timo

Yes. But perhaps you should be a bit more careful. After all, another relation which looks very similar, [math] ( {\bf a} {\bf b} ) {\bf c} = {\bf a} ({\bf b} {\bf c})[/math] does usually not hold (left-hand side is parallel to c, right-hand side is parallel to a). Partly. You have shown that the eigenvector [math]{\bf \lambda}[/math] is an eigenvector to the new matrix, too. You have to at least prove that for all of the other eigenvectors, also. Strictly speaking, you'd then have to go on and prove that there is no eigenvector of the new matrix, which is not an eigenvector of A. You have already found that the eigenvalue to the eigenvector [math]{\bf \lambda}[/math] is zero. You will find that for the other eigenvectors, the eigenvalues are the same as they were for A during the proof that all other eigenvectors are the same for both matrices.

You should try to write in complete sentences. Also, you should make your point clear rather than posting "a chicken crosses the road. think about it!". Expecting people to invest time into your thoughts and at the same time putting less than five minutes of effort into it yourself is just rude.

You're very much welcome. I would try it heads on. Take a vector [math] {\bf v}_i [/math] which is an eigenvector to the i'th largest eigenvalue, and show that it is an eigenvector to both matrices (of course you don't have to explicitly show that for matrix A since that was the condition that you start with). Hint: [math] \left( {\bf vv}^T \right) {\bf x} = {\bf v} \left( {\bf v}^T {\bf x} \right) [/math] (you might want to verify that first, if you don't know why).

The electron is attracted to the proton and not to the neutron because the proton has an electric charge while the neutron has not. In the quark image, just add up the individual electric charges of the quarks making up proton or neutron to see that. Not completely sure if that was your question, though.

From the viewpoint of efficient gathering of data, I don't think it is a very smart strategy to insult people in whose opinions you are [allegedly] interested in before a discussion has even started. EDIT: guess I'm being proven wrong by this thread.

That depends on the type of gas molecules you put in. If you add gas molecules of the same temperature as your gas had, then the temperature will remain the same and P will increase proportional to n. If you you put in n2 mole of the same gas at a different temperature T2, then the resulting temperature will be different. So mathematically, you are then asking what happens to P if V and R are constant, and n and T vary. And the answer is: that depends on how n and T vary. Or put in other words: you simply don't know. That is the mathematically correct answer. In the real world, say working as a scientist, you are usually not given such abstract questions, but will often encounter such questions in the context of an actual problem, i.e. have a few more information than you gave here. For example, the equation you gave is some state equation for the ideal gas. So in many problems where you use it, you could expect that the equation [math]E= \frac f2 nRT[/math] also holds (don't worry about the f, just assume f=3 if you don't know the equation in this form). So if now I add n2 mole of the same substance with a temperature T2, I get a total energy [math]E' = \frac f2 R(nT + n_2 T_2)[/math] and a temperature T' of the mixture which is [math] E' = \frac f2 R(nT + n_2 T_2) = \frac f2 (n+n_2) R T' \Rightarrow T' = \frac{nT + n_2 T_2}{n+n_2}[/math]. So with this not-too-far-fetched additional assumption, you can calculate the new temperature, and therefore know how P will behave. In case you are too lazy to proceed from that point yourself: unless the temperature of the new particles is zero, the pressure will increase (assuming constant volume in all of the above, of course). I hope it's not too confusing for you. The problem here is that people without a rigorous training in a scientific field tend to formulate problems and solutions in very general ways but at the same time take a lot of assumptions for granted which are not necessarily true. The 1st paragraph of this post is the correct answer to your question. The 2nd paragraph is more likely to be in the line of what you actually wanted to ask, and probably more helpful.

