uncool

I consider it to be correct, not sacrosanct, because I understand how it is proven from the postulates, and why the postulates are assumed. If light velocity is always c in the vertical direction It is not. It is always c in total. It is only c in the vertical direction when there is no other component to the velocity. =Uncool-

The pulse will hit the target, at speed c. SR is real. The laws of SR are real. The correct answer is that the light will hit the target, moving at c on a diagonal path relative to O1, but the person in the ship will still see it moving at c in the y-direction. You are saying that that is impossible because you have an unstated assumption about how velocities add, which contradicts SR. Except that your "analysis" of light emission is making assumptions itself that you aren't noticing. Note all of your assumptions whenever you make a conclusion here - you are assuming that there is no time dilation and coming up with a contradiction. =Uncool-

Ahh. That is what I misunderstood. Only if you assume Galilean relativity. In other words, you have made two contradicting assumptions; one must give. Unless "ballistic" means the same thing as "Galilean" to you. Relative to which frame? The original frame (O1) or the new frame (O2)? Only once you make the assumption that there is no time dilation (or sideways spatial dilation, although that happens to be true even in special relativity); in other words, you've proven (again) from the assumptions that there must be time dilation. Only if you discard one of the key components of special relativity - time dilation. In other words, you are specifically assuming that special relativity has contradictions to prove it has contradictions.=Uncool-

Let me make sure that I have your initial ideas correct. So you start in a spaceship which has velocity (0, 0, 0) relative to an inertial frame O1. This spaceship lets out a light pulse in the x-direction (relative to O1) towards a stationary (still relative to O1) detector at (x, 0, 0), right? The spaceship then acquires a y-velocity of v, giving it a velocity of (0, v, 0). Then there is another inertial frame O2, relative to which the spaceship is stationary (but then the detector is moving at (0, -v, 0)). This spaceship lets out a light pulse in the x-direction (relative to O2) Is that the situation so far? ETA: Or is the detector inside the spaceship? =Uncool-

So it is a given that at least 51 of the villagers are honest? =Uncool-

Here is the way to think about it under special relativity: Let's say that we have a spaceship which emits photons in the x-direction. Then in the frame of the spaceship, the path of the photons is the set of all points (t, ct, 0, 0) for positive t. We now take a look in the frame in which the spaceship is moving at a rate v in the y-direction. If we have an event (t, x, y, z) in the frame of the spaceship, the event becomes [MATH](t', x', y', z') = (\gamma (t + v/c^2 y), x, \gamma (y + vt), z)[/MATH] So the event (t, ct, 0, 0) becomes [MATH](t', x', y', z') = (\gamma t, ct, \gamma vt, 0)[/MATH] If we scale t, we get [MATH]t*(1, c/\gamma, v, 0)[/MATH] We then get that the speed is [MATH]\sqrt{(c/\gamma)^2 + v^2} = \sqrt{(c^2 - v^2) + v^2} = c.[/MATH] So no problem with relativity. In short, the answer is that it moves with the same transverse velocity as the source, but the parallel velocity is changed. =Uncool-

5 = 6*1 - 1 7 = 6*1 + 1 7 - 5 = 2. The twin primes are of the form (6k - 1, 6k + 1) for some k. =Uncool- First, you have to define what you mean by the "average distance between prime numbers". Do you mean: [math]\lim_{n \rightarrow \infty} \frac{p_n}{n}[/math]? If so, then you are wrong, as that limit is infinite. To see why you are wrong, consider a 210-sectioned spiral. Then there are a total of 48 (mod 210) possibilities, meaning that according to your argument, the average must be 210/48 = 35/8 = 4.375, which is substantially larger than your number. Not only that, but it is always possible to keep going up (next, try 2310, which gets you 4.8125). =Uncool-

Actually, there are only 6 quarks (and the corresponding 6 antiquarks, but in nearly every system you will have to include corresponding antiparticles). The other members of the "particle zoo" are additional leptons and force carriers. Neither gluons nor Higgs bosons are quarks. They are other parts of the Standard model, but they are not quarks. That is most certainly not the definition of elementary. When colliding electrons, many other particles are found. Quarks do not "fall apart on their own". They decay into other kinds of quarks, just as electrons can. The fact that electrons generally don't is simply due to the fact that electrons are the lowest energy state. Do you know what the definition of an elementary particle is? Err. Why not? The quarks can rather easily react with each other. The total charge will be 2 the entire time. Additionally, the theory is not that protons collide and form 6 separate quarks. The theory is that two protons are 6 separate quarks. The theory of quarks is equivalent to the idea that the down quark has charge -1, the up quark charge +2, and the proton +3. Fractional charges are equivalent to non-fractional charges, simply by multiplication. Actually, what quarks have going for them is a solid, decently-understood theory. Not including gravity, the theory is well-understood, and has provided many predictions borne out by experiment. Additionally, there is physical evidence that quarks exist from experiments that showed that nucleons were composed of three pointlike particles. If you want to criticize quark theory, you should learn more about it. Do you know quantum field theory at all? =Uncool-

Let's consider an event that happens on Earth (x1 = 0) at time 0 (t1 = 0). Then consider an event that happens on the sun (x2 = 1 AU) 10 minutes later in the Earth's frame of reference (t2 = 10 minutes). Then the spacetime interval will be: (c(t2 - t1))^2 - (x2 - x1)^2 = (10 lightminutes)^2 - (1 AU)^2 = 9.97562483 × 10^21 m^2. Now, the claim is that in all frames, the spacetime interval will be the same. =Uncool-

If both are measured in Earth's FOR, c. If the 0.02c is measured in the frame moving at 0.98c (as is assumed in the relativistic velocity addition argument), then slightly less than c. =Uncool-

