swansont

But not in this case Yes. And when the sun is overhead, it compresses the atmosphere, owing to the radiation pressure. Not sure where this is coming from. 780 nm is not high frequency light. No, it’s not. No Of order 1 mW/cm^2 for Rb This is not a thing, in this process

Scientific terminology is sometimes different from lay usage And? You mentioned gamma radiation, for some reason. What is “it”? This was a comment on gamma radiation?

No mention of “saturated” here. And I was asking if you knew what it meant, not if you could google, and you’ve confirmed that you don’t. Light with a higher frequency has more energy regardless of whether it’s interacting You don’t keep them stationary. You’re trying to slow them down, so they are moving. Vapor is invisible, so how could you tell? Gamma radiation has nothing to do with our discussion, save for your irrelevant tangents.

You have no idea what saturated means in this context, do you? Energy and intensity are not the same thing. Absorbing from multiple directions destroys the cooling effect. You want an atom to absorb a photon that opposes its motion. As I said above, it does matter what direction the light comes from Of course it happens. A free-electron laser emits photons. Of course it does. Failing without learning from your failure is pointless.

What’s the connection to your claim that “The more photons they get, the colder then get.”? Fact of the matter is that when we do the final cooling stage in our fountains, we turn the intensity down to get colder temperatures. “goes into stimulated emission”? Stimulated emission has a velocity? WTH are you talking about?

The attitude from some engineers that they don't need to understand physics has always scared me.

Interesting that they don't seem to break that down by jobs lost vs people voluntarily leaving a job (a second job) because they don't need to work the extra hours. (I recall that was part of some analysis for the ACA, for people who worked solely to have insurance)

I don't know about elsewhere, but COVID vaccines are not approved in the US. They are being used under an emergency use authorization (EUA) The issuance of an EUA is different than an FDA approval (licensure) of a vaccine, in that a vaccine available under an EUA is not approved. In determining whether to issue an EUA for a product, the FDA evaluates the available evidence to determine whether the product may be effective and also assesses any known or potential risks and any known or potential benefits. https://www.fda.gov/news-events/press-announcements/fda-takes-additional-action-fight-against-covid-19-issuing-emergency-use-authorization-second-covid

This is something you could estimate. This does not follow. The photons have to be the right photons (correct frequency, direction, etc.) in order to cool. This would seem to be yet another non-sequitur That's one of many. X-rays tend to be absorbed. Even dealing with UV light is problematic. Magnetic fields do not slow the atoms. And "fastest" is a dubious claim if you can't actually confine them because they aren't cold. I'm not aware you've arrived at a valid conclusion. Despite being told that this will not work. The problem is you never stop to question why things won't work, when you are told your ideas won't work. You just toss out a new idea to get shot down, and ignore the feedback. Have you ever considered learning physics?

That’s exactly the arrangement we used. The larger picture is you need space on your vacuum chamber for windows, and need ports to look in and to get atoms into (and possibly out of) the chamber. More windows means a larger chamber, which might be a luxury (you might have space constraints, and it costs more money)

Yes If you shine light toward a cloud of atoms, and scatter the light with a filter, how does the scattered light get to the atoms? You have a collimated source, aimed at a target. If it scatters, it’s no longer directed at the target. That what lenses are for. You can expand and re-collimate the beam. Or one laser and beamsplitters, or one laser and a bunch of mirrors to redirect and retro-reflect the beam, or three lasers and retro-reflection. I’ve done all of these options. X-rays won’t work. If you understood the atomic physics you’d see this, or be able to ask a pertinent question, rather than just tossing out random ideas and/or repeating the same flawed suggestions. It’s not a matter of them going in the same direction, but good luck with that; x-ray optics are notoriously difficult to work with, and inefficient. Or just use mirrors to expand the beam. That’s not the sticking point here. md65536 was not the one spouting nonsense about spin. You can technically cool the atoms with 4 beams (I’ve done that, too) but it’s not very forgiving, and is “leaky” so you don’t get as many atoms. 6 beams is overconstrained so any small misalignment doesn’t hurt you - it’s fairly robust.

You have no clue what the myriad problem with x-rays are, even though the information has been presented. Some of it is in the blurb you posted about laser cooling. The steps are not optional, and are fairly specific. If you don’t follow the instructions, you don’t get the result.

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

The fact that you are mentioning anything other than the D2 resonance suggests you don’t understand how it works. Yes. If you have to ask, or think a filter would work, you don’t understand what’s going on.

http://www.worldstartech.com/what-determines-the-lifetime-of-a-laser-module/ Laser distributor disagrees Manufacturers, too, that I’ve talked to in DARPA program reviews. They were working toward 100k hour lifetimes

Then you need to come up with the math to quantify this, and a model for why behavior depends on it

780 is in the infrared, and flashlights are not single-frequency Specific frequencies over a wide range. The absorption spectrum is discrete. A couple of wavelengths in the visible, at best. The ionization energy of Rb is a little over 4 eV. x-rays would remove an electron. What rotation? Yes, and nothing you’ve presented thus far addresses this. As I said, I’ve been doing this for a while, so I hope you didn’t post this to educate me. Now, the big question: did you understand any of this, and can you apply it? Can you see how it doesn’t mention rotation? Can you see how the slowing takes a half a meter, so a lower scatter rate or momentum of the photon by more than a factor of 2 means you can’t do this in a 1 meter system? (so a momentum a factor of tens of thousands smaller means this is impossible for RF) It does not matter. The atoms are not rotating. I would say they have spin, but it’s quantized, and not physical spin. It’s not something that is being reduced, so fredreload bringing it up repeatedly is based on misconceptions

The ones I use last 1-2 years, typically, and the failure is that it mode-hops, which is likely because if degradation of the AR coating. If you aren’t worried about the exact frequency of the light, the diode itself would last much longer. Well, no. It won’t lase below threshold, but there’s no reason to assume you’d normally operate it there. If threshold is e.g. 25 mA and you normally operate it at 100 mA, there is plenty of room to lower the current.

My objection is not based on my having questions about the inverse square law. No need to explain it to me. You could have just said you were calculating the surface area of a sphere at the distance of the earth. But when you just throw numbers up, without an equation or units or explanation, the method is not always clear.

Yes, we both posted information attempting to show this But (150 x 10^9 km)^2 isn’t an area with any physical meaning, so it makes no physical sense to multiply it by the power per unit area It does make sense ( sort of) to calculate the power per unit area using the 1/r^2 nature of the radiation, but if that’s what you did you’ve skipped some steps

I don’t understand your calculation. 150 million km is the distance to the sun, not the size of the earth. Why are you multiplying by the square of this, to get an energy from an energy per unit area? You should multiply by the projection of the area of the earth facing the sun, which would be pi* r^2, where r is the radius of the earth. And that’s the energy hitting the earth, while to OP talks of the energy emitted by the sun

Does this involve a property that be quantified and measured?

Stellar aberration’s basic explanation was known since ~1725, which precludes us being at rest with respect to the aether. We had to be moving, but the M-M experiment showed that to be wrong. https://en.wikipedia.org/wiki/Aberration_(astronomy)

Profitability wasn’t the point under discussion, nor was powering Europe, or job creation.

Because electricity is famous for being used at the location where it’s generated?