Energy put into a system will in some sense spread evenly over all possible energy contributions. I.e. whatever form of energy you give your new particles, after a while all particles will have certain average translational energies and certain average vibrational energies (assuming your gas particles can vibrate, of course). What actually pushes of course is the motion of the particles, not their rotation or vibration. None. By what you said so far, n and V increase in a not-further-specified way. Hence, pressure and temperature can still do pretty much anything. In practice, you might often have the situation that T or P are constant (say because you have your gas in a container through which energy can be transferred (-> system will reach same T as outside) and a mobile container wall exerting no forces by itself (-> same pressure as outside).

There is an extension of the space-time symmetries which is called "supersymmetry", or SUSY for short. If SUSY exists in nature, then for every elementary particle, a similar particle, a superpartner, must exist. The elementary particles currently known are not superparter to another, so if SUSY exists, quite a lot of yet-unknown particles must exist. These superpartners of known particles are called sparticles (and also the superpartner of say the electron is the selectron). SUSY is by far the most prominent model for exotic physics, i.e. extensions of the Standard Model of Particle Physics. It is not the same as superstring theory. String theory is often considered to need SUSY, but SUSY does not require string theory, and is more general. And yes, the relation particle <-> anti-particle also holds for sparticle <-> anti-sparticle. People don't speak of sprotons or sneutrons, though. Those are not elementary particles, and an object being composed of squarks and gluinos (sometimes, the superpartner is labeled with an ending -ino instead of a leading s), provided it could exist, is not likely to resemble a neutron or a proton.

Yes. Past, present, and future are like behind you, at your place, and before you. That's not three dimensions in the same sense as length, width, and depth. It's only different values of depth.

When I was a 1st year math student a professor in a seminar asked me if I knew what a proof is. I told him something along the lines of "series of logical steps ultimately tracing back to the basic axioms". He said "well, actually it's an argument that people understand and agree with". So whether your proof is correct, complete or even a proof at all somewhat depends on whom you want to convince. Anyways, you're not asking for anecdotes but for more specific replies. So a few comments that come to my mind: - It better be [math]| \lambda_2 | > |\lambda_3|[/math], not [math]| \lambda_2 | \geq |\lambda_3|[/math] for your statement to work. - I'm not sure what you need the proof for. I'd try a more specific definition of what convergence towards eigenvector [math]{\bf v}_2[/math] means. Or in other words, saying that [math] \lim_{k\to \infty} A^k {\bf x}_0 / \lambda_2{}^k = c_2 {\bf v}_2[/math] is fine, but I'd motivate more why you look at [math]A^k {\bf x}_0 / \lambda_2{}^k [/math] in the sense of why the convergence of this term towards the 2nd eigenvector means that [math]A^k {\bf x}_0 [/math] converges to (a multiple of) [math]{\bf v}_2[/math]. - Strictly speaking, it is possible that [math]{\bf x}_0[/math] is orthogonal to the eigenvector of the 2nd largest eigenvalue. - What if there is more than one eigenvector to the 2nd largest eigenvalue? I'm not sure to what extent my points above are relevant for your actual problem, so decide that for yourself.

Perhaps that's never been thought of before!

Having a substance somewhere is not a binary property, but a gradual one. You are just going to end up to have more water on the right side side with the solutes, i.e. a higher density.

Yes [math]\alpha_0[/math] (called [math] c_1 [/math] in the snippet you posted) is zero. It had been cooler if you had added the reason for that in your post, since it sounds a bit like random guessing this way. But well, don't bother now. Just make sure you understand why that is the case.

Let [math]\vec e_0[/math] be the eigenvector with the largest eigenvalue. What is [math]\alpha_0[/math]? I think that helps a lot.

Assume you have a vector expressed as a linear combination of the eigenvectors of A, i.e. [math] \vec v = \sum_{i=1}^{12} \alpha_i \vec e_i[/math]. What is [math]A \vec v[/math]? What is [math]A^n \vec v[/math]?

