timo

[math] E(v) = m(v) c^2 = \frac{m(v=0) c^2}{\sqrt{1- v^2/c^2}}[/math]. No acceleration there, only velocity(v).

And when the photon and the anti-photon meet they vanish, leaving only an energy-burst consisting of a photon and an anti-photon?

Which kind of exam (which field, which qualification, ...)? Speaking of physics exams in germany, post-examinatory notes of the examinees for pre-diploma and dimploma exams are not really wanted (but tolerated) by the universities and the examinators. Hence, they are not available for download publically but can usually be obtained/copied from the student body of the faculty. Oh, you could also state whether you speak of oral examinations or written exams.

@1: You need more energy to pick up the hammer. That's reflected in the hammer hurting more when you drop it on your foot, because it has a higher momentum at the same speed. @2: The attraction that the feather and the hammer extert on the moon is very small, ignored in the calculation claiming they both drop within the same time and in fact small enough that there's no significant (measurable) change if it was considered. Ignoring part of the mutual attraction is not valid in your planet/planet or ball/ball example.

Hm, might be formulation-dependent, i.e. depending on what you demand from the operation in the first place. However, my algebra book seems to agree with your professor, so listing closure explicitely as a group axiom might indeed be uncommon. Wikipedia imho doesn't count as a reliable source. Articles on relatively "everyday topics" like this one (or alternatively topics mentioned in popular science books" do, according to my experiences with the german Wikipedia, have the tendency to be written or at least heavily influenced by interested pupils or undergraduate students at best, which often lack the overview over the topic and only know the particular formulation that they learned in their classes.

Sure that's defined by above rules: [math] \infty + \, -\infty = \infty, \ -\infty + \infty = -\infty[/math]. To what extent this definition makes sense is another issue and probably only decidable in a given context (e.g. you now will run into problems applying this to limits of series or functions). But to come back to my original statement being "don't write infinity minus one equals inifnity": I see this kind of proven as an analysis of the statement did indeed cause confusion as plainly demonstrated - not even the two definitions of the people who said "sure, no problem to include the infinities" do match. In fact, I've additionally asked a friend of mine who did his diploma thesis in a related field (set theory) and he spontaneously gave me a sketch of a 3rd way of including the infinities. EDIT:@Tom: [smart-ass]Afaik, closure is a group-axiom.[/smart-ass]

If you're replying to "energy isn't conserved under changes of coordinate systems" with "show me the math showing this" and then reply to the numbers with "that cannot be true because energy is conserved under changes of coordinate systems" we're stuck in a loop, I think. Yes, the two values [math]E_A[/math] and [math]E_B[/math] are the values for total energy in the respective systems (you described). Yes, they are unequal. The \sqrt{1/0.19} comes from gamma = \sqrt{1/(1-v²/c²)} = \sqrt{1/(1 - 0.9²c²/c²)} = \sqrt{1/0.19} and is the Gamov-factor. See it this way: Energy is conserved for any physical processes. Changes of coordinate systems are not physical processes, but merely switching to a different description.

[math] E_A = m_A c^2 + \gamma m_B c^2, \ \gamma = \sqrt{1/0.19} [/math] [math] E_B = \gamma m_A c^2 + m_B c^2, \ \gamma = \sqrt{1/0.19} [/math] I still don't see any energy released, but the total energy (before any potential crash) is different, already.

It's the rest-mass. Your equation lacks a ")" which supposedly is a typo. [math]E_{\text{kin}} = E - E_0 = m_{\gamma}c^2 - m_0c^2 = \gamma m_0 c^2 - m_0 c^2 = (\gamma - 1) m_0 c^2[/math] where [math]m_{\gamma}[/math] and [math]m_0[/math] are relativistic mass and rest-mass, respectively.

There's a 4-line implementation of a quicksort for Python on Wikibooks. It might not be the best possible solution due to relatively large stack usage, but you could try that ... *looks it up* ... http://en.wikibooks.org/wiki/Algorithm_implementation/Sorting/Quicksort#Python Considering that and the relatively large programming overhead in creating a usefull tree-structure I think doing a sort on the list would be option #1.

