timo

On a sidenote (the term cross-product was already mentioned - I'll spare you of more formal terms that I'm not too familiar with, myself): The sine-term has an important disadvantage: Assume I use the proposed method and come up with sin(angle) = 0. What's the value of angle? You just don't know - and since I'm I let you figure out the "why", yourself. The value of 0 was just an arbitrary choice, in fact one where you immediately see the result for angle. But I could as well have picked 0.623, where the result is not so obvious, anymore. EDIT: I'll give you some time to think about my comment some more (or not) and post the solution later - tomorrow. It's not a big thing anyways. It's just a very suggestive pitfall that you already stepped into and that you'll hopefully remember (and avoid from thereon) after having seen what I meant.

Which needs a few corrections . Depends on what you call "really good". Realisticly, the original purpose of creating a scientific encyclopedia that's more accurate and detailed than other Wikis out there was pretty absurd from the very start . I considered more as a mixture as a blog and an editable FAQ, for which it did indeed serve well (at least I linked to my articles there when writing forum replies once or twice and had done so some other times after WiSci was gone). But it seems that was "after your time" (considering I wrote the articles after the letter project was already dead). Afaik, like all -to my knowledge- sfn-related projects started by sfn non-staff members, the letter was never finished and sent.

Ok, I have to think about it some more, but what I can already say: You might have misunderstood the proof of the infinite number of primes. It's a proof by contradiction, similar to this (not being a mathematician, I think it's necessary to add the "similar to" part - I don't exactly know all the subtleties involved): 1) Assume there is a finite number of all primes (it doesn't really matter if they are "known" or not). Call the set p. 2) Then, the product P of all those primes exist. 3) P+1 does not divide by any of those primes. Hence either 3a) P must be prime. 3b) Or, there must be some missing prime P' that divides P. 4) If a prime P+1 is not in p (which seems obvious, dunno what the exact reason is), or a P' exists, then you have a contradiction to the assumption. 5) To clearify: This then means that there is no finite set of all primes and that therefore P+1 does not exist. Ok, what do we learn from that: - There is a non-finite number of primes. Hence, you cannot contruct new primes in the exact way you tried in the proof. - Your next-best guess would be that you just take any finite set of primes and try to pull off that trick. Of course, you quickly find a set of test-primes where this won't work (namely at least all that don't contain 2). - Next-best guess is taking the N lowest primes. This might work, but I don't really see a reason why the N+1'th prime shouldn't divide that product+1 (or -1). Might be interesting to plug that into the computer, it might spit out an example quite quickly. EDIT: Product of 6 first primes plus 1: 2*3*5*7*11*13 +1 = 30031. 30031/59 = 509.

Darn, my post got deleted. So you get the sketchy version: - The accelerator you are looking for is (most likely) the LHC at CERN. - It has not been run in full-power mode. Dunno what the tests already done included. - The section is not too unfitting. If some moderator thinks it belonged somewhere else, he can move it, anyways. - http://en.wikipedia.org/wiki/Large_Hadron_Collider , especially the last paragraph about "contruction accidents". - http://press.web.cern.ch/press/PressReleases/Releases2007/PR06.07E.html for the seemingly official statement about LHC shedule. EDIT: [nitpick] CERN is not a particle accelerator [/nitpick].

Because the do? Drop a piece of paper from a height of 2 m and stop the time. Then do the same with your laptop . Pieces of papers and laptops falling down is much more common than metal balls that are later melted into a single ball.

