photon propeller Posted January 6, 2014 Share Posted January 6, 2014 Two spheres of equal size, starting at the same position A, take to different paths to the same destination B. The path of sphere 1 is curved. It is two perfect semicircles. The path of sphere 2 is a straight line. Sphere 2 travels at 10 mph. How much faster does sphere 1 have to travel in order to arrive at point b within the same time period as sphere 2? If sphere 1 and 2 simultaneously accelerate to light speed, is it possible for sphere 1 to keep up with sphere 2? If electric and magnetic fields are oscillating on curved paths orthogonal to the direction of propagation, how do they reconcile the necessary difference in speed needed to keep up with the wavefront moving linearly at light speed? Link to comment Share on other sites More sharing options...
Sensei Posted January 6, 2014 Share Posted January 6, 2014 If sphere 1 and 2 simultaneously accelerate to light speed, is it possible for sphere 1 to keep up with sphere 2? In Special Relativity you cannot accelerate massive object to light speed. You would need infinite energy. Link to comment Share on other sites More sharing options...
photon propeller Posted January 6, 2014 Author Share Posted January 6, 2014 (edited) If sphere 1 and 2 are an electron and positron, their rest mass is converted to energy when they annihilate, becoming the photon, which travels at light speed. So it follows one can accelerate to light speed if one's mass is completely converted to energy. The underlying paradox concerning this example is that sphere 1 can reconcile the necessary speed difference until sphere 2 reaches light speed. Even if they both travel at light speed, the curved path of sphere 1 prevents it from keeping pace with sphere 2. Edited January 6, 2014 by photon propeller Link to comment Share on other sites More sharing options...
imatfaal Posted January 6, 2014 Share Posted January 6, 2014 Two spheres of equal size, starting at the same position A, take to different paths to the same destination B. The path of sphere 1 is curved. It is two perfect semicircles. The path of sphere 2 is a straight line. Sphere 2 travels at 10 mph. How much faster does sphere 1 have to travel in order to arrive at point b within the same time period as sphere 2?.. v=d/t v_2=4r/t v_1=2.pi.r/t v_1= pi/2 * v_2 so S_1 travels about one and half times the speed of S_2 ...If sphere 1 and 2 simultaneously accelerate to light speed, is it possible for sphere 1 to keep up with sphere 2? If electric and magnetic fields are oscillating on curved paths orthogonal to the direction of propagation, how do they reconcile the necessary difference in speed needed to keep up with the wavefront moving linearly at light speed? Per sensei for the first bit. Second bit what makes you think anything is keeping up with wavefront and moving linearly at the same time? The fields oscillate such that the propagation of the oscillation is orthogonal to both fields and moves at c - nothing in the idea moves both at c and linearly; there is nothing going along with the light wave that is moving linearly If sphere 1 and 2 are an electron and positron, their rest mass is converted to energy when they annihilate, becoming the photon, which travels at light speed. So it follows one can accelerate to light speed if one's mass is completely converted to energy. The underlying paradox concerning this example is that sphere 1 can reconcile the necessary speed difference until sphere 2 reaches light speed. Even if they both travel at light speed, the curved path of sphere 1 prevents it from keeping pace with sphere 2. Talking about the acceleration of a massless particle is fraught with danger - they travel at c. And in fact if S_2 was going at just below .7c then a particle following your hemicircular path would not be able to keep up - that of course is based purely on the speed limit Link to comment Share on other sites More sharing options...
photon propeller Posted January 6, 2014 Author Share Posted January 6, 2014 Per first bit, close answer, sphere one has to travel 5.7 mph faster. Per second bit, wavelength is measured straight across from peak to peak, not along the curve of the wavelength itself as the curved path of the oscillating fields. frequency x wavelength =c.The rate of change in amplitude follows the curved path as the wavefront moves straight forward at c. Change is occurring at the curved rate plus the linear speed. .57 times faster than c. This should require proportionally the same difference in speed as the above problem. Meaning the curved path of oscillation should require a speed increase of 95,820 mph to keep pace in the same time period. Link to comment Share on other sites More sharing options...
