Strange

Until you can show, in mathematical detail, that "electromagnetic loop fields" are able to explain all the evidence, then you might as well invoke unicorns. Currently, only dark matter as a form of non-interacting matter is able to meet all the requirements.

Are you saying that there is a difference between substituting the values φ=π2λ=0α=π in three dimensions and two dimensions?

It would be if you look towards it. But, fair enough. I was just trying to draw a distinction between a filter that just absorbs some rays (so you can still see the source, just as you can without the filter) and something that absorbs all the rays (so you can no longer see the source).

It has to explain, consistently, the motions of galaxy clusters, the velocity curves of galaxies, gravitational lensing, the Bullet cluster, the patterns in the CMB, large structure formation, and so on. Until you can show, in mathematical detail, that your ... erm... "model" is able to do that, then you might as well invoke unicorns. Currently, only dark matter as a form of non-interacting matter is able to meet all the requirements.

You can always see the light source. It just becomes dimmer (and safer in the case of lasers or the sun) with a filter. But the filter doesn't make it visible. And if you block all the rays, then you can't see the light source at all. This is why you can't see then sun when you are down a coal mine. Or at night.

The "sky" rays have been (partially) absorbed, so the sky appears black. (Because of the dynamic range of the sensor.) The sun rays have also been (partially) absorbed, to exactly the same extent, so the sun is darker than it would have been without the filter. If "all the sun rays" had been absorbed then you wouldn't be able to see the sun, would you? If you took the same picture without the filter, but with the exposure adjusted to get the same brightness image of the Sun, then the sky would still appear black. (Because of the dynamic range of the sensor.) --- But do not do this as you are likely to damage the camera.

Good point!

But does it, that's the question.

And that may be a valid interpretation. I don't know whether there is a general solution to the equations in those terms but as the Schwarzschild solution can be applied to a body like the Earth, as well as black holes, then there are at least some cases where it is valid. P.S. The Baez pages explain why, in one case the equations describe expansion and in another they describe "attraction". P.P.S. Remember Joan Baez? Cousins.

Well, the field IS space. (And time.) Distances and times. Surprisingly. BTW You might find John Baez's pages on the Einstein Field Equations helpful. They do go into the maths in a little detail but most bits are understandable even without fully understanding the maths. http://math.ucr.edu/home/baez/einstein/

Yes. Space-time coordinates.

Space isn't "stuff" so it doesn't need to be annihilated or compressed. We are just talking about coordinates "flowing" into the black hole. As the Gullstrand–Painlevé coordinates are exactly equivalent to Schwarzschild coordinates (or any of the other possible choices) it can't give results that are different from any other choice.

Well, it does "feel" like it should be true. But that is why we use science! And, as studios says, it will probably be true for oil and grease.

I think the Gullstrand–Painlevé coordinates show that it is a valid concept, at least in the case of a Schwarzschild solution (a spherically symmetrical, non-spinning object). Whether it can be applied more generally, I don't know. ... Apparently, it can: https://arxiv.org/abs/0905.3244

Read that as it being either a BH (with an EH) or a neutron star (no EH). Surely, if you have an event horizon, then it is a black hole, by definition?

I'm fairly sure I read an article recently about research showing that hot water was no more effective than cold at removing germs. I will see if I can find it again... Here http://www.bbc.co.uk/news/health-40118539 As they say, it is more important how you wash them.

It doesn't. Very small black holes can evaporate. But, as far as we know, black holes like that don't exist.They could, hypothetically, have been formed in the early universe and some could still be around. People have looked for the characteristic radiation or gravitational lensing that we would expect to see and no evidence has been found. MACHOs are one candidate for dark matter. But as with other candidates they have not been detected. They would also be insufficient by themselves, so some amount of non-baryonic dark matter would also be required. And there is no evidence to say they are the source of the Big Bang. Interesting paper. But it doesn't, as far as I can see, say anything about the internals (inside the event horizon).

Not necessarily. Space doesn't have to be Euclidean, and isn't in the presence of mass.

As there are dozens of alternative gravity (and other) theories being explored, that doesn't seem to be true. Unfortunately, one reason that there are so many is that none of them work: they just don't fit all the evidence. An unknown form of matter does, which is why it is still the favoured explanation. As Katie Mack puts it, "Dark matter: still the worst explanation. Apart from all the others."

Both the laser and the Sun are always visible. The filter just makes them dimmer. Great. Let us know the results.

OK. Maybe we are getting somewhere now. This relates to the difference between your use of 5 lasers and photographing a general scene. So you are comparing the loss of the background with the addition of the filter, to the loss of some of the lasers? This is not a valid comparison. In the case of the Sun you have one bright source surrounded by an illuminated sky. Both the Sun and the sky radiate light in all directions. (Unlike the lasers.) The Sun is many times brighter than the sky (a few million times brighter, I think). This means that if you put a filter in the way, the sky will rapidly drop below the level of detectability, while the Sun is still visible. (This would be equally true if you achieved the same exposure by using a very short exposure time and a very narrow aperture - it is not related to the filter itself, just the relative brightness of the Sun and the background.) In the case of your 5 lasers, there is almost no other source (I am assuming we are doing the experiment in the dark, as in the video you posted). Therefore the only beams are the (unidirectional) beams from each laser, creating the images of those sources. The brightness of these will all be changed equally by the filter. So there is nothing that will disappear or appear. It s just the brightness of the image that will change.

The beams will be partly absorbed and hence dimmer, and so will the image formed from them. It won't change the path the beams take and so they will still be blocked by the diaphragm. An image of the sun with or without a filter will be identical other than brightness. The filter is only used to reduce the brightness (so the film doesn't catch fire!) BTW that is a very odd picture of the Sun. Where did you get it from?

What is the difference between a beam being visible and the light source being visible? The light source is only visible because there is a beam of light from it. And why would the presence of a filter change whether the beam was visible? It could only make it dimmer. (Are we back to polarisation, by any chance?) Indeed. Feel free to do that. Just in case (and feel free to be offended by this) ... you do realise that the reason you can see the beams from the lasers in the video you posted is because they are shining on (and reflected from) the white surface that the experiment is being done on? You can't see laser beams in the air. But you knew that, of course...

No. Projecting five parallel laser beams through the system is NOT the same as "any other scene". Several people have explained why these are very different scenarios. In the case of your five lasers, closing the diaphragm WILL block two of the beams and so only 3 sources will be visible. Adding a filter will not change that. In the case of "any other scene" then the entire scene will remain visible if the diaphragm is closed. Adding a filter will not change that. It can also be confirmed by theory. The value of your prediction is: false.

https://en.wikipedia.org/wiki/Pi#Definition That page has a neat animation showing the definition: