Everything posted by md65536
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Moon shadow
I thought this might be incorrect reasoning, because you could change the east/west direction of the shadow simply by changing the rotation rate of the Earth, without affecting how the moon moves relative to the sun. But I think your reasoning must be right. It seems then that if the Earth were spinning much faster, then even though day-to-day the moon appears to have lagged behind the sun, during a single day the moon would appear to be overtaking the sun at lower latitudes. This would be due to parallax. The animation at https://en.wikipedia.org/wiki/Solar_eclipse_of_August_23,_2044 shows how the shadow can go "backwards" at high latitudes. It seems that in this case, the sun and moon are "to the North", eg. during "night time" where there is midnight sun. The sun still appears to be overtaking the moon, but they're both moving in a west to east direction that late in the evening! This should happen anywhere at high latitudes when there is an eclipse in the evening after the sun has passed the westernmost point in the sky and begins moving eastward again before setting, or at dawn before the sun reaches the easternmost point.
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Four marching bugs
- How does ChatGPT work?
You can ask ChatGPT about itself. It probably knows a bit more about it than scienceforums.net users. Parts of its answers: Much, but not all, of what it says is just based on what other people have said. For example if you ask it a science question that is generally misunderstood repeatedly across the internet, and is not something some higher-weighted "expert system" would describe correctly, you might likely get the wrong generally misunderstood answer. Months ago I asked it if the Twin Paradox required general relativity, and it told me that it does, sounding just like a post you might see anywhere across the internet. Now if I ask it, it says it doesn't. I don't know if that's better data, better model, better weightings, or if it is actively being corrected on answers it commonly gets wrong. There is a small army of humans providing it with specific data, not just scraped stuff. It is not as simple as taking everything that's been written and scraped, and generating something similar. It is programmed with a bunch of different specific abilities. For example, it can multiply 2 numbers no one has ever multiplied before, with no data in existence to suggest what the probable answer is based only on what's been written before. It could conceivably learn the steps to do something it's not programmed to, but in this case it has specific programming for it. The general and most basic functionality is to mimic what it has seen before, and that alone can answer a lot of questions correctly and carry on a conversation similar to ones that have happened before. But there is a lot more additional programming that it hasn't just "learned" by itself. Certainly, additional programmed capabilities will be added over time. An AI like this can create new ideas. For example, if someone somewhere has associated A with X, and someone else has associated B with X, it's possible for an AI to associate A with B, even if no human has ever done that. When people say an AI doesn't, or never will, "understand" something like a human does, I wonder how they define understanding, or feeling, thinking, etc., without using "like a human does," in their definition. How do we know that human understanding is more than just the learned connections between a very large set of concepts and knowledge?- Another missing area
I can't imagine ever noticing that, but now I see why it works for other triangles with the same areas. If you skew the triangle eg. to make a right-angle triangle, you don't change the areas. If you uniformly scale the triangle vertically by r (preserving the ratios between the areas), and then scale horizontally by 1/r, each region's area is scaled by the same factor.- Find the perimeter
- Another missing area
- Find the missing area
That makes it a lot easier! Is it- Find the perimeter
I see there's other ways to figure this out, but I noticed that there are lots of ways the DE line can be chosen... Anyway the answer I get is- A grid and a shape
Intuition is that rotation will complicate things because it allows for more ways in which a shape can be moved to dodge the grid, but it seems not to matter because it can still be placed off the grid without rotation.- A grid and a shape
No fair asking for something that's impossible in the original problem 😠 "Proof": I think there might be a problem here talking about the area of points, like with the Banach-Tarski paradox.- A grid and a shape
We're trying to prove 2 opposite things. Does trying to prove that an arbitrary shape can be placed without touching a point, lead to a contradiction?- The sum of the series
How did you solve it? I noticed a pattern in what each subsequent term was "leaving out", and figured out a formula for the sum of the first n terms, then used induction to see that it works as n goes to infinity. Is there another way?- ... paradox help needed ...
