md65536

1 hour ago, MigL said:

Light does not stand still, even inside the EH, so there is no 'catching up' to the light emitted by your feet.
As soon as your feet cross the EH, light emitted inexorably moves towards its future, the center of the BH.
There is no path backwards to your eyes, and you cannot catch up to it.
So what do you see ???

Regardless of what happens inside the BH, the original scenario doesn't involve any light from inside the EH whatsoever. It doesn't say anything about what happens after some moment when the astronaut's head is less than a body's length from the EH, still outside, still only receiving light from outside the EH.

But, I think you're also wrong about what happens inside the Schwarzschild BH. Light inside can still move in different directions. Draw some light cones on a diagram. Yes, all light cones inside the BH will be tilted so that all light paths head toward the singularity, but the light cone from an event inside the EH can still intersect the worldline of an infalling particle. The lightcones do not instantly tilt an extra 45 degrees or narrow and degenerate into a single line, the instant you cross the EH.

56 minutes ago, geordief said:

Is the tiny distance between the head and the feet a misleading example?

Would whatever is seen,(or not seen) be  essentialy the same as if the distance was the same as that between any object and an observer  say 1,2 or 1000 light seconds away?

So the object crosses the EH in its own timeframe  and the observer sees it up until it crosses the EH in the observer's timeframe.

Basically. Any observer outside the EH will see the object approach the EH but never appear to cross it. The exact measurements will be different for different observers. My example was basically two observers who both fall in to the BH.

But sure, you could make the head and feet two ships arbitrarily far apart, and just make the BH big enough. Actually this thought experiment started from trying to figure out what would happen if two ships at relative rest were flying far from a BH so large that they couldn't detect it, such as J.C.MacSwell suggested, but if one ship ended up inside the EH and one was outside. I think either both ships enter the BH and don't notice, or the ships necessarily separate (head rips off), as the outside ship has to escape the inside ship if it is to escape the EH.

It causes problems thinking of the BH as a giant sphere whose event horizon can hover harmlessly between the two ships, or between head and feet. It is a lightlike surface. If the head and feet see each other, or the ships are in communication, the EH between them is moving as fast as they can communicate.

I agree with Halc. When the feet cross the EH, the photons making up the image of that, and directed outward, forever remain at the EH, which is a lightlike surface. It's stationary in the very distant observer's coordinates, but locally moves at the speed of light. The astronaut notices nothing unusual because the EH and image move past its head at the speed of light, just like photons from the feet do in usual circumstances.

An astronaut falls into an extremely large Schwarzschild black hole, so large that they don't notice any spaghettification-like effects. Their head can see their feet the entire time, and their helmet is constantly sending information to Earth, say. At some point, the astronaut sends a message basically saying "My feet are now inside the event horizon", and Earth eventually receives that message.

Is there a mistake with this scenario? What prevents the astronaut's head outside the event horizon from seeing their feet inside the horizon, and how does that look to them if they're watching their feet the entire time?

I have an answer in mind but I'm curious if there's any room for experts to disagree on a resolution, and also curious if fellow amateurs can resolve it with about a pop-sci level understanding. I'm interested in how different the answers might be, so please share if you have an opinion!

6 hours ago, dimreepr said:

That's why I believe Einstein could become a prophet like deity.

Yes, I think you're proving your point. If you don't understand something, it can seem god-like. If you read what Einstein wrote and understand even just parts of it, it's easy to see that it's basically a set of assumptions that match observations of reality, and some mathematical consequences of those, whose predictions also match observation. There's nothing god-like about it. But if you don't read it, it's an unknown, and it's already been discussed that people tend to attribute what's unknown to gods.

So it sounds like you're assuming that Einstein's work will become an unknown, but that the stories of Him or His work will survive and be passed on (with language, I guess?). But the same can be said about anything that's known. Science can explain lightning, but if everyone ignored and forgot that, then Zeus could regain popularity.

2 hours ago, fredreload said:

You are beaming the Near Infrared Laser of 780nm to inhibit sodium atoms from emitting infrared energy as heat energy while having it absorb light of a different wavelength so it keeps getting hotter and hotter until it turns into a BEC or possibly a black hole.

Why do you think it would turn into a BEC if hot? A BEC involves the atoms being in the lowest quantum states, not highly energetic ones. The laser cooling described in the thread is done to take energy away from the atoms.

