kjitta

Yes, the midpoint on a chord is the point closest to the centre, therefore a line joining the two points will be perpendicular to the chord.

I must admit this thread has been an amusing read. I am starting to think that this is a windup, in that light this thread makes perfect sence. If that is not the case and the original poster trully beleives he is on to something new, then I would encourage the poster to try and formulate his arguments with mathematics and leave the handwaving arguments. If you are thorough and achieve the same result you may publish your thoughts in any journal you wish. But you wont catch me holding my breath till that happens.

1) x^2 - x^2 = x^2 - x^2 2) (x + x)(x - x) = x(x - x) 3) x + x = x 4) 2x = x 5) 2 = 1 !!! Going from line 2 to 3 I devide by x - x witch is an illegal operation since x - x = 0 for all x and one cannot devide by zero.

Bloodhound, Your'e quite right. Of course it doesn't work that way, and that is exactly my point. Ignore simple rules and you get nonsence, just like wespe's supposedly proof of relativity being wrong. He is ignoring experimental evidence, simple rules and making up new ones as he goes along.

By God, you are right! I always knew they were fibbing with us at school and at university. I have known all along since I could proove that 1 is infact not 1 but 2! x^2 - x^2 = x^2 - x^2 (x + x)(x - x) = x(x - x) x + x = x 2x = x 2 = 1 !!!

I usually live in a moderate climate, but served 1 year in the army way north in my country. At one point it was -40°C outside, cold enough for a cup of hot water to freeze before it hit the ground, when it was tossed into the air. It wasn't any harder or easier to breathe there than on a warm day.

Alt, the wave definition you quote above is true for mechanical waves. Compression and relaxation of particles with mass, mass here being an important point. That is why sound energy is not transmitted through a vacuum. Photons have no mass, they are pure energy and that is how the energy is transported. Once you stop comparing em radiation with mechanical waves you will maybe be able to answer your own question.

EM radiation, photons, do not need a medium to propagate through, or any other means of transmission. You cannot compare EM radiation to mechanical waves. 3D space, as you metnioned earlier is not a medium as such and thinking of it as the medium for particles to propagate in is inaccurate and misleading. Mechanical waves need a medium because they are indeed manifested as compression and relaxation of matter. This same analogy does not hold for EM radiation. "I bet photons have their own EM fields, but the strength of the EM field coming from a photon is so intense, its registration seems equally or more important than the physical entity itself, in the eyes of scientists." The EM field is to fast to detect, and radiation is detected in the form of absorption of photons (photo electric effect). The EM field does not originate from photons as with a charged particle where the electric field is infinite at an ideal point charge. I prefere to think of EM radiation as a stream of photons whos statistical behavoir assimilate wave properties.

Cummon Tess, are you suffering from a mental block or do you just refuse to lissen to what I and others are trying to explain to you. 1) It is possible possible to single out single photons with certain limitations but nothing that can't be described by some basic statistics. We do this in quantum cryptography all the time, and it works. Hell we can now even teleport the quantum state of a single photon onto an other photon. You might want to check your sources becasue experiments on the single photon level has matured a lot lately. 2) mirrors reflect single photons more than 1-2 times, no question about it. As I tried to point out some will reflecte many times some not even ones. A photon will have a mean lifetime in the cavity, and it can be calculated. This is simple, just think about how a laser operates. Many photons are traped in a cavity and undergo gain as they pass through the gain medium. Each photon undergo an average number of cavity roundtrips before they excit. Given this, what makes you think it is different with a single photon. Why sohuld it only experience 1 or two roundtrips in the cavity? 3) a photon moves 30 cm every 1 ns make the cavity long enough and time of flight between two mirrors will be several ns giving you enough time to move a mirror. It can be done.

As I said, don't think of it of a one of hit or miss experiment. you can make a cavity with one mirror as close to 100% reflective as possible, the second say 98%. If you have a train of pulses prepared so that single photons arrive at the cavity every minute, say. statistically 2 in every 100 potons will enter the cavity and bounce back and forth with a certain average life time of which it will continue bouncing back and forth before it is destroy by some mean or it is transmitted by one of the mirrors. The bottom line is, it is possible to prepare a cavity with a single photon within it and the photon will have a certain life time in the cavity before it is absorbed (as you were saying) or it is transmitted through a mirror. It is just as with an excited atom. It will decay again at some average time after the excitation, but you will find if you repeat the experiment that some excited atoms stay excited much longer than most.

I can contribute something on single photon issue in this thread. It is possible to seperate single photons from many, it is used in quantum cryptography as one example. but there is one catch, you are never able to say for sure that your pulse containy 1 and only 1 photon. The common method is to use a pulsed laser and attenuate it so much that each pulse contains on average only .01 photons. In this casse you have a train of pulses of which many pulses will be empty, but some of them will contain only a single photon. There is also a chance that some pulses have more (poissonian statistics if you want to look it up) but you limit the chance of that happening by chosing a low average photon number per pulse as said. Single photon sources are still some time away but they are in development. A group at the neighbouring university I know are working on these and we aim to characterise them over a 67 km fiber link within a few years. As for trapping a single photon in a cavity. It is possible, but dont think of it as a one of experiment. you would have to carry out thousands and build up a statistical picture of what you wish to characterise.

This is a very simple experiment to perform: You have a polarized source incident on a mirror along the normal. Between the source and the mirror you have placed a polarizing beam splitter (PBS) in such a way that it transmits the light from your source. Now place a half waveplate between the PBS and mirror in such an orientation that it rotates the lights polarization 45 degrees when passing through it before it hits the mirror. On the way back from the mirror (assuming light is reflected back along the normal as you were questioning) the polarization of the light is rotated an other 45 degrees. When the light reaches the PBS the polarization is normal to the incident ligth, meaning the light will reflect at the PBS instead of transmitt. Place a detector at the reflecting output off the PBS, and hey presto, you will find that you measure the same intensity as the output of your source, showing that light, indeed, reflects back along the normal. (This assumes all components do not scatter or absorb any of the light incident on them)