question, it is horribly incorrect: it avoids the core answer, which is that there is nothing in consuming Siberian Tigers that affects longevity differently than consuming basically any other mammal.)Similarly, you do not need invariant mass to become a black hole. Momentum does suffice.From the point of view of an external observer, as long as they cannot precisely measure the velocity in barycentric coordinates, they cannot really tell whether the approaching object curves spacetime because of their mass (energy), or because of their momentum.Part of the difficulty I'm facing here is the lack of valid examples geared toward a layman. Any event that could impart sufficient momentum to a single point-like mass to become a black hole is necessarily energetic enough to make the entire point moot. (Based on known ways of transferring momentum, we can immediately say it is only possible in a system that is already a black hole.) It is completely unrealistic. So, whenever momentum is discussed wrt. black holes, it tends to be about angular momentum, which causes interesting measurable properties on black holes even when so small its effect on spacetime curvature is neglible.Yet, contrary to what one might think by reading Dr. Baird, how much curvature a fixed invariant mass spaceship causes to spacetime, most definitely depends on its velocity. Sourcecharge used his text to refute my points. I do not make much of a difference between "typical reader gets the wrong impression and understanding" and "writer being wrong", because the effect is the same, and physics is about describing reality, not intent.

author: Non-Abelian
author_rank: Contributor
timestamp: December 20, 2020, 03:49:48 am
content: QuoteThe spaceship is at rest in its own frame, so its velocity is zero. It makes no sense to refer t the velocity of something without referring to what that velocity is relative to, in which case the situation is recipricol and you can take either to be at rest or neither to be at rest relative to something else.QuoteI'm not a layman. I have a PhD in nuclear physics. Feel free to do this using differential geometry.

author: vad
author_rank: Frequent Contributor
timestamp: December 20, 2020, 04:20:14 am
content: Quote from: nctnico on December 20, 2020, 01:30:03 amNo, no, no.You can even use Newtonian mechanics, not to mention SR, for accelerating object / object moving in gravitational field. And it will yield very accurate results. Accurate enough to send men to the Moon, New Horizons to Pluto, or ICBM to another continent with pinpoint precision.GR starts playing major role only where there is significant curvature of spacetime caused by energy-momentum tensor (let say vicinity of black holes, neutron stars). In all other places classical mechanics and SR are good enough for most models, and GR only gives tiny adjustments of the second and higher orders.

author: MIS42N
author_rank: Frequent Contributor
timestamp: December 20, 2020, 09:49:43 pm
content: Quote from: vad on December 20, 2020, 04:20:14 amBut not accurate enough for GPS satellites. I found this "A calculation using General Relativity predicts that the ground-based clocks will run slower than the GPS satellite clocks by about 45 microseconds per day.". If that were not taken into account, it would produce an error of about 12km/day rendering the GPS system useless.

author: vad
author_rank: Frequent Contributor
timestamp: December 20, 2020, 10:54:36 pm
content: Quote from: MIS42N on December 20, 2020, 09:49:43 pmNewtonian physics is more than accurate enough for orbital mechanics of GPS and other satellites. In fact that’s the physics that is used in that field. Neither SR nor GR is needed.GPS system, on the other hand, requires extremely accurate clock. GR-related effects introduce tiny 0.00054 ppm error that is too big to be left unadjusted in software.

author: Nominal Animal
author_rank: Super Contributor
timestamp: December 21, 2020, 06:01:01 am
content: Quote from: Non-Abelian on December 20, 2020, 03:49:48 amIf you had read my post, you'd know that I frequently and explicitly talk about "barycentric coordinates" and "barycentric coordinate system".Why do you so obstinately refuse to read the entire thread? I find that gross, selfish, and pretty much a signature of an idiot. Prove me wrong.Quote from: Non-Abelian on December 20, 2020, 03:49:48 amI'm not trying to show examples to you; I do not doubt you know this stuff. That is why I asked you to point out any errors in my initial post; if you were to describe any, I would definitely trust you enough to seriously examine my claims, and either admit error and fix it, or explain why I believe I am not in error. I do not extend that trust to Dr. Baird, because the plain reading of their text indicates they are either completely wrong, or being nitpicky about the terms used ("relativistic mass", which is just an useful analog for those working with a Newtonian mindset) and failing to point it out and therefore leading non-physicists completely askew in their understanding (in particular, that momentum does affect spacetime curvature; a plain reading of Dr. Baird's question and self-answer makes non-physicists believe the opposite, which is the error that bugs me).I am trying to explain things to the original asker, and anyone else who reads this thread and sees things the same way.The reason I am using the words I am using, is because I'm trying to help them understand the situation in terms they can understand. I am directing my words to laymen. That does not mean I think you are a layman.You just swooped in from the side, didn't read the thread from the start, and just started to nitpick on me, without realizing I am the one here trying to describe the model that currently best matches reality. Then you get offended and demand I start using math notation you're familiar with, because you think I'm offending your sense of grandeur. Get real, and read every message in this thread, before posting again; you're making yourself look like a poser who knows the lingo but not much else, not a scientist.As to myself, I don't have a PhD yet, and probably never will (the university politics have become too crazy for me), although I do have taken the courses needed for a PhD in physics (my specialty being computational materials physics), plus probably enough in CS for a side MSc. (PM or email me, and I'll let you know my real name, and provide you my course records. It isn't magnificent - the GPA isn'tthatgood, except in the deeper, specialized courses, but shows I am being honest here.) I prefer to stick to classical models (because they are lightweight enough to compute complex enough systems to have practical use cases I'm interested in, like material degradation in fusion reactor walls and raw natural gas turbines), but have done enough research into general relativity to understand the issues of properly simulating e.g. galaxies without resorting to rough approximations. I do my best work when supporting other researchers, maintaining computing clusters and developing simulators, visualisation, and sonification of (especially massive) simulations. I am not a theorist, because I am not that good in math (I really need Maple, Sagemath, or Maxima to do proper work with field equations, or really, anything to do with differential calculus), but I have in the past successfully described complicated physical systems and processes to laymen, giving them a solid base for correct understanding; and that is what I am trying to achieve here. Not precise technical correctness, but to provide the laymen – not you, the asker and others reading this thread, you oaf! – a better picture of the reality involved here.I have also ran an IT company as part-owner and CEO for a few years, but found am not suited for the scammy world of IT business. That, and previous unfortunate workplace choices have broken me utterly. I suffer from recurrent severe depression, and although it is pretty well under control now, I still cannot handle stress at all, physiologically. I get immediate painful rashes from stress (histamine reaction from stress hormones), and there is still something seriously wrong with my brain pathways, as stress measurably decreases my problem-solving skills.I am a broken, burned out husk of a man, and not something you should see as a competitor or challenge to your PhD.Now, please, re-read all posts in this thread, especially focusing on my first post (fourth in the thread, I believe). Then, if you still believe I have made any mischaracterizations or errors in my descriptions that leads laymen astray – again, I don't care if those who know this stuff disagree with the terms I use, because the end goal is not technical correctness, but a robust overall description at the layperson level that leads tocorrect understanding–, post them or send me a PM; I will consider them throughly, and highly appreciate any suggested corrections, and publically admit any error and correct them if you show any. (Just look at my posts in other threads; I do that all the time. I am often wrong, and will gladly admit to being wrong when shown proof, and be grateful for letting me know. Truth isthatimportant to me; way more important than my own ego. How about you?)