Lorentz Jr

Nothing better to do, and most of this thread was a good exercise for me. Relativity, TeX, Inkscape, all good skills for me to work on. It can also be helpful for beginning readers to see the real theory and calculations. 🙂

I love the way Otto sneaks his little insinuations into innocent-looking questions. "If it's not 0.866 c", as though anyone in this thread has said anything to that effect. And of course his imaginary premise naturally leads into damning speculations like "Are you going to make up some random velocity now?". He's very good at knocking down straw men. 😄

Then why did you make all those comments about "Do you seriously think it's scientifically acceptable" and "Either we know the velocity of the clock or we don't, which is it?" 0.4641 c is from your calculation, not mine or anyone else's, and the light beam only travels ten meters in ten seconds, not 18.66 meters. This "stretched out light beam" idea is just another one of your little fantasies.

Exactly. +1 for that, and any comments to the contrary are best forgotten.

Where do you get 0.4641 c from?

Why is the video with the shorter answer so much longer? 🙄 All the first one says is what they all say (e.g. at 2:30 in Dr. Lincoln's video): that for the standard version of the paradox (with only one spaceship), acceleration is required to change the spaceship's velocity. The second video clarifies that acceleration is only involved indirectly. With two spaceships, you can get the same result with no acceleration.

Well, I can't stand this "conversation" anymore, so here's another pretty picture for anyone who's interested, just for fun:

Not new neurons, but neurons that already remember relationships between the network's inputs in existing memories (e.g. visual characteristics of complete images). They will activate other neurons that remember other aspects of the same memories (e.g. various images that share properties represented by that neuron, testing to see what other properties the new perception shares with those memories). (a) The state of each neuron is represented by a dimension in the net's (very high-dimensional) phase space, (b) each memory corresponds to a point in that space, and (c) the net is designed in such a way that its state will approach (i.e. be "attracted" by) the point that corresponds to a memory if a new perception puts the net in a state that's close enough to that point, i.e. within the point's "basin of attraction". An initial state of the network is the point in phase space, i.e. the collection of individual neural states, that a new perception (e.g. image) puts the net's neurons into. This is the flow map of how the net's state can go from various initial states, through other intermediate states, to the state for the specified memory during the net's recognition operation. This is the set of states that will eventually lead to the point for the memory. The endpoint (attractor) is the "lowest-energy" point in the basin, and the boundary of the basin is a "ridge" that separates the basin from other basins. So you can think of the net's phase space as a hilly terrain where each valley corresponds to a memory.

Lincoln mentions adding the field generated by the electrons (i.e. by atomic polarization with quasi-stationary nuclei) to the incident wave.

The fact is, you have been either unable or unwilling to solve the problems in this thread correctly, and all of your claims and suppositions have turned out to be incorrect. Your "analysis" of these problems is equivalent to saying that a car will collide with a slightly slower-moving car ahead of it, simply by reaching the point on the road where the leading car was when you started your analysis. Even in the old physics of Isaac Newton, that's not how moving objects work. The leading car has moved on, so the trailing car still needs to catch up to it.

Hey, you're right! The first wave moves so slowly at 7:05, I thought it was supposed to be stationary. Dr. Lincoln also repeats the "combined wave" idea at 8:30.

I suggested that, but I tried not to be too factual or authoritative about it. Dr. Lincoln said there's a light wave moving at c and a standing electron wave (i.e. not moving at all). Although he does say (at 7:05) that they produce a more slowly moving wave as a "result", whatever that's intended to mean.

The distance of the horizontal beam's trip only depends on L and v/c, both of which you've kept the same, so it still has to be 37.3 meters. And t = v/c, so the time is 37.3 seconds. After five seconds, the front of the car is at 9.33 meters and the beam is at 5 meters. The beam still hasn't reached the front. In other words, you've made exactly the same mistake you made in the c = 10m/s example. And the same thing applies for the diagonal beam: The distance is the same as before and the time is ten times longer: [math]\displaystyle{ x = \gamma L \left(1 + \sqrt{2} \frac{v}{c}\right) = 2(10m)(1 + .866\sqrt{2}) = 44.5m }[/math] [math]\displaystyle{ t= \gamma \frac{L}{c}\left(\sqrt{2} + \frac{v}{c}\right) = \frac{2(10m)}{1m/s}\left(\sqrt{2} + .866\right) = 45.6s }[/math]

[math]\displaystyle{ x = \gamma L \left(1 + \sqrt{2} \frac{v}{c}\right) = 2(10m)(1 + .866\sqrt{2}) = 44.5m }[/math] [math]\displaystyle{ t= \gamma \frac{L}{c}\left(\sqrt{2} + \frac{v}{c}\right) = \frac{2(10m)}{1m/s}\left(\sqrt{2} + .866\right) = 45.6s }[/math]

No, Grenady and I both doubt they would. My point was that all their data would still extrapolate out to c, even if they can't reach it experimentally.