imatfaal Posted January 6, 2014 Share Posted January 6, 2014 Per first bit, close answer, sphere one has to travel 5.7 mph faster. Per second bit, wavelength is measured straight across from peak to peak, not along the curve of the wavelength itself as the curved path of the oscillating fields. frequency x wavelength =c.The rate of change in amplitude follows the curved path as the wavefront moves straight forward at c. Change is occurring at the curved rate plus the linear speed. .57 times faster than c. This should require proportionally the same difference in speed as the above problem. Meaning the curved path of oscillation should require a speed increase of 95,820 mph to keep pace in the same time period. With respect my answer was exact v_1= pi/2 * v_2 You are fundamentally misunderstanding how a wave propagates either electromagnetically or even a water wave. The curved path of maximum amplitude is an artefact - nothing travels along it; the graph showing the variation (normally as lovely sine curves not semicircles) of amplitude of the E and B field both orthogonal with direction of propagation is a just a plot - there is no physical path along the points of maximum amplitude. The fields oscillate in two orthogonal directions the wave proceed in the third mutually orthogonal direction. Rates of change are a mathematical function they do not follow the path of an amplitude time plot. and where that second number is from I do not know Link to comment Share on other sites More sharing options...
photon propeller Posted January 7, 2014 Author Share Posted January 7, 2014 The curved path is the surface of the water. Do the fields themselves not travel the curved path, or their associated particles? If light moves linearly forward at light speed, how can its path oscillate without an increase in speed? The rate of change by the fields is the curved path plus the linear distance.The second number is the associated speed increase is found by 3.14/2 = x/c. 282,820-186,000=95,820. Link to comment Share on other sites More sharing options...
swansont Posted January 7, 2014 Share Posted January 7, 2014 What curved path? An EM wave propagates along a straight line as defined by E x B, just as imatfaal said. Link to comment Share on other sites More sharing options...
imatfaal Posted January 7, 2014 Share Posted January 7, 2014 The curved path is the surface of the water. Do the fields themselves not travel the curved path, or their associated particles? If light moves linearly forward at light speed, how can its path oscillate without an increase in speed? The rate of change by the fields is the curved path plus the linear distance.The second number is the associated speed increase is found by 3.14/2 = x/c. 282,820-186,000=95,820. A demonstration - the tokens merely move up and down the y axis ( 0 -> 1/2 -> root3/2 -> 1 -> root3/2 -> 1/2 -> 0 -> -1/2 -> -root3/2 -> -1 -> -root3/2 -> -1/2 -> 0) I am pretty sure that you will recognise the wave, you will note at the beginning the wave front as well. Nothing- absolutely nothing follows the root of the curve of maximum amplitude; the coins move up and down and the wave moves perpendicular to the coins Coins in sine.wmv Will attach a youtube link when they sort themselves out www.youtube.com/watch?v=vN_Iwei9j2Y there is a reputation point prize for anyone who can identify the tokens - especially the fourth one counting from the left Link to comment Share on other sites More sharing options...
photon propeller Posted January 7, 2014 Author Share Posted January 7, 2014 (edited) If a dc current (electrons) moves straight from point a to b at light speed, how does the magnetic field, which spirals around it, produce a spiral line as fast as a straight line. Wouldn't it would require the spiral to be produced faster than the straight line as in the above scenario? Are the spirals falling behind at the rate difference? Is space time compressing longitudinally as frequency goes up and rarefacting as it drops ? Is the coin a ww2 era one cent ceylon? Edited January 7, 2014 by photon propeller Link to comment Share on other sites More sharing options...
swansont Posted January 7, 2014 Share Posted January 7, 2014 If a dc current (electrons) moves straight from point a to b at light speed, how does the magnetic field, which spirals around it, produce a spiral line as fast as a straight line. Current doesn't flow at light speed. Link to comment Share on other sites More sharing options...
