GETTING RID OF EINSTEIN RELATIVITY
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From: Pentcho Valev on 12 Sep 2007 10:41 On 12 Sept, 16:06, "Paul B. Andersen" <paul.b.ander...(a)hiadeletethis.no> wrote: > Pentcho Valev wrote: > > On 12 Sept, 13:10, "Paul B. Andersen" > > <paul.b.ander...(a)hiadeletethis.no> wrote: > >> Pentcho Valev wrote: > > >>> The emission theory gives the equations c'=c-v and c'=c+v whereas > >>> special relativity gives c'=c. Which equations: c'=c-v and c'=c+v or > >>> c'=c, are relevant in the interpretation of the Sagnac experiment? > >>> Pentcho Valev > >> The Sagnac experiment: > >> - Given an inertial frame which is the reference > >> for all speeds mentioned below. > >> That is, all speeds are relative to this non-rotating frame. > >> - Given a stationary circle with radius r. > >> - Given a light source moving at the speed v around the circle. > >> - Assume the light is moving around the circle (infinite number of mirrors). > >> - Let tf be the time the light emittet in the forward direction > >> uses to catch up with the source. > >> - Let tb be the time the light emittet in the backward direction > >> uses to meet the source. > > >> Prediction according to SR: > >> --------------------------- > >> The speed of the light emitted in the forward direction is c. > >> The speed of the light emitted in the backward direction is c. > > >> So we have: > >> 2*pi*r + tf*v = tf*c > >> tf = 2*pi*r/(c-v) > > >> 2*pi*r - tb*v = tb*c > >> tb = 2*pi*r/(c+v) > > >> delta_t = tf - tb = 4*pi*r*v/(c^2 - v^2) > > >> Setting w = v/r, A = pi*r^2, g = (1 - v^2/c^2)^-0.5 > >> we get: > > >> delta_t = (4Aw/c^2)* g^2 > > >> The g^2 will obviously be unmeasureable different from 1 > >> for any practical Sagnac experiment. > > >> So SR predicts delta_t = 4Aw/c^2 which is in accordance > >> with enumerable practical experiments. > > >> Prediction correct, SR confirmed. > > >> Prediction according to the emission theory: > >> -------------------------------------------- > >> The speed of the light emitted in the forward direction is c+v. > >> The speed of the light emitted in the backwards direction is c-v. > > >> So we have: > >> 2*pi*r + tf*v = tf*(c+v) > >> tf = 2*pi*r/c > > >> 2*pi*r - tb*v = tb*(c-v) > >> tb = 2*pi*r/c > > >> delta_t = tf - tb = 0 > > >> So emission theory predicts delta_t = 0, while enumerable practical > >> experiments shows delta_t = 4Aw/c^2 > > >> Prediction wrong - emission theory falsified. > > >> Paul > > > Andersen Andersen there are so many sites containing calculations of > > the Sagnac experiment performed by people much cleverer than you. Why > > didn't you refer to some of them instead of demonstrating your zombie > > reasoning? See this for instance: > > >http://www.newtonphysics.on.ca/faq/invalidation.html > > > Find the mistake Andersen Andersen! > > << > However, we see above, that the velocity of Earth is responsible > for the change of time light takes to go around the Earth. [SIC] > >> > > .. to mention one. > > Paul Happy zombie finds mistakes easily, many mistakes finds happy zombie very easily: http://www.newtonphysics.on.ca/faq/invalidation.html "However, we see above, that the velocity of Earth is responsible for the change of time light takes to go around the Earth....Equation 9 is identical to Sagnac equation (Sagnac M. G. J. de Physics., 1914, 4, 177-195). Sagnac equation gives a double amount as expected, because it represents by definition, the difference of time interval between two beams traveling simultaneously in opposite directions, corresponding to a change of angular velocity from -w to +w . It has been clearly measured and demonstrated (see question 2-E above), that the Sagnac equation is in perfect agreement with observations." Pentcho Valev
From: bz on 12 Sep 2007 10:26 sean <jaymoseley(a)hotmail.com> wrote in news:1189444302.990016.237750(a)d55g2000hsg.googlegroups.com: > Post 363 > > On 6 Sep, 17:33, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: >> sean <jaymose...(a)hotmail.com> wrote >> innews:1189089230.764471.53320(a)22g2000hsm.googlegroups.com: >> >> > On 30 Aug, 23:55, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote: >> >> sean <jaymose...