Sorry. My fingers were going faster than my brain. 🙄

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And even if they did know about relativity without having access to a vacuum, they might be able to infer the real value of c from the electromagnetic constants and other phenomena, such as how muon decay rates depend on their speed. Maybe that's a stretch, I don't know. Their access to good experimental data might also be limited.

That can't happen in relativity, so your question would be better off in the Speculations area. Otherwise, as swansont explained, the answer is no, there's no analogy between exceeding c and reaching c.

A slightly higher energy, but only because the external potential makes that the new ground state. So it's a complete transition, not just partial mixing, and it's to the equivalent state (i.e. S to S, P to P, etc.). I don't think transmission can be based on any mixing or transitions to excited states. As you said, that's how absorption works.

Yes, but we're talking about transmission. There's a classical theory for calculating the index of refraction. I think the idea (to put it into quantum-mechanical terms) is that the applied potential from the light distorts the electron wave functions (i.e. their energy eigenstates) "smoothly" (so the photon energy must be less than the band gap) so the electrons oscillate without jumping out of their ground states.

He says at 7:05 that the sum of two waves moving at different speeds is a wave moving at some intermediate speed. Electrons can't escape the glass, so their wave is a standing wave, like the vibration of a guitar string. Its speed is zero. You can think of it as a combination of two waves that move in opposite directions and keep getting reflected internally at the surface of the glass. I'm not sure what the speed of those electron waves is. Offhand, I would guess that it should be less than c because of the mass of the electrons, but don't quote me on that. Come to think of it, maybe the speed of the "electron" waves is just c/n, where n is the medium's index of refraction. That's the speed of "light" in the medium. Physically speaking, the best way to think about it may be that there isn't a pure "light wave" going through the glass, but a combined light+electron wave that has its own properties. So the overall wave is still the sum of two waves moving in opposite directions at speed c/n (or at least it can be broken down that way mathematically), but the one going in the direction of the incident light has a larger amplitude than the one going the other way.

Good! I'm glad you figured that out yourself, Otto. Only gravity or an observer in a noninertial reference frame can make a light beam appear to be curved. Difficult? Yes, if you don't know the rules. Ridiculously? Not really. You just have to forget about common sense for a moment, focus on events, and use the Lorentz transformations. You use 45 degrees inside the train (because that's something you know), and then you apply the Lorentz transformations to the end of the diagonal light beam's path. Inside the train, [math]x' = L[/math]. The distance traveled by the diagonal beam is [math]s' = \sqrt{2}L[/math], so [math]\displaystyle{ c = \frac{s'}{t'} = \frac{\sqrt{2}L}{t'}}[/math] [math]\displaystyle{ t' = \frac{\sqrt{2}L}{c}}[/math] Now transform to the ground frame: [math]\displaystyle{ x = \gamma (x' + vt') = \gamma \left(L + v \frac{\sqrt{2}L}{c}\right) = 2(10m)(1 + .866\sqrt{2}) = 44.5m }[/math] [math]\displaystyle{ t = \gamma \left(t' + \frac{vx'}{c^2}\right) = \gamma \left(\frac{\sqrt{2}L}{c} + \frac{vL}{c^2}\right) = \frac{2(10m)}{10m/s}\left(\sqrt{2} + .866\right) = 4.56s }[/math] The end of the trip is when the light beam reaches the front of the car. You're not supposed to use anything except the Lorentz transformations and whatever specific information the problem gives you. The process can be a bit tricky and counterintuitive, but it's not horribly bad once you get used to it. 🙂

Yes, there's definitely something going on in space that goes far beyond the static nothingness of the vacuum in Newtonian physics. But the principle of relativity states that space doesn't have any "parts" or "constituents" that could have a detectable velocity, which is a property that's normally associated with things that are "made of something".

Why is this thread in the Relativity section? The question is profoundly speculative, and there are nontrivial answers associated with other theories. If anything, relativity seems to inherit the classical model of space as an empty void that particles move through, only acknowledging that it can be curved and not that it can be made of anything that has a state of motion. The definition of spacetime is mathematical. It's a four-dimensional space that includes time along with the three spatial dimensions, and it's based on the Lorentz transformations (which are supported by scientific evidence). The relativistic four-interval between two points (i.e. the "distance" between them) is equal to the proper time along the shortest path between them. (The proper time is the time that would elapse on a clock as it moves along the path.) That's a straight line in special relativity (Minkowski space, no gravity), and it's a minimally curved path called a "geodesic" in general relativity (curved spacetime, with gravity).