md65536

2 hours ago, swansont said:

The ISS is only about 400 km above the earth, which has a radius of about 6400 km.

 

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?

14 hours ago, Arthur Smith said:

Well, I guess a histogram would do it. Start with a graph of horizontal scale 100 km and rectangles of one millimetre. Height of each will be proportional to the fuel used over the distance interval. Play with scales and you have a representation of fuel consumption over distance as the area under a line following the height of the rectangles. Why do I feel I'm differentiating?

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?

3 hours ago, swansont said:

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The falling sand is exerting a force on the bottom of the hourglass. This decreases as the sand build up, as the falling sand has less momentum when it hits

 

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.

4 hours ago, TheVat said:

Well, the cosmic ray's clock is moving more slowly relative to an observer on, say, a nearby planet or space station.  So if the ray were passing through an object of negligible mass, then it would observe that's object's clocks as moving faster, therefore cooling faster.

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.

10 hours ago, DimaMazin said:

We should create law of addition of non-smultaneities. We can image any velocity(and bigger than c)  as non-simultaneity.

9 hours ago, Genady said:

I still have no idea what the question, "What is velocity of non-simultaneity?" means.

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.

56 minutes ago, Steve81 said:

Seems a great deal like my blanket and weights idea 😀

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.

4 hours ago, Genady said:

I don't know if there are data to support this statement, but I'm ready to believe it. I am ready to believe it, because this phrase is repeated so often, that if they remember anything about black holes, then they remember it.

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.

4 hours ago, Genady said:

Anyway, I think this phrase is just a popular cliché.

 

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.

34 minutes ago, Genady said:

Aha! To somebody, it does:

 

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".

24 minutes ago, Genady said:

What is unclear to me is a role of the word "even" in it.

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."

On 7/31/2023 at 2:30 PM, Genady said:

My question is, why they use the phrase "even light" as some kind of extreme, as if light is expected to escape from everything and everywhere? What is it about light that if IT cannot escape then NOTHING can? (I am not asking about the physics of it, but about the use of this phrase in the layman descriptions.)

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.

3 hours ago, geordief said:

The first pair of physical events is  a massless object fired,* from the origin through a vacuum  and leaving a physical mark ** at the site of the second point on the Minkowski  map.

The second pair of physical  events is a spaceship travelling at c/2 and arriving at the site of the mark made by the  massless object.(at a different time and spatial distance)

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.

5 hours ago, Genady said:

ChatGPT has handled it very well:

Surely you're joking.

What do you call the female equivalent of an American?

On 6/11/2023 at 5:23 AM, Genady said:

If Teresa's daughter is my daughter's mother, then what is my relationship to Teresa?

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.

On 6/11/2023 at 1:52 PM, npts2020 said:

 

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Seems to me they could start a pile and keep adding to it until one of them will accept it as their 1/5 share and repeat the process until the last two where; "one cuts, the other chooses". (if 2 pirates want the same pile and there isn't equivalent items to make an identical pile, repeat process using different combinations of items until only 1 accepts)

 

6 hours ago, Genady said:

What is the minimum amount of loads they will need to get to the next system?

6 hours ago, Genady said:

Is there a limit to the distance between systems that they can cross?

On 6/4/2023 at 4:26 AM, Jez said:

1. Rough out 1/5 of the loot, and ask if anyone wants it, if no-one wants it then keep adding to the pile until someone takes it, or if everyone wants it, keep removing pieces until there is only one pirate asking for it, then they get that.

What do you do if there are 2 pirates who want it, and then neither wants it after the adjustment, or vice versa?

On 6/4/2023 at 4:26 AM, Jez said:

2. Roughly halve the remaining pile and the two pirates standing next to a given pile will spilt it like for two people. If there are 3 or more pirates standing next to a pile, move the treasure piece by piece to the other pile until each pile has two pirates who want to share that particular pile.

What if there's 3 pirates on one side and 1 on the other, and you move some treasure and two (or three) pirates move to the other side? How many adjustments will it take to guarantee there are 2 on each side?

4 hours ago, Genady said:

Here is the first step of the procedure, if you're interested:

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A pirate takes from the treasure chest and put out a pile that he is willing to take as his share.

(I.e., he does not split the treasure into five parts.) 

 

7 hours ago, Genady said:

Aha, got it. Still looking for a counterexample. Let's try this scenario:

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B wants U or W.

C, D, E each want U or X.

Seems OK for A and B to take U and W between them. C, D, and E now need to split X, Y, and Z. 

But X+Y+Z might be < 3/5, and somebody will not be satisfied.

Am I right?

 

Yes, that's right. I'm assuming that if someone wants a share, and someone else takes it, and the former doesn't also get a share that they're satisfied with from that same division of shares, then they're dissatisfied. Like a child saying they wanted the purple-flavour yogurt that their sibling took and the parent opening a new purple-flavour yogurt and trying to argue that it's the same amount... it's hopeless! Only their own judgment can determine if they're satisfied.

13 minutes ago, Genady said:

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Let's try the following scenario. The pirates are A, B, C, D, E. A divides the loot into 5 shares: U, W, X, Y, Z.

A is willing to take any of them or let go any of them. B is willing to let only U go. C is willing to let only W go. D is willing to let only X go. E is willing to let only Y go.

Seems to me that there is no "a subset of n pirates for which n shares can be divided to each of their satisfaction, with the remaining pirates satisfied to let all of those n shares go." Do I miss something?

 

11 hours ago, Genady said:

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I still think that the conclusion,

is correct.

What I wrote is ambiguous...

The solution I'm working on seems to work, but I'm having trouble proving it:

2 hours ago, md65536 said:

Therefore I think that no solution that involves the piles being split up, and then allowing one of them to be kept, can work. I think any solution that involves one person splitting up the loot, would require that the shares be adjusted before guaranteeing that they all agree.

Nope, I'm wrong again! This is bad reasoning.

Lol, this puzzle has broken my brain. Every time you say I'm right I disagree. Every time I realize something I just said was wrong, I forget some other detail and say something else wrong, all the time flipping back and forth between 2 incorrect positions.

What I forgot is that is that it doesn't matter if someone is not satisfied with letting some other share be taken, IF they're also taking a share that they're satisfied with at the same time. So it might (should? but I don't want to give any more answers unless I figure it out) be possible to have someone split up the treasure, and have either everyone satisfied with the shares everyone else is taking, OR they're taking a share they're satisfied with. (However, my previous solution still doesn't work, sometimes not fulfilling either condition.)