-
Posts
3451 -
Joined
-
Last visited
-
Days Won
2
Content Type
Profiles
Forums
Events
Everything posted by timo
-
I am not completely sure why you think that a proton decay would cause neutron stars to evaporate. I do not think that neutron stars collapse to a black hole either. I think they are pretty stable (but I am not expert). Anyways, let's talk numbers: a) The prospected lifetime of a proton vs. the estimated age of the universe: ~10^30 years vs. ~10^10 years. In other words: The amount of protons that did decay since the start of proton existence is zero for most practical purposes. b) The total energy of a nucleon is 1 GeV. You're not going to get more energy out of a nucleon decay (actually less but that will depend on the reaction mechanism you have in mind). Compare that to your "highly energetic cosmic rays" you had in mind: Is that enough?
-
Since "cube is meaningful" presumably means "all of the sides and edges are meaningful" (what about the points in the corners, btw?) this seems trivially true. No, it's the other way around. If MT is meaningless then the whole cube cannot be meaningful. Of course, part of the cube can still be a meaningful statement despite the cube as a whole not being so. Example: Assume MT is a meaningless expression in any context imaginable. Then, the cube as a whole does not consist of meaningful expressions only. Still v*t is part of the cube and can be a meaningful expression in some contexts (namely being a distance traveled during a time-span t for an object with constant velocity v). I'd always chose "my ideas are great" if I had the choice. Sadly, it's not for you to decide except in your fantasy. While A is a correct statement, it also implies the possibility that the cube as a whole might simply not make sense. I am not sure that you understand/-stood that.
-
1) In TeX, so the term can be read better. You mean [math] \partial_i [ \lambda \delta_{ij} \partial_k U_k ] = \partial_i \lambda \partial_k U_k + \lambda \partial_i \partial_k U_k[/math], right? 2) First off: The expression is not correct. On the left-hand side there is an open (=non-contracted) index j and on the right-hand side there is an open index i. EDIT: The position of the indices (upper or lower index) sometimes (e.g. in relativistic physics) is important. I have dropped that here - writing the terms with correct positions of the indices is straightforward. 3) I am not completely sure about the convention what to derive. I'd expect that [math] \partial_x AB := \partial_x (AB) [/math] but the parentheses in your example seem to indicate otherwise. So I think it is the chain rule (plus the correction of the indices) that took place there: [math] \partial_i [ \lambda \delta_{ij} \partial_k U_k ] = \underbrace{\partial_i \delta_{ij}}_{=\partial_j} [ \lambda \partial_k U_k ] = (\partial_j \lambda)(\partial_k U_k) + \lambda (\partial_j \partial_k U_k)[/math]. Since you say lambda is a constant that term would then simplify to [math]\underbrace{(\partial_j \lambda)}_{=0}(\partial_k U_k) + \lambda (\partial_j \partial_k U_k) = \lambda \partial_j \partial_k U_k[/math]. 4) I do spontaneously not see what the expression are supposed to represent. So keep in mind that what I said above is purely algebraic, not physical.
-
If people you consider idiots feel that what you consider idiotic points are validated by that: so what? If you bother about other people reading the debate: Just imagine that those people are intellectually capable of forming their own opinion, irrespective of who had the last word.
-
Pushing things so that they do not stop moving is nothing special. It's how motors work, for example. I am not sure that I really got your point, though. As a comment on form and style: Approximately proper spelling, capitalization and punctuation would rock.
-
There is actually no problem with this statement. However, you are used to relative velocities being constant when you switch to a moving frame: If you see two cars approaching each other with 50 km/h in opposite directions their relative velocity is 100 km/h. This holds true in relativity. But you are used to think even more, namely that then the cars' drivers see the other car approaching with a velocity of 100 km/h, too. That simple transferring of relative velocities to other observers does not hold in relativity, anymore. Oh, and welcome to sfn. According to my system clock and your user profile you are about to register here tomorrow .
-
The posters exist in German, Italian and French. The three main official languages spoken in Switzerland. I posted the French version because it is the language I simply assumed the most people to know at least a little. I am not from Switzerland at all.
-
Sure, you've got your fancy lab coats and your real diploma. But I am a logical thinker who questions things and can think out of the box. Also, I scored 154 in an online IQ test.
