timo

Same here.

I assume you know virtual particles in the context of Feynman diagrams: Feynman diagrams are at the very basic level nothing but pictorial representations of mathematical terms (addends in a Taylon expansion). So in that sense they are exactly as "real" as the x in the Taylor expansion of sin(x) = x + O(x²). They can also be tested in a similar way. Just plug in several small x and you´ll notice that the result is roughly x. So in that sense the x is real, necessary for sin(x) and can be tested.

My statement went further. What I said was that if A and B are square matrices, A is invertible and AB=1, then B is the inverse of A. So I was not assuming that A and B are each other´s inverse, I was showing it under the given conditions. What I was trying to show is that for an invertible A, there is only one B such that AB=1 and therefore that B must be the inverse.

The function used depends on the substance your are passing through and the velocity you´re using. For slow velocities, a function ~v might be sufficient. For higher velocities, you´ll need v² terms. Generally, all those function should better be regarded as an approximation only valid within a certain range. Computing friction for realistic scenarios is a complicated task and afaik an industry / branch of research in itself: The friction is not calculated by simple terms like above, but instead a full simulation which also includes turbulences and stuff like that is run. I don´t know how good these simulations are these days but I doubt that anyone would let a plane transport passengers when its aerodynamical properties have only been calculated on a computer and not really tested in an experiment - the computer simulation could at most be used to reduce the amount of experimenting because effects can already be estimated from the simulation. Another interesting thing in this context is that such experiments are not scale-invariant. You cannot just model your new plane with a size reduced by a factor of 100 and expect it to have the same aerodynamic properties as the real plane - a reason being that you cannot scale down the size of an air molecule by a size of 100. In short: A realistic calculation can be arbitrarily complex, I think. For simple things like shooting someone in the back, effective models like ~v² can be sufficient.

Prelude: I am not sure if it´s correct but a math prof of mine once said "a proof is an argument that everyone understands and agrees with", so in this case: I don´t understand the "for any vector b" part. Had you said "for any vector x", I had understood it as you can multiply any vector x with the matrix B. What I don´t understand why any vector can be constructed by multiplying another vector x with B. There definitely is an additional restriction on B (just let B=0 and try to construct and [math] b \neq \vec 0 [/math]) which you didn´t mention. Sadly, you need the "any b"-statement here: So the "any b" statement must be made more explicit. But I have the bad feeling that it´s only true if AB=BA=1 ....

No but almost. A and B additionally must be square matrices. I think in most cases it´s easier to check if they are square matrices. That is because there is none. 2x2 is square and hence AB=1 => BA=1. I´ll sketch a proof in below but I cannot guarantee that it will be too helpful to you as I don´t know which things I can assume known and which not; for a formal proof, better check a book on linear algebra. To my own surprise, it is. Assume A and B are their inverses, respectively. Assume there was another matrix C=B+X such that AC=1. Then, AB = 1 = AC = AB + AX and therefore AX=0. You might already guess that AX=0 => X=0 (and therefore C=B) if A is invertible but let´s make it explicit: [math] AX = \left( \begin{array}{c} A_1^t \\ \dots \\ A_n^t \end{array} \right) \left( X_1 \dots X_n \right) [/math], where I have written the matrices in row-vectors [math] A^t_i [/math] and column-vecotors [math] X_i [/math]. Since A is an invertible matrix, the vectors [math] A_i [/math] must be linearly independent. Now, since we want [math] C \neq B \Rightarrow X \neq 0 [/math] it can be concluded that there is at least one k, such that [math] X_k \neq \vec 0 [/math]. But from AX=0 we know that for all j: [math] \left< A_j | X_k \right> = 0 [/math] where <...|...> denotes the scalar product. Now, the crucial point is that the A_i form a base for the n-dimensional vector space they and the X_i belong to because A is invertible (This is the point where I assume knowledge that might not be there. They must be linearly independent for det(A)=0.). Hence, to satisfy [math] \left< A_j | X_k \right> = 0 [/math] for all base vectors [math]A_j[/math], [math]X_k [/math] can only be zero, which was explicitely excluded in case that [math] B \neq C [/math]. So far it was shown that if AB=1 for an invertible A, B is unique. Since updating showed me Zareon already replied, I´ll first read his post before making any further thoughts.

Perhaps it was just a demonstration of why the advice is good, especially the "don´t let anyone find out"-part

No, I was mostly joking. I don´t think your name is so prone to that as mine and it was actually only one or two PMs in the last 2.5 years.

You might regret using your tag from computer games here as soon as you start to get private messages telling you "hi, do you really like slaughtering? I like war, too" .

It´s not so easy to see, especially since you either made a mistake or there is a typo in above.

So what was the issue? Was it really that google adopts very quickly to changes (at least I have the feeling that the number of posts has increased within the last week, I just cannot tell what is cause and what if effect) or some incorrect setting/similar (what exactly)?

