timo

Preparations: A) Figure the message you want to convey. Sadly, that's "we did the same thing as the other guys, but with slightly different parameters/systems" in many cases. Possibly even more sadly: that's enough to get it published. B) look through the results of your calculations/simulations/experiments. How do the data support the claim, and which of them do you want to show. Sadly, cherry-picking data (i.e. showing only the data that fit best and ignoring that some of the data not shown wouldn't even support the hypothesis at all) is very common there. Luckily, you might not need to because one can get away with rather dubious data anyways (a colleague recently got a paper accepted in which the referees said that the data are not entirely convincing, but probably the best that can be done with today's technology). Certainly, a lot can be said about actual writing, but (cliche!) the process can often be cut down to realizing that B) is void. No offense meant, btw. I'd gladly say a few words about the actual writing once you figured out B). It's just that before that it is a waste of time.

I don't really understand the question, because I could only guess which variables correspond to "rating" and "power". Anyways, perhaps the identity "log(2x) = log 2 + log x" helps you? Another related and helpful one is [math] \log a^n = n \, \log a [/math], btw. Should both work in all bases.

I would not expect that "3^3*x^9*y^12 over (2x)^3*y^6" is a consistent rearrangement of "(3x^3y^4 over 2xy^2)^3" in the alternative reading - that's how I came up with my interpretation, at least. I agree that unambiguous presentation of terms would be cool, though (but still think "not saying what's actually asked for" is a more serious mistake).

I don't really see what you're missing CaptainPanic? "If an Atom is kept away from any form of external influence then how long it will live ?" seems to be a well-defined question to me (and I am pedantic). The answer is, as Tom said, that it depends on the atom. While some atoms decay very fast (->radioactivity), there are atoms that will live forever - at least as far as I know.

You're incorrect in assuming that $6000/$3000 = $2000, for instance.

The raw_input() is supposed to stop exactly this immediate closing of your program. It should be the last command/line in the program file itself, i.e. a test program (stored in a file ending on ".py") could be print "Hello Windows ", 3+4 raw_input("Hit <enter> to ->exit") If that doesn't do the trick, then you'll either have to play around with some Windows settings (there might be something like "close window after program finishes") or learn how to use the Windows console.

I agree that khaled's buddy was likely to refer to something like relativistic space-time intervals, but I disagree that he has any point at all. The existence of a path from A to B with a length of X does not imply that the distance (in the sense of "any properly defined distance") between A and B is X.

Or add the line raw_input() as the last line of your program (it will wait for you to press <enter> before finishing the program), and try to run your program by double-clicking it. It you're lucky, files ending on ".py" are automatically opened with python. While Cap'n is right that Windows has a console, he's wrong in assuming that Windows users use it .

To make matters short (and not explaining what you did wrong - it only really matters if you are not using Windows, which seems rather unlikely). I'm not a python expert, but you could try import myfirstprograp.py or execfile("myfirstprograp.py") @Capn: My suspicion is that dragonstar might simply not have a console.

Create a homepage.

Sometimes you have to throw only once (which is half of the time where you immediately get heads), sometimes you have to throw more often. So obviously the average is more than one. The proper way to calculate a statistical average is to sum up all possible results weighted with their (absolute) probability. For example: let's assume you'd throw the dice twice and want to know the average number of heads showing up: each possible combination (heads-heads, heads-tails, th, tt) has the same probability 1/4. The number of heads in the combinations is 2, 1, 1 and 0, respectively. The average number of heads is [math]\bar h = 2\frac 14 + 1 \frac 14 + 1 \frac 14 + 0 \frac 14 = 1[/math]. Getting the average number of coin flips is very analogous, and only slightly more complicated because the number of relevant combinations is infinite (but the result will still be finite).

@moo: I know and understand that not everyone is such a pedant as me who wants (their) students to understand why they are doing what and to be able to communicate this to others (of course I am not completely clueless what the task might have been). But actually giving the answer with all steps is not exactly the idea of the homework section, and a bit too much - see Janus' post for what I guess everyone would consider a reasonable reply.