-
I found the Taliban woman on that poster somewhat more inappropriate. A skyline full of minarets at least had something to do with the question (in principle, not necessarily in scale). Anyways, my impression is that blaming the result on a questionable campaign, narrowed down to a single poster, is a lame excuse. It is in fact also an offense to the voters, too. I claim that the average voter did not base his decision on that poster. They (can't remember exactly who, perhaps some newspaper, perhaps some official statisticians) did ask the German people about what they had voted if they had had the choice. Result: about 50:50. Consider people are more afraid to admit politically incorrect attitudes (disallowing building of minarets would be a total no-go officially) when directly asked than to secretly give their vote for such endeavors in the voting cabin => I think there would be quite a solid majority for banning building of minarets in Germany, too. And, and this is my point, that is without any muslims-come-to-assimilate-our-glorious-culture campaigns at all. Assuming Germany being somewhat similar to Switzerland: The result of this vote is not due to questionable campaigns, it is a representative statement of the people's will/fears/racism/atheism/whatever. EDIT: Since most people will not know the poster, here's a picture:
-
No way. I believe in natural selection. (An explanation for this post, since I assume not everyone has studied math: A set is said to be countable if there is a mapping of every element onto one unique natural number, i.e. a labeling of every element via a unique natural number. Integers are countable but the real numbers are non-countable ("over-countable", though I am not 100% sure if that is the correct English term). IOW: There is no way to label the reals with natural numbers.)
-
Using anything other than integers and their fractions is irrational, at best.
-
No. A sequence that is constantly equal to 0 does converge to a limit (zero) and does not grow over all bounds (i.e. diverge to infinity). Same goes for the series that results if you add the terms in the sequence up. A sequence that does not grow over all bounds does not have to converge. It still can.
-
I mean that a sequence that alternates between the values 0 and 1 does not "diverge to infinity" and does not converge, either. Therefore, a sequence that does not diverge to infinity but does not converge either exists. I certainly did not say that a sequence that does not grow arbitrarily and does not converge must always alternate between two numbers. It can obviously alternate between three numbers, too. And it's hopefully obvious that alternating between two or three numbers does not sum up all possibilities, either. EDIT: Could it be that some of your "sequence" is supposed to be "series" or "sum of"? Because on 2nd reading I really wonder whether your opening post "Assume A. Now, if not A ..." perhaps was supposed to make sense?
-
No. Just imagine a sequence alternating between the values 1 and 0 as a counter example.
-
Or you just claim that with saying [math] \sqrt{x(x+1)} = \sqrt{x^2+x} \approx \sqrt{x^2} = x[/math] you throw away a (large) term x under the root while with [math] \sqrt{x(x+1)} = \sqrt{x^2 + x} = \sqrt{ (x+1/2)^2 - 1/4} \approx \sqrt{ (x+1/2)^2} = x+1/2 [/math] you only throw away a small constant so the latter approximation is better.
-
No. Counter example: The first derivative of velocity with respect to time is acceleration, not velocity. ok. No. When you set the velocity equal to zero and solve for time you get the time at which the velocity equals zero. It might be interesting to think about what that physically means (and how the whole motion look, physically).
-
It's really not my intent to piss you off or something like that but a) "because the derivative is velocity" is not an answer to any of the three questions I asked and b) the only sensible answer to your question whether to set it to zero or not (that is in agreement with the forum rules) is again "why so?". I mean, you must have an idea why you want to do so? If you feel like that you have an idea but find it hard to express: Finding something hard to explain to others is often a sign of not having understood it properly. So it really can be worth poking exactly such points.
-
Why have you taken the derivative and of what and with respect to which variable? If you know why you did so then you're probably close to the solution.
-
As a layman I would expect that the answer is "yes" for most modern laws and that the lawyers even have a name for this type of law. As a personal comment: I find the idea than health care for everyone was against general welfare somewhat alien.
-
Science talks: Board & Chalk, White Board or Projector?
timo replied to ajb's topic in Other Sciences
Of course in theory that depends on field, topic, audience, message you want to get across, ... . I don't see me doing anything but beamer (slides, perhaps some live simulations) or transparencies (if there is no beamer), soon. For math where the (details of the) calculations might be more interesting to the audience blackboard is probably also appropriate. I do not really see the difference between a black- and a whiteboard except that on a whiteboard there's better chance to have a readable hand-writing. If you're doing a statement->proof->next statement-> next proof->... type of talk then having the statements on a transparency shown during the whole talk and doing the details on the blackboard sounds like a promising idea. Particularly if some of the earlier results are re-used later on (speaking for me, I am never able to follow a talk in which a result from two slides ago is used unless I knew that result before the talk, already). Just an idea of mine, though. -
Apart from a few typos (p1=-p2, not p2) and potential problems with your denominator I have no real objections, yes . There's two objections I have to the last step but I can't put my finger on it, atm. Anyways, those two objections are: 1) You derived an equation in the cms (=center-of-mass = system where the total 3-momentum is zero) frame. In that equation you have something that transforms like a Lorentz scalar on the left side and something that probably transfers like inverse of the 0-element of a vector. Your equation would probably not survive a boost into a different system. 2) I don't think there is such a system. In the cms frame you have the two daughter particles flying apart back-to-back. To which direction would you want to boost that both daughter particles have zero momentum?