I think the ancient Greeks thought that a body's natural state was being at rest and that keeping in motion did require energy. It is generally attributed to Newton (again, if I recall that correctly) to realize that this is not true and that the natural state of an object is to keep moving at the current velocity unless somehow affected from the outside. So to answer your question: it has been thought of (by the ancient Greeks). It is generally considered not being the case since Newton (~300 years). Incidently, the idea that nature could be understood by rational thinking alone was also the Greek's attitude, I think, and the attitude that ideas alone are insufficient and should be tested in experiments is rather new (though possibly slightly older than 300 years).

The oxygen occupies the whole volume before you remove the CO2, and it does so after you remove the CO2. A gas occupying a volume does not restrict another gas to occupy the same volume (in fact it has to for the whole thing to be a mixture). Keep in mind that thermodynamics is about the large-scale effects (the "thermodynamic limit"), not about atomistic "but in reality, the atoms cannot occupy the same place at once"-details. EDIT: An alternative statement applying in this special case and which more people will like: Ideal gases don't interact. So your ideal-gas oxygen doesn't feel the presence or absence of the idea-gas CO2, anyways.

I have not looked at the questions. Neither do I know the sample size, nor the sample size that would be required to make accurate statements. The problem with distrusting statistics to deal with properties of large quantities of people is that there's no real other option to trust instead. Perhaps prejudice, since that's a fast and easy method. But prejudice is not very scientific. So in a scientific sense you've got to chose between statements you do not trust for whatever reason, or no statements at all. For the validity of these data: the results are pretty much stable over the years, so I believe the sample size is large enough for differences not being completely due to random fluctuations. Since even I can think of this possible problem, it's not too surprising that the experts seem to have done that correctly. If you are talking about the PISA test that the articles skyhook linked to talk about: it's not an IQ test at all. It's a test that evaluates the skills of 15 year-old pupils in the fields of reading comprehension, math, and natural sciences (and possibly others that I am not aware of).

Why would a test designed to test the average student not be an indication of the general population, but only of the top students? One possibility would be that a larger part of the general population already dropped out of school at the age of 15, but that's purely speculative. I think you should bother to add why you think so.

Book to read: Ben Goldacre: "Bad Science". Thingy to build: a LEGO robot controled by a simple neural network that drives around without running into walls. Hobby to start: vegetarian cooking with/for your girlfriend/family. Program to program: I do have a few ideas for projects of B.Sc. level in statistical physics, the CUDA one probably being the best-suited for someone without access to serious CPU power. The idea is pretty much guaranteed to work, but you'd basically not get anything out of it. Book to write: "Establishment in the box": the best conspiracy theories of the last 5 years of sfn.

I wish my research was as valuable as the cited "questionable" examples.

I tend to agree with D_H on all points expect for Garret Lisi. 1) You (D_H) presumably refer to his Internet-shaking "exceptionally simple theory". It's certainly been well known for its interesting title. I'm not too convinced about its actual impact, though. 2) According to Wikipedia, he was on a 77k$ research grant while writing this paper (which 2 years later was extended by another 77k$), and teaching physics at a university before that. I don't think I'd technically count him as an amateur.

I meant "initial current" (or more abstract: initial state), not "initial oscillation". Corrected. Don't take that sentence too serious, anyways. It's just a rough idea that's supposed to lead to an Ansatz, not a well-formulated result.

The corresponding homogeneous differential equation (deq) to your problem would be [math] L \frac{dI}{dt} = -R I [/math], which is well known to have solutions of the type [math] I = e^{\lambda t}[/math]. The deq you present is inhomogeneous, as there is an additional driving term of type [math] \sin( \omega t) [/math]. The solutions of an inhomogeneous differential equation are the solutions of the corresponding homogeneous deq plus an arbitrary solution of the inhomogeneous one. From physical intuition, what I'd expect is that due to the damping term RI, the initial oscillation current is suppressed over time, and only the oscillation driven by the [math]\sin(\omega t)[/math] term remains. To make my point: I'd try [math] I(t) = e^{\lambda t} + \alpha \cos ( \beta t ) [/math] as an Ansatz (possibly even chosing [math] \beta = \omega[/math] from the very start on). No guarantee that it's going to work, though - I didn't test it.

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