timo

I am not sure what exactly you are talking about but in case it is not obvious to everyone: The guys from the "International Institute for Biophysics" pretty much meet all crackpot criteria I can spontaneously think of. Might be more obvious if one speaks German, though. EDIT: To be fair I should say that it is only the impression that the homepage of the IIB leaves.

You are at least missing a) the units of alpha, b) mentioning what L is (stricly speaking you also didn't mention L0, L1, T1 and T2 but one can guess what they are supposed to be), c) the result you got and called senseless.

I'd spontaneously not try to approach it by using contradiction. I'd show that if there are two inverses (or neutral elements) a and b, then a=b. But thinking about it that might be what Bignose meant with a proof by contradiction (the result it is kind of a contradiction to the assumption but I do not get where a violation of the axioms would come in).

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That is quite an extrapolation to say that one pair of particles does not annihilate (whatever "annihilation" exactly shall be) therefore no other pair does either. Z-bosons do annihilate. Depending on whether by "annihilation" you mean that the reaction happens even when there is little to no kinetic energy involved or if you just mean any reaction that changes the particle types then photons either do not annihilate because of conservation of energy or they probably do. I see no reason why electron+positron -> 2 photon should not work the other way round given sufficiently high-energy photons (at least in principle).

There's a (written or unwritten, I am not sure) policy on sfn that others should not do your homework for you. Try it yourself. If you get stuck or your attempts fail, then ask for help and present what you did/tried so far.

Why do you think that was the case?

That's not a (complete) argument: If planets orbited the sun as they do they would be giving off gravitational wells. The energy must be conserved, so as the gravitational waves are emitted, the planets would lose energy. So, as you can probably see, if planets orbited the sun, then they WOULD crash into it! You are at least missing a time-scale in your argument (alternatively you can of course deny the existence of gravitational waves). I am not sure to what extent it is a good idea to mix a non-QM model of atomic structure with a quantized electromagnetic field (->photons).

Much better for me.

But you have to admit that it is the best quantitative answer to the question so far; even though that is because it is the only one. Nice. I wasn't aware of that. So how about the rest? It is a reasonable approximation to split earth in two pieces, core and mantle, and assume sphericallity and uniformity for both of them? That would still lead to a relatively simple answer to the question (sum of two forces).

The weight decreases linearly until it reaches zero at the center; so it would be W*(R-1000 mi)/R, with W being the original weight and R being the radius of earth.

But of course an annual increase by any constant percentage is an exponential function

Not really Since mass is a form of energy, you cannot have mass without energy. ??? Dark matter is considered having mass. Personally, I often consider things that are directly proportional (with the proportionality constant being a natural constant, not something like F and a in F=ma) as being the same thing measured in different units. In fact, particle physics even uses units where your E=mc² would read E=m.

Just out of interest: What's that lecturer's field of research?

Energy, kinetic energy in particular . A bit of velocity too, but at some point the increase in velocity becomes negligible. Curved spacetime has nothing to do with the question I can assure you (spacetime has, but only the non-curved one is relevant). Sadly, my experience teaching/explaining relativity, particularly to non-physicists, is pretty much non-existent. I could tell you that the relation between energy and velocity for a particle with mass m you know, [math]E_{kin} = \frac 12 m \vec v^2[/math], is simply not correct and that the correct relation is [math] E_{kin} = \left( \frac{1}{ \sqrt{ 1 - \frac{\vec v^2}{c^2} } } - 1\right) mc^2[/math] (where m is non-increasing with velocity in this formula). But that will not really help understanding anything I'm afraid. At least from that you can verify that for any kinetic energy however large, the velocity leading to this energy is smaller than c. And that any kinetic energy, however large, can be achieved even when the speed is limited to be <c.

According to modern mainstream physics there is no theoretical limit for the kinetic energy of a particle. Any particle will still move with a speed <= c regardless of kinetic energy. Saying that an increase in mass was responsible for this effect (v<=c regardless of kinetic energy) is a popular explanation amongst laymen; and at least not completely wrong (I personally dislike it, though). Particles are either stable (will not decay or transform into something else without the influence of another particle) or unstable (spontaneously transform into something else). The kinetic energy or speed of a particle does not influence that.

