Moving Mirror Proves Einstein Wrong
in [Relativity]
|
Prev: LHC marries Aunt-Al
Next: SR and a lightbulb
From: YBM on 2 Jan 2010 13:03 Henry Wilson DSc a �crit : > I'm as interested in what SR says as I am in the Koran. But you pretend to have proven that SR is self-contradictory... Can't you see who's contradicting himself here?
From: PD on 2 Jan 2010 13:13 On Jan 1, 8:57 pm, xxein <xx...(a)comcast.net> wrote: > On Dec 31 2009, 1:04 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Dec 30, 10:03 pm,xxein<xx...(a)comcast.net> wrote: > > > > On Dec 29, 6:51 pm, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: > > > > > Henry Wilson DSc wrote: > > > > > This experiment involves a light source and a mirror. > > > > > Note: I am not really an "Einstein supporter". Rather, I > > > > am a physicist. That is, I STUDY physics and use the theories > > > > which best model the physical phenomena of the world. For this > > > > sort of question by far the best model is Special Relativity > > > > (no other theory comes close, except for theories > > > > indistinguishable from SR). > > > > > Tom Roberts > > > >xxein: 'That is', that you refuse to recognise or distinguish other > > > theories. You'd bleed to death over your belief of 'your' physics. I > > > hate to explain to brick walls and the physic hates it also, but the > > > physic is just the physic and someone other than the 'pontificating > > > TR' can certainly understand it better than you. > > > > Can you describe the physical mechanism (not a/some physics) of how > > > velocity addition works to our perception and measure of it? NO. The > > > only thing you have is a math formula designed to describe the math of > > > how it eerily conforms only to subjective observation. > > > Objective observation, where the objectivity is determined by the > > common result obtained by independent investigators using > > complementary methods. > > > The second you start saying that observation is suspect and that > > objective truth can be determined without reference or verification > > from observation, you have stopped doing science. > > > > You have > > > discounted the objectiveness of the physic. > > > > Einstein died with a healthy doubt of his own very successful > > > theories. I don't see you relinquishing yours.- Hide quoted text - > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text - > > xxein: On the contrary. The invesitgators are subjective to the > objective affects (however hidden from thinking in their mode of > investigation). If you doubt that, then dismiss the Doppler effect. The Doppler effect is objective, since it can be measured by independent means by independent investigators. Are you thinking that physical quantity that is frame-dependent is subjective? How about velocity? Is that a subjective physical quantity? > > You can make a math about it but then you have to explain how the > physic works to support it. A man-made physics is just a guess and > never correct enough. Otherwise we would have TOE's running rampant > because there would be no ultimate qualification. That's what current theories do. I get that you don't understand the physics that supports the math, but that doesn't mean that it's just math. It just means that you don't understand it.
From: Inertial on 2 Jan 2010 17:07 "eric gisse" <jowr.pi.nospam(a)gmail.com> wrote in message news:hhod94$nht$3(a)news.eternal-september.org... > ..@..(Henry Wilson DSc) wrote: > >> On Sat, 02 Jan 2010 12:19:51 -0800, eric gisse <jowr.pi.nospam(a)gmail.com> >> wrote: >> >>>..@..(Henry Wilson DSc) wrote: >>> >>>> On Sat, 2 Jan 2010 10:21:39 -0800 (PST), PD <thedraperfamily(a)gmail.com> >>>> wrote: >>>> >>>>>On Jan 1, 2:16 pm, ..@..(Henry Wilson DSc) wrote: >>>>>> On Fri, 1 Jan 2010 10:21:35 -0800 (PST), PD >>>>>> <thedraperfam...(a)gmail.com> wrote: >>>>>> >On Dec 31 2009, 8:09 pm, ..@..