Earth's surface is a good approximation to an inertial frame in this case. The correction due to the gravitational field that you stay in is in the order of [math]10^{-8}[/math]. Correction due to earth's rotation is probably in the same order (too lazy to calculate it). Corrections for the house are even smaller (and counter the corrections for the earth frame, anyways). That depends on the point of view. From my point of view, the equations are the same in every frame: The eigentime tau an objects experiences during its path is: [math] \tau = \int_0^1 \| \frac{d\gamma (x)}{dx}\| dx [/math], where [math]\gamma(x)[/math] is the curve the object takes through spacetime (not the Gamov factor!), x is the curve parameter running from 0 to 1 and [math] \| v \| := \sqrt{g_{\mu \nu} v^{\mu} v^{\nu}}[/math] (I know ppl who would kill me for this notation ). You might want to keep that formula in mind for your lecture for that it plays a very fundamental rule (of being the definition of eigentime and also the starting point to derive the GR equations of motion - we even had an article on WiSci about that). What I wrote was just the specialization of this for a certain choice of spacetime geometry (approximately flat) and coordinate system (time-orthogonal cartesian, resting with the guy on earth). As said, you can chose any coordinate system in principle. If you wanted to take into account earth's gravitational field, the rotation of earth and make the simplification that there are no other spacetime-bending bodies, then you could take the Schwarzschild metric and have both bodies moving in that CS accordingly (that's the assumption I have approximated the [math]10^{-8}[/math]-correction from). The big point in GR is: You can chose any coordinate system you want, the result for certain types of values (scalars like lengths of trajectories, depending on point of view possibly also vectors and tensors) will, and must, always result in the same value. I don't really understand what he said. Perhaps if you quoted some passages it would be easier for me to understand what they/you mean. Have fun with your GR course. Personally, GR was by far the most interesting course (topic-wise, the lecturer was no too good) I took in university. Yes and no. SR is just an approximation. I'm not sure if I agree with the statement "extra term" for that the calculation approach is simply a bit different. But of course you can always define "extra term" = "GR result" - "SR result". That depends on where you draw the line between SR and GR. Some people say SR is flat spacetime and non-accelerating frames. Others count accelerating frames to SR (the step to accelerating frames is not much of a deal) and draw the distinction to GR when non-flat geometries are involved. Personally, I tend to go one step further and make the distinction at the point where the influence of energy on spacetime is considered (that is, when the equations of motion of objects within spacetime are completed by the field equations that describe the "reaction" of spacetime on the particle content). Either way, and that probably clears things up a bit: The equation I posted was derived by GR methods -using some approximations- in the first place. I just (and not too surprisingly) turned out to approximately result in the equations you'd expect from SR in the first place. I barely think about relativistic problems in the SR context (quantum theories being the big class of exceptions) because conceptually I find GR easier to understand.

Sometimes, the words around the equations are important, too. "In the inertial frame you stay in" meant earth's frame. Tau is the time measured by the ppl in the house, t is the time measured on earth, v is the velocity of the house as seen from earth. I did not claim the equation to be true in the other frame. I have not stated the integration boundaries (because they were irrelevant for my statement). Certainly, both objects (house and you) experience a certain time between the NASA theft and the reunition. These times cannot depend on a chosen frame of reference. So it is possible to explicitely calculate the in one frame (like the one for which the equations are true) and know other possibly more complicated results would give the same. Assuming a specific acceleration profile, the results could look something like this: , with T being the total time measured in earth's frame.

Yep. And personally I don't really see the point of this thread. In the end, it just seems to use a different meaning of time travel than I'd expect. No, but you probably know that yourself and just used sloppy language. Not exactly sure what you mean. Yet, I think a fitting reply is "no". I think here I can safely say "no". It's a frame-dependent statement at best. I'm even more convinced of a "no", here. I'll pass on this one. In the inertial frame you stay in, [math] d\tau = dt/\gamma(v), \gamma \geq 1.[/math] No acceleration term in there. For me, that's more convincing than the next-best google hit. The idea works, except for the badly chosen numbers (your friends will most likely not want a beer after being smashed by the enourmous acceleration required for these numbers) and NASA not having the technology to do it - but that's nitpicking. I just don't really see the point of the example. The result seems pretty much a triviality (like said, it's just the twin's paradoxon).

Yeah, I had that suspicion, too. Hence, the disguised attempt to find out if that point distribution is a global phenomenon.

I bet against that. You can see which posts you were given rep points somewhere under "User CP". Speaking for me, it's all been spam posts or one-liners that people gave me reputation points for . Reputation points are given by members for whatever reason they have giving them. This can be thinking a post was a valuable contribution (by whatever criteria) or for simply thinking "nice post, it made me laugh". Assuming that people have the tendency to remember people they gave points to better, the term "reputation points" seems to fit. I don't get your point about primes and quantization. Integer numbers are obviously quantized.

Probably the signature. No, not really.

