uncool

You can't simply take the inverse transform; the quotient of the transforms isn't the transform of the quotient. However, I'd recommend that you look at the equations just before you take the quotients. You have a full equation for X(s): X(s)[s^2+s+a+c+sa+sc+sb+ba]=0 You should know that X(s) isn't 0 for all s; so that should allow you to "solve" for s. I would note, however, that in this context, the Laplace transform technically shouldn't be applied.

When asking whether it's faster, you'll have to define how "fast" each operation is. Please note that the following is from memory: I'm going to approach this from two assignment of "speeds" to each operation. The first assignment is that multiplication takes "infinitely long" compared to addition. In other words, the way we "rate" any algorithm that only uses addition and multiplication is to check how many multiplications it uses. Using the usual factorial algorithm (put 1 in memory for the factorial, put 1 for the index, start the loop, multiply the factorial by the index, add 1 to the index, end the loop), you need n multiplications to find n!. Using your algorithm, you will also need n multiplications, as follows: Assuming n is odd: 1: Find (n + 1)/2 (this doesn't actually count as a multiplication but rather a shift; multiplication and division by 2 is a special case). 2: Find ((n + 1)/2)^2 (this does count; you are multiplying (n + 1)/2 by itself) - 1 multiplication 3: For each i from 1 to (n + 1)/2 - 2, find ((n + 1)/2)^2 - i^2 (one multiplication), and multiply that in (another multiplication) - 2 multiplications for each i from 1 to (n + 1)/2 - 1. Step 3 gives us a total of n - 1 multiplications, while step 2 gives us one multiplication, for a total of n multiplications. If n is even, it takes (n - 1) + 1 = n calculations, since you reduced it to the case that n is odd. The second assignment is a bit more interesting. When doing circuitry, it turns out that the algorithm for multiplication takes 3 cycles, whereas addition only takes 1 cycle, so to evaluate how many cycles it takes, you "assign" any algorithm 3 times the number of multiplications it does plus the number of additions it does. The "naive" factorial algorithm just adds 1 each time and multiplies, for a total of 4n cycles. I think that your algorithm uses 2 additions (one by adding 1 to i, and one for subtracting the squares) for each i from 1 to (n - 1)/2 and one more (to get (n + 1)/2 in the first place); it does use 2 multiplications for each i from 1 to (n - 1)/2 and 1 more, for a total of n multiplications and n additions. So it turns out that as long as all you care about is additions and multiplications, your method will take exactly as long as the naive method. I would be surprised if there were a "fast" method to calculate factorials exactly without using some kind of lookup table.

Not quite sure I know what you are saying here; length contraction is direction dependent. In the direction of travel, there is contraction, while perpendicular to the direction of travel, there is no contraction.

To be exact, it means no acceleration of frames. Objects can accelerate, but in SR inertial frames are in constant rectilinear motion with respect to each other.

Take f(x) = 2x + 2; f^{-1}(x) = x/2 - 1. d/dx ((2x + 2)(x/2 - 1)) = d/dx (x^2 - x - 2) = 2x - 1 f(x) = 2x - 1; f^{-1}(x) = x/2 + 1/2 d/dx ((2x - 1)(x/2 + 1/2)) = d/dx (x^2 + x/2 - 1/2) = 2x + 1/2 For affine functions (f(x) = mx + b), you will always get that d/dx (f(x) f^{-1}(x)) is also an affine function; then if we call it m' x + b', then m' will always be 2, and b'/m' will always be equal to -1/2 b/m.

