timo

Yes, I´m pretty sure. You get an equation for the velocity which (directly taken from your post; didn´t check it) is v = sqrt(G*m1/r). Any change in m2 will not alter the value of v. You can also leave m2 in if it makes you feel better: v = sqrt[(G*m1*m2)/(m2*r)]. Still: Changing m2 doesn´t change v (replace m2 with 5*m2, for instance). The mass cancels out at a much more basic level, actually. The force excerted by a gravitational field g on a mass m is F = mg. Now from Newtonian mechanics, you know that F=ma => a = F/m. Hence, the acceleration of a body caused by a gravitational field is the same for all bodies, regardless of their mass: a = F/m = m*g/m = g. Two bodies which have the same initial positions and the same initial velocities at a given time and do always experience the same acceleration move exactly the same path. So if your heavier satelite starts at the same height and with the same initial velocity as the lighter one, it will have exactly the same orbit. EDIT: Yes, I know it sounds illogical (unintuitive fits better, here). But it´s also counter-intuitive that a feather falls down as fast as a hammer if there´s no friction due to air resistance.

I´m pretty sure. What do you think is wrong? And why?

There is no gravity in special relativity, that´s my point. Newtonian mechanics violates special relativity because of it´s instantaneous changes in the gravitational field (the time between cause and effect is not limited by the speed of light).

or 4) The spacetime of special relativity? (just added this so that you have a correct answer to chose from, even though the statement Bettina found doesn´t make much sense, then).

Wikipedia is often a surprise considering what´s in there and what´s not. One day you might find a small note about exotic physics like the susy partner of the graviton, the next day you aren´t able to find any information about some basic term of mainstream physics (can´t remember what it was, something spinor-related, I think). The next day you find a very detailed and very good article about a german terrorist group. Always good for a nice surprise but nothing to rely on. Sorry, I don´t know any good resources for electricity. You did try a google search, did you?

I might have an advantage as I am european and I do have a friend who´s gay with whom I sometimes hang around in gay clubs (I think he doesn´t even go to non-gay clubs anymore). But I really didn´t find it hard. I got 7/8 right (not counting the "trick question"). One thing that would interest me, though: Did the creators of the game actually ask the people whether they are gay or not or is the right/wrong only based on their thinking? Cause in the latter case that game´s not really "educational" but only about checking whether you´ve got the same prejudices as the creators.

"Anyways" erscheint mir passend. Well, "Wert"="value". The strict translation would probably be something like "That´s not valuable to me". Such a sentence will be usually used in the sense of "I don´t need it"/"I don´t care about it"/"It doesn´t matter for me" but often with the unspoked addition ".. so don´t do it". Such a sentence would appear in the sense of "Ich lege keinen Wert auf lange Herleitungen der verwendeten Formeln" ("students: You don´t need to derive every equation you use explicitely") or "Er legt keinen Wert auf teure Kleidung" ("He doesn´t care about/need expensive clothings"). "Unterredung" = "parley"/"discussion"/"talk" "brennen" = "to burn" but "auf etwas brennen" = "to be keen on something" I´m not very good in translating but I´ll give it a try: "Our last discussion wasn´t in such a way that I´m very keen on continuing it" ot losely speaking (I am better at this, certainly) "Last time we talked you really pissed me off so I´m not going to discuss with you any further".

You actually did finish the question already: You get the velocity by setting the centrifugal force F = Mv²/r to equal the attractive gravitational force F = GmM/r². Wheter you replace M by 3M (M is the mass of the satelite; m the mass of earth, here) or not doesn´t play a role. It still cancels out in your equation for v. So the velocity needed to keep a satelite in a stable orbit does not depend on the satelites mass. Given a certain orbit it´s equal for all masses M of satelites. This should not really come to much of a surprise. In fact, a key feature of gravity is that the movement of a body due to a given gravitational field is always independent of that bodies mass. "... i think that the mass would be the same no matter how heavy it is." ^^ That was a typo, wasn´t it?

If you´re smart enough to take a second attempt if the first building you randomly chose is a nuclear bunker, then yes. Most houses I´ve seen which I´d classify as "average house" do indeed have more or less the same number of windows per meter of wall and roughly the same height of the rooms. As several people already pointed out: These kind of questions are about rough estimates, not about making statistics about the window-density distribution in US buildings.

Are there no houses with windows close to where you live?

No. Wavelength is the wavelength of a photonic state contributing to a packet. A packet with only a single definite photonic state contributing has infinite size as the probability density for such a state is constant with regard to position. Frequency and energy of a photon is the same; it´s just expressed in different units.

You got a vector space V and you got a reziprocal vector space D (also called dual vector space). Covariant vectors are physical vectors. Contravariant vectors are elements of the dual space. But since the dual space of the dual space is the original space' date=' and since one can obtain corresponding vectors by contracting with the metric tensor, one usually doesn´t pay much attention. So: In first principle there is a huge difference. For practical purposes there´s usually none. Same applies for tensors, of course: A contravariant index works on an element of V and a covariant index works on D. I know this sounds stupid, but what do you mean by "graph out" ? Actually, I have never thought of that before. By definition, a tensor of rank(a,b) is a multilinear map [math] V^a \times D^b \rightarrow \textbf{R} [/math]. So there´s not much connection to be seen in the first place. However, the two typical tensors I´d think of now -the energy momentum tensor and the curvature tensor- are in fact related to derivatives. So one could find an analogy, there. I would be careful to restrict yourself to "it´s just an extension of the Jacobian" but it might be an interesting analogy from time to time. Of course, the meaning of "extension" is arbitrarily extendable .