(a)hotmail.com> wrote in news:1188512224.511353.237820 >> >> @i13g2000prf.googlegroups.com: >> >> >> > If you bothered analysing your sean planets sim, youd >> >> > see that its the only way to have light propagating away >> >> > from any source at c. >> >> >> One of the most fundamental laws of physics is that an object in >> >> motion continues in that motion unless acted upon by an outside >> >> force. >> > You ignore several things here, First of all if emmision theory >> > predicts that light always is at c relative to a source >> >> That only applies at the moment of emission. If the source changes its >> motion after emission, the light does not know or care. > Maybe in your own personal version of emmision theory. But Im saying > that if one can model emmision theory as having light propagate away > from any source always at c relative to any source, then,.... > one can explain MMx and sagnac > >> > then it would be impossible for light to be effected by gravity under >> > these terms. >> >> Once light has left the source, gravity, mirrors, lenses, etc can >> effect its motion. The 'leaves its source at c' only applies at the >> moment of emission and then only when measured with equipment that are >> co-moving with the source and hence in the same inertial frame of >> reference. If the source suddenly accelerates at right angles and goes >> flying off, into the night (you have just thrown your flashlight away) >> the light that has been already emitted from that flashlight does NOT >> suddenly change its motion in order to maintain a velocity of c wrt the >> source that it has left far behind. > How do you know. > Wheres the proof ? You have none. I discussed this earlier in this > thread with George and I pointed out that it is technically almost > impossible to test these claims. At least there is no experimental > setup > that exists that can test this clai you make. I disagree. > If a lab was 10 meters long and a rotating light source on one end > was observed from the other end. The actual displacement > due to this `path drag` would be minute. Something like a > 0.000003 mm deflection over 10 m. Let us put a LED (light emitting diode) on the rim of a ultra centrifuge rotor that is 10 cm in diameter and rotate the disk at 50,000 rpm. The circumference will be 0.628 meters. The speed of the diode will be 523.599 meters per second or 1.7 micro c. Now, we spin the rotor clock wise for one test and counter clockwise for another test. We pulse the LED near the point where it is closest to us (thus traveling at max speed perpendicular to the line of sight) From a distance of 10 meters we use a long focal length microscope to view the position of the LED at the moment that it pulses. Our microscope gives us a 100 to 1 magnification of the image of the LED. We use a ccd camera to display the image on our computer screen. What do I predict and what does your theory predict? I predict that the image will appear in the same location, non rotating, or rotating clockwise or counterclockwise at 50,000 rpm. What does your theory predict? Your theory predicts that during the 0.033 us it take the light to travel from the LED to our tele-microscope 10 meters away, the light will have moved 0.017 mm in the direction of rotation. As there are LEDs that are only 20 um in size, http://www.patentstorm.us/patents/6410940-claims.html so, I calculate that if your theory were correct that the image would be displaced by 873 diameters in the direction of travel. I see no reason that such a test can not be performed, inexpensively. > Youd never be able to measure that. > Nobody has yet. Therefore no proof exists that tests this possibilty > one way or the other >> > Secondly you ignore what I say repeatedly in the post >> > youve just replied to. >> >> I ignore many things because I am trying to focus your attention on one >> small point. > And Im trying to focus your mind on one point. That is that the MMx > source > is moving yet the light that leaves the source still has a speed of > c relative to the moving source. Contrary to what you claim taht it > isnty > moving at c in the lab. And I have yet to see anything else that works that way. The golf ball does not bobble as tiger completes his follow-through. It is contrary to all data we have and it is easy to test, as I have just shown. >> > That emmision theory cannot have light as a particle with mass. >> >> I don't care if light is a wave or a particle at this moment. It >> doesn't matter. The light, once emitted, goes its merry way. Unlike the >> teenager, leaving home for the first time, the light does not 'write >> home' and ask for more money. The light doesn't care what happens to >> the source after it has left home. >> >> [snipping the rest of the stuff that is unrelated to the point I am >> trying to make and snipping your insults. If you want to continue our >> discussion, you will avoid insults. I don't bother with those that lack >> enough self respect to treat others with respect.] ..... > My apologies. I get non stop sleazy insinuations, character smears > and > insults from the rest of your peers. I forgot to check if you werent > up to > the same. A quick check shows me that so far you seem to have been OK. > Thanks. I appreciate that. Accepted. > > JUst dont forget that Im argueing here for a wave only emmision model > to explain the various observed phenomena. So please dont expect me > to incorporate particle theory in this model. Like for instance.. > electrons. Electrons are sources of photons as they