photon propeller Posted January 7, 2014 Author Share Posted January 7, 2014 correct, current flows as slow as syrup but the associated em energy flows at light speed. In the case of the photon, what are the dynamics of field position relative to the axis? Are particle representatives present and oscillating about the axis as the photon propagates? Or are fields merely oscillating and the particles are existent only in the form of energy not matter? Link to comment Share on other sites More sharing options...
swansont Posted January 7, 2014 Share Posted January 7, 2014 correct, current flows as slow as syrup but the associated em energy flows at light speed. In the case of the photon, what are the dynamics of field position relative to the axis? Are particle representatives present and oscillating about the axis as the photon propagates? Or are fields merely oscillating and the particles are existent only in the form of energy not matter? No, the EM signal flows at less than light speed. The speed of a signal in a coaxial cable is around 2/3 c. In unshielded wire it's about 0.95c. Link to comment Share on other sites More sharing options...
photon propeller Posted January 7, 2014 Author Share Posted January 7, 2014 So in the case of the photon, in place of the cable we have its axis, the straight line, which is traveling linearly at light speed. Keeping in mind nothing happens faster than c. Yet in a circularly polarized photon, e and b fields are transversing and rotating around their axis as the fields relative point positions at the wavefront also transverse and rotate. If their common axis point at the wavefront is moving linearly forward at light speed, how do the relative field points keep up? Link to comment Share on other sites More sharing options...
swansont Posted January 7, 2014 Share Posted January 7, 2014 Keeping in mind nothing happens faster than c. That's a vague statement and ultimately not true. Information cannot be transmitted faster than c, and massive objects cannot move at c or faster. But lots of "things" "happen" faster than c. This just came up in another thread: if you shone a laser dot on the moon and then moved it, the dot could move faster than c. The E and B fields move along with a photon, so they are moving at c. That their strength is varying is not additional motion, much like the dot on the moon is not physically moving. Link to comment Share on other sites More sharing options...
photon propeller Posted January 7, 2014 Author Share Posted January 7, 2014 Doesnt any change, including varying field strength, produce a trail of information? Does that trail equate to the production of information? Is it that information may be produced faster but not transported faster than c? How does the image,( the information transported), relate to the relevant point wavefront positions? Does the change in amplitude contribute to a change in the topography of the wavefront? Hence, is the information produced by the change in amplitude simultaneously revealed at and as the wavefront propagates? Link to comment Share on other sites More sharing options...
swansont Posted January 7, 2014 Share Posted January 7, 2014 Doesnt any change, including varying field strength, produce a trail of information? No, not just any change. Does the change in amplitude contribute to a change in the topography of the wavefront? No. There is no change in amplitude of the light. Hence, is the information produced by the change in amplitude simultaneously revealed at and as the wavefront propagates? There is no change in amplitude of the light. Link to comment Share on other sites More sharing options...
photon propeller Posted January 15, 2014 Author Share Posted January 15, 2014 sure it does. amplitude changes as it oscillates (e&b field strength varies)from the inertial line to its maximum. The manipulation of that maximum is the basis of modulation for am signals. That is how a carrier wave transports information. Frequency may also be modulated as in the case of fm signals. If an electron and positron represent these fields within the photon, and adding more electrons in phase modulates the amplitude, how can you believe nothing physically travels the curved path? What paths do the electron and positron follow? What are the dynamics of their motion which propels the photon linearly? Link to comment Share on other sites More sharing options...
