I've already answered that for the specific case of your gendanken, but I assume you want a more general solution. I could talk it through, but the least ambiguous way is to present an equation. The equation in question is the vector form of the relativistic velocity addition equation. I haven't derived this myself (I imagine it would be an interesting exercise), I pinched it from John Baez's site. If frame S is moving at velocity v = (v<sub>x</sub>, 0, 0) in frame O, and a thing has velocity u = (u<sub>x</sub>, u<sub>y</sub>, u<sub>z</sub>) in frame S, and if the axes of frame S and O are oriented in the same way, then the velocity w = (w<sub>x</sub>, w<sub>y</sub>, w<sub>z</sub>) of the thing in frame O is given by: w<sub>x</sub> = (u<sub>x</sub> + v<sub>x</sub>) / (1 + u<sub>x</sub>.v<sub>x</sub> / c²Please Register or Log in to view the hidden image! w<sub>y</sub> = u<sub>y</sub> / [(1 + u<sub>x</sub>.v<sub>x</sub> / c²Please Register or Log in to view the hidden image! gamma(v<sub>x</sub>)] w<sub>z</sub> = u<sub>z</sub> / [(1 + u<sub>x</sub>.v<sub>x</sub> / c²Please Register or Log in to view the hidden image! gamma(v<sub>x</sub>)] In the case of your gedanken: S is the rest frame of the starship. O is the rest frame of the Earth. The thing is the emitted photon. v = (0.9999c, 0, 0) is the velocity of the starship with respect to Earth. u = (0, c, 0) is the velocity of the photon with respect to the starship. w = (0.9999c, 0.014142c, 0) (derived from the SR equation) is the velocity of the photon with respect to Earth. It is not difficult to see that u and w (the velocity of the photon in the two frames) have the same magnitude, but different directions.
CANGAS, If in the gravitational interaction like poles attract and negative poles repel then a galaxy composed only of negative gravity would behave exactly the same as one composed only of positive gravity. Since opposite poles repel and like poles attract in the gravitational interaction, this type of interaction would eventually result in a universe with seperate gravitationally positive and negative areas.
Pete: you are so full of s'it that you remind me of the most recent time I went to the elephant exhibit at the zoo. Elephants have even you beat at spewing out s'it. But don't feel disheartened; an average elephant outweighs you by several tons. Your rebuttal will be posted very soon. You will read it and weep, as we say in serious games. A long time ago, you should have learned how far to bluff with a pair of deuces.
Prosoothus: I get it. I just wanted to be sure I understood correctly what I thought you wanted to communicate. You may be on the right track. But there is still a long way to go.
You're starting to make a habit of this, CANGAS. If your schedule for the dreaded rebuttal is like your schedule for your relativity killer, I guess I'll be waiting a looooong time... But that's OK. It is much easier to posts insults than to engage in rational debate, after all, so feel free to carry on as usual.
For those who may have missed some physics regarding light motion, please consider thefollowing: 1. The motion (velocity) of light is independendent of th emotion of the source of the light. 2. Once emiotted, the light will move in a straight line. The motion of the ship moving perpendicular to the motion of the emitted light is irreelevant to the future motiuon of the light. Soon after the light is emitted the side of the tube covering the light pulse will intercept the light pulse. If the side of the tube absorbs the light then no light pulse will strike the detector at the opposite end of the tube. If the tub is reflecting, then the light pulse will be deflected in a zig-zag course down the tube. The Light wiull then strike the detector after making a number of reflections as the pusle bounces off the sides of the tube wall. The responses I reviewed that assumed the light would have a velocity component in the direction of motion of the moving ship are ignoring the basic laws of light motion: light moves at a constant velocity c = 3 x 10^8 m/sec. If one adds the velocity of the ship into the scenario then using simple pathogrean(sic?) maths, the speed of light comes out much greater than c. This is so basic chaps. Why do you insist on altering the laws of physics? Is it to maintain your dogmatic belief system (scientific religiosity) in the special theory of relativity? It surely seems so, sadly, for those in such an abject state of ignorance. Geistkiesel