@1) There is a set [math] \{ a_i \}: v = \sum_i a_i v_i [/math] by the definition of a spanning set => [math] 0= 1*v - \sum_i a_i v_i [/math] which is linear dependence (factor 1 explicitely written for clarification). @2) there is an a>0 so that 0 = a*v4 + x*v1+y*v2+z*v3 => v4 = (x*v1+y*v2+z*v3)/a. Left for you: Why is there such an a? what if v4=0?

That was more or less my point. In the given context you cannot tell whether one model is more correct than the other so you´re just practical and take the easiest one.

I have no idea what you mean. Perhaps you should clarify that point. Cosmology assumes a mostly uniform distribution of mass in the universe. A black hole assumes all mass being at a single point (or at least a small area with a lot of void around it). So it does not seem the same to me. I dunno. Since it was you who said something about something being infinite, it should be you who knows what´s meant by it.

"Locally" rules out any effect of an [again clasically speaking] inhomogenous gravitational field within the experimentation apparatus. Time dilatation would be the same for all clocks positioned anywhere in it and the heavy side would not point towards earth since the net force on the heavy and the lighter side of the device would experience the same force (none in free fall). Of course it is just an idealization and theoretically, for any non-zero extension you have a finite inhomogenity in the gravitational field of a homogeneous massive sphere. Imagine the restriction "locally" as the statement as "you can continuously shrink the measuring apparatus down to a size where the effect caused by the inhomogeniousity of the gravitational field is below a measurable amount". How well locallity is true for an actual realitstic system is a practical issue.

You should keep in mind that "locally" means "sufficient close to avoid any impact of a finite extension". So if you had something in mind like [description follows in terms of classical Newtonian mechanics] dropping two stones and seeing that during freefall their horizontal distance slightly decreases (because both are attracted to the center of earth), then such a test is explicitely ruled out by "locally". If you had something in mind like "the closer you get to earth, the bigger the attraction will be - I could measure that", then the question would be: How? @Zareon: Apart from some nitpicking one could do (e.g. the motion of an object B is not independent of its mass if it gravitationally influences other bodies G around it which causes a change in the gravitational field caused by G), that all sounds fine.

Gimme a hint then: What is 10 ft and 12 ft ?

I don´t know much about string theory but I am pretty sure that a significant share of string research goes into checking a proposed scenario for what observations it describes and ruling out models which give incorrect (non-compatible with nature) predictions.

Plug it in?

I think the name is "Tsunami" : http://www.magicarsenal.com/productimages/178890.jpg Since you asked in the Relativity subforum you get a relativistic answer: There is a theoretical limit on what events can be related. Two events (an event is described by a position and a time; it doesn´t matter what the actual event is) that cannot be connected by a particle travelling from event A to event B with a speed of c or slower are not causally connected and cannot influence each other. If the events can be causally connected, then there is still a unique order: Assuming the particle travels from A to B, then A always happens before B. This means that A can influence B, but B cannot influence A. A is called the cause, B is called the effect.

Assuming "measurement" is what I assume it to be (0.5*a*b, where a and b are the two shorter sides of the triangle), then additional assuming a<b (meaning that a is the shortest side), you can get the length of a by substituting b=4/3*a in above. Knowing the length of a, you know the lengths of the other sides (because you know the ratios). EDIT: Forget what I said in above. If the "measurements" are the area as I supposed in above, then it should be square-feet, not feet. They can, however, also not be the lengths of two sides (because their ratios are 6:5 which is none of the two possible ratios 5:3 and 5:4). What is it?

I don´t know the function so perhaps you should give some additional information. How is it obvious that there´s more than one f(z) for a given z? Did you possibly mistake the plot The Tree posted for an "f(x) over x" one? Judging from the labels I´d assume it is the complex values of the line f(x) : x -> Complex, f(x) = Riemann(1/2 + i*x) for x in some interval [x0,x1].

I think the idea of creating a scientific wiki is pretty much dead and I give it neither much chance, nor do I see much point in trying to revive it in the way it was supposed to be. Other than that, I think you have a free hand to pretty much do anything you want there at the moment - except for oviously unvanted stuff like illegal activities or abusing it as a cheap data storage for your private use (it´s just what I think, not the opinion of the people in charge of WiSci - I don´t even really know who that is). So if you see a way to put it to good use, just do it - and perhaps check back with the SFN admins in case you are really starting to invest a huge amount of time. EDIT: Wow, more than 10 minutes to type that few sentences.

Nice brides. Expect for imho being a bit too fast, the slideshow seems to work from my Laptop (System is Win XP home, in case that matters) i.e. all pictures showed up during the roughly 1 minute I ran the slideshow. EDIT: Oh yes, seeing that Klaynos mentioned the browser: It works with both: Internet Explorer and Firefox.

The same thing that possessed you? I would assume that strictly speaking such a scenario might be impossible. Not because of physics -I see no problem there- but because of the new definition of a planet which, if I got it correctly, requires an object to have no other gravitationally relevant objects in its area to be called a planet. But that´s only pedantry about definitions, of course (not even sure if the definition of a planet really is a hindrance).