swansont

A laser is not a blackbody, doesn't have a well-defined temperature and the Stefan-Boltzmann equation doesn't apply. It's not a situation where energy is being transferred due to a temperature difference, as you note. So it's a non-sequitur. I'm not sure how you came to the conclusion you wrote. The forms of spontaneous heat transfer include radiation. That's not the same as saying all radiation is spontaneous heat transfer (i.e. caused by a temperature difference)

You talked about contradicting the wave-particle duality. It's what I quoted. That's what I was responding to.

Or running through a forest, blindfolded. The concept here is called the mean free path — how far can you go before you hit something. Target density (which depends on pressure) is one of the factors.

We weren't discussing the complications that arise when you treat this relativistically.

I really like that. Merged post follows: Consecutive posts merged The wave-particle duality is introduced to wean physicists from classical notions — we get to hang onto the concept that there are still particles. But we say particle when we mean localized and quantized behavior. Just because you've stopped looking for the wave behavior doesn't mean it's not there. Merged post follows: Consecutive posts merged Yes, it is. One is expected to support arguments, and limit the debate to the science. If one believes a statement to be incorrect, one needs to address why they believe it to be so.

The answer is subtle but there's a clue in what Sisyphus explained. There is no perfectly rigid object, so there is a slight bend as you move outward. That means there is both a force toward the center (centripetal force), required for circular motion, but also a tangential force, required to speed the arm up. This tangential force does work, and because the path is longer as the radius increases, it does more work as r increases, and being a maximum at the tip. Proximity to the energy source does not mandate that the energy transferred be a maximum there. Energy is transmitted through forces (i.e. doing work) and at the center the force is entirely centripetal. And centripetal forces do no work.

Yes, it includes radiant energy. The net amount it radiates depends on the temperature of the surrounding reservoir, or objects. That's why there are two temperature terms in the Stefan-Boltzmann equation, and they have opposite signs. An object at the same temperature as the surroundings radiates no net energy, because it absorbs the same amount that it radiates.

Cites of books really need page numbers. And no, it's not patently obvious (to me) that you can recreate the exact trajectory of an electron where no measurement has been made.

Well good for you, Pete. Not everyone is so capable. Here's where you and I differ. A lot of people find science to be inaccessible, and one reason is the attitudes of scientists who can't be bothered to explain science at a level that the general public can understand. Those 12-year-olds who get the brush-off never get interested in it. I've taught and tutored, I do public outreach at work. And I spend time here, because I like explaining physics to people interested in physics, even if they aren't physicists or studying to be physicists. That's the primary audience at SFN. Keeping answers at the proper level of discussion — appropriate to the question — is one thing to keep in mind here. Merged post follows: Consecutive posts merged Someone familiar with science and statistics should be able to recognize that this is a bad argument. 40% of textbooks may not represent 40% of physicists, because a lot more students use the lower-level texts, and fewer use any particular specialized, higher-level texts. Where did that claim originate, anyway? I doubt that it is inclusive — the majority of texts wouldn't bother with it because they don't go into relativity. Is this 40% of relativity texts, making it a minority even within that sub-field?

One definition that has a tiny bit of relevance — meaning it mentions wavelength of an emission — is when the meter was defined in terms of a Krypton transition: 1,650,763.73 wavelengths of a particular transition in Kr-86. That was the definition from 1960 until 1983.

jsaldea12 has been suspended for three days (automatic suspension for accrual of infraction points). Argument by persistent repeated assertion = trolling.

This passed ridiculous about three stops back. jsaldea12, you have been told several times that this is a scientific discussion board, and so discussions have to contain science. Argument by repeated claim does not qualify. You had your chance. Thread closed.

We aren't talking about whether it will overhwhelm me. I've met a lot of people whose eyes would glaze over by such an explanation. And our goal is to not have that happen.

Atoms are neutral. The nucleus isn't, electrons aren't. The nucleus is held together by the strong nuclear force. There is no positive and negative there.

All the light that is reflected will contribute. It's all heat. The sun is at 6000K. That's your limit for passive heating.