Density does not directly affect air resistance but it does indirectly. Frictional force is a function of the object's geometry (=shape) and velocity - physicists sometimes write this in a shorthand notation [math] F_F = F_F(\text{geometry},v) [/math]. Now let's assume some geometry and velocity are given and look how mass will affect the acceleration: [math] ma = F = F_G + F_F(\text{geo},v) = G \frac{m M_0}{r^2} + F_F(\text{geo},v) \Rightarrow a = \underbrace{G \frac{M_0}{r^2}}_{a_{\text{free}}} + \underbrace{ \frac{F_F(\text{geo},v)}{m} }_{a_\text{frict}} [/math]. The term [math]a_\text{free}[/math] is independent on the mass of the object. The term [math]a_\text{frict}[/math] (which has a different sign than [math]a_\text{free}[/math], meaning it's magnitude has to be substracted from [math]a_\text{free}[/math]'s magnitude - I just dropped the vectors in favour of readabillity) however depends on the mass. Therefore due to the extra term [math]a_\text{frict}[/math] increasing mass will lead to increasing acceleration once you consider friction. Now to come back to how density affects acceleration: Keeping geometry the same and considering the same current velocity, higher density will lead to a greater mass, decrease the magnitude of [math]a_\text{frict}[/math] and increase acceleration a. EDIT: Or in short and trying not to overcomplicate things: Yes, that (your previous post) is more or less true.

I think the part of dropping stuff from the tower of Pisa is just a myth. Doesn't matter for the physics, though. The common view is that density does not play a role, either - you could as well do the thing with lead and gold (I propose gold nuggets rather than gold cannonballs for economic reasons). What the experiment is supposed to show is that when you can sensibly neglect air resistance, the masses of the objects do not matter for the time it takes them to hit the ground. More specifically, the mass-terms in the gravitational force acting on an object (heavy mass) [math] F_G = G \frac{m M_0}{r^2}[/math] and the mass term resisting acceleration (inert mass) in the equation of motion [math] F=ma [/math] cancel: [math] ma=F_G=F=G \frac{m M_0}{r^2} \Rightarrow a = G \frac{M_0}{r^2}[/math]. This does not hold true when [math] F \neq F_G[/math] like when a non-neglectible friction force works on the object (i.e. [math] F = F_G + F_F [/math] with [math]F_F[/math] being friction-force). .The plastic ball would hit the ground later than the cannonball, but that's because it's affected by friction stronger than the cannonball. Similar experiments as the original one have been repeated often in the past dunnohowmany-hundred years. The most interesting ones imho are the ones performed under conditions where there is not friction, that is in vacuum conditions. It's a relatively widepsread school experiment, actually: Take a glass cylinder with a feather and a stone inside. Pump out the air and then turn the cylinder upside down -> both objects will fall at the same speed as above calculation suggests. Same thing has also been performed as a fun experiment on the moon; you might even find the video of it somewhere on the net, but the video quality is quite bad. EDIT: Considering the follow-up question: The air resistance of a leaden cannonball is ... errr ... more negligible than the effect of air resistance on a plastic ball. Like I said the thing about throwing the balls from a tower is perhaps just a myth. It seems more practical to roll the balls down a ramp to keep the velocities (air resistance increases with velocity) in a reasonable range.

You could try hitting the "report post" option and ask for some moderator to hide the reply with the spoiler tags. Explicit note: That's not what the "report post" option is for, but I suspect the current usage of the option is limited enough so that you don't cause overly much bureocratic trouble. Often, when you see someone using a feature that you don't know how to use (the spoiler tag is used quite frequently in this very subforum, so just look around for it) you can quote the post and look at the source you get within the quote tags. EDIT: Not as frequently as I thought; I was fooled by someone using the spoiler tags in his/her signature ;|. Whatever, I found a post using it, quoted it. The code is [ hide] secret message [ /hide] or [hide] secret message [/hide]

I don't exactly know what time is meant by "in the very beginning" but I suspect you're talking about "right after the photon is created". Anyways: If you stay in the classical picture, then when a particle is created it must start with some velocity (zero is also a velocity). Why not the speed of light in the case of light? The initial velocity is actually determined by conservation of energy and momentum. The force that governs the creation of a photon from de-exitation of an atom is the electromagnetic force. Yet, there is no force acting on the photon. Actually, the photon is the "force particle" (that's not really the term, but forces in the sense of F=m*a are not really a concept of modern physics, anyways) of the electromagnetic force itself. I can understand why you have problems understanding that, but the point is that the physics you are probably familiar with misses two important aspect necessary for understanding photons: - It does not describe massless particles. - It does not describe the quantization of the electromagnetic field. The closest you get to a photon is an electromagnetic wave like a radio wave. Admittedly, that's already very close to the actual concept of a photon. So perhaps it's easiest to think of it as a normal EM-wave.