It has been boldly claimed by a Relativity devotee that photon speed does not matter in in the thread gedanken. We reexamine the matter in new light to determine if the bold conjecture is a lucky guess or is absurdly wrong. And we more thoroughly explore the Special Relativity mandate that a moving object must be considered to be a stationary reference frame while the rest of the entire universe is actually moving instead. We employ two observers. One really is a stationary observer on Earth, the Stationary Observer A, or simply the SOA. The second is the usual Relativity moving stationary observer riding on the moving object and claiming it to be stationary, The Stationary Observer B, or simply the SOB. Suppose a train car is traveling through the landscape at .999 MPH. A lightning bug is creeping straight across the floor from one side to the other at 1.0 MPH. The moving stationary observer, riding on the train, the SOB, says that the bug is simply moving in a straight line across the car at the usual speed of lightning bug, 1.0 MPH, while the landscape is whizzing past the stationary car at .999 MPH. The stationary stationary observer, the SOA, standing on the embankment, sees that the bug is moving in two directions simultaneously. It has two perpendicular component velocity vectors which add to a resultant velocity vector which is a trajectory practically diagonal to the train movement direction. The SOA solves a right triangle with sides .999 MPH and 1.0 MPH. Long side equals square root of of sum of .999 squared plus 1.0 squared: roughly 1.41. The lightning bug, relative to the SOA, is is traveling faster than the speed of lightning bug; it is moving at 1.41 MPH! In the spaceship, the moving stationary observer, the SOB, claims to observe that the photon is simply moving in a straight line across the pipe length at the usual speed of light, 1.0c, while the universe is whizzing past at .999c. IF the SOB's claim about the photon is true, the stationary stationary observer, the SOA, standing on the Earth, observes that the photon is moving in two directions simultaneously; it has two perpendicular component velocity vectors which add to a resultant velocity vector which is a trajectory practically diagonal to the spaceship movement direction: it is moving straight through the pipe length at 1.0c while simultaneously the pipe is moving in a perpendicular direction at .999c. The SOA solves a right triangle with sides .999c and 1.0c. Long side equals square root of of sum of .999 squared plus 1.0 squared: roughly 1.41. The photon, relative to the SOA, is traveling faster than the speed of light; it is moving at 1.41c! If we refuse to believe that a photon moves at 1.41c, we must question the claims of the SOB. If the SOB is wrong and the spaceship is actually moving, and if the photon is not magically swept along and so is quickly struck by the sideways movement of the pipe after it exits the Xaser, rather than traveling through the entire pipe length, then the SOA observes the photon to travel in only one velocity vector at 1.0c, perpendicular to to the spaceship velocity vector. The photon can only travel all the way through the pipe if it has a practically diagonal trajectory in the observation of the SOA. If the SOB waves its arms and shouts that the photon velocity does not matter, we must stubbornly insist that we want the SOB to explain the result ( light going 1.41c ) of his claims. We want the SOB to explain how Special Relativity can be true while it produces such an outrageous contradiction.
I don't think so. Please feel free to prove me wrong be quoting the actual words of the person concerned. Correction: a moving object may be considered to be a stationary reference while the rest of the entire universe is actually moving.[/color] In special relativity (and galilean relativity), you can use any inertial reference frame at all... the only must is that if measurements are taken in different frames, then they must be transformed appropriately before being used together... and also that if you want to determine what SR predicts, then you must use the SR model. Correct. In this case the addition of velocities is very closely approximated by the galilean model. The SR model would give a correction for the through-the-pipe velocity component in the SOA frame in the order of 10<sup>-18</sup>mph, which (as I'm sure you'll agree) is insignificant. Not correct. In this case, the addition of velocities is not closely approximated by the galilean model. The through-the-pipe velocity component in the SOA frame predicted by SR is in fact 0.0447c, not 1.0c. Note that this means that the time taken for the photon to traverse the pipe is 22 times longer in the SOA frame than in the SOB frame. So, the SOA solves a right triangle with sides .999c and 0.0447c. Long side equals square root of of sum of .999 squared plus 0.0447 squared: roughly 1.0. The photon, relative to the SOA, is traveling at the speed of light; it is moving at 1.0c! CANGAS, Rewording your original gedanken is not a rebuttal, and showing that Galilean relativity predicts the photon to have FTL speed isn't relevant. If you simply want to disbelieve the postulates of SR, that's fine... but when you assert that it predicts something it doesn't, I will correct you.
