timo

Are you sure he rotates the 2nd filter at an angle of 45° and not adds a 3rd filter at an angle of 45° between the previous two? The latter experiment is much more impressive. How it really works: The free electromagnetic field (i.e. light) can be mathematically cast into a sum of different addends each having a wavelength, a direction of travel and a polarisation vector that is perpendicular to the direction of travel. The square of the amplitude is the intensity (for that respective wavelength for that respective direction). Let's assume the travelling direction was the z-direction so that the relevant degrees of freedom for the amplitude are all vectors of the xy-plane [math]\vec A = \left( \begin{array}{c} A_x \\ A_y \end{array} \right).[/math]. A filter projects the amplitude onto some direction (i.e. cuts out all shares in the perpendicular direction). A filter on the x-direction would mathematically be applying the matrix [math]P_x = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right] [/math], correspondingly a filter on the y-direction would be [math]P_y = \left[ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right] [/math]. So if you apply both filters on the incoming light, the resultant amplitude on the other end of the filters would be [math]\vec A_{\text{after}} = P_y P_x \vec A_{\text{before}} = \vec 0[/math] and hence the intensity would also be zero. A projection on a 45° rotated direction would be [math]P_{45} = \frac{1}{2} \left[ \begin{array}{cc} 1 & -1 \\ -1 & 1 \end{array} \right] [/math]. Now the rather boring thing is that when you rotate one of the two polarisation filters to a 45° position you get [math]\vec A_{\text{after}} = P_{45} P_x \vec A_{\text{before}} =\frac{1}{2} \left( \begin{array}{c} A_x \\ -A_x \end{array} \right)[/math] which will not always be zero (meaning that there's some part of the light that shines through). The much more interesting thing happens when you put a 3rd 45°-rotated polarisation filter between the two perpendicular ones: [math]\vec A_{\text{after}} = P_y P_{45} P_x \vec A_{\text{before}} =\frac{1}{2} \left( \begin{array}{c} 0 \\ -A_x \end{array} \right)[/math] That is usually also not equal to zero.

I don't think uncertainty really add difficulty into obtaining the molecular structure as a WF except in the sense that you cannot pin down a single electron with two 3-tuples position and momentum but have to describe it with a whole R³->C function. I am not sure if I really would say that describing an object with a WF rather than two 3-tuples is a result of uncertainty. It's tied (at least for the uncertainty between momentum and position) but I personally feel that "it's due to uncertainty" gives uncertainty too much credit. In practice, QM calculations for more complicated objects use simplifications (the most straightforward being to treat the nuclei or the ions with the innermost shells classically) and/or some sophisticated methods like Density Functional Theory. I think I have a relatively simple approximation that shows that even for a very simple molecule (H2, iirc) a straightforward solution is not achievable due to the huge amount of data (was something like "if you burn it to CDs you'd need all the mass of the visible universe to create enough CDs"). Sadly, the lecture notes that it's in -assuming I still have it at all, I threw away quite a lot of stuff lately- are in my other flat at which I won't be until next week. I'll post the example here next week if I happen to still have the notes and if I have access to the internet. It might show why and how the amount of data explodes. If someone else has a similar or the same example calculation (I imagine it being quite a standard example) please post it here - saves me time for searching In short: I don't see uncertainty as problematic when it comes to finding a wave function. I'd rather see the fact that you describe an object via WFs (and not via R³ vectors) as tied to uncertainty.

1) Mass is not inversely proportional to velocity. Unless you specify some constraints they are almost completely independent. What you probably had in mind is that for a fixed amount of kinetic energy mass and velocity were inversely proportional. This is not the case as E=mv²/2 => m = 2E/v², i.e. mass is inversely proportional to the square of the velocity, not velocity. Note that this example is just a guess of what you might have had in mind and not a general rule (e.g. there is no reason why a car and a truck should have the same kinetic energy). 2) Some people say that mass increases with velocity and call that definition of mass the "relativistic mass". It is not wrong to say so but in most cases it is not advisable. That's a bit off-topic, though. 3) If some value increases when another variable increases it does not necessarily have to do in a proportional fashion. Proportionality means that if one of the values is multiplied by x then so is the other. There's an arbitrary amount of other relationships that have value A which is dependent increasing as B increases without being a proportionality, e.g. A=B³.

If an object moves with a velocity v<c in some coordinate system then the ratio between time passage in this coordinate system and the time passage for the particle equals the Gamov factor [math]\gamma = \frac{1}{\sqrt{1-v^2/c^2}}[/math] (which is the same gamma as the one mentioned by Swansont). Gamma is neither proportional nor anti-proportional to v.

