timo

Well, the inverse square law follows from the assumption that (1) no light is absorbed, (2) the area A(d) covered by light increases with d² as a function of the distance d from the light source and (3) the area A(d) is uniformly illuminated. Because of (1) the total light shining on A(d) must be the same for all distances. because of (2) the area A(d) goes with d² so because of (3) the ammount of light shining on any unit area at distance d must go with 1/d² (that´s because this value times A(d) must be the same like said in the previous sentence). So your inverse square law is made up of three assumption which -as I will show in the following- are usually not completely true: - No light absorbtion: That is of course only a simplification. Light will be absorbed by air, scattered, converted to other wavelenght, whatever. A 1st order approximation of the absorbtion would be adding a factor exp(-X*d) with X being a material constant that describes the absorbtion. But for your 1 meter or 5 meters distance from a laser the light absorbtion from air is probably neglectible. - Area uniformly illuminated: Afaik that´s not true for lasers. Laser intensities rather have a gaussian shape. - Area covered increasing with d²: Well, you can just measure the area covered. If you neglect absorbtion the total ammount of light shining on any area A(d) must be the same. So by dividing by A(d) you´d get an average intensity (average, because lasers do not light the area uniformly as allready said). An alternative to measuring all A(d) would be to make the assumption that the laser beam is conical with a small but finite area at the opening of your laser. If you measure A(d) for two different distances and create a formula for A(d) out of these measurements and above assumption you´ll end up with a formula of A(d) ~ (d-d0)². That´s close to A(d)~d² for sufficiently large distances but not exactly the same. An short: The inverse square law is almost allways a (very usefull) simplification. Hope that helped EDIT: Before someone jumps in to start nitpicking I´ll add the following myself: - (2) is formulated a bit fuzzy but it should be obvious what I mean - (3) is not nessecary for an invere square law but it greatly simplifies any discussion. - a (4) "light rays propagate in a straight line" seemed obvious so I ommited it for the sake of readability.

Care to explain what you mean by "inverse square law"? The rest of your post also seems a bit messed up. What are you talking about?

1 second roughly equals 3*10^8 m / c, yes.

1a) The Einstein Equations are the basic equations in General Relativity. They´re very complex diferential equations and there´s no know general solution to them. 1b) The first analytic solution found was that for a spherical symmetric mass distribution - The Schwarzschild solution. Even though "spherical symmetric mass distribution" might sound like a strong restrictment it´s quite usefull as planets and suns can be considered spherical symmetric to a good approximation. As a sidenote: The Schwarzschild solution is only valid outside the mass distribution. 1c) The Schwarzschild solution has a certain area where the coordinates are not defined. Since this area lies inside the mass distribution for suns or planets noone initially paid much attention to it. 1d) Someone (I think it was Chandraseka) was able to show that suns that run out of fuel will collapse to being so small that this "undefined area" lies outside the mass distribution if their mass exceeds the Chandraseka mass. That´s what you´d call a Black Hole: A spherical symmetric mass distribution with it´s undefined area (which is enlosed by the Event Horizont) outside of the mass distribution. 2) Einstein (General Relativity), Schwarzschild (1st analytic solution), Chandraseka (showing that suns will collapse beyond the event horizont), Kruskal/Penrose (there´s a coordinate that´s called "Kruskal Koordinates" or "Penrose Coordinates" that´s able to map the area within the Event Horizont. Can´t really tell those people´s contributions to it, though). 3) None has been directly observed, afaik. Methods would include looking for "jets" that are emitted by them, observing the movement of suns to find centers of gravity where no mass can be seen and perhaps light deviation. 4) Not completely sure what you mean with that question. The event horizont is just the surface that seperates the area where the Schwarzschild-coordiantes are defined from that where they aren´t (or at least where they don´t make much sense). Sources: That was off my head. I´m neither interested in experimental physics nor in history of physics too much, but I hope that still helped you.

The problem with conservation of energy in GR is not that energy is a frame-dependent concept - you could still assign a definite value in any coordinate system that would be conserved if the stress-energy-momentum tensor was. The problem -if I remember that correctly- is rather how to define conservation of energy. You also have a continuity equation here (that´s your 2nd formula in your 1st post) but problems occur when you want to derive the integral form of it (sorry for being so vague, I´d have to look up a few things to be more precise and correct). It´s funny that you posted research papers in your first post without any comments on them so I couldn´t understand their connections to your statement and that you now post a link to an introduction to special relativity and call it "rather advanced stuff" . Perhaps it would help if you said what in particular your problems with the text are so maybe I or others can clear things up for you.

Sorry, I didn´t understand your answer. So with that "gravitational field energy momentum tensor" you are not talking about the matter-side of the Einstein Equations? And what do you mean by "The gravitational field energy momentum tensor is a unique pseudotensor because actually there is none in general relativity"?

Why is the energy momentum tensor a pseudo-tensor and not a tensor? Not only that this was completely new for me I´d furthermore doubt it since the energy momentum tensor forms one side of a very basic tensorial equation called Einstein equation. That´s, at least, if a pseudo tensor is what I currently have in mind for this term: A tensor-like structure with non-tensorial transformation laws for it´s entries. What makes you come to the conclusion that because of local conservation of energy "gravity must draw upon some energy/power source in order to do work"?

