md65536

For example, in Schwarzschild coordinates, the rate at which light propagates at the event horizon is zero, but still the local speed of light at the horizon is c. Obviously both things can't be referred to as "the speed of light" and be used interchangeably. I used "rate" instead of "speed" because I don't think the latter is the right word to use here. What's the correct term for what you're describing (or for what Bjarne-7 is having a problem with?). I wouldn't use "speed of light" when talking about curved spaces unless whatever d/t that's referred to is the same as the local speed of light over d. Anything else I think needs to be described more specifically than just "speed of light."

If Bob made no mistake then he or she understands that a local clock is ticking at a different rate than a clock in empty space, and that light traveling through empty space at a speed of c according to a clock in empty space, is not going to give the same result when using clocks that are ticking at different rates. You measure the time of light's journey according to conditions all along the journey, not just those at the end. (In SR this would be different, because with everything assumed to be empty space, the conditions at the end are the same as conditions throughout the journey.)

It sounds like the question is only about gravitational time dilation, not about inertial frames or relativity of simultaneity. Bob and Alice agree that the local speed of light is c. They don't say "I can measure the speed of light in empty space using measurements I'm making locally in a gravitational well." Bob errs in doing so. Another way to say it is, "the coordinate speed of light isn't always c", however I don't know if "coordinate speed" (as in, measuring d and t using a distant observer's coordinates) is a standard term.

c is a universal constant that refers to the speed of light in a vacuum. It's not shorthand for "speed of light" in all possible contexts or coordinates. In a medium, light generally propagates at a rate lower than c, it doesn't change its value.

Look up those things and let me know what you want to discuss. I can't teach it.

I came to that same conclusion a decade ago. In figuring out that it's wrong, I learned the definitions of speed, time, and c. The speed of light doesn't have to be measured, because it is defined. There are other measures of rate of motion where the rate of (or approaching) light is infinite, such as rapidity and celerity. These also have definitions.

Right, which means that it's in the sun more than someone on the ground is. How much more is what I'm wondering. At 400km high, the ISS can be seen from 2300km away. That's more than a timezone at the equator, and more than 2 at the highest latitude the ISS passes over (51.6 degrees). So for example, the sun may have set for you an hour or two ago while the ISS above you is still lit by the sun. Or to think of it another way, if you were on the ground and had 12 hours of daylight, someone in a 400km-tall tower would get in the ballpark of 2 to 4 more hours of sunlight that day, which means 4 to 8 more hours of light than dark (very rough estimate). That's 58% to 67% of the time in sunlight, rather than 50%. I'm leaving out some important details, but how different is the correct answer from this estimate?

All the sources I see, repeat that the ISS spends on average about 45 minutes of a 90 minute orbit in darkness. I can't fathom why and I can't find any more accurate numbers. Only at equinox, would half of every orbit be above night on Earth. In summer and winter the orbits can vary over the day, as some orbits are in longer days and some in longer nights. However I'd like to ignore that variation and consider only the average times. First, the ISS is raised above the surface of the Earth, so it is in the sun more than the surface is. The higher a satellite is, the less it will be in the Earth's shadow. For example, the moon is in sunlight about 100% of the time. Second, the ISS orbit is tilted off the ecliptic. If a satellite is in a polar orbit (or better, orbiting over the edge of the polar circles in an optimal way), it can spend 100% of the time in sunlight for part of the year. Combined, using one very rough measure using a satellite tracking website, the ISS can spend about 5.25 minutes in the sun while the point on Earth below it is in darkness, on every sunset and sunrise. I think this measure must vary? But if this were the correct average measure, in a 93-minute orbit the ISS would spend 57 minutes in daylight and 36 minutes in darkness, or 61% of the time in light. This seems like an important difference from "about 50%" that I see quoted everywhere. Am I getting something wrong? Is there a more accurate calculation available? To me it seems like the 50% estimate is more misleading than a useful simplification. Why wouldn't they use a more accurate number, even if it's too complicated to explain?

Because you are. Drop down one dimension and the problem is similar to finding the area under a curve. Your questions about the "instantaneous fuel consumption" might be equivalent to asking what is the area of an infinitesimally wide line under the curve. I believe it was thinking along these lines that lead Newton to develop calculus, and that if you look into Newton's reasoning, it might help make sense of the fuel consumption idea without apparent paradoxes.

What is the cross section of a pipe filled with fuel, that you consume by moving along it instead of moving the fuel through the pipe?

However, the sand's momentum is gained while in free-fall, not contributing to the weight of the hourglass. As the sand builds up, the falling sand has less momentum, but there's also less sand falling. When the hourglass is started, there is sand falling weightlessly, and no sand hitting the bottom, and the hourglass briefly weighs less. At the end, there is sand hitting the bottom but no more of it falling, and it is briefly heavier. The start and end are similar to standing on a scale holding some mass, letting go of it, and then catching it lower down. https://demoweb.physics.ucla.edu/content/110-weight-hourglass But I see other links claiming experimental verification of different results.