circulate in the synchrotron. The fact that the electron is a particle is NOT important. The fact that we KNOW the path that the electron is taking as it emits those photons IS important. The electrons are moving along a curved path at a high velocity. The electrons are the SOURCE of the photons. The source is moving along a curve. By your theory, the photons are 'tied' to the source and must mirror its motion. The photons do NOT do that. This is experimentally verified, as I showed. > >> >> The light emitted consists of 'objects'. Once emitted, they must >> >> travel in straight lines at a constant velocity unless acted upon by >> >> an outside force. >> >> >> What force would 'tie' the light, once emitted, to the source so >> >> that it must follow the source? >> >> >> There is no evidence of any such 'tie' between light, once emitted, >> >> and its source. >> > You jump to conclusions here. What "evidence" do you have that light >> > doesnt >> > travel at c in any direction from a source? >> >> I made no such claim. It DOES travel at c in all directions from where >> the source was at the moment of emission. But once emitted, it doesn't >> care where the source goes afterwards. > This is an odd statement . You say you never claimed that light > `doesnt travel > at c relative to a source`. And then you contradict yourself in the > same > sentence above and say that in fact it `doesnt travel at c relative to > the > source`.?? > So what do you want to claim? That it does or it doesnt travel at c > relative > to the source? It travels at c in all directions, as measured in an inertial frame of reference co-moving with the source at the moment of emission. If the source changes motion or even ceases to exist after the light has been emitted, the light neither knows nor cares. >> > My evidence that they do is MMx. Notice here that the MMx source is >> > rotating >> > around the earths axis yet we also know that the light is travelling >> > at c on >> > both paths. This would not be possible if light travelled away from a >> > rotating >> > source at different speeds in different directions as you erroneously >> > suggest.above >> >> You misread my suggestion. > How did I misread your sugggestion? YOu said... > > "The light emitted consists of 'objects'. Once emitted, they must > travel in > straight lines at a constant velocity unless acted upon by an outside > force." > > In other words you say above that once light leaves the MMx source it > must > travel in straight lines at c + the speed of earths rotation etc, > relative > to the source. They travel at c as measured in an inertial frame of reference that was moving at exactly the same velocity as the source at the instant of emission. > THat means that you think that light travels at different speeds away > from a source relative to that source. They travel at c as measured in an inertial frame of reference that was moving at exactly the same velocity as the source at the instant of emission. Light travels at c as measured in ANY inertial frame of reference, irrespective of the velocity of that frame of reference wrt the source. [with the caveat that no inertial frame of reference can ever be defined so as to move at c with respect to any object that has mass]. >> >> In a cyclotron, particles traveling around a circular path emit >> >> beams of xrays. >> >> Those xrays travel in straight lines from the point of emission. >> >> They do NOT follow, in any way, the subsequent motion of the >> >> emitting particle. >> > Two things here.. Firstly,.. whats your proof these xrays >> > travel in straight lines?? You have none. >> >> You put three lead plates with small holes in them along a straight >> line leading away from the source. The xrays only reach the detector >> when all three plates line up with the holes in a straight line. If the >> xrays were traveling along a curved path, the holes would need to line >> up along a curved path. >> >> > Show me the part >> > of an experimental setup that checks the direction relative to the >> > known >> > source and the speed of the xray. You have neither nor do both tests >> > exist. >> >> You have never worked with an xray beam line at a synchrotron. >> visit a facility such ashttp://www.camd.lsu.edu/and you will see that >> they use the synchrotron to generate xray beams. BEAMS of xrays >> traveling in straight lines. If your idea about the emission following >> the movements of the source were correct, CAMD would not work as it >> does. >> >> > This is a fabrication on your part. >> >> I don't need to fabricate anything because I know what I am talking >> about. >> >> > Secondly,.. you make the assumption that the xrays are produced by >> > moving particles/ie ...rotating sources like the MMx or sagnac >> > source. But what proof have you that the sources in a cyclotron >> > is rotating?? NONE >> >> The