@6431hoho: I´m not completely sure if you´re aware that the ammount of Dollars (or any other currencies) is not regulated by the ammount of bills someone prints but by the base rate set by the federal bank (<- hope that´s the correct institution; I´m neither an economist nor interested in american politics) since most of the Dollars that are around only exist as a code of bits in a computer. Changing the base rate is something you cannot do without anyone noticing (or if you do: Nothing will happen as noone noticed you changed it). Afaik, creating new money is rather common for states with money problems and is a sign for a breakdown of it´s economy.

you need a special cable (crossover cable) to connect two computers directly. Try to pick a better name for your thread next time.

I never said you did. It was just an (seemingly wrong) assumption why the v suddenly appears. As I´ve told you before: Please show your calculations step by step (with comments). That´s the only way to find the mistake you made.

You steps from "h²f²=(mc²)²+(pc)²" to "m=[h * sqrt(v²-c²)]/[lambda * c²]" are a bit mysterious and -because of the 2nd equation being wrong- obviously flawed. Note that none of the c's in the 1st equation is the particle´s velocity.

You did. Dying and being reanimated (hey, that really hurts) probably counts as "times of trauma".

What´s CMB ?

Count the number of times you run the loop, convert this number to a string, use the string as the name for the output file.

Actually, they are simultaneous in any frame of reference. Reason: If the two points have the same space and time coordinates, they are the same point in 4D. There´s only one tuple of coordinates for any one point in 4D in any coordinate system.

You´ll find that everywhere. What he´s trying to say is that he knows better than our current theories.

EDIT: Wow, so much action in a rather long-dead thread. Seems that I should add that the following message is in reply to mydadonapogo: Momentum isn´t p=mv, that´s only an approximation for small velocities. If it was, any mass could easily reach and succeed lightspeed for example, as the nessecary energy to reach lightspeed would be given by your E² = 2(mc²)². For small particles these energies can be achieved relatively easily (energy would be 1 MeV/c² for electrons - modern acellerators operate in the range of 1000 MeV/c², I think).

Didin´t follow the rest of the post (mainly due to the broken TeX). My main point is the following anyways: E² = 2(mc²)² is not the correct expression for a particle´s energy. How did you come up with this? Did you assume pc = mc²? If so, why so? Well, I do have a guess but I´d like to hear how you got to E² = 2(mc²)².

Where does the hf = 2(mc²)² come from? EDIT: And please, for the future: Add the steps that led you to a conclusion. You cannot expect any better answer than "you are wrong" when you don´t.

Actually it does. But a sidenote here: The statement that clocks of moving observers move slower is a bit incomplete. Actually, observers moving with a relative speed to each other have a different understanding of what space and what time is. In the twins example the part of spacetime that the rocket considers time -it´s movement direction- is seen as a mixture of space and time by the twin that stays behind. That depends on the form of Special Relativity you´ve seen. The most common -and in your problem pobably sufficient- would be: The problem is not symmetric. The planet´s (or whatever is left behind) rest frame is an inertial frame for the whole process. The spaceship´s rest frame isn´t an inertial frame because it has to acceelerate to change it´s direction. I doubt that anyone knows the "why". Afaik it´s just an experimental fact that our modern theories were built upon.

Well, what would you expect to be the command for a dot? .... it´s "\dot" . two dots is "\ddot", btw

No, I´m obviously no match for your knowledge in physics if I don´t even know that there´s a rest frame for photons.

a) A frame where photons are at rest does not exist. b) The (rest)mass of photons is zero. c) I made a mistake. E=hf applies to all particles. But pc = hf only applies to massless particles. Simple reason: (hf)² = E² = (pc)² + (mc²)² => [math] pc = \sqrt{(hf)^2 - m^2c^4} [/math]. So pc = hf exactly if m=0.

you mean mp² instead of mc²? No, a simple check of the units will show you it cannot be mp².

I rather think your problem is that you think the energy for particles was E = hf. That´s only true for photons. The energy is E = sqrt[ (pc)² + (mc²)² ] just like you got it. Step-by-step derivation: - Start with a plain wave: [math] \psi = \exp \left( -i(Et - px) \right) [/math] - I´ll leave it 1D to save some time typing. Also I set hbar to one. It would cancel out in the next equation anyways. - Plug it into the Klein-Gordon wave-equation: [math] \left[ \frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2} + m^2 \right] \psi = 0 [/math] (note that due to my lazyness in typing I also set c=1). => [math] \left[ \dots \right] \psi = \left[ -E^2 \psi + p^2 \psi + m^2 \psi \right] = \left[ -E^2 + p^2 + m^2 \right] \psi = 0[/math] Since the wavefunction is nonzero the term in brackets must be zero => -E² + p² + m² = 0 => E² = p² + m² Putting back the c´s to get the SI units thus gives: E² = (pc)² + (mc²)² or E = +- sqrt[ (pc)² + (mc²)² ] Note that you also get negative total energies. If I had used the Dirac equation (which is a bit more complicated) in above those solutions with negative energies would be associated to anti-particles. To be honest I´m not completely sure if this can also be said in case of the Klein-Gordon equation.