That's not how SR works. The observations are reciprocal, not absolute. If the planet observes the cosmic ray's clock ticking slowly relative to its own due to relative motion, the cosmic ray observes the planet's clock also ticking slowly relative to its own.

Earlier in the thread, it was used like this: Given a set of events, the velocity of non-simultaneity is the distance between pairs of subsequent events divided by the time between those events. So for example if you have a train at rest with a bunch of synchronized clocks along its length, the events "noon" at each clock are all simultaneous and the "velocity of non-simultaneity" is undefined. In other frames, the events would propagate along the length of the train at a rate greater than c (the pairs of events are space-like). SR can handle particle velocities greater than c just fine, you just can't accelerate something from slower than c to c or faster, or vice versa. I guess a frame of reference moving relative to another at v>c doesn't make sense? So a composition of velocity formula for v>c doesn't make sense? Anyway, DimaMazin didn't you already calculate the rate at which events would propagate along the x-axis for a given reference frame velocity? You would just use the existing composition formula for that velocity, which is less than c, and find the "velocity of non-simultaneity" using that.

I think it's better to stick to "rubber sheet" instead of blanket, because the latter doesn't suggest as well that stretching is involved. If there's no stretching, there's only extrinsic curvature, and the analogy isn't as good. I think you could probably even demonstrate intrinsic curvature using a stretched rubber sheet without requiring extra dimensions.

Here's a few people you could ask if that's why they used the phrase: "within a certain region of space around it — nothing can escape its gravitational pull. Inside what's known as the black hole's event horizon, not even light itself can escape from a black hole." -- Ethan Siegel https://www.forbes.com/sites/startswithabang/2020/03/24/sorry-stephen-hawking-but-every-black-hole-is-still-growing-not-decaying/ 'Physical objects (those that move at or more slowly than the speed of light) can pass through the “event horizon” that defines the boundary of the black hole, but they never escape back to the outside world. Black holes are therefore black — even light cannot escape — thus the name.' -- Sean Carroll https://www.preposterousuniverse.com/blog/2020/11/26/thanksgiving-15/ "Ultimately, when the star has shrunk to a few tens of kilometers size, its gravity grows so enormous that nothing, not even light, can escape its grip. The star creates a black hole around itself." -- Kip Thorne https://www.its.caltech.edu/~kip/scripts/PubScans/BlackHoles-Thorne-Starmus.pdf Incidentally, I first looked up Misner, Thorne, Wheeler - Gravitation thinking they might have used the phrase there, and they didn't, and I couldn't find a case where Hawking used it either. The older descriptions of black holes introduced them as "collapsed stars", I guess because that was their only expected existence? They introduced them by describing the formation of the horizon, with a lot more to think about than just "nothing can escape (not even light!)". So I can see how that phrase is at least disappointing. However, I couldn't suggest an improvement, and I think simply omitting "even light" would be even less helpful.

I think it's more useful to include it than not to. It's not just aimed at people who expect light to escape from everything, but also those who wouldn't even consider that light might be related at all. That light--specifically--can't escape a black hole is a huge part of an average lay understanding of black holes.

The expectation that light ALWAYS escapes comes from "in my experience" here, not from the statement "Even light cannot escape!" At best the statement acknowledges that it might be expected that light wouldn't escape, not that it must be expected. But that justifies inclusion of "even light". I think it's more likely that someone already expects that light should escape, and the statement corrects that common(?) misconception, rather than that someone doesn't think that light is expected to escape until after reading the statement. This doesn't really matter though. The statements including "even" are correct, and succinct. I think it's a good way to describe black holes. I think it's useful for beginners to understand that BHs involve spacetime curvature, and I think it's unlikely that anyone who already understands is going to be misled by the word "even".

Typically laypeople would think of inability to escape in terms of strong gravity and its effect on masses, so it is reasonable to emphasize that it also applies to light. I don't think the writers needed to worry, "this word might confuse people who already understand this."

The phrase "not even light can escape a black hole" is correct. Do you find it confusing? The statement doesn't imply that light is expected to escape from everything, or that if IT cannot escape then NOTHING can, and I doubt many others have that confusion.

You wrote "I feared as much" but I think you got it close enough to not fear. Here, Genady has drawn the world lines of the different particles, onto the same Minkowski diagram. The diagram, and the coordinates on their grid, represent the measurements for one particular inertial observer (aka. inertial reference frame). You can draw the same worldlines on a Minkowski diagram for a different observer, and the subluminal "time like" world lines will be at different angles. If you make a physical mark at some fixed location in a this particular Minkowski diagram, and extend it through time, you get a vertical line like the Moon's. So this diagram represents the rest frame of the moon. If your mark is at rest, all events at that mark have the same spatial component (but they have different time component), and the spatial distance of the pairs of events you described will be the same in this particular frame's Minkowski coordinates.

Surely you're joking. What do you call the female equivalent of an American?

There are several answers, but why wouldn't "You're Teresa's daughter" be the simplest assumption? It minimizes the number of people and relationships being talked about. "Mother in law" adds an assumption of marriage, which at least is a cultural bias an AI would need to be trained for.