No offense meant, but I don't see a problem in the first place. Your re-arrangement [math]\left( \frac{3x^3y^4}{2xy^2} \right)^3 = \frac{3^3x^9y^{12}}{(2x)^3y^6}[/math] is a mathematically correct step. But what is it supposed to be (good for)?

Wikipedia sais [math] \log_b r = \frac{\log_a r}{\log_a b} [/math]

I'm not sure if that's what I said - I don't understand it. I'm not even sure if at least one of the sentences is grammatically correct.

You misunderstood that statement in at least two respects: (1) More a detail: I am talking about a free electron. The only energy that a free electron can have is kinetic energy (momentum), which is not quantized in this case. Speaking about "energy levels" does not seem like a useful concept in this case. At least I don't know what you mean by it. (2) The presumably crucial point: The term "state" in the context of QM is a technical term, which may not exactly be what a QM-layman expects from it (which is why I wrote "note that 'state' is a QM term"). In classical physics, the state of a free particle is determined by its position and momentum. In QM, that is not the case. In particular, an object's state might not uniquely determine its momentum (which is why I wrote "the final state is not a state belonging to a definite direction"). I understand that it might be difficult to understand statements that use abstract QM concepts. But since this sub-forum is called "Quantum Theory", I think it is appropriate.

If you're seriously interested in a discussion and/or in learning something then I'd appreciate if you gave my posts and your replies more than ten minutes of thought - and subsequently be more accurate when referring to my statements (it's a huge difference to measure an object as not having a certain property and not to measure the property). At the moment, I consider this as a waste of my time.

Probably got to do with the energies involved.

I don't see why measuring a photon as not having gone a certain direction seems to be such a catastrophy to you.

I don't understand what you are saying to trying to say with that. There should be plenty of texts to be found about the double slit experiment, though.

It doesn't. Consider seeing with your eye as being the measurement. You either measure the photon as having gone your way (then you see it) or as not having gone your way (you don't see it). sidenote: QM has the reputation of often being counter-intuitive. Hence, it should not be too surprising that it often works differently than you'd intuitively expect.

That's not how QM works. What happens is that if you shoot a photon at a free electron at rest (no need to complicate things by making it a bound electron), the electron and the photon will be in a definite state (note that "state" is a QM term). This state, let's call it "final state", is entirely determined by the state the electron and the photon were in before ("initial state") and the rules of physics. There is no randomness in there. However, the final state is not a state belonging to a definite direction, but a mixture of different direction. When you try to figure out which direction the electron went, you have to measure its direction. At this measurement process, randomness then comes into play, and a random direction will be picked (with the probabilities being dictated by the properties of the final state). For this simple experiment, it seems like a rather cheap excuse to claim that everything was deterministic, and only the measurement process destroys determinism. It seems more natural (in the sense of being in the lines of everyday thinking) to assume that the transition from the initial state to the final state is a random process completely fixing the outgoing direction. However, in more sophisticated experiment (you don't measure the final state immediately but let a 2nd reaction happen first), one can tell the difference.

Oh come on. You've been working long and didn't figure out [math] {\rm Tr} (cA) = \sum_{i=1}^n ca_{ii} = c \sum_{i=1}^n a_{ii} = c \, {\rm Tr} (A) [/math] ? 1b, 1c work the same way. Question 2 very analogous.

I don't know what you mean by "deterministic as confirmed by absorbtion and collision (Compton's effect)". If you mean that a collision of a photon is like the collision of two billiard balls, then "no". Loosely speaking, if you were shooting a single photon at a single electron at rest, you cannot know what happens in the sense that you don't know the momentum of the electron after the measurement. However, simply saying that the process is not deterministic is not appropriate to the complexity of the situation/scenario. Ultimately, it boils down to the problem that the part of quantum mechanics which breaks determinism, the process of measurement, is not fully understood (certainly not by me, but that's also what our mathematical physicists tell me). You can directly write down a wave function for a free photon. The basis is simply plane waves.

My first thought reading your post is "geology". Not sure if it's offered as a university degree in <whatever country you are from>, tough. You should also check how big the market for planetary science jobs is (and could at the same time have a look at the typical requirement for those jobs); planetary science does not sound like a big field with countless positions; you might want to arrange for some backup plan.