timo

The answer is probably "no". I don't know what you mean but you possibly meant that if photons had mass the energy of the photons coming from the sun would be significantly larger. It'd be the same.

I was referring to yourdadonapogos' location box saying "earth". I've in fact already heard of that weird claim that earth wasn't flat and implicitly referred to that.

Here on earth?

I think I found the woman of my dreams. But just to be sure: That is a diamond and it is yours?

Not exactly violated but the concept of relativistic mass isn't really used any more as it is misleading. It is generally considered that there is ONLY rest mass which is called mass (normally) and is a Lorentz invariant. This has been discussed indepth in another thread including references and also mentioned in this very thread.

Not sure if I really understand the question but the approach I'd try would have been something like that: - Energy per turbine required: 99 MW/66 = 1.5 MW/turbine. - Since each turbine has 50% effectivity you want 3 MW/turbine to be transferred by the wind. - Energy transferred to the blades is the kinetic energy of the wind: E = mv²/2. - The mass of wind hitting the rotors each second equals A*v * 1 kg/m³ (volume passing through times density), where A is the area of the blades. This is the point where some information, the size of a blade, seems to be missing. Or I don't really understood the question. - Anyways, from this you get E/1s = Av³/2 * kg/m³. This shall equal 3 MW => 3 MW = A v³/2 kg/m³ => 6,000,000/A Wm³/kg = [math] \frac{6,000,000}{A} \ m^5/s^3[/math] = v³. Please note that I might have misunderstood the question, but I think the three blades were supposed to be on one rotor and that some width for them is missing in what you posted.

Makes more sense than I expected . I would assume that the Canadian teacher association in charge has taken care of the problem, e.g. by starting the tests at the physically same time everywhere (meaning earlier hours in the east, later hours in the west) or making different tests. But it was certainly kind of you if respected that possible loophole in the system and waited a few hours (e.g. number of time-zones minus time it takes to take the test) and then post the questions - you won't have that many time-zones, I'd think. I'd bet that the chances that you'll meet someone on this board who took the same test are more or less zero. But on the other hand judging from the questions from the last years I am pretty sure there is a lot of people interested in the questions (and trying themselves on them) and also several people that can answer them or answer additional questions about them. So please post the questions here at some time that you consider appropriate. EDIT: Oh, and welcome to sfn, btw.

Why would you think you were not allowed to post the questions after the exam was taken? Are you doing some secret nuclear weapon research with your 11th-graders ?

I don't know that and I think that is wrong or at least incomplete. In theoretical physics (and those branches of experimental physics that I am familiar with) phases mean that a substance can be in different states for which some (macroscopic) property significantly differs. For the transition between fluid and gas, for example, a suitable property would be the density. There's other phases and phase transitions that are just less commonly mentioned but just as valid as phases as the classics. Examples: - Transition between ferromagnetic and paramagnetic state in iron (-> magnetisation) - Transition between isotropic and nematic phase in liquid crystal systems (-> nematic ordering parameter). - Different lattice systems (relative positions of the atoms) in crystals (-> dunno). - Transition from superconducting to non-superconducting state by change in temperature (-> electric resistivity).

http://en.wikipedia.org/wiki/Lorentz_transformation

What would that have to do with a "girls' clique" ?

I'd imagine that in addition to possible loss of currently-written data (but you wouldn't switch of your computer while you write important data to the HD, would you?) temporary data (-trash) might not be correctly cleaned up. Personally, I think "damages your HD" or "messes up your filesystem" sounds like an urban myth. It would really surprise me if the manufacturers of HDs and designers of file systems were not aware that the average computer gets hardware switched-off (by power failure, system crash, ...) somewhere between once a week to once a year.

Unlikely that someone know that off their head but I see no problem converting it: 1 = 3600 s / 1 h 1 = x kcal / 1 J (look up the number) 1 = 1 J / 1 Ws => 1 W/m² = 1*1*1*1 W/m² = 1 W/m² *3600 s/h * x kcal/J * 1 J/Ws = 3600x kcal/hm².

It's basically because [math]c \cdot 0 = 0[/math] for any value of c. Have you tried what I proposed? What did you get and/or where did you get stuck?

What exactly is your problem? You don't understand what the question means? Just add the components of the vectors [math]\vec x = c\vec v + d\vec w[/math] (i.e. [math]x_1+x_2+x_3[/math] keeping c and d as variables and see that the result will equal zero - regardless of the values of c and d.