I wouldn't phrase that as saying that time is "hyperbolic" and the spatial dimensions are "circular"; the way the transformations work deals with the relationships between the dimensions. So for example, if instead we had 2 temporal dimensions and 2 spatial dimensions, the transformation between the two temporal dimensions would be "circular", the transformation between the two spatial dimensions would be "circular", and the other transformations would be "hyperbolic". In other words, you'd get the following transformations: t1 -> (cos s) t1 + (sin s) t2 t2 -> (- sin s) t1 + (cos s) t2 x1 -> x1 x2 -> x2 OR t1 -> t1 t2 -> t2 x1 -> (cos s) x1 + (sin s) x2 x2 -> (- sin s) x1 + (cos s) x2 OR t1 -> (cosh s) t1 + (sinh s) x1 t2 -> t2 x1 -> (sinh s) t1 + (cosh s) t1 x2 -> x2 (where you can switch x1 and x2, or t1 and t2 for the last transformation). So as you can see, it's not an aspect of the "temporal dimensions"; instead, the "hyperbola" comes (in some sense) from the interaction between the temporal and spatial dimensions. Between two temporal dimensions, you would still get a "circle".

My point is that you are using standard symbols to mean something non-standard and expecting others to understand what you meant without explaining. What do you mean by your symbols '1', '+', '=', and '2'? I can answer precisely what all of those mean to me and to scientists. Don't blame us when you have yet to explain precisely what you mean.

You said this: This can only be correct if by "synergetically", you mean something entirely different by + and = from what standard mathematics says.

md65536, it seems that you have confused simultaneity with synchronization. Einstein's paper says that two clocks are synchronized if the delay is equal.

You may want to restrict the second one to the right-handed limit, i.e. [math]e=\lim_{n\to 0^+}{\left[n+1\right]^{\frac{1}{n}}}[/math] Then, we can easily see that the two are equivalent by looking at the first limit and letting [math]n = \frac{1}{n'}[/math].

Depending on what waitforufo meant precisely, "voting themselves raises" doesn't necessarily require violating the 27th amendment. First, "themselves" could refer to Congress as a class, rather than as its individual members; then they can vote themselves raises, although it takes effects in the next cycle. Second, given the huge rate of incumbency in Congress, there isn't that much difference between cycles. So if they vote for a raise, most of them will survive the next election to receive the raise in the next cycle. =Uncool-

Where do you get the idea that "scientists are going to impose a punishment" for anything? How would scientists "impose" this punishment? Do you think that scientists control the government? "Scientists do not like for transitions of cesium isotope atoms to be considered in this way" because they have shown it to be inaccurate, just as we could say that scientists do not like for the Earth to be considered as flat. Once again: you are free to be wrong. Just because someone tells you that you are wrong doesn't mean that they are going to punish you (unless you consider laughter to be punishment). =Uncool-

Correct - the clocks on the airplane showed different times because they measured this particular trip in a slower or faster way. As I said before, they would show a different amount of time for a different plane (while they stay on the same plane). That is because of the x-dependence of t'. I have no idea what you are trying to say with this. It seems like nonsense. Except that's not a prediction, it's a postdiction. But further: please do so. Find out exactly how much change you actually get. I'm guessing that the amount of change will be far off from what was actually found (note that I am referring to the amount of change, not the actual values; relativity predicts the amount of change accurately, so you need to as well). You have the order wrong. The prediction was the combination of the Lorentz equations and GR; that prediction came before the experiment. And in doing so, you will fail to predict. Yes, they do; that is the point of t = t'. You are adding another transformation. Which is fine - but it's not the Galilean transformation equations; it's your expansion. And as I've said before, you are free to use them incorrectly; no one has to accept that your idea is correct (and it is not). =Uncool-

Yes, and that's the point - you can't simply say x^(1/2), since that is ambiguous. You can't just say (-1)^(1/2) must be i. =Uncool-

You completely ignored what he wrote in blue. No, a/i is not both ai and -ai; it is -ai, period.