xrays are produced when a beam of particles is forced to CHANGE >> directions. When this is done, they emit photons. This is known as >> cyclotron radiation. >> >> > You can only assume the xray `sources` are rotating theoretically. >> > IM not familar enough with the technical setup of a cyclotron >> >> I am. >> >> > but my reference refers to the `particles` not as atoms but as >> > electrons >> >> Electrons or positrons or protons or heavier particles can be >> accelerated. When the beams are bent(forced to change direction of >> motion) photons are emitted. >> >> > As far as Im aware electrons are emr. >> >> You are quite mistaken. The motion of electrons generates electro >> magnetic radiation but electrons are NOT electro magnetic radiation. >> But, all of this is beside the point. > Sorry Im not sure why I said this .Yes the electrons in wave theory > too , arent the emr but they generate the emr . That is ok. We all make mistakes at times. Wisdom comes from making mistakes and from learning from them. >> The point being that once emitted, light doesn't care what happens to >> the source. >> >> [snipping the rest of the side issues] >> >> .... >> >> > Lets see if for starters you can supply the lab setup in a cyclotron >> > that measures the speed and dierection of the observed xrays.. >> > You cant, because no such test was ever made. >> >> Wrong. The direction is easily determined by three lead plates with >> holes in them. > OK fair enough. This is the first time anyonse asked me to define a > cyclotron in wave only theory so I need to get uo to speed on how they > are constructed. Its not quite clear in my reference how this is done. > For instance it shows the defelected `electrons` but not where the > xrays are generated. I assume the whole cavity generates them or is it > just a point at where the electron is deflected at the small outlet > hole? Its very vague my reference. Xrays are only emitted where the beam of electrons is bent. Xrays ( or lower energy emr) can be emitted whenever electrons are made to slow down suddenly, such as the beam of electrons from the cathode strike the anode in an x-ray tube or when the electron beam strikes the screen on your TV tube. That is why the voltage on the TV CRT is prevented from exceeding a certain threshold voltage (somewhere around 30KV if I remember correctly) to avoid excessive xray production)[but this is again beside the point]. >> The speed is easily determined by measuring the time between the moment >> that the batch of particles gets forced to change direction of motion >> and the time the xray pulse hits a target along the beam line. Also, >> the beam line can contain two detectors, a known distance apart and the >> time for the pulse to travel between those detectors can be measured. > You obviously know a good deal about cyclotrons. I admit need to find > out more about cyclotrons . For instance my reference > shows the electrons being fired out the small hole? The electron beam may start from an electron gun similar to that in your TV CRT. > Or is that the > xrays. The xrays are allowed to exit through a series of small holes so that the beam is of precise energy and its direction, energy and area of illumination are known. > Its not clear. Any ways in a wave only theory I have to > explain electrons as waves only so its going to be difficult > for you to argue that the source is particles. It doesn't matter if electrons are waves (they DO have wavelike properties) or particles (they do have particle like properties) or just best described by a series of equations and not labeled with 'wave' or 'particle' at all. > For instance > I notice that the setup of the cyclotron bears close resemblance > to the radio transmittter. That is no accident. > Both have oscillating magnetic fields > as the `driver` of the radiations and both in the standard model > describe these as streams of electrons. But wave theory as > I point out in my "redshift without expansion " describes the > oscillating field in the transmitter as oscillating magnitic > fields of the atoms in the wire which produce the emr radiation. There are instruments called 'ESR' and 'NMR' (electron spin resonance spectrometers and Nuclear magnetic resonance spectrometers) that can be used to precisely determine the properties of the magnetic fields from the particles in the atom. You need to study how these work and what they tell us about the environment around each atom in a complicated molecule. If your conjecture were right, these instruments would show much different results than they do. You must be able to explain, for example, why each peak on a 2d 400 MHz (1H, 1H) COSY-45 spectrum of a particular compound shows up at the exact location it occurs at. Here is a place to start. It will tell you 'how it works'. http://www.cryst.bbk.ac.uk/PPS2/projects/schirra/html/theory.htm > No electrons needed. So Id say the cavity in the cyclotron is > the source and it is the oscillating magnetic field in the > cavity that generates the xrays. Wrong. If you were correct, then my microwave oven would generate xrays. My ham radio transmitter would generate xrays. We would all be dead because the cell phone towers would generate xrays. > Just as the the wire `cavity` > in the radio transmitter generates the radio waves. The 'cavity' does NOT generate the radio waves. Oscillations of current (electron flow) in the antenna cause electromagnetic radiation to radiate in the form of radio waves. > I assume > the cyclotron cavity oscillates at higher frequencies hence the > higher frequency emmision.Thats whats suggested in my ref anyways. Higher frequency, yes. Xray frequency, NO. > As with the transmitter the cavity as a whole is the source. No to both. If I fail to hook an antenna to my transmitter, there are no radio waves emitted. > And it doesnt move in the lab. It is in the same frame as the > observor. Hence contrary to what you thought no bent lines in > the x-radiation are predicted by wave only emmision theory > because the source isnt moving in the lab The sources of emission are the electrons. If the microwave accelerators are TURNED OFF, those electrons still in the beam will continue to circulate until the electrons loose energy and the beam loses coherence. At the 'bends' xrays would still be emitted. It is NOT the microwaves that produce the xrays. Anymore than it is the gunpowder that produces the bullet. The gunpowder propels the bullet but the bullet is a piece of lead that is crimped into the cartridge. >> >> He did NOT say 'break apart', he said 'distort'. >> >> In the MMX, all it takes is moving part of the apparatus by a small >> >> fraction of the wavelength of light in order to create a detectable >> >> indication. >> >> >> THAT is why he said 'distort'. Spinning the apparatus would cause >> >> changes in the instrument that would make it useless for the purpose >> >> you propose to use it for. >> > I assume by this that you and george imply that a rotating MM >> > experiment >> > physically stretches one arm moreso than another and doesnt give a >> > null >> > result? If so please supply a reference for such an observation, >> > because >> > I dont believe you have any proof to back up this claim.(notice that >> > will not be able to supply this as none exists. Youve just fabricated >> > this evidence) >> >> If you ever spin anything (a car tire for example) you will notice that >> getting everything 'in dynamic balance' is NOT trivial. Any >> experimental apparatus consists of many different parts, each with >> their own mass, density and, in this case, distance from the point >> about which the apparatus is to be rotated. >> >> The MMX apparatus must maintain all dimensions within fractions of the >> wavelength of light. >> >> ANY slight imbalance or difference in the distances from the center >> would produce different torques and invalidate the experiment. Have you >> ever seen an MMX apparatus? >> >> And one final point. The MMX apparatus IS spinning at a slow rate of >> spin, 1 revolution per day. > I dispute these points on several counts. First MMx DID spin the setup > and it didnt create distortion in the arms. WRONG. The movement of the apparatus from one position to another 'produced fringe shifts of up to 7 fringes' and that is after the apparatus was allowed to stabilize. You do realize that the entire apparatus was floating on mercury in order to reduce the stresses of rotation, don't you????? > So it is possible.Second, > if this experiment hasnt been tried in your opinion, it doesnt mean > that > one can assume that the outcome would be that light wouldnt travel at > c > relative to the spinning MMx source,.. which leads to my third point; > As you point out above. A spinning MMx setup has been tried where > no distortion occurs and it does show that light from a spinning > source > still travels away at c relative to the source. > Its called the MM experiment. And it spins 1/day. And it shows a negative result. >> > And secondly assuming the remote chance that it does distort such an >> > apparatus and not allow us to check what I propose may be observed >> > (ie whether or not a rotating MMx that isnt distorted does still >> > have light at c on both paths)...THen how is it that you or george >> > can say with certainty that a rotating MMx source does not allow >> > light >> > to propagate down both arms at c? After all, youve just claimed >> > that any experiment to test this is impossible. >> >> The math is not subject to the physical limitations of the equipment. >> >> [snipping the rest because it has nothing to do with **the point** that >> >> once light is emitted, subsequent motion of the source can not and does >> not effect the