Sensei Posted January 15, 2014 Share Posted January 15, 2014 If an electron and positron represent these fields within the photon, and adding more electrons in phase modulates the amplitude, There is no electrons and positrons in photon... Electron and positron have billions time more energy than radiowave photon. Link to comment Share on other sites More sharing options...
swansont Posted January 15, 2014 Share Posted January 15, 2014 sure it does. amplitude changes as it oscillates (e&b field strength varies)from the inertial line to its maximum. The manipulation of that maximum is the basis of modulation for am signals. That is how a carrier wave transports information. The amplitude of a wave that's a pure tone is the peak value, not the value at some point. A carrier wave transports information by modulating it, making it the sum of waves, but this does not arrive any faster than the carrier so it doesn't exceed c. If an electron and positron represent these fields within the photon, and adding more electrons in phase modulates the amplitude, how can you believe nothing physically travels the curved path? What paths do the electron and positron follow? What are the dynamics of their motion which propels the photon linearly? As Sensei notes, the premise is not true, so any conclusion is irrelevant. Link to comment Share on other sites More sharing options...
photon propeller Posted January 15, 2014 Author Share Posted January 15, 2014 (edited) There is no electrons and positrons in photon... Electron and positron have billions time more energy than radiowave photon. E=hv. an electron and positron annihilate to form a gamma photon. gamma photon is absorbed then emitted as radio photon. both are still combinations of e and b fields, one less energetic than the other. 2 oscillating point vectors represent each field and a single point vector represents the axis. Do these field vectors not act like point particles that transverse the axis and rotate with it? Do their relative positions at the wavefront not change as the wave propagates? Isnt the vector information converted to raster format,(rasterization), converting the visual image to an electrical stream of proportional intensities? (TV)Doesnt the variance of these point vectors, with each frame, outline the image(information) transported? Edited January 15, 2014 by photon propeller Link to comment Share on other sites More sharing options...
swansont Posted January 15, 2014 Share Posted January 15, 2014 E=hv. an electron and positron annihilate to form a gamma photon. Two, actually. Or sometimes three, and on rare occasions even more. gamma photon is absorbed then emitted as radio photon. Absorbed and emitted by what? A 511 keV photon will ionize any atom it encounters. You can eventually get radio waves, after several steps of various interactions. Because, as Sensei noted, radio waves have much less energy than the electron or positron. both are still combinations of e and b fields, one less energetic than the other. 2 oscillating point vectors represent each field and a single point vector represents the axis. Do these field vectors not act like point particles that transverse the axis and rotate with it? Do their relative positions at the wavefront not change as the wave propagates? What's a point vector? Anyway, that seems to be in rough agreement with an EM wave. Isnt the vector information converted to raster format,(rasterization), converting the visual image to an electrical stream of proportional intensities? (TV)Doesnt the variance of these point vectors, with each frame, outline the image(information) transported? Huh? A modulated signal contains information, that can be decoded. The specifics of the information doesn't matter all that much. But it's not a vector that contains the information, it's the modulation. Link to comment Share on other sites More sharing options...
Sensei Posted January 16, 2014 Share Posted January 16, 2014 (edited) there is a reputation point prize for anyone who can identify the tokens - especially the fourth one counting from the left From left: 3rd, 5th, 8th, 10th and 11 are 10 groszy (ten penny), Polish coin, equivalent to 3.333 USA cents current rate (ajb should have them in the pocket). 4rd one is some penny from New York. All of them are 10 penny but from different countries? Are you collecting 10 pennies? Edited January 16, 2014 by Sensei Link to comment Share on other sites More sharing options...
imatfaal Posted January 16, 2014 Share Posted January 16, 2014 From left: 3rd, 5th, 8th, 10th and 11 are 10 groszy (ten penny), Polish coin, equivalent to 3.333 USA cents current rate (ajb should have them in the pocket). 4rd one is some penny from New York. All of them are 10 penny but from different countries? Are you collecting 10 pennies? Well spotted on the groszy! I had a handful of zlotys and groszys after a night out in soho (long story) and the 10 groszy coin was the right sort of size so I searched for a few other interesting ones (none are still current) two dutch cents pre euro an english halfpenny two spanish 5 pesatas a new york subway token I have always collected coins - now it is pretty lackadaisical hobby in that I just keep a pocketful from whereever I have been and add them to a large glass pot. The NYC subway tokens are a favourite of mine - I even converted 4 into a set of cufflinks Link to comment Share on other sites More sharing options...
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