As I've explained, no, they are not faster or slower. That is your addition to relativity. And as I've explained before, with relativity the clocks aren't "slower" or "faster". That is not part of relativity. They are different, in a way that is determined by the Lorentz equations. That is what relativity says. If you wanted to respond to what relativity actually says, you could claim that it is physically impossible for a clock in K' to be neither slower nor faster than a clock in K. You'd still be wrong, but you'd be dealing with what relativity actually says, rather than your imagined version of relativity. The problem isn't with relativity itself - it's with your imagined version of relativity. What you are saying is that you couldn't determine it from your equations directly, meaning that you can't predict the experimental values. On the other hand, GR and SR did predict these values, and pretty accurately at that. And in science, you take the theory that predicts over the one that doesn't. That's not a prediction. That's a postdiction. Prediction comes from the Latin dico dicere, to say, and pre, meaning before. You are making pronouncements after having seen the answers. John Cuthber was asking you to get the answer without knowing the outcome. That's the problem - your equations don't predict the experiments. SR and GR, on the other hand, predicted the values - they figured out what SR and GR implied the values would be, and then ran the experiment, and found them to match up well. That's the sign of a good theory. He's saying that your idea doesn't predict the results of the experiments, and you have just confirmed it. Why? The request to use them properly was a further request, not a "Use them properly or not at all" request. He hasn't "forbid [you] from using them". No one has. He's pointed out that you've been using them wrongly. Galilean transformation equations assume absolute time. That's what swansont's been saying. There's been no "forbidding" - it's a simple fact about the Galilean transformation equations. They are, but that doesn't mean that you're using them correctly. =Uncool-

Well, that is fine. We were told that the moving clocks were slower, which is the way I always thought of it. Then you were relying on a depiction of relativity that is meant to be shown to people who have poor understandings of math and physics - the idea that there isn't necessarily a "faster" and "slower" that can be applied to all clocks is confusing, and less likely to help people understand. Well, scientists have to agree on something in the end, and they are going to agree on whatever seems most advantageous to scientists. If they are getting large sums of money from governments to do "research" using Einstein's theory, it seems to me they are going to be influenced to some extent by that. But you've gone far beyond that. What you said was that they would do "whatever it took to say the experiments matched." That's not just "being influenced"; that's fraud. No, I was just pointing out that events with regard to a light ray directed in the -x direction do not agree with what scientists have been saying. Oh? You have an experiment that shows this? Again, every experiment that we have tested shows that the speed of light is the same in every direction relative to every frame. That is what scientists have been saying; that is what experiment shows. I do not agree with scientists about relativity of simultaneity either. I realize that. That's irrelevant to what I was saying. The point I was making was that you are saying that relativity has a problem because you have added a law that relativity doesn't assume to it and the pointed to a contradiction. The problem isn't with relativity, it's with your addition to it. On a side note: the way you are currently using quotes mixes your text with mine. Please try to use [ / quote ] and [ quote ] (remove the spaces) to distinguish. Example: [ quote ] [ quote ] I say blah [ / quote ] You respond blahblah [ / quote ] I respond blahblahblah to your response. Becomes You respond blahblah I respond blahblahblah to your response. Having that distinguishing feature makes conversations much, much, much easier. =Uncool-

And as others have told you, that is a falsehood about scientists. You have a very, very distorted view of science and scientists. No, they have not. They have disputed your beliefs about the Galilean transforms. That is not the same thing, and goes to illustrate the distortion I mentioned earlier. Then don't discuss it. That's your problem, not one for scientists. Do you want me to make a list? I'll start with the neutrino scientists at OPERA, who within the past 3 years published a paper that demonstrated results that seemed to dispute relativity. In science, things that conflict with the mainstream are much more likely to get money - as long as they are done correctly. This doesn't follow from the above. It also is blatantly false and defamatory. There are plenty of scientists who still use the Galilean transforms. Just not in the realm of high speeds, where special relativity is much closer. As you've been told, it's not prohibited - it's just incorrect when dealing with high speeds (and no, 30 mi/s is not a high speed). Yet another example of distortion. And to match the experiments. Which, as you've repeatedly be shown, you don't do. =Uncool-