A layman explanation of a BEC (and possibly completely wrong, others please correct me) is that when a particle's speed is very close to zero (ie. very low temperature), you know the momentum of it to high certainty, so due to the uncertainty principle, you can only know the position of it to lower certainty. With a cloud of particles acting as if the particles are each spread out, they overlap each other (their wave functions overlap) and are indistinguishable, and the cloud behaves in ways that are like it's one macroscopic fuzzy particle (for lack of words), instead of behaving like it's made of many microscopic particles. If the atoms are hot (in aggregate), I don't think there's a practical way to force the positional uncertainly that you need for these effects.

6 hours ago, dimreepr said:

You're kinda proving my point...

I'm not seeing your point at all. Maybe you can explain, why would a religious person not believe in atheism? Is that similar to what you're asking? Can you describe the analogous point to the one you're making from that perspective, or if it's not analogous, what's the difference?

18 minutes ago, dimreepr said:

Most athiests only attack the concept of a god, rather than a critique of the human understanding of what is written, and understood at the time a religion is inspired.

Most? Do you have data on that?

Why would we not believe what neanderthals believed? We could try to understand what they did in the context of the world they lived in, but does that mean we should share their beliefs?

I lot of the beliefs of a religion have a god as their core. If ancient people ate raw pigs and got sick, but they didn't know why, they may explain it as "God does not want us to eat raw pork!" Now we understand why you would get sick, and don't need the concept of a god to explain it. It's not like atheists say "I don't believe in god, so I won't believe there's anything wrong with eating raw pork!" Atheists and religious people and neanderthals share a lot of the same beliefs. However, if you believe in something, and can't separate the concept of god from it, but are an atheist, I think that means there's a contradiction in your beliefs. As it is, there is nothing in reality that is inconsistent with the belief that no supernatural god exists.

8 hours ago, studiot said:

Certainly my thoughts are that if you allow scattering before the impact of the photons ther is no point in using a laser

A laser is practical for other reasons, like being single wavelength, high intensity, easy to direct, and needing a small window into the vacuum chamber. If you had some other setup and a single-wavelength light source that directed photons into the atom cloud, but the photons came from different directions, that would work too. I think all that's needed is enough photons with a) right wavelength b) aimed at the atoms c) symmetric or balanced distribution of directions.

8 hours ago, studiot said:

Further I don't see that scattering after the impact is relevent since any change to the target's momentum will already have occurred.

So why is scattering not a red herring ?

If you're talking about fredreload's ideas, I don't agree that any of them improve on what's already been done. If you had a way to scatter light into the cloud to get a distribution of directions, that's great, but using mirrors and multiple laser beams seems much easier.

Scattering after the impact is basically how the atoms are cooled. As per the quote provided earlier by fredreload, from https://www.sciencedirect.com/topics/chemical-engineering/laser-cooling

Quote

The absorbed radiation is quickly emitted by the atom, either through stimulated emission or spontaneous emission. Stimulated emission occurs in the same direction as the absorption, and the recoil effect accelerates the atom back to its original velocity so that the absorption and emission effects cancel. Spontaneous emission, on the other hand, results in a random emission of a photon (and hence recoil of the atom) in any direction, with the net effect that, on average, the atom slows down in the direction of the absorption.

So to recap, "in the direction of absorption" is why you need photons coming from multiple directions, and you need a wavelength slightly longer than the atom's resonant frequency so that the photons are only absorbed by atoms moving toward the light source (blue-shifted to the resonant frequency). Contrary to what fredreload seems to suggest (I may be misinterpreting), you need the atoms to release that energy to get them to a lower quantum state. If they only absorb, I suppose you could still cool them, but they won't form a BEC by being in their lowest quantum state.

1 hour ago, studiot said:

How do you address my point about two of these being irrelevant ?

Which two are irrelevant?

Do you mean when you wrote, "Do you not realise the basic mechanics that momentum directed along 2 of those 6 directions cannot affect spin?" I'm not talking about affecting spin, I'm talking about slowing of atoms that are moving toward the laser, ie. laser cooling. If you have 4 lasers aimed at +x, -x, +y, and -y, then atoms with movement in the z direction will not be cooled effectively.