Only if m := e/c²

Abskebabs, the problem with your questions is that a serious answer would probably be too long for a forum post (and that my article explaining at least the very start of it was lost with WiSci). So a very compressed reply: Kind of "yes". GR bases on "differential geometry", hence thinking of vectors as arrows (tangential vectors on spacetime, actually) imho greatly helps in understanding them. A "contravariant vector" strictly speaking is not a vector but its coordinates in a given base (which is usually not mentioned explcitely). The name comes from that for the vector to remain the same under a base transformation, the coordinates must transform inversely to the base. Note: If the coordinate base is given or obvious, then of course the four numbers of the "covariant vector" are a sufficient description for a vector. Pretty much the same as with contravariant vectors except that they are the coordinates in the reciprocal base (which is pretty much the same as the reciprocal base in solid state physics except for the factor 2pi and a different metric). I don't exactly get the question, but the pseudo-scalar product (which is often just called "scalar product") of two vectors is an invariant scalar, regardless of being the scalar product of the vector with itself or with some other vector. That's similar to the angle between two lines drawn on a piece of paper being independent of the coordinate you use to describe points on this piece of paper (in fact it's even independent on whether you use a coordinate system at all or not - you just can't calculate anything without a coordinate system). The result of the product of a momentum 4-vector with itself is the rest-mass squared: [math] \left< p | p \right>_{\text{pseudo}} = \underbrace{E^2 - {|\vec p|}^2}_{\text{requires a coordinate-system to make sense}} = \underbrace{{m_0}^2 }_{\text{always true, no cs needed}}[/math]

The common view is that this additional mile per hour is unattainable because of the "inconceivable amount of energy" required. As long as v<c (which is the constraint for massive objects), the energy is always finite => no infinite energies.

If you take a close look at Dak's code then you'll see that he finds a duplicate ("if dicnohash[a] == dicnohash:"), correctly copies it to the list of duplicates ("dups.append(dicnohash)") and later checks it again over the range [a'+1,len(list)] when a'=b.

Things you could try: 1) eliminate words from the text that were already found as duplicates so they're not checked again later. That's little effort with probably little gain; in fact, the question whether you get a performance gain from that will be strongly dependent on how the text structure is stored and how it performs deletion of intermediate words. 2) read-in the whole text in a better-suited structure (e.g. alphabetically sorted binary tree where each node counts the number of occurences of the word). That changes complexity from O(n²) to O(n*X)+Y, where X is the complexity of writing to the structure (i think that typically log(n) for trees) and Y is the complexity of reading out your data. A lot of effort with questionable outcome but could be worthwhile investigating - even if it's just for getting a feeling for the advantages and disadvantages of the different approaches.

I'm not sure how you want to calculate the "energy released", not even what it's supposed to be (and admittedly I don't want to think about it right now). But energy is not a conserved quantity under changes of coordinate systems - you can easily see that when you consider a particle in its rest-frame where it has no kinetic energy and a frame where it moves (where it then has kinetic energy). There's a combined structure of energy and momentum called "4-momentum" that is conserved, but in a slightly different manner: It's a vector with 4 entries. The entries themselves change but the vector itself is considered unchanged - imagine describing an arrow on the wall with two different cordinate systems like {"up", "left"} and {"down", "right"}: The entries of your descriptions are different (the respecitve negative of each other) but the arrow is the same in both cases. btw.: It seems you used this coordinate transformation on the 2nd letter "n" in Swansont's name

A moving relativistic particle has momentum [math] \vec p [/math] and energy [math]E[/math]. The energy modulo c² (speed of light squared) is sometimes called "relativistic mass" [math] m_{\gamma} [/math], more precisely [math] m_{\gamma} = E/c^2 [/math] (note that this is the same as [math] E = m_{\gamma}c^2 [/math]). Momentum and energy are related to each other via [math] E(\vec p) = \sqrt{|\vec p|^2 c^2 + m_0^2 c^4}[/math], where [math] m_0 [/math] is a constant that depends on the particle type. [math]m_0[/math] is called the "invariant mass" or "rest mass". In constrast to the relativistic mass which depends on the particle's momentum (remember that [math] m_{\gamma} = \sqrt{| \vec p | + m_0^2 c^2}/c [/math]), the rest mass is a property of the particle type (meaning e.g. all electrons have the same mass). The name rest mass can be understood from taking the special case in which the particle is at rest (meaning [math] \vec p =0 [/math]). In this case [math] E = \sqrt{0+ m_0^2 c^4} = m_0 c^2 [/math] meaning that for the frame at which the particle is at rest [math] m_{\gamma} = m_0 [/math]. In short: - Relativistic mass is another word for energy. - Invariant mass is a particle-type dependent term in the equation that relates energy and momentum (dispersion relation or energy-momentum relation). - For particles at rest, invariant mass and relativistic mass are equal. Otherwise, relativistic mass is greater than invariant mass. - When only speaking of "mass" without stating whether it's the invariant or the relativistic mass, you need to determine which one is meant from the context. If it's a particle property (someone saying "protons have a mass of ...") then it's the invariant mass. EDIT: And when Swansont said "rest mass" is not a property of everything that exists he (hopefully) meant that it can be zero for some things (like the photons).

