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I think you may have forgotten the length of the interval in calculating its area. =Uncool-

Please explain. For one thing, that's going too far. It's not that they can't be answered; for what we know right now, they can't be answered now, but will be answered in the future. Actually, 38. There are 36 normally numbered spaces and then two spaces labelled 0 and 00. That actually isn't true; there are ways of predicting which compartment the ball will be in based on mechanics. Often computers are necessary, but they have been used. It is precisely as futile - not at all. =Uncool-

I'm curious, exploration, what do you hope to do by spamming? =Uncool-

And all of us are inviting you to actually answer any of the responses in this thread, which you have consistently refused. Whether you reject our invitation or not is another story; at least we tried, and our conscience is clear. Your "ideas" will get nowhere until you actually sit down and think about what other people are saying. =Uncool-

False. Quantum mechanics and relativity are both things which have mathematical analogs that are taught at the undergraduate level. Any math major worth their salt should be able to understand relativity and quantum mechanics. Unsubstantiated. False. It is impossible to determine position and momentum simultaneously, but it is easy to determine momentum and mass. Not quite false, but only because of relativistic effects. A positively charged nucleus has the same effect on electrons as a sun does in a solar system according to Newtonian physics. But some of the effects of Maxwell's equations say that the electrons will radiate energy away, eventually causing the electrons to fall into the nucleus. False. It indicates that there are discrete energy values for electrons. False. There are infinitely many orbitals. Again, any math major worth their salt understands it. Citation, please. This is not just going to pass without being examined. If "sub quantum kinetics" is more accurate than the current quantum theory, then that is a measure of the "aether". So no, it would not be "something that they can't measure". False. Aether was the original theory; the point was that relativity explained everything in the relevant fields so well that aether was unnecessary. "Science" at that time was hostile to relativity until it could prove its usefulness. False. It was shown that Newton's theory didn't explain the data and that relativity did. False. There has been plenty of criticism of special relativity. Citations, please. And yet only change in energy ever matters, so you can set the minimum energy to be exactly what you want it to be and have the equation be true. The two are therefore entirely equivalent. False, with a large misunderstanding of how to determine error bounds. A small number times a big number can still have a small error bound. Actually, it is possible to separate into isotopes so that we actually know which isotopes we are talking about. False. Deuterium has mass 2.01410178 amu, while helium 4 has mass 4.002602 amu - which is a lot less than double the mass of deuterium. This difference is caused by special relativity. That's it for now. =Uncool-

This has absolutely nothing to do with what I asked. I asked you specifically: What about quantum electrodynamics does not "interface"? You are talking about quantum mechanics. I am asking you about quantum electrodynamics. =Uncool-

Dovada, you claimed that How exactly does quantum electrodynamics not "interface" with classic electrodynamics? Alternatively, how does quantum electrodynamics not "interface" with the "quantum atomic model"? =Uncool-

An indeterminate form is what happens when you take a limit, where the result cannot be determined by separating the stuff inside the limit into two parts, and taking the limit on each. So saying that [math]1^\infty[/math] is indeterminate means that there are 4 sequences [math]a_n, b_n, c_n, d_n[/math] such that [math]\lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} c_n = 1[/math] and [math]\lim_{n \rightarrow \infty} b_n = \lim_{n \rightarrow \infty} d_n = \infty[/math], but [math]\lim_{n \rightarrow \infty} a_n^{b_n} \neq \lim_{n \rightarrow \infty} c_n^{d_n}[/math] We can see that this is true with the following sequences: [math]a_n = 1, b_n = n[/math] which has limit 1 [math]c_n = (1 + \frac{1}{n}), d_n = n[/math], which has limit e. =Uncool-

Please make a specific, testable, falsifiable prediction from your not-yet-hypothesis. It becomes a hypothesis when you have such a prediction; it becomes a theory when that prediction is shown to be correct. =Uncool-

There are two reasons why this is wrong: 1) Only changes in the electromagnetic field travel at c; you could have a field that is constantly speeding the particles up. 2) It actually is possible to pull a faster object with a slower object. You just have to be creative in doing it. =Uncool-

Which equation? Schrodinger's? Schrodinger's equation is an equation that works in any number of dimensions. This line reads like gibberish. What mysticism are you talking about? There is no mysticism inherent in quantum physics. Naive mystics may try to pretend there is, but they would be wrong. I'm not seeing how this statement contradicts anything in any standard quantum mechanics textbook, for example. What exactly are you disputing here? =Uncool-