I concur. This sort of thing has cropped up before, and it has always been due to relativity of simultaneity (usually). Forget about the fluid for now. Consider the frame where both Vs are traveling toward each other at the same speed. They're length-contracted to an identical shape, and should therefore collide at all points along the V simultaneously. Therefore neither is there an enclosed volume to trap fluid nor extra space for the fluid to escape. The same events happen in all frames, ie. all points along the Vs must collide. The main difference is that the events aren't all simultaneous in other frames. Yes, in one frame the point will contact first, and in another the ends collide first. When you add the fluid back, in all frames the outcome is the same: either the fluid is moved early enough so it can escape in all frames, or it doesn't have enough time to move in any frame. You can maintain a paradox by stipulating some impossible requirement. Eg. have the 2 Vs come to mutual rest instantly in multiple frames. Eg. accelerate a perfectly rigid body (including a volume of incompressible liquid).- Momentum and spacetime curvature
I think that applies to all elementary particles in Newtonian gravity. I think I've been tricked into showing that I was wrong all along. Even with a Newtonian approximation of GR, adding energy somewhere does not necessarily simply add a Newtonian gravitational effect there. Even when the added energy adds rest mass to a system, it might not behave as if the mass was added where the energy was. Back to the original problem of 2 masses, it's possible it works approximately well when the masses pass, ie. with the system treated as a particle that gains rest mass as measured from a distance. However, when adding energy to each of the 2 masses, the behavior is not the same as if you increased their individual rest masses. To try to get away from the Newtonian view, mass is not some kind of "stuff" that exists independently, it's just the measurement. Of course I'd rather have simple answers but I suppose that's best, especially when bad intuitive explanations can stick around for decades, being repeated in pop sci media etc.- Momentum and spacetime curvature
Definitely a Newtonian approximation has an error amount and it seems completely useless in some of these cases. An example I was thinking of is if you have 3 equal masses in an equilateral triangle say arranged in a v shape, and a test mass balanced in the center. Then you increase the masses of all 3 with the mass at the bottom very slightly more than the others. With Newtonian gravity the test mass would be expected to accelerate downward. But suppose that the increases of mass on the top 2 were from rotation, with each spinning opposite of the other so that the test mass would instead fall directly upward due to frame dragging. That might work fine or not??? because as explained in this thread the effects are non-linear. You can't just add up or overlay 2 separate frame-dragging effects. So there may be some additional energy in the interaction of the 2 spinning masses (it might even be that to get the test mass to move upward, it still needs greater mass increase "upward"???), or some other thing I can't imagine. I guess either "there is an answer given just the information above, but it's more complicated than suggested" or "the answer changes depending on factors other than just which of the 3 masses is given more rotation or energy." I have no idea which. This is an example of how we're talking about different things. Mass in Newtonian gravity is based on how it's measured, such as how a mass fits gravity equations. It's not based on a prediction of what Newton's laws say it should weigh. For example we wouldn't say "The Newtonian mass of an atom is a lot less than the measured mass of an atom because it doesn't include the contribution of binding energy." Geons have mass, and Newtonian gravity can approximate its effect. Suppose there's some elementary particle that has a measured mass, and behaves approximately as Newtonian gravity predicts. Say, for argument only, along the lines of Wheeler's speculation, that it is discovered the particle is actually a geon. We wouldn't say, "According to Newtonian gravity, this particle can now be considered massless and is predicted not to gravitate, contrary to measurements." Mass in Newtonian gravity necessarily includes contributions of things that are only explained by GR. Meanwhile mass in GR does not have a single definition that works perfectly in all cases. So I guess if we're talking about mass and assuming that GR is the best model of spacetime, we're necessarily mixing concepts that are not purely GR. But, we still use mass and can make sense of statements about it, I guess because there is no alternative that can better convey the same information as concisely and meaningfully. Well, this is all rather pointless, except that I think the concept that energy in the COM frame is equivalent to mass is generally useful even if it can be problematic. Sure, there's always energy in any given frame (even where it's too low to measure), but for beams with the same direction, none of it can be defined as rest mass, because there's no frame where the net momentum is zero. However... if the beams aren't infinitely long, adding a single massive particle "to the system" then lets you define a rest mass that includes the beams. What would happen if you simply included a mass? My first intuition about it is certainly wrong. With a mass, the COM frame doesn't depend on where the mass is. Would adding a small mass arbitrarily far away change the behavior of the beams? That certainly seems counter-intuitive. Or instead of a mass, you could just add another beam or pair of beams aimed in the opposite direction. If those were added very far away, would that make the 2 beams aimed in a common direction start attracting each other? I think the answer is no, even though the system now has a rest mass. Indeed, if you have some beams close together in one direction, and some far away in another direction, that's approximately a description of the original experiment, except with a change of scale. The 2 beams aimed in one direction should bend toward the beams moving in the opposite direction, not toward each other. So either intuition has broken down here, or the intuitive explanation is that it must be that when separate bits of energy combine to make up a rest mass, the density distribution of that mass is not necessarily the same as the density distribution of the energy that makes up it. The mass is located in some way between the oppositely directed beams, not inside the beams themselves. (By this point, I don't know if it's a better argument that intuition fails, or that intuition can be improved to make sense of GR, but the standard explanation involving how the pp-waves of the different beams superimpose differently, might be a more intuitive and more useful way to think about it than this, and might better suggest what kind of mass distribution should be expected.)- Momentum and spacetime curvature