But my answer works without the equation. So Mr. Boltzmann need not attend the party — the very definition of temperature tells you that any collection of molecules at a temperature will have the same average KE. It's a question one should be able to answer after 15 minutes of thermo class, before any of the equations are derived/given. Going through the equations is a nice exercise, though.

There have been some nice scientific contributions in this thread. Sadly, none by the original poster. Repeated assertion is not a discussion. If at some future the the poster has some evidence to back up his assertions, PM a staff member and the thread can be re-opened

And will have a wavelength shift of a couple of picometers, which isn't something close to being able to be discerned by the human eye, which is the topic of the discussion.

IOW, use the data that agrees with you, and discard/ignore the data that disagree. Wonderful science, that.

I think I have no clue what you're talking about. Equilibrium of positive and negative what? What does "on mass of nucleus" mean? And you keep using "unerring." I do not think it means what you think it means.

Because you're the one belaboring the point. I think the point Martin was trying to make (and apologies if it isn't) is that SFN isn't a bunch of physicists sitting around talking to each other, and delving into such details is just going to confuse the heck out of the person asking the question about basic relativity. So while relativistic mass may be a perfectly fine and useful concept to use for someone who's going to spend (or has spent) time studying General Relativity, and reads the texts you list, this describes almost none of the people asking questions here at SFN. IMO, one of the big problems I see in getting people to connect with science is the inability or unwillingness for some people to put the conversation at the level that the audience can understand (though there are certainly some who are great at it). But the simple fact is that you can't give a graduate-level answer to a neophyte and expect them to comprehend, and if you insist on doing that anyway, it's not fair to be disappointed when they don't. I understand frustration at having to give a simpler answer when you think (actually, know) the full picture is more interesting, but the full picture is also overwhelming, and that's to be avoided. What we're left with is giving the useful part of the picture, and not worrying about the part that's has to be left under wraps until later. The useful part of this picture is that mass is rest mass. Yes, you can define it to be other things, but what's the point in further confusing someone who has a question (or worse, a misconception) about the basics of relativity by giving them the answer they won't be ready for until they've studied for a few more years?

Right and wrong attract? Males aren't occasionally attracted to males, and females to females? (in about 10% of the population?) Bogus. No such law.

Cite? i.e. support for this statement?

And you understand this violates the 2nd law, but don;t see how it violates it. [quote name=CaptainPanic;501046 The radiation originating from any object includes' date=' as was already mentioned, pretty much all wavelengths. Let's assume a perfect black body (see: graph), which does radiate all wavelengths. I still don't see why with a perfect lens you cannot concentrate the light so much that it will heat up an object to a temperature higher than the source of the photons. Just focus it on a really really tiny spot. The flux of photons should become so great that a higher temperature can be achieved? I understand perfectly that the energy balance will not be breaking the 2nd law of thermodynamics. The total energy of all photons originating from the source of photons will always be greater than on the "really really tiny spot". But why wouldn't it reach a higher temperature? That does not necessarily go against the 2nd law (look at heat pumps and your fridge: it's possible). I think that for determining the temperature that is achieved somewhere we must look at the flux (photons / receiving surface area) and the wavelength. This flux can be concentrated immensely using lenses. As insane_alien has noted, it's because you're adding energy. The objection of the 2nd law is for spontaneous processes only, and passive optical systems. And here's the issue: you insist that you focus the light down. You can't. You can see this with ray tracing: in order to get a small focus, you need parallel light, but you don't have that. You can try to get it with two lenses - one to capture light and the other to concentrate it. Put an object at the focal point of the other lens, and it will give you parallel light. But that means the source has to be infinitely small, and you've already said that it was bigger. Put another way: only one point on the source can be at the focal point of the lens. The rest must be away from the focal point, and these all map to different points near the focal point of lens #2. The spot you get will be larger than you have assumed. (This all ignores the fact that you can only collect half the light with a lens, but adding mirrors doesn't solve the problem.)

Because, as I said, GR reduces to Newtonian gravity (as it must) for weak gravity cases, and Newtonian gravity explains the tides quite nicely.