To be fair, he isn't claiming to be a "great scientist"; he apparently thinks that scientists are frauds. You can feel free to use the Lorentz equations to determine that what I am saying is correct. You may be interested in the fact that scientific reporting is very poor. A better way to say it would be that the moving clock was slower relative to the frame of the Earth. Which is what relativity predicts. That is an accusation of fraud that you really need to have a lot of evidence to show. You are accusing scientists of effectively making up data, which is one of the most heinous things you can do as a scientist; scientists are trained from the start to avoid it, and with an experiment this high-level, there is a huge amount of intense scrutiny involved. But it does make further predictions - namely, that the speed of light is not isotropic relative to every frame. Which has been experimentally tested. But as you've already said yourself - light travels both farther in more time (in the -x direction) and shorter in less time (in the +x direction), which means that the clock is neither faster nor slower. You are requiring relativity to follow a rule that it explicitly not only lacks, but violates. In other words, you are still setting up a strawman version of relativity. Once again, special relativity says that clocks that are moving at different velocities are neither "faster" nor "slower" than each other. This is a part of what is meant by the relativity of simultaneity. By requiring that there be a "faster" and "slower", you are adding in a contradiction to relativity that doesn't exist without your additional assumption. =Uncool-

Not quite. What it shows is that less time has elapsed on this flight. In other words, the clock that was on a plane (call it plane 1) measured this trip in a "slower" way. It would have measured a different trip in a different way, that is, the clock on plane 1 would have measured the rate of speed of a clock on another plane, say plane 2, in a different way, possibly faster. As I've said, position dependence of time is important here. That's what relativity of simultaneity is all about. As I have said, the Lorentz equations don't have one clock being "both slower and faster" than another clock, what it has is one clock being neither slower nor faster than another clock. And that is not a physical impossibility. The dependence of t' on x is of extreme importance. It implies that at some positions, you will have t' > t, and at others t' < t. And there is no reason for that to be a physical impossibility, but it does contradict your version of Galilean relativity. =Uncool-

Well, I guess you have enough concerns about this to use the quote button this time. I already showed that if the speed of light is c in both frames of reference for a light ray in the -x direction, then the Lorentz equations require a faster clock in K' instead of a slower one. If you think you can show something else, go ahead and try. t' in the Lorentz transformation equations is equal to t'=(t-vx/c^2)gamma. I don't recall saying anything different. And as I've already explained, the clocks for the Lorentz transformation aren't "faster" or "slower" than each other - that's a comparison that simply makes no sense. Which is why I asked you a question that didn't involve "faster" or "slower". Would you mind simply answering the yes or no question, namely: Are you claiming that my statement that " The speed of light is isotropic in every frame when using the Lorentz transform. " is false? =Uncool-

I'm only using your logic.

I want to make sure that I understand before I go on; please answer yes or no to the following question. Are you claiming that my statement that " The speed of light is isotropic in every frame when using the Lorentz transform. " is false? Which is precisely what the Lorentz equations predict, as I can show (if you want). Do you want me to do so? As I've said, there is another part that is important - the mixing of time and space. Namely, t' is not equal to \gamma t, but \gamma (t - v/c^2 x). As I've said before, you can't simply drop terms because you don't like them. What concern? Again, which part of which post here are you responding to? This is why I've asked you to use the quote button. =Uncool-

No, it can't, since i is neither positive nor negative. You can't say "the positive one" except when dealing with real numbers directly, and the square root of a negative number is not real. You did say (in effect) that (-1)^3 means the same thing as -1^3, by saying: Further, you haven't answered the question I've been pointing out. According to your logic, don't you have that: root((-1)^3) = ((-1)^3)^(1/2) = (-1)^(3/2) = ((-1)^(1/2))^3 = (root(-1))^3? =Uncool-

This doesn't seem to respond to anything anyone has said. The closest I can see is what swansont has said with 'The "Galilean transformation equations as I use them" are not the Galilean transformations. ' And your response has nothing to do with that; his point is that what you are calling the "Galilean transformation equations as I used them" are not the actual Galilean transformations. What do you mean by "flipping of frames of reference"? The objection that I have stated is that the speed of light is not isotropic except in at most one frame, if your idea is correct, which is absolutely not true for the Lorentz transform. The speed of light is isotropic in every frame when using the Lorentz transform. If you are stating that either part of what I have said is false, please demonstrate it; if you are saying something else, please try to explain what you are saying in a different way, since "flipping of frames of reference" doesn't mean anything to me. =Uncool-