Whatever inverse proportionality is supposed to mean I imagine problems for [math]\vec v = \vec 0 [/math]. -------- Start incorrect information ----------- The relation between space-distance and time-distance travelled during an eigentime interval is like [math]v_t^2 = \vec v^2 / c^2 + 1[/math] for massive particles. ------- End of incorrect information --------- Edit note: I incorrectly mixed [math]\vec v = \frac{d\vec x}{dt}[/math] and [math]v_t = \frac{dt}{d\tau}[/math] above (not the different variables against which the differential is taken).

You want to show that every (a,x) has an inverse (A,X), i.e. (a,x)*(A,X) = 1 = (A,X)*(a,x). It does (usually) not help you to find inverses for arcane combinations of elements ( (gh^e,ed) rather than a simple (g,e) ) - you'd have to proof that all elements of D can be written as such a combination. It is not sufficient to show that all (a,+1) have inverses; the (a,-1) need to have inverses, too. If e=1 then you are saying that (1/h,d) is the inverse to (gh,d). That's wrong for e.g. h=1, d=1, g!=1: (1,1)*(g,1) = (1*g,1) = (g,1) [math]\neq [/math] 1. Sidenote: 1 refers to the neutral element of the appropriate group (D, G or {+1,-1; *}). You show that (a,x)*(b,y) is an element of D for all (a,x) and (b,y) in D. In this case the closure deduces directly from G and {+1,-1; *} being groups (i.e. also being closed). Not if "it" is D. Integers are a group under addition not under multiplication (there is no multiplicative inverse of 2, for example) so integers under multiplication do not qualify as G. You seem to be thinking of multiplication in your example. You seem completely confused about the topic. For your own clarification, please try to construct complete statements and do not just throw in results or guesses. It is almost impossible to understand what you are trying to say. Remarks (possibly helpful; you seem to have problems understanding what the notations in the question mean): - [math]a \, \in \, G: \ a^1 = a; a^{-1}[/math] means the inverse of a. 1 is the identity of G. - Sample proof for the identity being (1,1): [math] (1,1)(a,x) = (1a^1,1x) = (a,x), \ (a,x)(1,1) = (a1^x,x1) = (a1,x) = (a,x)[/math], where [math]1^{-1}=1[/math] has been used in the 2nd equation (proof: 1*1=1=1*1 => 1^{-1} = 1). - Be very careful about the notations. They are not what you are probably used to. If G was the integers under addition then 1*x, x*y and x^{-1} (for x,y in G) would read differently in the notation you are used from school math where addition is written with a "+". Namely 0+x, x+y and -x, respectively.

I think Pong was console so that won't fit. But it's well known that Tetris was the most successful Sowjet blow on western economy.

- You should add and write out the proofs for the identity and the inverse, e.g. (a.x) * (1,1) = (a,x) = (1,1)*(a,x) for the neutral element. You did not state what (1/h,d) is supposed to be the inverse of. Again: Do and -if you want feedback- post the full proof, not the result (the result that D is a group exactly if G is abelian is already given in the question, anyways). - Closure is the simplest of all the steps, so try that next.

http://www.scienceforums.net/forum/showthread.php?t=30990 Hah, that blew the cover of your 2nd account, Klaynos!

Improper timing while typing. Probably caused by being influenced by a malfunctioning watch for too long.

The photon is a particle. The particle of the free electromagnetic field.

Not really .

I have not read any post in this thread but you should probably still try (a+bi)(a-bi) = a² + b².

I usually don't have a watch; there's plenty of clocks all around and I usually carry my mobile phone with me, anyways. The one occasion that I had a watch I had was during the two years that I worked as a paramedic and it was an analog watch. I did not explicitly chose it because it was non-digital but I imagine pulse checking to be slightly (bordering to negligibly) more complicated because with an analog watch you can see the 10, 15, 20 or 30 seconds that you count pulses visually while with a digital watch you have to compute and remember the stop time during counting which, being two arithmetic operations at once for the brain, sound more complicated and error-prone than one arithmetic and one geometric/visual operation.

1) Publish in a journal (already mentioned) or deposit suitable documents at a notary. Alternatively, set up a homepage and put the stuff there. Don't forget to pay the bill for the HP. 2) Consider doing arts rather than science if acknowledgement for your creativity is what you are striving for. Music, filming, painting, sculpturing, ... . 3) Reconsider making money with your creativity. Your points sound like a strange synopsis of "Ghost in a Shell" and "Final Fantasy".

From x=at²/2 => v = at (assuming your v means dx/dt - mine does), not v = at/sqrt(1+a²t²).

