swansont

The separation velocity of any two particles will be constant in time; absent any interaction, the velocity of each particles is constant. There’s nothing proportional with distance.

Two particles traveling in opposite directions, each at v wrt to the explosion, will separate at 2v (for v<<c) and this will not change with time, and therefor not change with separation. For v to increase in such a scenario you’d violate both conservation of energy and Newton’s first law

Basically the point I made as well. Antibiotics use in children is rampant, a third to half of kids 4 and under are prescribed each year in the US, and in some countries the average is 5 per year “Children in low- and middle-income countries (LMICs) are receiving an average of 25 antibiotic prescriptions during their first five years of life” https://news.harvard.edu/gazette/story/2019/12/high-rate-of-antibiotic-use-in-low-income-countries-alarming/ https://meps.ahrq.gov/data_files/publications/st35/stat35.shtml You’d expect to see a bigger effect in countries where more antibiotics are used. Do we see such an effect?

They checked the engraver but no mention of checking the obituary listings for Roger?

mar_mar2 banned as a sockpuppet of mar_mar. We do not need a sequel

I don’t think lasers are used to measure anything outside of our solar system. Certainly nothing where relativity is a factor. We get distances from the light that comes from the entity being measured, and those aren’t lasers.

I don’t understand. Are you claiming that collision avoidance doesn’t exist? Can you provide a link to what you read? https://en.m.wikipedia.org/wiki/Traffic_collision_avoidance_system https://www.faa.gov/about/office_org/headquarters_offices/avs/offices/afx/afs/afs400/afs410/airborne-collision-avoidance-system-acas

! Moderator Note No. This is a discussion forum. Soapboxing and promoting an agenda are against the rules. Stop it.

No, it reflects, but the emissivity is only going to reduce the blackbody power by a factor of 2 or so. This isn’t going to save your conjecture. Pick another solid that’s a better blackbody. But consider that you could paint a block of copper matte black, and this would make it a much better blackbody, yet this only changes the surface and not the bulk property of the copper. Negligible change in the thermal energy content. A model allows you to do calculations, and accounting for the photons is a critical part of your conjecture. If you can’t quantify the effects, you don’t have a model.

Then you don’t have a model. You’re proposing a new description for temperature and you need new physics to make it happen

Yes. Fermion number is conserved. (Anti-fermions have a negative fermion number, which is why we get matter and antimatter in pairs)

You won’t provide a calculation of your own. How do you get the number of photons your model requires? If you emit at the rate required for the blackbody radiation the number inside the material is negligible compared to the thermal energy.

I didn’t assume they emit at the same time. I assumed they emit at the same rate, because why wouldn’t they? They are identical atoms. And if they don’t emit and absorb at the same average rate, then some area would be emitting more or less than another, which would heat or cool it. But we have thermal equilibrium, so that can’t be the case. A photon that is absorbed by an atom ceases to exist. There is no more oscillating EM field. The energy and momentum are transferred to the atom, but there is no more photon. Boson number is not a conserved property. You can create and destroy bosons.

iNow has a point. You have a couple of threads where you start out in chicken little mode, and raise an issue that you could research yourself - which you’ve done here to some extent, but only after being challenged with regard to your assumptions and framing The evidence presented thus far is that there’s an effect (which nobody really challenged) but any possible permanent effect is not common.

Doesn’t really matter; you still can’t reconcile the emitted radiation with the photon gas density you need to account for the thermal energy, since the residence time of the photons is so short.

You would have to establish that it is, by quantifying the behavior and comparing it with experiment. You haven’t done this. A model is supposed to make predictions. I took your model and predicted how a solid would behave if it were true. If the established physics I used is wrong, you could point this out, but where? Calorimetry is pretty well-established, as is the value of speed of light. You’re basing your idea on the Stefan-Boltzmann law, so you obviously don’t have a problem with it. The rest is just pretty straightforward math. And a competing model has to be consistent with all the evidence the existing model covers. You can’t just cherry pick one or two data points. This is fallacious reasoning. Instances of physics being wrong in some aspect does not mean any arbitrary aspect of it is wrong. A model is wrong if it disagrees with observation. And your model disagrees with observation.