Hi geist. That particular assertion is what this thread is about. Do you have any arguments or references to support it? The motion of the ship once the light is emitted is irrelevant. But the motion of the ship at the instant of emission determines the direction of emission. I disagree, as I'm sure you're aware. You're making the same mistake as CANGAS in assuming that the time taken for the light to traverse the pipe is always the same - SR says that it is not. This is indeed basic. Why do you insist on being ignorant of the laws of physics?
According to what theory? Perhaps we should name this strange new theory CAR - CANGAS Anti-Relativity. It certainly is not a prediction of SR. It doesn't produce such an outrageous contradiction. That much is clear by the simple fact that one of the postulates of SR is that light always travels at c. -Dale
Hi Pete Can you give me an equation or a link to an equation that links the velocity of a light source and the resultant change in the light's emission vector. Thanks Please Register or Log in to view the hidden image!
Sure, look a few posts up. It's the relativistic vector velocity addition formula: If frame S is moving at velocity v = (v<sub>x</sub>, 0, 0) in frame O, and a thing has velocity u = (u<sub>x</sub>, u<sub>y</sub>, u<sub>z</sub>) in frame S, and if the axes of frame S and O are oriented in the same way, then the velocity w = (w<sub>x</sub>, w<sub>y</sub>, w<sub>z</sub>) of the thing in frame O is given by: w<sub>x</sub> = (u<sub>x</sub> + v<sub>x</sub>) / (1 + u<sub>x</sub>.v<sub>x</sub> / c²Please Register or Log in to view the hidden image! w<sub>y</sub> = u<sub>y</sub> / [(1 + u<sub>x</sub>.v<sub>x</sub> / c²Please Register or Log in to view the hidden image! gamma(v<sub>x</sub>)] w<sub>z</sub> = u<sub>z</sub> / [(1 + u<sub>x</sub>.v<sub>x</sub> / c²Please Register or Log in to view the hidden image! gamma(v<sub>x</sub>)] If the thing is a photon, then both u and w will have magnitude c. Tom2 posted a different version of the formula early in the thread: w=[gamma.v + u + ((gamma-1)/v<sup>2</sup>)(v<sup>.</sup>u)v] / [gamma + (gamma/c<sup>2</sup>)v<sup>.</sup>u] w= velocity vector of the photon in the observer's rest frame. u= velocity vector of the photon in the emitter's rest frame. v= velocity vector of emitter relative to observer. gamma = 1/√(v²/c²Please Register or Log in to view the hidden image! Tom's formula is more general, because it works directly on vectors rather than components. In particular, this means that v can be in any direction. In the first formula, v must be parallel to the x axis.
My recent posts have consisted of very clearly worded expressions of very clearly explained logic and simply expressed mathematics, which have proved beyond any doubt that Special Relativity gives grossly incorrect predictions. Always has. Always will. Dissenting recent posts have ubiquitously consisted of pointless whining with NO SUBSTANTIAL LOGICAL OR MATHEMATICAL ARGUEMENTS. Special Relativity has been proven dead. The only thing left is the wake and the burial.
ROFL Please Register or Log in to view the hidden image! ! You certainly have shown that your CAR theory gives grossly incorrect predictions. Unfortunately for your pointless rant, you actually have to use the predictions of SR in order to show that SR gives incorrect predictions. Try using the Lorentz transform to get a velocity of 1.41c. Until then you are not even talking about SR, let alone showing that it is wrong. SUBSTANTIAL LOGICAL ARGUMENT: SR asserts (by postulate) that c is frame invariant. Your analysis obtains a frame variant c. Therefore your analysis is not SR. QED. -Dale
You claim to be an expert mathematician yet you do not understand grade school trigonometry. 1.41c is not equal to 1.0c, Dale.
Hi Pete I did not specify an observer. Therefor there is no w. There are only S and u. If an observer is required then how can you tell where the apparent change in the light's emission vector takes place. IE does the vector change at the source or at the observer. Please Register or Log in to view the hidden image!
The "observer" is whatever the light source has velocity v relative to. w, the velocity vector of the emitted light, is a realtive thing - it depends not on the source or the observer, but on the relationship between them.
Hi Pete Light travels in a straight line so logically, light's calculated vector change must happen either at the source or at the observer. Please Register or Log in to view the hidden image!
I don't know what you're thinking, Montec. Can you expand your logic? Putting in in mathematical terms would make it clearer.
