timo

Dunno if that was a typo. But "quantum chaos" of course is the term for chaotic quantum systems. EDIT: Ok, it seems like that´s what you actually wanted to say. I have some problems with the assumption that non-deterministic <=> QM. If fact, I have problems with both directions of that equivalence. Neither do I agree that QM => non-deterministic, nor would I see why non-deterministic => it´s a QM system.

I don´t know much about that but at least I remembered a name associated to steering of satelites, so maybe start your search from here: http://en.wikipedia.org/wiki/Control_Moment_Gyroscope

Dunno that guy. But let´s assume he did: Do you think the guy on the posters is Sergio, then?

Generally I think using the "preview post" button is enough for all testing. But perhaps not everyone (especially not people new to this forum) will come up with the idea of using it. So: I like the idea.

That´s not Fermi, the guy´s name is Rovelli.

Perhaps at this point it would be fair to explicitely point out that of course it is possible to report posts by staff members if you feel attacked or to PM an 'independent' admin if you feel you were awarded unjustified warning points. You just volunteered to go through the next twenty "I have found a flaw in General Relativity" threads and have a nice discussion with the poster about possible flaws and possibilities of his/her groundbreaking new theory .

For me it´s a non-issue. Why should there be / is there a difference between an admin and a non-admin? At least I´m trying to do so quite often in the physics forum. I´m not sure what I should do but I would do something like this: http://www.scienceforums.net/forums/showpost.php?p=226349&postcount=16 http://www.scienceforums.net/forums/showpost.php?p=216732&postcount=11 (I´d like to point out that in the first case IIRC we were simply talking about two different things and sorting it out would have been a mess which hadn´t given rise to much new insight. In the 2nd case, Dudde indeed did PM me and we had the problem cleared very quickly and easily.) Or to say it simple: If something seems not worth discussing, don´t discuss it. In the end, it boils down to this: http://www.darkfire.net/~mrb/images/retarded.jpg (Hint: Above picture is not about disabled persons but about arguing in internet forums so unless given a good reason -feel free to PM me- I consider "that´s so offensive"-comments as political correctness hypocriticism). EDIT: @Martin: On 2nd thought it might be that I didn´t exactly meet the topic you had in mind here. As usually, I have a hard time understanding what you were trying to say/ask/whatever.

Those with an 8 or 9 as the last digit seem highly suspicious to me.

I do look here quite often and it´s especially nice to get a few more details about people I´ve already met in the physics section.

I think it can be understood with high-school knowledge, but it´s amazing for any level. Welcome to the wonderful world of simplifications. Above statement is true if you know what it means and pretty useless (or even incorrect) if you don´t know how to interpret it. Let me try to explain what actually happens: Let´s assume the photon (=light) moves in z-direction. As you hopefully know, it´s polarisation is perpendicular to the movement direction. So if we -for simplicity- rescrict ourselves to linear polarisation, the polarisation of the light can be described as a 2D vector in the x/y-plane (x,y). Vectors within this plane can be descibed by a linear combination of any two perpendicular base vectors e1 and e2. Example: e1 is the x-axis and e2 is the y-axis, then (x,y) = x*e1 + y*e2. Now what a polarisation filter does is projecting the vector onto a certain direction. Let polarisation filter 1 project on the e1-direction (x, right-left), polarisation filter 2 prohject on the e2-direction (y, up-down) and polarisation filter 3 project on the e1+e2 direction (x+y, right/up - left/down): Now let´s put this into the language of above projections. The beam of light arrives with any polarisation (x,y) = x*e1 + y*e2. The first filter projects this vector on the e1-axis so after passing this filter the polarisation is (x,0) = x*e1. The 2nd filter then projects the polarisation on the e2-axis, then. So after passing the 2nd filter the polarisation is (0,0) = 0. Now at this point I realize that I should have told you that the magnitude of the polarisation vector is connected to the intensity of light. I don´t want to rewrite all I´ve written so far so just believe me that a zero polarisation vector means "no light" and that a nonzero polarisation vector means "some light" - at least within the context of this post. I hope so. Yes, it is surprising. But let´s do the math. As before, after passing the 1st polarisation filter, the polarisation will be (x,0). Now, instead of arriving at the 2nd filter, the light comes to the filter 3 which projects onto the e1+e2 direction. To calculate the projection onto this direction, let´s rewrite (x,0) as a linear combination of e1+e2 and a perpendicular vector (e1-e2): (x,0) = x*e1 = 0.5*(x*(e1+e2) + x*(e1-e2)). So after projecting on the e1+e2 direction by polarisation filter 3, the polarisation will be (x/2,x/2) = 0.5*x*e1 + 0.5*x*e2. Already see what will happen now? I guess so but let´s complete the calculation anyways: This polarisation will then arrive at the polarisation filter 2 which projects it on the y-axis. (x/2, x/2) projected on the y-axis simply is (0, x/2) which is not zero (unless x was zero, of course).

