timo

You should at least change the mail address, too. It's not particularly polite storing data of/about people that explicitely said they don't want you to - to say the least.

The gauge bosons of the Standard Model are vector bosons, so yes. I refers to the transformation under Lorentz-transformations (as does "scalar"). So the answer is probably yes. I assume you meant to write "the same quantum state". Yes. @T_man: Don't be afraid that we went a bit off-topic in the last posts (last posts have been more about physics at subatomic scale than atomic physics) - it was just too tempting. You can probably ignore the last few posts and just stick to post #2 (the 1st reply by foodchain) if you have further questions.

Not necessarily. In the Standard Model of elementary particle physics, all "force carriers" happen to be bosons (more specifically, vector-bosons, which simply means spin-1 bosons). But there are bosons which are not force carriers, e.g. [math]^4\text{He}[/math] (and all the other atoms which Bose-Einstein condensation is done with).

I doubt that "there is an electrically positive nucleus consisting of neutrons and protons with a radius of ~10^-15 m surrounded by an electron cloud with a radius of ~10^-10 m" really is a better reply. Some comments on yours, though: Pretty much by definition if they are the building blocks for an atom . yes Spin is the only property defining whether a particle is a fermion or not (a boson, then). Not sure what you meant by "genera's" and "their status". They are simply three different particles. One of them is elementary (the electron), the other two are not. I'd go as far as to skip the word "typically". A different count in neutrons. A different count in protons would give a different chemical element, not an isotope of the same element. I think they actually are the only factor describing chemical properties (although the state of the electrons is determined by the nucleus, especially its charge, of course).

You're probably not properly taking into account that velocity and acceleration are vectors, not just some positive scalar numbers. In case you are familiar with linear algebra, look up the matrix form of the Lorentz transformation and verify [math]\Lambda(-v) \Lambda(v) = 1 \neq \Lambda(2v) [/math].

Yes, e.g. relativity and quantum field theory. Note that while they are incompatible with a discrete approach (due to appearing differentials), they are not necessarily completely incompatible with discrete space - they could be effective theories for large scales ("large" meaning something like the size of a nucleus or larger). Bjorken, Drell make a related statement in the 1st chapter (chapter 11.1 ) of their book "Relativistic Quantum Fields": Lose translation from german done by me, buying the book highly unrecommended (due to its age).

For finding out how where the values of a function are heading to.

No offense, but it looks like a bunch of semi-arbitrary statements to me. What does ""space is expanding" mean? Will the electrons obey the rules of QM and keep their allowed orbits or do the orbits also expand? In the sense of "composed of elementary particles"? In what relation does "everything scale with time"? Linearly? Quadratic? Cubic? Exponentially? Is that scaling equal for all observables or different for different ones? If the volume occupied by five matter-points is doubled, where did the five new matter-points from? No. Try using more precise statements, e.g. define the scaling behaviour for different relevant observables. You seem to be saying "if everything scaled such that every observable was unchanged, then there would be no observable difference".

The angular velocity does not equal the frequency (the rev/min). Think of it: If something goes in a circle at 1 rev/sec, then the magnitude of the velocity equals one circumference per second, not one radius per second.

What are your solutions and what is wrong with them ?

What's your ideas? Have you tried approaching the problem? How did you do that? What was the problems you encountered?

And see if you also get >100% for seemingly exclusive statements .

Instant transfer of momentum comes to my mind.

The letter m stands for mass, not for matter.

No, not at all. "Where" was rather meant as "where within the causal chain from the bouncing of two nitrogen molecules to the stormy weather next weekend", i.e. not in the sense of spacetime-coordinates. I am not sure to what extent the question makes sense, though.

I tend to agree, but problems arise when you take this thought one step further: - Assume the unpredictability comes from arbitrary small uncertainties in the initial conditions when given sufficient time (that's actually pretty close to the premise of at least non-qm chaos theory, I think). - Then, given at some scale you have qm behaviour and assuming something similar like a measurement process happens at some intermediate scale, you inherit the qm randomness into your system, hence making it random, too. - ... and somehow I wrote this oven an hour ago, just got back to my still-running computer and dunno what exactly wanted to say next. Had something to do with that QM is completely non-random except for the measurement process and that the question where the observer is arises.

You'd better go with a book.

Sounds like engineering, especially the "creation"-part. That sounds ... well ... impossible. I can almost assure you that the number of people thinking about improving q-tips is way lower than the number of people fantazising about black-hole based time machines. I am not so convinced. I think physics involves much more learning what other people found out and much less own creativity than engineering. But not being an engineer, that is just prejudice. Inventing new or improved detectors for particle physics experiments sounds a lot like improved q-tips to me. Most graduate physicists are keyboard-stroking nerds . Btw: Welcome to sfn, you nerd.

If you think the answer is important, then I have a very simple experiment in mind you could perform: 1) Write down your current postcount A 2) Post in a thread in the politics section 3) Write down your current postcount B 4) Calculate C=B-A 5) Hit "report post" on your own post and ask a mod to remove it (not necessary in case your post was actually sensible). 6) The anwer is [math]\left\{ \begin{array}{rcl} \text{no}&:& C=0 \\ \text{yes}&:&C=1 \\ \text{you idiot}&:& \text{otherwise} \end{array} \right.[/math]

You can probably get the rules I meant from those. Just see how far you get with your proof.

I'm not exactly sure what you mean by "why"? Of course you can just plug it in and verify it: [math] D_\mu g_{\alpha \beta} := \partial_\mu g_{\alpha \beta} - \Gamma^i{}_{\mu \alpha} g_{i \beta} - \Gamma^i{}_{\mu \beta} g_{\alpha i} = \dots = 0.[/math] (note that I have implicitely used the symmetry of g and the two last indices of the Christoffel symbols - iow: I have not bothered on the exact position of the indices). If "why?" was meant in the sense of "what is the physical meaning?": I dunno at the moment. It would probably be something like that relations (scalar products) between vectors remain unchanged under parallel transport.

I wouldn't assume anyone knowing what a matrix and a determinant is not knowing what the transpose of a matrix is, either. And I've never seen a different notation for the transpose, either. So yes, I strongly assume it was assumed knowledge. Anyways, - your transformation is correct. The k^t is redundant. The transpose of a scalar is, depending on point of view, either not defined (in which case you wouldn't get the k^t in the first place but just k) or equals k. Either way, you end up with -M + kI. - I later edited my post to ask what (M-kI)^t is, rather than (M+kI)^t. Of course, figuring that out goes exactly the same way as (M+kI)^t (alternatively, you could just replace k with -k). The reason why I later edited it to (M-kI)^t of course it that (M-kI) is the expression that appears on the left-hand side of your original question. - Since you didn't know the notation of the transpose: Do you know what det(...) is and do you also know some calculation rules for it (you will need two of them)?

Yes, but it's not taking something to the power of t but taking the transposed. If M^t = -M and I^t = I, what is (M-kI)^t ?

Your first step is wrong, already. M^t = -M, not M.

Yes. E.g., Neutrons bound in a nucleus don't decay at all (the extent to what you can still sensibly call that Neutrons being a different question) . If moving them around quickly also counts as "enviroment" (you could possibly move them around in a circle), then that's another option (see muons produced in the upper atmosphere). Dunno. Do you have a specific idea in mind? Not sure what you mean. Pion masses are ~140 MeV/c².