That's a boring and an interesting result at the same time. The boring part is that you're effectively saying that f(x) and f(y=2x) are equivalent. It is interesting in the sense that it seems to show where and how the quantization criteria interplay and kind of give scale to physics (well, that's at least what I do think; it's not that it's something I ever thought about before or have some authorative 3rd party opinion). In other words: I think the point you're looking for is the quantization criteria. [math]\phi[/math] and [math]\psi[/math] cannot obey the same quantization criterion because the commutator of two [math]\psi[/math]-dependent (where "-dependent" shall mean "expressed as [math]\psi[/math], the two variables are of course just multiples of each other) operators will differ by a factor of [math]\lambda[/math] from that of the [math]\phi[/math]-dependent ones. I'd have to read up a bit on it, but I think that's what it boils down to: As soon as you fix the quantization condition for your field, you cannot (or at least should not) abritrarily rescale it, anymore.

dunno how to answer that. You can see it this way but that doesn't mean the photon acutally was part of the atom. What happens is that the electromagnetic field interacts with the atom with the result that the atom gets de-excited and the electromagnetic field is excited by one photon. From the interaction process with the atom. No. We're past that idea for at least a few hundred years by now. In fact, if something moves and no force acts on it, it will continue to move unhinderedly. Its special relation between wavelength and frequency.

As soon as the volume in which the particles are spread out is much larger than the area in which the particles were randomly distributed, the original position ditribution can be ignored. What I find more interesting is the question to what extent switching on elastic collisions drives the effect. With collisions you should get an effective pressure from the inside to the outside (where there are no particles that could bounce you back). So I'd think that switching on the collisions would quicken the "Hubble's Law"-effect. It should even be possible to make thermodynamical predictions that could then be tested by your simulation.

Your english is fine, don't worry. I like the idea of such a little simulation but hope you're not getting too excited about having found the sage's stone of cosmology. There's several issues with your simulation, one being that I am pretty sure that you were using non-relativistic calculations and only restricted your initial conditions to be v<c in some frame of reference. One thing you could try out: Does the clustering have any influence on the velocity distribution at all? You can get (using a non-relativistic model) get the result of relative velocity being proportional to distance very easily: Start all points (radius=0 here) at the same point in space with some random velocity and no colissions. This will trivially exactly reproduce Hubble's Law. I think your results come from these conditions (which are relatively similar to your program).

Seen with a bit of distance it should be obvious what's meant. If you're critizising how logging of ant choices is done, then your experience would qualify you as "at least know what he's talking about". If you're talking about how scientific research is handeled in the international university research, then one way to get the "at least knows what he's talking about"-label would be actually having worked in that enviroment. Similarly for citizising how research is handeled on industry level. I could just go along and say a lot of breakthroughs in industry are prevented by incompetent superiors that are afraid of losing their jobs to the smarter people having the idea - but that statement is just pulled off my lowerback. Not having that label that or another way (I'm not claiming that working in a field is the only way to know something about it, but it's the most obvious) it strongly depends on people's personality and mood how seriously they take you and how much time/effort they put into giving you replies. Some take more time to explain their point of view in a detailed way, some just say "you obviously don't know anything, so there's no point even adressing your point" (which is a reply at least) and some will simply not bother replying or even reading what you wrote.

Both better than sitting in front of a computer in your air-conditioned house, it seems. Especially if you go there by bike. It would (for several reasons) be nice having a link to where you found the chart, though.

He surely didn't mean that since you copied his premises incorrectly. Other than that, if you agree that [math] -x = x e^{i\pi} [/math] you can indeed make a similar statement (I strongly assume saying 2*(-1)=2 was just another tyop of yours): [math] 2 \cdot (-1) = 2e^{0 i \pi} \cdot 1 e^{1 i \pi} = 2 e^{(0+1)i \pi} = -2. [/math]

Personally, I'd drop the "Internet explorer". Too me it sounds like "can count to 100". You should check the correct capitalization of the words, I saw some words that I'd have capitalized (e.g. "Internet Explorer").