swansont

He has a fairly popular blog and is a professor in the same state. Oh, and the little matter of the American Atheists conference, so his appearance might have been expected.

It doesn't take energy to push you into the gravity well, it takes a force. And the curvature acts like a force, because your motion is changed by the gradient of the surface, though one of the results of relativity is that a constant acceleration (i.e. freefall) is inertial.

"missing" as in we don't hit the sun.

Such delicious irony.

A classical vacuum has no temperature — there's nothing to measure. Reaching 0 K implies you were originally at some nonzero temperature.

This would occur if the particles could scatter off of each other and trade energy and momentum. E=mc^2 doesn't say anything like that. There is such a thing as gravitational potential energy. Gravity, however, is a force (or a curvature of space if you view it geometrically). but since [math]F = -\nabla U[/math], gravity isn't special in that regard.

I was comparing Mars to Earth. How did Venus get into the picture? Anyway, I was talking about optical depth, which does not mean that IR radiation does not escape — that would be in violation of thermodynamics. The radiation heat transfer ensures that radiation (mostly in the IR) will escape. The difference between "optically thick" and "semitransparent" is whether the IR is from the earth's surface, or if it is reradiated along the way out. And it's not just the visible portion of the spectrum, though there are plenty of time we can't see the sun. The issue is what does the atmosphere looks like at the wavelengths where the earth is radiating (which is somewhere out in the vicinity of 10 microns). This can be easily calculated from the absorption of gaseous carbon dioxide. See Phys. Rev. 41, 291 - 303 (1932) P. E. Martin and E. F. Barker "The Infrared Absorption Spectrum of Carbon Dioxide". Thomas and Stamnes (ref. 2, page 91) that shows 0% transmittance at 22 km and below for the 15 micron CO2 band. This section discusses the "opaque region" and also gives a very clear discussion of line broadening, which is an additional point that many people are unfamiliar with. Schneider, Kucerovsky, and Brannen (Appl. Opt. 28:5, 1998) give an absorption coefficient at 9.90 ± 1.49 cm-1 atm-1 for low concentrations of CO2 in a 1-atm nitrogen atmosphere at 4.2 microns. This works out to 376 absorbance units per km for 380 ppm CO2, which is about as close to 100% absorption as you can get. Heinz Hug, a global warming skeptic, measured a similar value (0.03 absorbance units/10 cm for 357 ppm at 15μm) from http://brneurosci.org/co2.html That's called optically thick, not semitransparent.

Almost. In a classical sense, you can't reach absolute zero, which is where nothing is moving, so there is motion. In a QM sense, you can't reach absolute zero, but this is where there would still be some small amount of residual motion from the zero-point energy.

How is this relevant to anything in this thread? ttowntom hasn't posted here, and the only link that has been presented did, in fact, provide only anecdotal evidence.

As opposed to the implications from fire, that can be (and has been) used by terrorists, and exposes people to radiation?

When do we get to the interesting part?

Exactly — they accelerate toward the sun but keep missing. (though the general form is an ellipse)

The sublimation temperature of water is around 200 K. At -30 ºC (243 K), the vapor pressure is around a few tens of Pa, and it goes up ~exponentially with temperature, so at 250 K it's 100 Pa. http://www.lsbu.ac.uk/water/images/phase.gif

As I said before, absolute zero from the second law of thermodynamics is a classical concept, before the development of the idea of a quantum-mechanical zero-point energy. Absolute zero was the removal of all center-of-mass kinetic energy of the atoms. The Feynman quote is accounting for the fact that when incorporating QM, you have to say that absolute zero is the removal of all available kinetic energy, and you will have the zero-point amount left. But as I demonstrated earlier, that amount is really small in macro systems. In any event, you can't reach absolute zero, and you can't bring atoms to a complete stop. This also points to the problem of simply quoting authorities on the subject. Feynman was correct, but his statement has to be taken in context in order to be understood, because even though it contradicts iNow's statement in post #11, iNow was also correct, taken in context. It was a comparison of quantum apples with continuous oranges.

Objects behave as waves, but sometimes the wave behavior is relatively unimportant, so you can assume particle-like behavior, e.g. ballistic trajectories. The energy is quantized (as are other properties) so certain interactions also seem like they involve particles. IOW, treating some things as particles is reasonable under many circumstances, and that often makes the math simpler.

Yes, unless your diet is rich in pitchblende.

Tell that to Fonzie. Cool/cold is just an arbitrary temperature threshold. That's temperature or thermal energy; heat is actually energy being moved around, or the process by which that happens, rather than a substance an object contains. An object with a uniform temperature in thermal equilibrium will have equal amounts of heat entering as leaving — there is no net heat flow. Also, one needs to be careful not to draw the conclusion from this that infrared and heat are the same thing. That's an important point — the body is a good sensor of heat flow, not temperature.

You do realize that human beings are inherently radioactive, right?

But if someone makes a point then we have a point and pointless existing simultaneously — the thread will be in a superposition of the two states. Which is proof that QM is right. Locking the thread will collapse the thread's wave function in the "pointless" state (assuming that wave functions are not bullshit)

Except you can't reach absolute zero. Let's take an ensemble of atoms and put them in some kind of confinement, (like a magnetic trap) of characteristic size 1 cm. The zero-point energy for that is of order 10^-21 eV for atomic masses around 100 amu. The equivalent temperature of that is around 10^-17 K. Quantum fluctuations in macroscopic systems are small. Not reaching absolute zero is a classical limitation, that is not contradicted by quantum mechanics, but neither is it driven by QM in classical systems. The oscillations in question — strings — are not thermal-atomic-motion in nature.

Tomato, tomahto, potato, potahto.

"The astronomers were surprised to find that the planet has more methane than predicted by conventional models for "hot Jupiters." " To paraphrase a comment I read elsewhere, astronomers must be the most-easily surprised of all the scientists. Almost every press release has them surprised at something-or-other. "Astronomers were surprised to find that extrapolating from a single data point can yield unusual results"

No, but the water surrounding it is, and will freeze at a lower temperature. Consequently, it can melt ice, with which it comes into contact, at a lower temperature. But I believe that's above the sublimation point, so there can still be loss of mass that increases with increasing temperature.

Or a wave taking up only a small region, as opposed to a long wave-train. Like this

I think what you'd have to do is account for the optical depth of the atmosphere for all of the wavelengths in the blackbody spectrum. IOW, is the atmosphere opaque at 10 microns, or partially transparent? If the latter, you'll couple to both the atmosphere and the CMB.