I have read your posts, and I have found no such thing anywhere. No, but it models reality surprisingly well. That is exactly what people have been doing here. You have only evaded their arguments. For example: what evidence do you have that the people you named are "wizards", and that they were as you say they were? =Uncool-

Alex, I'd like you to define exactly what you think x_1', x_2', t_1', and t_2' mean. So far, I've been defining the x_i' s as the distance between the origin (as seen in the second frame) and the walls, and the t_i' s as the times to get to those walls in the second frame. As far as I can tell, you are defining them differently. =Uncool-

EDITED: If you define x_1' as being the distance between source and wall, then that is true. However, in this frame, light does not have to cover that entire distance - the wall is moving towards the light already. If you're defining x_1' as the distance between the origin and the wall, then yes, I am contesting that. =Uncool-

Because we start with x_1 = x_2. We do not get that x_1' = x_2'. =Uncool-

At the moment, we're not talking about your idea. We're talking about relativity. You still have yet to show that there is a problem with it; I have shown you why your objection to it is wrong. Additionally, in context, your post makes no sense. There is no H; there is no W. =Uncool-

But relativity is more than just slowing time - again, space in one frame becomes time in another. Therefore, the fact that the spatial coordinates are different can change that. In this example: From the frame where the walls are standing still, the events we see are: Light rays emitted: [math](0,0,0,0)[/math] Light ray absorbed on the left: [math](\frac{L}{c}, -L, 0, 0)[/math] Light ray absorbed on the right: [math](\frac{L}{c}, L, 0, 0)[/math] From the frame where the walls are moving, according to the standard transformations: Light rays emitted: [math](0,0,0,0)[/math] Light ray absorbed on the left: [math](\frac{\gamma L (c - v)}{c^2}, -\frac{\gamma L (c - v)}{c}, 0, 0)[/math] Light ray absorbed on the right: [math](\frac{\gamma L (c + v)}{c^2}, \frac{\gamma L (c + v)}{c}, 0, 0)[/math] which can alternatively be written as: Light rays emitted: [math](0,0,0,0)[/math] Light ray absorbed on the left: [math](\frac{L}{(v + c)\gamma}, -\frac{cL}{(v + c)\gamma}, 0, 0)[/math] Light ray absorbed on the right: [math](\frac{L}{(v - c)\gamma}, \frac{cL}{(v - c)\gamma}, 0, 0)[/math] Which is exactly what you got, up to the factor of [math]\gamma[/math], which is there due to the spatial dilation (the barriers are closer together than in the non-moving frame). You have to include the effects of spatial displacements. =Uncool-

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This is gibberish. I think I can guess what you're saying, but I would like to make sure, so please try to reexplain. =Uncool-

No, it doesn't. It alleges that the times are identical in another frame. You forget that time is frame-dependent in relativity. You are mixing measurements from different frames. Relativity says that whether the transit times are the same is frame-dependent - in this frame, where the objects are moving, the transit time will be different. However, in the frame where neither are moving, the transit time will be the same. Relativity also gets that T1 = L/(c + v) and T2 = L/(c - v); but those are only applicable in this frame. In other frames, the time could be different. =Uncool-

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Why not? Which it shouldn't be, because light is traveling at the same speed in both directions, and is covering the same distance. Because you specified the same distance to be traveled. The time should not be different - because the speed of light should be the same in all directions in all frames. Remember, time and space mix - space becomes time and time becomes space. =Uncool-

This is still gibberish. What do you mean by "Time is not different"? =Uncool-

Where? Please demonstrate this. This is gibberish. Please quote the precise part of your post where you show that the equation is incorrect. I don't want an assertion; I want a specific proof. =Uncool- Again, Alex, your entire argument comes down to the assertion that changes in time-ordering are impossible. You have not given any reason to believe this. You have not shown that relativity violates causality in any way. As an analogy to Galilean mechanics that you might understand: Do you agree that you can order events - that is, points in space-time - according to x-position, given a choice of frame? Further, do you agree that this ordering changes depending on your frame? =Uncool-

No, you never have. First, under special relativity we don't see any two frames where one accelerates time relative to the other and the other slows down. Second, why does one exclude the other? You have yet to demonstrate this. =Uncool-