There are millions of mathematicans and physicists around. You´ll certainly find a few who question relativity. Even it it´s a vanishing minority: That doesn´t count because they are all brainwashed and don´t think outside the box :]. On topic: Didn´t understand what the author was saying. But it´s impressive how many "I found a flaw in relativity"-authors do this by trying to find flaws in papers which are a hundred years old.

Cause it´s a way to get attention. If he really had another opinion he had added an "...it´s actually due to <cosmological epansion/time dilatation/temperature/magical elves/...>", I think.

The question how to transform a tensor of course depends on how you want to transform ... Assume you have the following transformation for a covariant vector: [math] \var v_i = T_i{^j} v_j [/math]. For this matrix T the inverse [math] T^i{_k} \, : T^i{_k} T_i{^j} = \delta^j{_k} [/math] exists. Since scalars like the product of a covariant and a contravariant vector are invariant under coordinate transformation, one can now construct the transformation rules for a contravariant vector out of the trafo for a covariant one and the demand on invariance: [math] c = \bar v_j \bar v^j = \bar v_j \delta^j{_k} \bar v^k = \underbrace{\bar v_j T_i{^j}}_{:= v_i} \underbrace{T^i{_k} \bar v^k}_{\rightarrow v^i} = v_i v^i = c [/math]. Now that you know the transformations for covariant and contravariant vectors, the same method can be used to determine the transformations for a tensor. The key point is realizing that a tensors of n covariant and m contravariant ranks maps n contravariant and m covariant vectors on a scalar. The example of a rank-1 contravariant tensor is just the one from above. For higher ranks, you simply include more deltas. In the end you come up with the result that each covariant index and each contravariant index gets contracted by the matrix you´d contract a contravariant/covariant vector with. EDIT: Nah, that´s incoherently written, at least. Have to go to work now so I can´t edit it out. The key point still is that the tensor transformations follow from the vector transformation and the demand on invariance of a scalar.

Counter question: Why shouldn´t they be "allowed" ?

http://en.wikipedia.org/wiki/Einstein_field_equations ^^ now, that was not too hard, was it?

The physical description of a plain wave is f(x, t) = C exp(i(kx - wt)) with three -intially- arbitrary parameters C, k, w. Plain waves for massive particles exists and have a different relation between w and k than plain waves for massless -lightlike- waves (this relation is called "dispersion relation"; the name stems from optics, I think). In classical QM, particles are described by waves, which can be constructed by adding up plain waves : [math] \psi(x,t) = \sum_i C_i \exp [ i (k_i x - w_i t) ] [/math] (the sum goes to an integral, but that´s not my point here). The square of the wave´s amplitude is considered the probability density for the postion. So if you find an area A in which the sum over the squared amplitudes almost equals the total sum over the squared amplitudes, you say that the particle is in this area. In contrast to plain waves, where the amplitude squared is always |C|² for all times t and all positions x, the sum of those plain waves can produce almost arbitrary distributions. The area A which gives most contribution to the sum of the squared amplitudes can change with time. This is considered as "the particle moves". The area does not nessecarlily have to move at lightspeed. Might be an interesting weekend project to write few programs visualizing this aspect of QM. So if anyone knows of a suitable computer language for doing so, pls PM me. Python is not suitable as it seems too big to download, ISO C++ doesn´t seem good as I want to make a visualization w/o bothering about details like plotting the data or platform dependence (OpenGl is not an option), Mathematica isn´t free, dunno about Java. Inbuilt FFT, C-numbers and OOP structure would be greatly appreciated but are not really nessecary. And finally: Your question is rather unrelated to Relativity. So if you have further questions on this topic, do a search in the "Quantum Mechanics" section or open a new thread there if you don´t find answers to your questions.

To create a bijection between two sets, both sets must have the same number of elements. If the sets A, B, C, D, E, F and Z all have the same finite number of elements, then no bijection {A}x{B}x{C}x{D}x{E}x{F} <-> {A}x{Z}x{D}x{E}x{F} can exists.

No, a quantum mechanical wave is not nessecarily an electromagnetic one. But electromagnetic waves were the easiest example of a QM wave that came to my mind.

Yes. Radio programs are transfered by electromagnetic waves, for example. Are you spanish ?

v = k*f k-> k'= C*k f -> f' = f/C v -> v' = k'*f' = C*k*f/C = k*f

It´s no piece of garbage so far: The magnitude of 4-velocity is always 1 (or whatever you norm it to - c for example) for massive particles and 0 (regardless of the norm, of course) for massless ones as light. I just don´t undertand what "at rest with respect to time" and "moving throgh time at c" means. And I could bet you don´t know that either.

I don´t really get what you´re doing but maybe this helps: You seem to map (B,C) -> Z and want the inverse Z -> (B,C). As the number of combinations (B,C) |{(B,C)}| is about 1 million while |{Z}| = 1001 the map (B, C) -> cannot be injective. Therefore it cannot be bijective. In the case that Z is not limited to an integer in the range of 0-999 (1000 elements is easier to talk about and the extension to 1001 is not too hard) but to 1 milion elements, a bijection can be made rather easily by: (B, C) -> Z = 1000*B + C with the reverse Z -> (B, C) = (Z div 1000, Z mod 1000).

There are no boundaries of the universe (at least in mainstream physics). "The universe expands" means that the distance between two points in space increases with time. Consider the plane {(x,y) : x,y real}. This plane is infinite. Let the distance between two points P1=(a,b) and P2=(c,d) be defined by d(P1, P2) = t*sqrt[(a-c)² + (b-d)²]. The distance between those two fixed points increases as the time t increases. Therefore one could argue that the plane expands. Also, the distance between any two point is zero at t=0 (which is the big bang, then). But the plane still doesn´t have any boundaries. If you want to have an example of an expanding finite space without boundaries take a sphere with radius time.