-
I do not understand what you do there. Take positronium to 2 gamma as an example. In the cms frame the momentum of the positronium is [math]P^\mu = (M, 0, 0, 0)[/math]. Let the gammas fly out parallel to the z-direction. They then have the momentum [math]p_1^\mu = (M/2, 0, 0, M/2)[/math] and [math]p_2^\mu = (M/2, 0, 0, -M/2)[/math], respectively. This does of course conserve momentum as [math] P = p_1 + p_2 [/math]. The mass of a single particle is the pseudo-magnitude of its momentum, i.e. [math]M = \| P \|[/math] and [math] m_1 = m_2 = \sqrt{p_i^{\mu} p_{i,\mu}} = 0[/math]. Hence [math] M \neq m_1 + m_2[/math]. Of course it holds true that [math] M = \| P \| = \| p_1 + p_2 \|[/math] - that is just a weaker version of conservation of momentum (if a vector remains constant that naturally its magnitude also does). Perhaps the best question to understand what you do would be: What is M, m1 and m2?
-
Whatever you've been thinking of there, I think it is wrong. poss a) You you mean a decay where kinetic energy is negligible then the conservation of mass is only as true as the approximation you did (your equation would then be conservation of energy). Strictly speaking, I don't think it could ever be exact because you need some non-zero final-state phase space volume for a reaction to happen (excitation daughter particles might provide some phase space, though). poss b) If you were thinking of the pseudo-magnitude of 4-momentum to be conserved then the problem is that in general [math] \| \vec a + \vec b \| \neq \| \vec a \| + \| \vec b \|[/math]; in the letters of your example: [math] M = \| P \| = \| p_1 + p_2 \| \neq \| p_1 \| + \| p_2 \| = m_1 + m_2[/math]. By that I mean: The momentum is conserved, of course. So is its pseudo-magnitude which is sometimes even called the invariant mass of a system of particles. It is however not true that the sum of the individual decay products' mass was equal to this invariant mass; the simplemost counter example being Positronium (m roughly 1 MeV) decaying to photons (no mass individually). poss c) You might have meant relativistic mass above (I doubt it) but there'd be no reason not to call it "energy" in the first place.
-
No. It has energy pretty much the same way that it has some electric charge. Some mass-energy (it's called "mass" - I am just emphasizing that it is just a form of energy) and possibly some other energy, say kinetic energy. Particles can be converted into other particles. A photon is not energy either - although it is quite a common misunderstanding that it was. It is just another particle that has energy, the only difference being that the energy of a photon is purely kinetic with zero contribution from (zero) mass. Sidenote on different conventions: Some people call the total energy of a photon (or any other free particle) "relativstic mass". As you see, I don't. Like I said: You cannot convert matter to energy. You can convert mass (a form of energy) into kinetic energy (another form of energy) via converting particles into other particles. Assuming mass and kinetic energy to be the only energies possible: When the resulting particles have less mass than the initial ones, the total kinetic energy of the resulting particles must be larger than that of the initial ones such that total energy is conserved. The first process that comes to my mind here is nuclear fission. At the heart of the process, a neutron and a nucleus are converted into 2 (maybe also more?) other nuclei and some neutrons and possibly some other crap. The resulting particles have a high kinetic energy and can then heat water that drives a heat engine. You could in principle store electrons and anti-electrons in two different containers, let them combine and form two photons with high kinetic energy that you then harvest. The basic reason why this is not done that it's just not practical. Yes. In the case of the sun it even happens without significant risks due to radioactivity. Note: Since I can't edit my previous post anymore: Electric charge also need to be conserved under any process. I forgot that in my previous post. I am currently not sure to what extent conservation of angular momentum (which is conserved) is important, too.
-
(edited:) Whatever; see ajb's post which you'd probably like more.