swansont

Ole Roemer watched Io pass in and out of Jupiter's shadow, and noted that the time depended on how close Jupiter was to the earth, implying a finite speed. Bradley used stellar aberration. Wikipedia entry

Then it should have been posted in "Speculations"

"So what" is appropriate when you use logical fallacies. Did you not read beyond that? When you proceed from a false premise, or use invalid logic, any conclusion at all may be reached, so the conclusion is worthless. Hence the "so what." There is no scientific value in it. I've explained the science in some detail. What, specifically, don't you understand? The energy comes from the mass of the composite system.

Length contraction — the spatial dimensions do change. Time dilation has been observed/measured, over and over again. Thing is that scientific theories are accurate — they make concrete predictions so that can be tested and, in principle, falsified. IOW, you need general equations describing the behavior.

You are not accounting for the fact that it starts 2m off the ground, and will have more KE when it hits the ground than when you launch. The ground, where it has minimal PE, is where it has maximum KE, since the sum of the two is constant.

Einstein may have disagreed that constant gravity was curvature, but to that I say, "So what?" It doesn't matter how Einstein interpreted it. That's either argument from authority (Einstein also said "God does not play dice" and he was wrong) or equivocation, because it depends on whose definitions you use. What matters is the actual theory, which has progressed from the time Einstein introduced it. That paper states that gravity equating to curvature of spacetime is the current interpretation (p.15). It seems to depend on how you define your terms. Even considering all that, "There is no curvature" is wrong. "There need not be curvature, according to Einstein's definitions" would be correct, with constant gravity being one example, according to the paper cited.

If someone wants to discuss pushing on tachyonic materials, then all bets are off anyway.

Dismissing and not addressing are not the same thing. I have no stake in your metaphysical arguments, and have repeatedly pointed out that I have no desire to discuss them. But when you make a physics statement, and it's wrong, I will comment. If you will not be satisfied with "the argument uses the wrong concepts" then you'd better not use it to dismiss a counter-argument using accepted physics. Rest mass is a Lorentz invariant, and that's useful where you can assume flat spacetime. But two masses comprise a composite system, just like with a nuclear or electromagnetic attraction, and the rest mass of that composite system is smaller. I can measure it fairly easily with these two forces because they are large enough. I can't with gravity, because gh/c^2 is about a part in 10^16 per meter. If you can't do a measurement that shows the effect, because it's too small, then there's no need to worry about including in your calculations. As an example, in nuclear binding energy calculations the mass of H-1 is used for the proton and electron, even though that ignores the 13.6 eV of binding energy of the electron, and atomic masses are used for the product, even though they include a different amount of net binding for their electrons. It doesn't matter, because that's at least five or six orders of magnitude smaller than nuclear binding energies. If it were big enough, then people would worry about the term, just like they would note the rest mass having a gravitational potential term. They Feynman example shows that the mass terms have to be different by gh/c^2 to conserve energy.

Calculate the maximum speeds in both directions. (You should be able to determine that this is the value you get when the ball strikes the floor)

The forces involved propagate at c or slower. Infinitely rigid materials don't exist.

But there should be an interesting effect: because the momentum of the photons change there should be a force on the medium. If we do this with a long optical fiber, there should be a strain on it when the photons enter one end, but before they leave the other. Someone told me about a proposed experiment to measure this, but I can't recall who was proposing it, or find any results. But I know that Ketterle's group at MIT measured the recoil momentum of photon absorption in a gas to be [math]n\hbar k[/math] rather than [math]\hbar k[/math], which is the same idea.

You can observe wave properties, such as interference, yes. Whether you see particle- or wave-like behavior depends on the nature of the interactions I'm not sure what this means But the accelerations and bumps should have been comparable between the east- and west-bound clocks. Any bias they might introduce would tend to cancel. I suspect that from the prior experience of transporting clocks, the experimenters already knew the effect would be small.

The net momentum of an annihilation reaction is zero, so you'll be expelling energy out the front, too. To turn annihilation into propulsion will require some engineering.

I trap atoms for use in an atomic clock (Rubidium Fountain)

OK, I'll go with this. "Your view employs the wrong concepts" Are you willing to be dismissed this trivially?

Yes, light slows when it enters a medium with a higher index of refraction, but it can propagate through vacuum, whereas sound cannot. Light propagates through air, but the air isn't required. Note also that sound is a longitudinal wave, while light is a transverse wave. Light exerts pressure on things; it's the basis of laser cooling, which won the Nobel Prize for Chu, Cohen-Tannoudji and Phillips in 1997. Atoms are being trapped with light pressure in my lab at this very moment. It's not normally noticed because it's such a small force once you get above atomic mass scales.

They may be the same, but that does not mean they are derived.

The EM spectrum has some fuzzy definition. Most of the low-frequency stuff has general guidelines based on frequency or wavelength, though where e.g. RF stops and microwave starts depends on whether you talk to an engineer or a physicist and what their specific area is. But in physics, X-rays come from atomic interactions and gammas come from nuclear interactions. So it's possible to have X-rays higher in energy than some gammas. (In astronomy, AFAIK, they use an arbitrary energy division. IIRC it's 1 MeV) In general, gammas are considered to be high-energy, and there's no other name — nothing higher than gamma. Radio waves get subdivided, like high frequency, low frequency, very low frequency and extremely low frequency. But nothing lower. http://en.wikipedia.org/wiki/Electromagnetic_spectrum

Since when is smart defined as "able to divide by 1000" ?

Muddled and so difficult to always say exactly where it strayed. However Momentum transfer will be maxmized when the masses are equal [math]K_e=\frac {1}{2} mv^2[/math] doesn't "turn into" [math]E=mv^2[/math] [math]\frac {1}{2} mv^2[/math] is a nonrelativistic approximation, valid for v<<c

Probably. You need to be aware of the risks involved with playing with radioactive sources, though.

It's possible to make a homemade cloud chamber.

Yeah, and it was way cool. Here's a summary: http://www.physorg.com/news88439430.html

[math]E = \gamma m_0c^2[/math] Gamma is imaginary and also negative. So it all depends on what you want — if the energy is to be real and positive, then the rest mass must be imaginary and positive. If you want a real rest mass, that makes the energy imaginary and negative, or you could have negative and real rest mass, giving you an imaginary and positive energy. None of that currently has any physical meaning. I think that in some high-energy/particle physics, the relevant tems are mass^2 and energy^2 in some of the equations, but I think you still end up with a sign problem.

Into the EM radiation that you see, or don't see (outside the visible spectrum, or not hitting earth).