? = 2c. There is no problem having differences of speeds being >c. Velocities and therefore also differences of velocities depend on the coordinate system used (=frame of reference). The stringent from relativity is that every velocity must be <=c. So the difference of two velocities must be <= 2c. Two massive bodies travelling away from a common midpoint at 0.8c each will have a relative velocity of 1.6c. BUT: If you go to a frame of reference where one of these two bodies is at rest the transformation laws for velocities will guarantee you that in this system the relative velocity (which simply equals the velocity of the other body there) will be <c.

I actually wonder what your confusion is about Tom. To me it seems that we were all saying the same thing. I assume your confusion arises from the term "by definition" used by Matt? Not being a native english speaker I´d interpret "by definition" as both "is included in the definition" and "is a direct result of the definition". As I tried to show, the uniqueness of the 1 is garanteed by the set of conditions for a group I know. Adding an additional (superfluous) condition that there is a unique 1 will not change anything as both sets of conditions will still lead to the same definition in the sense that the set of all possible objects being a group doesn´t change. Or in other words: I don´t think that a mathematical definition for something must behave like a base for a vector space (minimal set of generating vectors) but more like a generating system (vector space is clearly defined).

No I can´t but I have the slight feeling that this is because an identity element is not what I imagine it to be. For me, 1 is an identity element exaclty if 1A = A1 = A for any A. For this, there should imho not exist more than one for each group: Assume A and B are two identity elements: A = AB = B.

I admit I didn´t go to the link you gave. But can you elaborate on the neutrino pressure (for example telling what it is) and the mechanism by which light loses energy to it here?

Perhaps I should note that I was thinking about those replaceable hard drives and storing the somewhere in a shelf when I talked about storing data on a HD. Not about HDs which are actually used all the time.

Stone carvings are still unbeaten in durability of stored data Does anyone know how long data can be stored on a regular hard disk? They probably won´t endure being put in a washing machine but I´d think that with proper treatment they can store data for quite a while. Plus, you can really store a lot of data on a hard disk. Also, HDs are relatively cheap.

That look good, yes. The reason I asked you to show your calculation (apart from that it´s always a good idea to do so) was that I assumed you might have forgotten the factor 9.81 m/s².

400/500 * 100% = 80%, not 8%

Don´t think so. Perhaps it´s best if you show us how you calculated your result so maybe someone can point out where your error lies.

I wouldn´t think so. The maximum change in speed a particle can have is 2c (c left to c right). Let´s make in c for simplification (massive particles could reach the 2c only asymtotically which complicates the treatment without giving any new insight - 1c is already sufficient here). Assuming a constant acceleration a which works over a time t, the change in velocity is a*t which shall be < c: a*t < c This equation can be satisfied for any acceleration -no matter how much- if just t is sufficient small, namely t < c/a

No force. Light, gamma and all other EM are all photons. Photons have the feature to always have a speed of c (regardless of their momentum) because they are massless. In lack of a better analogy: Grass is always green, whether it naturally grows in a grassland or it´s artificially planted in a city park. I am not sure if I understand that question. Purely classically: If a point A, which shall be fixed in space, emmits some other point B which then moves in a straight line: How can B have any other movement than "directly away from A"?

Your posts are moved to speculations for a simple reason: The physics section is for mainstream physics, not for speculations about other possible forms of physics. Calling moving your threads to the speculations section "banishing" only shows that you don´t take those "I know better that all the millions of physicists out there"-threads seriously yourself. I doubt that any of your posts were censored. True. And if you use momentum instead of velocity to describe a particle, all problems you might have with v=c disappear. So it´s better so regard velocity as an effect of momentum, not the other way round.

Agreed, except for the spelling. *hitsthereportbutton*

Assuming the "decrease in disorder" you have in parentheses is a relic from typing adn that we´re really speaking about "increase of entropy" now, then the two answers look more similar. Though, there is still some difference if you are really looking for one. Answer one leaves the possibility to have a decrease in total entropy, for example (because nothing is said about the quantities). I don´t think there´s much point in debating over this question without having the exact wording of the question and the answers at hand. Multiple choice questions (I assume it was multiple choice) often depend on nuances in the questions and answers.