I think my main confusion was taking this to mean that adding "mass" energy generally doesn't correspond with any "stronger" Newtonian gravity. I'm also not interested in Newtonian gravity except as an approximation of the predictions of GR. But as you pointed out, in my example the strong frame-dragging effect has no approximation in Newtonian gravity. It should be possible to contrive an example where the Newtonian forces are small, and the purely GR aspects dominate and give results that are the exact opposite of what Newtonian gravity predicts. Maybe, but in this example mass alone and the notion of energy having mass in the COM frame seems to work fine and fits with intuition. With 2 beams in the same direction, there's no COM frame. The energy of the beams is frame-dependent, and as total momentum approaches zero, so does the energy. Intuitively there is no energy that could make up "rest mass." With beams in opposite directions there is energy in the COM frame. True, that's not enough to guess how much they'd attract. I wouldn't even be certain the beams would attract (because of the uncommon geometry of the "mass"). As you've said such a notion of mass would be "problematic" in GR. But I think at least it's intuitive that the energy that doesn't vanish with a choice of coordinates, curves spacetime around the beams, similar to how any other equivalent mass would (and I don't know how to say that without using the word 'mass' even if it's just in a vague way).- Momentum and spacetime curvature
I re-read the thread looking for why this is not at all making sense to me. One thing I see is I was only talking about curvature far away from the masses, while OP mentions only the curvature at the point they meet. Other replies talk about both, and it's only the stuff about curvature far away that's not making sense to me. Another thing is I might not understand what is meant by a "source of gravity". I'll try again to rephrase OP's question to see if I can figure out where I'm going wrong. Say that you have two identical massive objects far apart from each other, and a test particle at rest at the midpoint between them. Then say you add a relatively massive amount of energy to one of the objects by spinning it. A very basic understanding of relativity is that the test particle will fall toward the object with more mass (energy), and that it is due to curvature of spacetime that it falls. Even though that additional energy is equivalent to mass, it is not what you're calling a "source of gravity"? Does this experiment make sense with what's already been said in the thread? The only other difference I can think of is that I'm talking about a huge contribution of kinetic energy and maybe other replies are talking about insignificant amounts? I'm also not sure if we're not in agreement about what would happen in an experiment like this, or if I'm just failing to understand the formal descriptions in GR.- Momentum and spacetime curvature
Then what is the correct way to describe their difference? Say you have 2 metrics, one for a system with little mass, and another for one with greater mass. The one with greater mass fits all intuitive notions of "stronger gravity", but how would that be expressed correctly? Greater mass corresponds with greater curvature, wouldn't the latter have greater curvature?- Momentum and spacetime curvature
I don't see how that is relevant so I wonder if we're even talking about the same thing? Where is my mistake in the following? All else being equal, a rotating object will have a higher invariant mass than the same object when not rotating. Curvature is related to invariant mass, in particular a higher invariant mass corresponds with a stronger Newtonian gravitational field. An object made of two massive bodies moving around each other, will have different spacetime curvature and "stronger" Newtonian gravity compared to the same bodies not moving (all else being equal). This answer's OP's main question with "yes." This also seems to contradict your statements. As far as I can tell, you're saying that because you can't plug the relevant values directly into the field equations, the result will be more complicated? But that won't change the statements above. That answer's OP's last question with "yes". I don't see how that matters. If you change the conditions that give you a different mass or different motion, you get different curvature, a different metric. Why would that imply anything about combining separate metrics?- Momentum and spacetime curvature
Yes, the rest masses don't change, but together the rest mass of the system includes the kinetic energy of the bodies in the system's COM frame. For example, the rest mass of an atom includes the (small) contribution of the kinetic energy of its electrons. This additional mass affects both the curvature around the atom, and the atom's Newtonian gravitational strength. No one's talking about doing that. I think a more appropriate analogy is comparing the Schwarzschild metric and the Kerr. If a non-rotating spherically symmetric mass (not necessarily a BH) is rotated, what changes? Its shape, and kinetic energy? But spacetime curvature is different, and so is its mass. Can you still say that the motion of the mass's components doesn't contribute to its mass or gravitational influence? It's similar for non-symmetric systems, such as 2 masses orbiting each other. Its mass and gravitational influence can include more than just the rest mass of its components. It's true for 2 orbiting bodies or for an atom or anything else that has a COM frame.- Momentum and spacetime curvature