I think you don't need lasers aimed in opposite directions, but merely a symmetric distribution of photon directions, to avoid propelling the cloud. 4 lasers arranged like a triangular pyramid might work, but probably not as well. Maybe it's better to use more than 6 lasers, surely with diminishing return. 6 is not "magic", but probably the simplest arrangement needed to get good results.

48 minutes ago, swansont said:

That what lenses are for. You can expand and re-collimate the beam.

Sorry, I (and fredreload it seems) thought you meant the photons had to have parallel directions, not just that they're lined up with the target.

2 hours ago, swansont said:

You want the light going to the atoms in a collimated beam, so what does scattering it in all directions get you?

Scattering seems fine to me. You want photons coming from different directions, because the atoms are moving in all different directions. As long as the wavelength is right, it can slow an atom moving in the opposite direction of the photon. Is that wrong? Why do you want it collimated?

Scattering would be less efficient if the photons aren't going through enough of the cloud. You'd use lasers for the single wavelength, focus, and high output. The light being collimated is actually a problem---you need the light coming from different directions---which is solved by using 6 lasers.

The spontaneous emission implies there's a lower limit to the average velocity of an atom that you can achieve, so a limit to the cooling. Even if absorption brought an atom to complete rest, re-emission would propel it in a random direction.

1 hour ago, fredreload said:

Well, visible light/flash light is 780nm, precisely the range for lighting up rubidium atoms. Although flash lights are not as precise and coherent than lasers. But you could create a magneto-optic trap with flash lights.

Why might a laser be preferred over a flashlight? What are some problems you might run into using a flashlight? By the way, how many atoms are you cooling? How dense is this cloud?

1 hour ago, fredreload said:

Alkali atoms also absorb a wide range of electromagnetic radiation including x rays.

You could also try a microwave oven, to bombard it with microwaves. You could add a toaster, and let it absorb infrared. But will that cause spontaneous emission like in what you quoted? And how do you get the atoms moving toward these heat sources to preferentially absorb the light? Why does resonance matter, in what gets absorbed and what is emitted, if the atoms can simply absorb light of many different wavelengths?

6 hours ago, studiot said:

I have absolutely no idea how this would happen, but isn't that exactly what the word condense means?

Isn't that what happens when a gas condenses to a liquid or a liquid to a solid ?

A new form of PE is aquired - surface energy.

It's a different state of matter, but the meaning (according to wikipedia) comes from, "Einstein proposed that cooling bosonic atoms to a very low temperature would cause them to fall (or "condense") into the lowest accessible quantum state". But even if it's analogous to condensing from gas to liquid to solid, those are typically going from higher energy states to lower energy states. Eg. water releases energy when it freezes, it doesn't absorb it. Anyway, a BEC involves atoms being in their lowest quantum state; they must get rid of that energy to form a BEC.

32 minutes ago, swansont said:

Yes, the doppler shift will be changed slightly. That’s why you need millions of photon scatters to slow an atom to close to zero velocity - the imparted momentum from a single photon is small. p=E/c, and c is a big number.

As a toy model just to describe where the energy's going, is the following a reasonable description?

Say you start with one million photons moving to the right, and one atom moving to the left.

After absorption and re-emitting, you end up with one million photons, scattered in a spherically uniform distribution of directions. The re-emitted photons would have variance in Doppler shift (because they're not necessarily immediately re-emitted?), but the photons re-emitted to the left would be blue-shifted on average, and the photons directly to the right would have no shift on average (basically, if a photon enters the atom from the left, and is re-emitted to the right, that photon hasn't changed the atom's final speed).

Also you end up with an atom with near zero velocity. The total energy of the system is the same. I would guess that nearly all of the kinetic energy of the atom at the start, goes into the scattered direction of the photons in the end, and a tiny portion of it goes into the Doppler shift of those photons?

2 minutes ago, studiot said:

Unless the cloud can convert KE to PE somehow, since temperature is a function of KE, not PE  don't you think ?

I assume it doesn't. If it did how would it do that? Atoms in a higher energy state? Even if it were possible, it couldn't be done forever. That would mean the more time spent cooling the cloud, the more energetic it would get.

1 hour ago, fredreload said:

Meaning they accept and release a photon so that uses some energy of the laser I think.