I've investigated the problem this night. But I am not very good at maths so I need someone that can help me putting my theory into mathematical terms. Here's my results (I am thinking about a rewrite to cut the long sentences a bit):

I can't really comment on helpfulness for you. Severian, being a teacher for that kind of stuff, is probably a better advisor on that. There is a lot more to the index notation than writing stuff in matrix notation [math] \vec a = M \vec b[/math] shows. I'll try breaking my thoughts down into tiny pieces that hopefully sketches why and when you can use matrix notation for Lorentz transformation, why I do not like to do so and why the original question strictly speaking had no answer. Be aware that the following is really basic and possibly not exactly helpful for your current class - like said, stick to thinking in matrices if your course does and it's easier for you. At a very basic level [math]\Lambda^\mu{}_\nu[/math] as such is just 16 real-valued indexed numbers. If you just want to write those numbers down you can either arrange them in a 4x4 square as a matrix (and for simply writing it down it doesn't matter at all which index you chose as row and which as column) or just in a linear vector-like style: [math]( \Lambda^0{}_0, \Lambda^0{}_1, \Lambda^0{}_2, \Lambda^0{}_3, \Lambda^1{}_0, \Lambda^1{}_1, \Lambda^1{}_2, \Lambda^1{}_3, \dots )[/math]. It's all just a way to write down the same 16 indexed numbers. At this point, it certainly is not clear which index is row and which is column; it is not even clear that your representation you chose to write down the 16 numbers on a piece of paper has rectangular form (a circular arrangement would also be nice ). Now, luckily you added that these 16 numbers are supposed to represent an LT, so the obvious usage of these 16 numbers is to express what happens to the entries of a 4-vector under a coordinate transformation. Assumption 1: In the most basic and most common version of expressing special relativity (there is a little ambiguity where exactly you draw the line between SR and GR) 4-vectors are symbolized as [math]v^\mu[/math] (upper index) where v0 is a number associated to time (e.g. time-position for position vectors, energy for 4-momenta) and v1, v2, v3 are numbers associated to the respective space direction (e.g. position in space or momentum). Assumption 2: Your [math]\Lambda^\mu_\nu[/math] means [math]\Lambda^\mu{}_\nu[/math] and the values of the 16 Lambda are given such that the transformation law for the numbers [math]v^\mu[/math] onto the numbers [math]w^\mu[/math] representing the same vector in a different coordinate system is [math]w^\mu = \Lambda^\mu{}_\nu v^\nu[/math]. That is also standard notation.Explicitely writing that out (i.e. expanding the compressed notation of the Einstein convention) yields four equations; one for each number of [math]w^\mu[/math]: [math]w^0 = \Lambda^0{}_0 v^0 + \Lambda^0{}_1 v^1 + \Lambda^0{}_2 v^2 + \Lambda^0{}_3 v^3[/math] [math]w^1 = \Lambda^1{}_0 v^0 + \Lambda^1{}_1 v^1 + \Lambda^1{}_2 v^2 + \Lambda^1{}_3 v^3[/math] [math]w^2 = \Lambda^2{}_0 v^0 + \Lambda^2{}_1 v^1 + \Lambda^2{}_2 v^2 + \Lambda^2{}_3 v^3[/math] [math]w^3 = \Lambda^3{}_0 v^0 + \Lambda^3{}_1 v^1 + \Lambda^3{}_2 v^2 + \Lambda^3{}_3 v^3[/math] You can write this more elegantly as [math]\left( \begin{array}{c} w^0 \\ w^1 \\ w^2 \\ w^3 \end{array} \right)[/math]= [math]\left( \begin{array}{cccc} \Lambda^0{}_0 & \Lambda^0{}_1 & \Lambda^0{}_2 & \Lambda^0{}_3 \\ \Lambda^1{}_0 & \Lambda^1{}_1 &\Lambda^1{}_2 & \Lambda^1{}_3 \\ \Lambda^2{}_0 & \Lambda^2{}_1 &\Lambda^2{}_2 & \Lambda^2{}_3 \\ \Lambda^3{}_0 & \Lambda^3{}_1 &\Lambda^3{}_2 & \Lambda^3{}_3 \end{array} \right)[/math] [math]\left( \begin{array}{c} v^0 \\ v^1 \\ v^2 \\ v^3 \end{array} \right)[/math]. From this, the question which index corresponds to rows and which to columns can be answered for that case. But: There were two assumptions that were made, one about how vectors are specified via numbers and one about how vectors and Lorentz transformations are combined. Intermediate result: As long as you are dealing with coordinate transformations given as numbers [math]\Lambda^\mu{}_\nu[/math] and vectors are given as numbers [math]v^\mu[/math] and the transformation rule is [math]w^\mu = \Lambda^\mu{}_\nu v^\nu[/math] and the numbers [math]v^\mu[/math] are arranged as column vectors, then using matrix multiplication in the sense [math]w = \Lambda v[/math] and equating the 1st index of Lambda with row and the 2nd with column is fine. If you feel like filling a few pieces of paper you could verify for yourself that [math]w^\mu = A^\mu{}_\sigma B^\sigma{}_\nu v^\nu[/math] can in the same manner be written in matrix formulation [math]w = ABv[/math] with the same identifications when the same assumptions hold true - I'm certainly not gonna tex that in here . Without the two assumptions, there had -as far as I can see- not been an answer to your question. In short: For practical purposes within your course you are possibly fine with thinking in terms of matrix multiplications. Problems: There is several terms that are syntactically correct that you can -in principle or in practice- encounter that do not satisfy the two assumptions above: - [math]w^\mu = \Lambda_\nu{}^\mu v^\nu[/math]. I just made that one up, but if you write that out as I did above, you'll see that 1st and 2nd index are exchanged with respect to the previous example. - [math]s = g_{\mu\nu}v^\mu v^\nu[/math]: One of the most important terms in relativity and one you probably already encountered (invariant pseudo-magnitude of a vector v). Obviously, if all numbers for g and v are given you can calculate s via the rules the sum convention dictates (if you don't see that, try it out - getting accustomed to Lorentz-indices is really worth it). But writing that as a matrix equation is not possible without previous transformations or additional assumptions. - [math]R_{\alpha \beta \gamma \delta}[/math]: An object with four indices (R is the common name for the curvature tensor in GR). Contraction of indices (e.g. [math] F_{\alpha \beta \gamma} = R_{\alpha \beta \gamma \delta} v^\delta[/math] is defined by the summation convention. But how would you even represent such an object in a matrix-vector style? You could use a 2D matrix with each entry being a 2D submatrix but there's little gain for practical purposes. Ok, that was a lot of text; hope I didn't make too many errors. Bottom line: Thinking in and expressing equations via matrices might suffice for your case. I don't like it. I don't do so (so perhaps it actually is a good idea and I am just not used to it). I think it's very prone to making errors unless you remember some additional rules of how the terms with Lorentz indices transfer to matrix equations. Considering that when you have understood the index notation it takes only a few seconds to deduce the correct form in case you have to write it out that way, I think it's better to understand the index notation rather than remembering stuff like "first index is row, 2nd is column". Some quick answers: No. The former is simply two collections of numbers (staying in the language of above) seperated by a comma and grouped by parentheses. The parentheses do not indicate any mathematical operation. The latter is a summation, though (-> Einstein sum convention). It doesn't. Especially not in the sense of object meaning a physical object such as a stone or an electron. I more or less mean "variable".