I figure it is in this case. (There was another topic about this...) If you have 2 masses moving relative to each other, then there should be a center-of-momentum frame for them. In that frame, the kinetic energy of the masses contributes to the total energy of the system, and its rest mass. It should contribute to gravity just like any mass would? In the case where the two masses are at relative rest with each other, they're at rest in the center-of-momentum frame, so the system has no kinetic energy to contribute to its rest mass.- Recording or perceiving the activity of an oncoming object
They use "apparently faster-than-light motion" to mean the same as "apparently faster than c", so my earlier distinction between the 2 was pointless. I wish they'd distinguish though, because "faster than light" can mean "faster than light measured correctly (measured differently than the object)" or it can mean "can beat light in a race (measured the same way as the object)", and they mean only the former. Then people get in arguments and it turns out they're saying the same thing but meaning 2 different things. If an object can appear to move faster than c, then light can appear to move faster than c, and "light can appear to move faster than light" is confusing unless it's clear that it's implying 2 different measurements.- Recording or perceiving the activity of an oncoming object
Mostly no. Any light inside the event horizon will "fall" toward the singularity, but not all at the same rate (light sent inward will get there faster than light sent outward). No light from inside can go outside to the spaceship. Or, light from the spaceship can't go to places that light going past it can't... if we can't see what's in the black hole then either the spaceship can't see it or we can't see the spaceship. The event horizon passes over the ship at the speed of light, locally. It can "see" light that's part of the horizon, but it can't send signals faster than the light on the horizon. However once the spaceship enters the horizon, it could see some light from objects from inside the BH that fell in before it. Basically, it can fall toward the singularity more quickly than light directed outward does. Or, everything including light is moving toward the singularity, but you can catch up to light that is aimed toward you.- Recording or perceiving the activity of an oncoming object
Yes, that's correct. The relativistic Doppler effect includes delay of light and SR's time dilation. The Doppler factor period_received/period_sent = sqrt((1+beta)/(1-beta)) where beta*c is velocity, and for beta=-.6 the factor is equal to 0.5. https://en.wikipedia.org/wiki/Relativistic_Doppler_effect Yes it appears as if it's moving faster than c, but not faster than light, because incoming light appears to arrive at the same time it appears that it was sent! To look faster than that (v < -c), you'd have to see it arrive before you saw it leave, see it in 2 places at once, etc. The incoming (outgoing) object would also appear stretched (compressed) by the relativistic Doppler factor, so if something could approach the speed of light you could see it arrive just after it left, appearing stretched along its entire path, as well as bright and "very" blue. Sure, like if you consider an object falling into a black hole and sending out signals. A signal sent outward as the event horizon passes, becomes a part of the event horizon. Then if another observer falls in, it could see that signal as the event horizon passes it. I think that if both observers fell in at similar speeds, a photon sent exactly at the event horizon's passing wouldn't be redshifted, because where would the energy go? (It's not like the horizon itself stretches out.) I may be wrong. It seems, if you were both far above a black hole, and one went in, the other would see the one in front slow and redshift to the point of appearing stopped and invisibly dark. Then if the other followed, they'd see the one in front redshift less and less until appearing normal speed at the event horizon, and I think they'd then see the one in front blueshift increasingly redshift again and eventually "end" as there's some point where any "outward"-directed light from after that point will reach the singularity before you can catch up to it. That is, if the interior of the black hole has Schwarzschild geometry. (I think it would be blueshifted because the light you could receive would be increasingly later in the first object's life, it would have to appear sped up???) Edit: On second thought it seems that light would get "stretched" in opposite directions on either side of the event horizon, and it would immediately increasingly redshift as soon as the event horizon passed you. However, I'm trying to figure it out intuitively and could be way off. This is related to a topic that came up before, and I found it useful to see the paths of light on a Kruskal–Szekeres diagram, when talking about what infalling observers might see. https://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates Edit2: Okay looking at those diagrams again, the paths of light are 45-degree diagonal lines, now I remember what the diagrams showed: The second infalling observer could see images of the first observer from when they were "above" where the 2nd observer is now (higher r value). The 2nd observer would hit the singularity while still being able to "see" the first, but could only see the 1st's life up to some minimum r, whose value depends on how long the 2nd went in after the 1st. That implies the second observer would see the 1st redshifted inside the black hole.- Lorentz transformations btw rotating frames of reference
The transformations are between the coordinates in the respective reference frames, and I think you can/must choose the coordinates for Earth here. For a spherically symmetric rotating Earth with nothing else in the universe, the Kerr metric should work (an exact solution), and Kerr–Schild coordinates is apparently a good choice. I think this would be coordinates where the Earth is rotating, not a rotating frame where Earth is completely stationary, but I suppose an additional transformation between those could be used. But I think you could choose different metrics, including ones without exact solutions, and different coordinate systems that I think could include or not some of the negligible relativistic effects. For the transformation, would you basically derive the Lorentz transformation using Kerr–Schild coordinates? - How does ChatGPT work?
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