1) The particle physicists definition: Standard Model elementary fermions and bound states containing Standard Model elementary fermions. Since the photon is a Standard Model elementary boson, it is not matter. 2) Cosmo-guys definition (afaik): Stuff that is well-treatened as massless is radiation, stuff with significant mass (compared to thermal energies I'd think) is matter. Since for massless photons, being massless is a good approximation, photons are not matter in the cosmo sense. 3) Definition of people who don't understand that "mass" and "matter" are two different words (even if they start with the same letter): A photon has energy, therefore by the only thing that I know of this Einstein guy (E=mc²) it is matter. In that definition it counts as matter. 1) Is probably closest to what you'd call a definition that everyone (in physics) agrees on - although mainly because there's not so much variety of elementary particles on a larger scale; it's all just quark-gluon combinations, electrons or photons there. 2) Is just a practical definition for calculations but I do not know how widespread it really is (I just happen to have read it in a cosmo book), 3) is probably the most widespread definition on the internet ... The idea that a photon was only energy is widespread and silly. A neutron has mass and energy. A photon has energy but no mass. A car has a color and tires. A ship has color but no tires. No one would call a ship being a form of color. But for a photon people suddenly think it was a sensible idea calling it "pure energy" or similarly.

The 1st google hit for "physics university lisbon" is a group for mathematical physics: http://gfm.cii.fc.ul.pt/ . I'd find it strange having a group for it but no opportunity to study there. Perhaps there's courses offered by the math department ? Or they take advanced students having completed studying physics engeneering? Or it's a different university?

I cannot give you an easy explanation for your question but I'd like to comment on something else: From the perspective of a theoretician that is fundamentally wrong. The orbitals are one-electron solutions. Strictly speaking they would not even be valid for many-electron systems but they seem to work fine. The reason why that the number of electrons in an orbit is limited is the Pauli principle, not the negative charge of the electrons. The electrons do not crash into the nucleus (whereas it is not entirely clear what that means in QM; in some sense they do) because there is no such orbit - but that is not a good answer, of course.

I find that a rather strange proposal considering that according to himself, Kip Thorne researches gravitational waves, not teleportation or time travel. @Enigma: Considering you know pretty much nothing about the topics that you definitely want to go into you should not rush it. If you feel you really need to go to the top university in that research area you can still do that after you got your BSc, masters or PhD. If you are interested in moving to a different country or even continent for studying to get some life experience that is fine. But don't do that because Internet rumors told you the best universities were there. If "their theory department has people working on the stuff I am interested in" is an argument to chose one university over another, then that is fine - particularly if they'd suit you equally well, otherwise. But don't make that the only criterion.

I think it is out of question that with N digits 0 and 1 I one can form 2^N different combinations. I would like to emphasize that still no one forces me to use a 32-bit integer to address memory just because I am on a 32-bit machine (at least in theory, in practice that is how they address their memory nowadays) - this is also what Pangloss meant, I think. Popular counter-examples: - My 1st computer was a C64: 8-bit registers, 2 registers for addressing the memory => 64 kB addressable memory. - Pretty much all 16 bit home PCs had by default a 1 MB addressable memory range (due to some completely moronic addressing system), more (called EMS and XMS, iirc) was accessible with loading special libraries. The 1MB limit mentioned above was, at least according to common belief, introduced because the designer foresaw that no computer will ever need more than 1MB of RAM. Considering scientific problems there is no limit of RAM that you could fill if you wanted to (provided you have the computing power to make use of the additional data).

That should carry some unit, probably Joule. "nx=1, ny=2, nz=1" probably is a typo and should read "nx=1, ny=2, nz=2"? A degenerancy of 1 is usually not referred to as degenerate. I did not check the numbers. Technically, a proof that all other combinations of nx,ny,nz would exceed the maximum energy is missing. Other than that and the (minor) comments above it looks fine to me.

Yes. You should by now have been able to come up with that yourself. It's merely the sum of energies of three degrees of freedom (x, y and z) in a 1D box. Let's just call it nx=1, ny=1, nz=1. You mean the right thing, but there simply is no variable called "n", anywhere. More generally: How it relates to the given energy. It's quite improbable that a quantized energy will exactly equal a given one. Yes. Now try to find all states whose energy is lower than the given. Then think about the degeneracy.

Assuming "cartesian space" means (x,y) where y=f(x) then the n-th degree polynomial [math]P^n[/math] running to the points [math](x_i, y_i), \ i=0 \dots n[/math] (with [math]i \neq j \Rightarrow x_i \neq x_j[/math] is [math] P^n(x) = \sum_{i=0}^n y_i \prod_{j\neq i} \frac{x-x_j}{x_i-x_j} [/math]

Is that the same n or is it three different n? You get faster with experience plus you have copy&paste which is a huge advantage over using pen&paper. In fact I sometimes to tedious calculation on latex rather than paper when they involve a lot of repeating the same stuff because I am a) sometimes a bit faster then and b) less prone to errors when copying something from the previous line to the new one.