It means to calculate the "length" of the curve in spacetime (where length is the common name but problematic due to the metric not being positive definite). You can see this as calculating the line of a curve in R^n which is a map from an interval I in R on R^n: [math]\phi : I \to R^n, \phi(s) = (x^1(s), \dots , x^n(s))[/math]. The length of such a curve in R^n between the points s=a and s=b>a is calculated as [math] L = \int_a^b \| \frac{d\vec x}{ds}(s) \| ds [/math], i.e. by summing up the lengths of small changes. Not much changes in your problem except that you must find a parameterisation for the curve and that the magnitude of a vector is a pseudo-magnitude: 1) A practical approach is to take t as the curve parameter in which case [math]\phi(t) = (1t, \vec x(t)) \Rightarrow \Delta \tau = \int_{t_i}^{t_f} \| \frac{d\phi}{dt}(t) \| \, dt[/math]. And [math]\frac{d\phi}{dt}(t) = (1,\vec v(t))[/math] is a 4-vector, a tangential vector on the world line (sidenote: Strictly speaking (t,x) is not a vector but a position - in many cases in SR it can be treated as if it was a vector, though). 2) The relativistic squared pseudo-magnitude of a vector v is [math] v^2 = v^i v_i = g_{ij} v^i v^j[/math] (imho writing "v²" for that is nasty but it's common convention). If v is the tangential vector of the world line of a massive object, then v²>0 (or <0, depending on the signature choice of the metric - ajb is the only one on sfn that uses <0, I think). You could therefore write [math] \| v \| = \sqrt{v^2} [/math]. Plugging that into the equation for the length above yields [math] \Delta \tau = \int_{t_i}^{t_f} \sqrt{1- \vec v^2/c^2} \, dt[/math]. So much for a practical calculation of the given problem. Now why is that called "integrate dtau"? Tau is the eigentime of an object. The eigentime of an object is simply the length of its world curve. Had you chosen tau as parameter for the curve then you had had [math] \Delta \tau = \int_{\tau_i}^{\tau_f} 1 \, dt[/math] which is quite a no-brainer but useless for the calculation here. Please note that factors of c in above are not guaranteed to follow some consistent usage, especially not the convention in your book. The reason is that those factors do not carry any "structure" but only give a scale. At least in theoretical physics it is common to completely omit them and only add them at the very end of a calculation in case a number is desired as a result (and only if that number shall be SI compatible).

Do you know what dtau is? What expressions for dtau (as a function of other parameters) do you know?

More things like that? Tachyon healing!

After unsuccessful searches for ET they now sell having found a radio transmission from YT as a great success?

Well, if in the link YT gave with "pure energy" they really mean some s-channel particle then that's closer to mass (if you were to define mass as cms energy which is a reasonable definition) than to kinetic energy. So if you took that definition of "pure energy" and transferred it on the photon then the photon was kind of an anti-thesis of "pure energy". I don't think the term "pure energy" has any use other than in "I think everything in fact consists of pure energy"-style crackpot ideas or "I don't want to bother with or explain the details here"-type of approaches/explanations.

- They do not call photon pairs produced in e+ e- annihilation "pure energy". What they supposedly mean by "pure energy" is some intermediate virtual particle with some energy and zero momentum (s-channel). - Assuming with "pure energy" they meant the intermediate virtual particle I still do not like calling it "pure energy". I think it's wrong. I think you are on the safe side if on the most fundamental level you take the stance of Klaynos seeing energy as a property of the physical object, not a physical object itself - or even worse: Something more fundamental than elementary physical objects. Yet, if you keep that in mind you can of course sensibly talk about energy being a physical object in an effective (=appropriate for the current topic) theory/framework. I find it perfectly appropriate to talk about "waste of energy" or "energy consumption" and stuff like that when it comes to technical stuff like power supply as long as you keep in mind what these terms really mean (conversion of a form of energy that you can use for your current application into a form that's useless to you).

^^ Probably started out as a work/company team of some arsenal (in or near London).

Yes, it is. As Klaynos said, energy is a property of some physical system i.e. something that describes the physical system/object. It is not a physical object by itself. A photon is not "pure energy" similarly to a car not being "pure red". Even worse: If you agreed that all red cars are "pure red" (in which case the "pure" would be misleading as it does not carry any additional information) then only red cars were "pure red". Not all cars are red. But for example all elementary particles have energy. So if you called all particles that have some energy "pure energy" then all particles were "pure energy".

Why would you think so other than that it spontaneously sounds like it makes sense? Example: If the photons created had momentum: Would they still be "pure energy"? If "yes, pure energy can have momentum": Why wouldn't they be pure momentum that can have energy? Why not pure photons that have both, energy and momentum?