! Moderator Note I’m not sure what this has to do with ethics Soapboxing is against our rules, as is spamming.

It’s not opinion. Your idea is wildly incompatible with established physics. Shopping for a more credulous audience won’t change that.

It’s not a “discovery” as such. It’s a model based on tired light and varying fundamental constants, without the experimental support one needs to support those ideas. Not a good foundation for a model. The tone of the article suggests that this is somehow a credible experimental result, when it is very far from that.

Radiant heat. There is also conduction and convection. Planck was modeling the radiation spectrum, not the behavior of atoms in a solid. Copper is not transparent Photons disappear all the time. Ever turn off a light at night and notice how a room immediately gets dark? Both sethoflagos and I have pointed out that these photons don’t have enough energy to cause atomic excitations. You can’t just wish these reactions into existence What you think isn’t nearly as important as what you can show, both theoretically and experimentally. You talk of photons passing through materials like copper to transfer heat, when empirically we know copper is opaque. All you’ve done here is regurgitate some stuff from wikipedia. There’s no actual analysis to see if the claims are reasonable, and match up with what we observe. The thermal energy content of the copper in my example is around 1 kjoule. The blackbody photons average around 0.1 eV, meaning there need to be around 6 x 10^22 photons in existence at all times in the material. (this is not an emission rate, this is a population) But the block is only 1 cm on a side. A photon can travel, at most, 1 cm before it either is absorbed, or leaves the block. It takes less than 0.1 ns to travel that far. An atom has to be emitting 10^10 photons a second to have just one be existing at any given time. That’s an emission rate of 6 x 10^32 photons a sec for there to be a kilojoule of photon energy in the material Atoms near the surface are the ones that can have photons leave the material. If that’s only the monolayer at the surface. 10^16 atoms. Half will emit in a direction that leaves the material. That’s 10^38 photons a second. 10^18 watts. Slightly higher than what the Stefan-Boltzmann law predicts. Your idea doesn’t hold up to scrutiny. It’s not even close.

Because you are presenting a pet theory of yours.

But the issue is the energy content. You’ve provided no analysis to show that the thermal energy of the object is from the photons. Just assertion.

Why should it be? The 2.75 W is dependent on the surface area. Show your work if you disagree. Thermal equilibrium means the surroundings are at 300K, and 2.75 W is also being absorbed. But that’s at the surface. The kilojoule is the thermal energy content (mass* specific heat capacity * temperature), not the radiated power. They are two different things. The radiated power will tend to reduce the thermal energy by reducing the temperature.

Not all that much. In copper it’s about 0.25 nm. Light would take ~10^-18 sec to traverse the distance If you had a block of copper, 0.01m on a side, it’s going to radiate 2.75 W at 300K At 4 atoms per nm, there will be 4 x 10^7 atoms along one dimension. 6 x 1.6 x^10^15 atoms on the surface, so perhaps 10^17 atoms near the surface can radiate outwards. at 300K, the photon energy peak is about 0.1 eV. 2.75 W needs about 2 x 10^20 of these photons per second. So each atom is responsible for about 200 photons/sec, or one every 5 milliseconds. But we know these photons can only live for a time that’s around 10^14 times shorter. These are estimations, so there might be a factor of 2 here and there that could be off. But you’re ~14 orders of magnitude short 1 cm^3 of copper is about 9g, so the block is somewhere around 10^23 atoms, or 2 x 10^25 photons/s for 10^-18 sec. 2 x 10^7 photons let’s call it 10^8, just to be safe, at 0.1 eV. Which is around 10^-12 J. Compare with the heat capacity of 0.385 J/gK, and we’re at 300K, so we have around a kilojoule of thermal energy in our 9g block. No, it’s not photons.

This discussion was originally about a solid, and my posts were in that context, but go ahead and calculate the amount of EM energy. The S-B law gives you the radiated power from a surface. What’s the surface area of the interior of a gas? How do you ger an energy content from it? If the inside of a solid material the system is at the same temperature, there is no net radiation. Any photon emitted is absorbed, and since c is a big number, the time between these events will be small