Sure, but what is meant by "use"? The energy is not reduced through use.

Naively I can think of it like, photons are absorbed by atoms moving toward the laser, and the kinetic energy of the two partially cancel each other out. But really I think it's more like, an atom absorbs a photon with energy E in the atom's rest frame, changes speed in the process, then emits the same energy E (as one photon? or several over time?) in its new rest frame. In the lab frame, the energy absorbed is less (on average???) than the energy emitted due to the atom's change in speed. --- However, if the light is re-emitted in a random direction, I'm not sure this idea makes any sense.

The issue is, the real cooling corresponds with energy coming out of the cloud, not energy going in.

1 hour ago, swansont said:

But it’s probably all for naught, because a 1 meter cloud of atoms will be optically thick, meaning the laser light won’t penetrate, and the re-radiation would cause heating. You would likely not be able to get a 1m cloud of cold atoms to begin the process of forming a BEC.

So basically, if you can start cooling the edges of the cloud, more laser light penetrates further into the cloud, but the further in, the less chance the re-emitted light has of escaping the cloud? The cooling effect occurs because of absorption, but it still requires the energy being put into the cloud to escape? The initial absorption reduces heat energy, but since you can't just increase the non-thermal energy of the atoms indefinitely, it's going to have to escape the cloud or end up increasing the heat.

Does this mean that the laser cooling process is effectively using light to "poke" the atoms just right so that they emit more light energy than they absorb, so that more light energy goes out of the cloud than what goes in?

PS. as a spectator up till now, I appreciate the patient informative responses. Replies that focus on discouraging amateurs don't help others who are reading.

8 hours ago, Commander said:

There are 2 Solutions known. You need to find both !

It's a bit confusing because it's not a valid sentence with all of those question marks, but if they're changed to commas, one solution is

Is this the easier solution?

Oh I see, just by trying some things out:

I was curious about how many answers there might be, so I wrote code. I gave up before trying to deal with parentheses, but got the following:

(11 - 17) * 13 + 2 + 19 + 101 = 44
(11 - 13) * 19 - 17 - 2 + 101 = 44
(13 + 101) / 2 - 11 - 19 + 17 = 44
(11 - 13 - 17) * 2 - 19 + 101 = 44
(13 - 17 - 19) * 2 - 11 + 101 = 44
(17 - 19 - 2) * 11 - 13 + 101 = 44
(17 - 19) * 11 * 2 - 13 + 101 = 44
(13 - 11 + 19 + 101) / 2 - 17 = 44
(11 - 17 + 19 + 101) / 2 - 13 = 44   (my answer)
((17 - 19 + 101) / 11 + 13) * 2 = 44
((13 * 17 - 19) / 101 + 2) * 11 = 44
((19 - 11) * 17 - 13 - 101) * 2 = 44
(13 * 17 - 19) / 101 * 2 * 11 = 44   (Commander's answer, ignoring order)

Sensei's answer isn't here because it's not left-to-right order of operations. I wouldn't doubt this is a small fraction of the possible answers, but neither would I bet that it is. (I also manually culled duplicates so I may have removed too many.)

I got (101+11+19-17)/2-13=44.

On 1/20/2021 at 1:23 AM, Markus Hanke said:

As I said previously, my response was based upon my own understanding of the OP. If it completely missed the point, then it is the OP’s job to clarify things.

Okay, but I'm trying to understand a statement that you made, because it makes no sense to me and I'm trying to figure out where the misunderstanding is.

You wrote, "He’s essentially using an inertial coordinate system to show that there is no acceleration - which is trivially true." That makes NO sense to me, because you can describe a particle that's accelerating, using an inertial coordinate system. The coordinate system has nothing to do with whether the particle is properly accelerating or not.

Eg. Consider an inertial observer on a train bank. A train accelerates from rest at a rate of 1 m/s^2, relative to the observer. The observer's coordinate system doesn't trivially show that there is no acceleration. Am I using the term "coordinate system" incorrectly? What I don't understand is if one can refer to a particle, and its own coordinate system, as the same thing, or if you and Anamitra are talking about different things. If I have an inertial observer and a train accelerates, I don't think I can sensibly describe it as "an accelerating coordinate system". I'm trying to find out what you're saying is trivially true.