already emitted light > This is wrong and unsubstantiated. Ive pointed out many times > that the MMx results contradict this claim you and others make. > Because we know that when light leaves the MMx source it still travels > in all directions at c relative to the source. THis is only possible > if > the subsequent motion of the source did effect the already > transmitted > light. Sean, I have given you the truth, as best I know it. I have suggested simple experiments that YOU could do to test your ideas. If your conjecture were correct, the CAMD synchrotron would not work as it does, the beams of xrays would follow a curve rather than a straight line. Neither waves on water nor anything else behaves as you propose for light. There is no reason to suppose that the light can know or care what happens to the source of the light once the light has left it. NOTHING else behaves that way. There is no evidence that light behaves that way. -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+spr(a)ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
From: bz on 12 Sep 2007 11:15 Pentcho Valev <pvalev(a)yahoo.com> wrote in news:1189598024.812605.162630 @g4g2000hsf.googlegroups.com: > http://www.newtonphysics.on.ca/faq/invalidation.html > Whoever wrote that site has many errors. To start with, they conflate 'absolute velocity' and 'absolute velocity of rotation'. They claim that Einstein's theory of relativity says that there are no absolutes. That is incorrect. [quote] Einstein's Theory claims that there is no possible way to detect the ABSOLUTE velocity of the Earth. The very use of the expression RELATIVITY comes from Einstein's hypothesis that parameters, like VELOCITY, are relative so that any absolute motion (like absolute velocity) is meaningless. [unquote] If they had read Einstein's paper and understood it, they would have known that the term 'ABSOLUTE velocity of the Earth' as used in that context has an implied 'along a straight line'. Einstein was speaking about things in uniform states of linear motion. Things in what we now call 'inertial frames of reference'. It should be clear to anyone that any object that is rotating is not in an inertial frame of reference. Rotation IS absolute. If you are in a room which is within another room and the room you are in is absolutely not rotating, it doesn't matter if the outer room is rotating, you can tell with simple test equipment that your room is stationary. for example. "This optical gyroscope detects the absolute velocity of the rotating Earth from any location on Earth, even if Einstein's theory claims that light travels with respect to the observer. " The optical gyroscope does not detect 'the absolute velocity' of anything much less 'the rotating Earth'. It detects the velocity of rotation of anything. Rotation IS an 'absolute' as is acceleration under special relativity. The author of the page does not have a basic understanding of physics. That paper is not worth the paper it is not printed on. -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+spr(a)ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
From: bz on 12 Sep 2007 11:35 bz <bz+spr(a)ch100-5.chem.lsu.edu> wrote in news:Xns99A95FF903997WQAHBGMXSZHVspammote(a)130.39.198.139: > If the source changes motion or even ceases to exist after the light has > been emitted, the light neither knows nor cares. In fact, as the 'source' is most often an ion, atom or molecule in an excited state, and the act of emission makes the source 'cease to exist' [in the excited state] then there is no longer a 'reference' to which to tie the light, once it has been emitted, so there is no way for the light to know what it's 'mother' does, once it has left home. -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+spr(a)ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
From: Pentcho Valev on 12 Sep 2007 15:16
On 12 Sept, 18:15, bz <bz+...(a)ch100-5.chem.lsu.edu> wrote in sci.physics.relativity: > Pentcho Valev <pva...(a)yahoo.com> wrote in news:1189598024.812605.162630 > @g4g2000hsf.googlegroups.com: > > >http://www.newtonphysics.on.ca/faq/invalidation.html > > Whoever wrote that site has many errors. This is irrelevant since the problem is: Does Sagnac experiment confirm special relativity or does it refute special relativity by confirming c'=c+v, the implication of Newton's corpuscular theory of light. The authors claim Sagnac experiment confirms c'=c+v so only errors which, when fixed, reverse the situation so that Sagnac experiment starts confirming special relativity are relevant. Find those errors, change the respective equations and prove that Sagnac experiment confirms special relativity and refutes c'=c+v! If you can do this in a convincing way, you will automatically become the Albert Einstein of our generation (Hawking and Tom Roberts will be degraded to silly zombies). Pentcho Valev |