Depends on the physics you are doing/using. But as soon as you seriously bother about quarks chances are that you are doing QCD or Standard Model physics. The former case has no electric charge, in the latter the electric charge is indeed modelled via two other charges. Not exactly on-topic, though. No, I don't think that's going to serve as an analogy. It's more like a coordinate transformation.

I'm not convinced that it's that simple (e.g. as soon as you go to electroweak theory the electric charge is just a composite of two other "more fundamental" charges, the weak isospin and the weak hypercharge). The elementary (electric) charge is currently defined as the charge of the electron (or position? not exactly sure). If a textbook sais so, then it's not because it was outdated - it's still the working definition of the term.

- matrices can be written by \begin{array}{cccc} 11 & 12 & 13 & 14 \\ 21 & 22 & ... \end{array}. There's also a more direct command named something like "matrix" but I am neither familiar with the command, nor does the "show/hide latex reference"-button in the advanced editing options work properly in my browser. - No, the indices on Lambda do not represent the number of a coordinate system but the index of the respective coordinate. Your trafo would be [math] \left( \begin{array}{c} t2 \\ x2 \\ y2 \\ z2 \end{array}\right) [/math]= [math]\left( \begin{array}{cccc} \Lambda^0{}_0 & \Lambda^0{}_1 & \Lambda^0{}_2 & \Lambda^0{}_3 \\ \Lambda^1{}_0 & \Lambda^1{}_1 &\Lambda^1{}_2 & \Lambda^1{}_3 \\ \Lambda^2{}_0 & \Lambda^2{}_1 &\Lambda^2{}_2 & \Lambda^2{}_3 \\ \Lambda^3{}_0 & \Lambda^3{}_1 &\Lambda^3{}_2 & \Lambda^3{}_3 \end{array} \right) \left( \begin{array}{c} t1 \\ x1 \\ y1 \\ z1 \end{array} \right) [/math] (sidenote@admins.sfn : could you increase the maximum number of characters in a tex string from 400 to something bigger?). - Yes, the inverse of an LT is an LT. Let LT L12 (1 and 2 not being indices but part of the name!) be the transformation that transforms coordinates from frame 1 to frame 2. LTs are the transformations from inertial frames to inertial frames so there is an LT for the transformation from frame 2 to frame 1: L21. Transforming coordinates from system 1 to system 2 and then back to system 1 should better give the original coordinates back, hence L21*L12 = 1 (to be understood as a matrix multiplication with the order of execution read from right to left). Sameways, transforming coordinates from frame 2 into frame 1 and then back into frame 2 should also yield the original numbers: L12*L21 = 1. Therefore L12*L21 = 1 = L21*L12 which is the definition for two algebraic elements (replace "algebraic element" by "matrix" if it makes it easier to understand for you) to be their respective inverses. If you prefer a more direct construction, multiply the LT matrix for a transformation belonging to a relative motion of v with that of a relative motion of -v. In short: Yes, inverses of Lorentz transformations are also Lorentz transformations. Their physical meaning is that they are the transformation that transforms from the original destination frame back onto the original source frame. - Thinking of the Lorentz transformations as matrices might help because it's a familiar concept. But the formalism with upper and lower indices is a bit more/different than that, so better stick with [math]\Lambda_{\mu \nu} v^\nu := \sum_{\nu = 0 \dots 3} \Lambda_{\mu \nu} v^\nu[/math] if in question. As I previously said, you can possibly get a representation of a matrix multiplication by comparison of the appearing terms. You should be on the safe side of representing the contraction of an object with two Lorentz indices with an object with one Lorentz index (contraction meaning that one index appears on both objects; once as an upper one and once as a lower one) as a matrix multiplication whenever the former object is given as [math]\Lambda^\mu{}_\nu[/math] (note the order and upper/lower position of the indices) and the latter object is given as [math]v^\nu[/math] (note the position and that it corresponds with the 2nd index of the "matrix"). [math]( \Lambda_\mu{}^\nu, v_\nu )[/math] also works (but is a tad less common). As soon as you get terms like [math]\Lambda_\mu{}^\nu v^\mu [/math] or [math] R_{\alpha \beta \gamma \delta} \cdot \text{something}[/math] () you run into trouble with your graphical "first index is line index, 2nd is column index"-rule. In short: Don't put too much into matrix multiplications even though they might be familiar to you. Better think of matrix multiplications as a visualization tool for certain types of mathematical operations.

I'm not sure if you can really tell without having specified a vector the transformation acts on. However, the standard notation is using [math]v^\sigma[/math] as the entries of a column vector which supposedly leads to [math]v'^\alpha = \Lambda^\alpha {} _\beta v^\beta[/math] and from comparing this with a matrix multiplication to the first index being lines and the 2nd being columns.

That's not even an equation, less a proof for something. It's merely a set of rational numbers without further explanation. I didn't understand your explanation. In fact, I fail to see any explanation at all. Please repeat it, possibly a bit more detailed.

If I understand the setup correctly, then 1) is correct but your approach to 2) is wrong. Draw that standard picture of what happens at the grating to see why (I mean this picture: http://en.wikipedia.org/wiki/Image:TwoSlitInterference.svg).

Who cares? If people believe that knowing that a year has approximately 52 weeks had something to do with intelligence, then you could make them believe the same for being able to calculate the approximate diameter of a neutron star. Possibly. People from cultures in which the event horizon ends at the territorial borders will have a harder time.