... that's where copy&paste comes in.

Well, the ~4 weeks are over. There's only one entry so comparison of the codes and discussion of the different approaches is skipped. @Aeternus: You can check your results and compare with the primitive BigInteger version yourself, so I'll skip that, too (I assume you did check your results, anyways). If you want to, I can have a look at your code and give some comments, but I'm not sure that there's much to get out of it for you.

You should probably state what programming language you are talking about. Additional useful information could be the operating systems used on your computer and the college computers.

I don't understand what you're asking. When you give a mathematical expression involving an operator like x+y=z, then I assume the operation to be defined for the x, y and z being members of some common set. It is very unlikely that someone adds a tree (or the tree ) to seven Volt. I assumed the "1" in your equation stood for an element of the real numbers or a subset thereof (the rationals or the integers). Infinity is not a member of those sets; it's usually not even considered a number at all. Therefore, for the statement [math]\infty + 1 = \infty[/math] to make any sense to me, I expect a set that consists of at least the two elements 1 and [math]\infty[/math] (the minus case makes it a bit more complicated but let's forget that for now). R [math]\cup \{ \infty, -\infty \}[/math] simply seemed like the most straightforward one of the possible sets. The notation simply means "all the real numbers and two additional elements called plus and minus infinity". Well, the rest is written already: Your calculation rules on this extended set didn't convince me to be obvious enough that they wouldn't need further explanation on (a) what exactly they are and (b) on where the big limits and differences to the calculation rules on the reals are (lack of some inverse elements under addition or no associativity, no closure at least under multiplication). Mathematical definitions possibly worthwhile looking up (for those who don't understand the terms "inverse", "closure", "associativity" and their relevance here): Group, Body.

I think you should rework the "site navigation" field on the main page, soon. 4 out of 7 links there are either pointless or dead (I came to the book discussions via the link there and hence didn't notice that it's in the archives - Ben probably came there the same way).

- Please try to be more precise (link) what you're talking about. I assume it's about http://www.scienceforums.net/forum/showthread.php?t=23782 . - Your low postcount is most likely not the reason you may not post there - I must not, either. I bet that either there's a registration list somewhere or (more likely) it's an incorrect setting that no one told the admins about, yet (there's currently a lot of dead links, outdated information, ...). - Chosing blike to send the pm to might have been the worst possible choice among the admins - I think he's online least of them. - The book discussions is just one of the many examples what happens when you try to get something started and running with people that you cannot kick in the lowerback physically (i.e. people from the net): Someone has a good idea -> everyone agrees that the idea is great -> everyone adds his ideas how to make it even better or alternatively comes around with a similar idea -> the volume of ideas increases faster than the people contributing -> the number of people contributing to a particular idea approaches zero -> end of story. Iow, I think the section is dead.

Personally, I'd accept addition as a semi-group operation, too. But I do not really know what the common stance is, there. Problem is, that the given rules (and I'm not talking about the [math]\infty = 0[/math] version here ) either violate closure or -if you define [math] \infty + \, -\infty [/math] to equal something- associativity. Hence, the operation defined can't be the operation on a semi-group, either.

I wasn't complaining about the part in parantheses in the term x^(x^x^...) to be infinite, I telling you that you shouldn't write "because [math]\infty -1 = \infty[/math]" as a justification because that does seem to need more justification than the original statement. Telling someone that an excess of mathematical formalism (not exactly sure what you meant by semantics) is blurring the mind seems counter-productive to the purpose of a mathematics section of a science forum to me.