9 hours ago, Markus Hanke said:

He’s essentially using an inertial coordinate system to show that there is no acceleration - which is trivially true.

No acceleration of what? It originally mentions and later clarifies that the acceleration of "particles" is being discussed, and that's not trivially true. If you're assuming it's reference frames being accelerated, where is that stated?

This topic is confusing from the start, with seemingly some explanation or details missing?? It's difficult for me to follow if I first have to guess at the same assumptions being made, even if they're reasonable.

9 hours ago, Markus Hanke said:

Ok, but then, what is his point? He starts with a metric in Cartesian coordinates, then manipulates it using relations that imply an inertial observer, and ends up with the conclusion that there is no acceleration...?

Yes, that seems to explain what is going on, thanks. I did not get that from the discussion so far. Basically is it: specify a particle that has no acceleration, conclude that it cannot be changing speed?

(I'm guessing then the main flaw in reasoning is that OP starts with equations of inertial motion, but is treating them as "the equations of general motion in SR"?)

8 hours ago, Markus Hanke said:

It’s how I understood it, based on the fact that the metric given is of a form that would generally be used by an inertial observer, so it is natural to assume that these are Minkowski coordinates, and not hyperbolic ones. The geodesics calculated from his metric ansatz are straight lines, not hyperbolas - unless the coordinate basis is not Cartesian, but the OP never indicated that. This is also consistent with the OP’s conclusion: “[...] we see that the particle cannot accelerate”, which is of course trivially true, based on that metric.

Then I still don't understand. Particles *can* accelerate, and their motion can be described in the coordinates of an inertial observer. Why do you need an accelerating observer to describe accelerating particles? It can be described in Minkowski coordinates, why suggest Rindler? Not all particles are observers; to me it looks like where OP is talking about particles, you're describing observers.

"in this article we see that the particle cannot accelerate." (emphasis mine) I understand to mean that we're talking about the case of constant gamma, ie. we're limiting ourselves to particles that don't accelerate.

If you two are really on the same page and understanding each other, then I apologize.

(Besides all that, a particle can accelerate and maintain constant speed relative to an inertial observer, thus constant gamma in that observer's inertial frame but only changing direction, while the particle undergoes proper acceleration, so I think I disagree with both of you however your statements are understood!)

14 hours ago, Markus Hanke said:

I was referring to the fact that, if an accelerometer co-moving with an observer measures something other than zero, then that observer cannot be inertial by definition. I understood the OPs comment to say that a particle can be in a state of proper acceleration, yet still be inertial - hence my comment.

That's not what OP literally said, and you've interpreted what OP wrote as nonsense, but still I don't know what OP really meant because it could be interpreted different ways. I don't think you two are talking about the same thing, at least in some cases like this, and I don't see how the problems can possibly be resolved if you're not even talking about the same things. I think OP needs to clarify first.

Eg. in this case, Anamitra were you talking about general particles, as they are measured in various inertial frames of reference? Or particles in their own rest frames, being inertial in some inertial frames of reference and accelerating in other inertial frames of reference? Or something else?

I think some consider a particle being "in" a given frame to mean that the frame is its rest frame, but I take it to mean the particle "as measured in" a given frame. Eg. a train can have a positive speed in the bank's frame. The train is in all the different frames, not just its rest frame. Is that at odds with your interpretation of OP's statement?

10 hours ago, Markus Hanke said:

Not true. If a frame is inertial, then by definition there is no proper acceleration. There can, however, be coordinate acceleration, but that is an artefact of how we choose to label events, and not something that a physical accelerometer would measure.

I'm not following this. An accelerometer that is measuring proper acceleration can be described in the coordinates of an inertial observer.

My reading of what you replied to, is "An inertial observer remains inertial as measured in any other inertial reference frame, but a particle can (properly) accelerate as measured in an inertial reference frame." It only mentions particles being able to accelerate. What are you referring to when you say "there is no proper acceleration"?

On 11/29/2020 at 2:12 PM, MigL said:

You are aware that you actually don't 'consume' water, but simply recycle it ?

If you 'consumed' 1 gallon of water everyday, would result in your weight gain of 2900 lbs after a year.

Actually humans create water. How much more you'd weigh depends on what part of it you keep. Plants consume water but they don't generally hold on to the oxygen.