Actually, you're not missing anything. One of the major points of all field theories is to analyze everything through the fields, rather than through the particles. =Uncool-

In that case, you don't have Can you make an explicit statement of the difference? In physics, a force is applied at any time when an object accelerates. Therefore, gravity is a force, and the string is a force. And forces have been mathematically analyzed. =Uncool-

Not that it's perpendicular to the Earth. That it's moving perpendicularly to the displacement from the Earth to the moon. And the moon does speed up and slow down relative to the sun. However, in most orbits, you try to use a frame in which the object around which you are orbiting is not moving - in which case, force is perpendicular to velocity. The string is accelerating (note: physics word - doesn't necessarily require the speed of the object to be increasing) the ball. That means that it is applying a force to the ball - in other words, "pulling" on the ball. Do you have a mathematical description of your theory? This theory is more than well-thought-out. =Uncool-

No. I think that ajb wasn't quite thinking about it correctly; this actually corresponds the notion of the difference between "spacelike" and "timelike" separations. It is true that for any pair of events, there is a frame where either the two can be said to be simultaneous (in which case the separation between the two is called 'spacelike') or there is a frame where the two are considered to have happened at the same location (in which case the separation between the two is called 'timelike'). Something that is true: for any pair of events, there is a third event such that the first and third event, and the second and third event, are separated by spacelike intervals. =Uncool-

Victor: would you mind posting, in one post, your entire proof? =Uncool-

The 32 elements are: 0, a b, a + b c, a + c b + c, a + b + c d, a + d b + d, a + b + d c + d, a + c + d b + c + d, a + b + c + d e, a + e b + e, a + b + e c + e, a + c + e b + c + e, a + b + c + e d + e, a + d + e b + d + e, a + b + d + e c + d + e, a + c + d + e b + c + d + e, a + b + c + d + e =Uncool-

Ahhh, I misunderstood. I thought you meant that it was trivial without assuming FLT; . =Uncool-

Err. No, it is not trivial. I'd like to see your proof of it, because proof of that is entirely equivalent to Fermat's Theorem, as follows: Assume a solution to a^n + b^n = c^n. Then let a' = a/b, c' = c/b. Then a'^n + 1 = c'^n. Therefore, if what you say is true, we have a contradiction, and we are done. =Uncool-

Giving the Cayley table would take a long time; writing the relations of the generators is enough. Additively: a + a = 0, b + b = 0, c + c = 0, d + d = 0, e + e = 0 a + b = b + a a + c = c + a a + d = d + a a + e = e + a b + c = c + b b + d = d + b b + e = e + b c + d = d + c c + e = e + c d + e = e + d fully describes the addition in the group. Multiplicatively: a*a = c a*b = d a*c = 0 a*d = 0 a*e = 0 b*a = d b*b = e b*c = 0 b*d = 0 b*e = 0 c*anything = 0 d*anything = 0 e*anything = 0 =Uncool-

That must be true. A spring is a partial ring which is not additively commutative, correct? If so, then the set of products cannot be the entire thing, as the set of products is commutative. See earlier: Additively, it's (Z_2)^5 - it's a ring without unity. To be specific: c, d, and e all have trivial multiplication. Then a*a = c, a*b = b*a = d, b*b = e. Then, as it is commutative, and only 3 elements do not eliminate anything, we can see that there are only 6 products, not the 7 that would be necessary (c + d + e is not a product). =Uncool-

Actually, I think there is a normal ideal here. The normal closure of I is the set of all elements such that their abelianization is generated by x. This is a normal subgroup, and as it contains the brutal ideal, it is an ideal. Therefore, it is a normal ideal. Also, the products are always a brutal ideal, if when you say "build" you mean generate. It is not true if you mean that they are a brutal ideal - there is a 32 element group in which the products do not form a subgroup (one is missing). =Uncool-

Interesting; seems like this could work for any homomorphism to the integers... Let G be a group; let f be a nontrivial homomorphism from G to the integers. Then let g be an element which does not go to 0 (which would mean that the subgroup generated by g is infinite). Then define x • y as g^(f(x) + f(y)). ETA: This also happens to work when it's going to any cyclic group. The natural version of the ideal in a pring is the normal ideal - that is, a normal subgroup which is an ideal. For those, the two equivalences are the same. =Uncool-

No problem; I like feedback on this. I'm trying to figure out whether these partial rings are the symmetries for any objects at all. Also wondering if it's possible to create pring-modules... =Uncool-

I would advise you to rethink this. I read your post very carefully. I thought that you were unclear. In that case, using |OB| as the bound of integration is probably not a good idea. Good; then my advise is to not name two different things in the same way. You should not use a variable as both the boundary and the bounded variable in an integral - it causes confusion. By the definition of limits, this is also 0. =Uncool-

Honestly, mishin, this gets old quickly. Assuming that everyone that disagrees with you is "self-assuredly silly" is a way to quickly get ignored. I have been quite honest with all of the reasons why I think that there are problems with what you are saying. Shouldn't you use dx rather than d|OB|, since |OB| is a constant? If anything, |OB| is a bound of integration, and dx is the differential that should be used in integration. You are using |OB| as if it were changing throughout the integral - as if it were something that were different for each of the "thin plates" that is summed to get the area of the rectangle. You should use a different variable instead. mishin, a major problem that you have is the fact that you use variables of integrations as bounds of integration. That is completely and entirely a mistake - it makes your statements nearly incomprehensible. In other words, your last equation should be written: [math] S _ {OFGB} = |OF |\cdot|OB | =\int\limits _ {0} ^ {|OB |} |OF|dx [/math] since the name of the variable of integration doesn't matter. =Uncool-