(Henry Wilson DSc) wrote: >>>> >>>>>> >>>>>> >I know the whole idea of observer-dependent quantities just throws >>>>>> >you for a loop. >>>>>> >But even Galileo noted that even the *shape* of the trajectory of a >>>>>> >falling body changes just because someone else looks at it. A >>>>>> >cannonball dropped from the mast of a ship falls in a *straight >>>>>> >line* >>>>>> >to the bottom of the mast, according to an observer on the deck. But >>>>>> >it falls in a parabola, according to an observer on shore. >>>>>> >Now, according to you, if a path is one shape, it can hardly be made >>>>>> >a different shape just because someone else looks at it. But it's >>>>>> >plain that this DOES happen. So what you think is impossible appears >>>>>> >to be possible after all, even though you don't understand how it >>>>>> >can >>>>>> >be. Same thing here. >>>>>> >>>>>> So why are some things frame dependent and not others? >>>>>> >>>>>> They must contain L/T or (L/T)^2 >>>>> >>>>>There are a number of things that don't, like electric field. So >>>>>obviously, it's not true that they "must". >>>> >>>> What are the dimensions of electric field? >>>> >>>> The dimensions of electrical units are pretty vague generally. They >>>> include >>>> things like 'M^1/2' and 'L^3/2'. epsilon a mu are also there...and >>>> they >>>> have (unknown) dimensions too. >>>> >>> >>>Wow, did you just argue that nobody knows what dimensions an electric >>>field has? >> >> Do YOU know? > > Force over charge. > >> >> Electrical phenomena cannot be expressed in just M, L and T. > > That's because you forgot about charge, one of the fundamental SI units. > >> >> mu and epsilon have to be present as well. > > Those aren't units. > >> >> If you accept that these two are fundamental dimensions then that might >> be >> OK.... but maybe the two quantities are themselves expressible in terms >> of >> other dimension about which we know nothing. >> >> So I would stick by my claim that the whole area is pretty vague, >> wouldn't >> you agree. > > What book did you study E&M from? I think we can extend the 'Henry doesn't know, doesn't care, and can't understand it" principle to many areas of physics, not just SR. He really needs a new hobby, one that a knows something about, cares about, and understands (if there is such a thing) because he is a failure at physics.
From: waldofj on 3 Jan 2010 15:08 > news:ea47baf0-a8f9-4f58-935d-812452ab275d(a)o19g2000vbj.googlegroups.com... > On Dec 30, 7:36 am, "Androcles" <Headmas...(a)Hogwarts.physics_q> wrote: > > > I don't care about physical explanations, there is no mathematical > > explanation. Anyone handwaving their precious gamma around > > needs to derive it first. > > > gamma = sqrt[(c-v)(c+v)/c^2] > > > actually gamma is the reciprocal of that function and deriving it is > easy. What's the big deal? > ======================================== > Go on then, show and tell. Just remember that the speed of light is > c in all frames of reference and so you can't use c+v or c-v without > being self-contradictory. Ill derive it two ways, the first is the way it was taught to me in college, the second is a way I came up with myself. First: compute gamma by analyzing a light clock, two mirrors with a light beam bouncing between them. The mirrors are oriented so the light is moving parallel to the Y axis and perpendicular to the X axis. I will call the time it takes light to travel from one mirror to the other from the point of view of an observer that is stationary w.r.t. the clock T and the time from the point of view that is moving w.r.t. the clock T. Gamma is defined as the ratio between these two times. Gamma = T / T The distance between the mirrors is D and the moving point of view is moving with velocity V along the X axis. From the stationary point of view the time for the light to bounce between the mirrors is just D divided by the speed of light. T = D / C From the moving point of view the light traces out a diagonal path and a right triangle is needed to analyze this motion. The light path is the hypotenuse of the triangle and its length is the speed of light times the time it takes for the light to bounce between the mirrors, CT. The vertical leg of the triangle is the distance between the mirrors, D. The horizontal leg of the triangle is the distance the moving point of view moves in the time it takes for the light to bounce between the mirrors, VT. So from the Pythagorean theorem we have (CT)^2 = (VT)^2 + D^2 (CT)^2 (VT)^2 = D^2 T^2(C^2 V^2) = D^2 T^2 = D^2 / (C^2 V^2) T = D / sqrt(C^2 V^2) So Gamma = T / T Gamma = (D / sqrt(C^2 V^2)) / (D / C) Gamma = C / sqrt(C^2 V^2) Gamma = 1 / sqrt((C^2 V^2) / C^2) Gamma = 1 / sqrt(1 V^2 / C^2) The second way I stumbled on to when I tried to derive the LTE from an idea I had. The first step was to find transformation equations that embodied the principle of the constancy of the speed of light (PCSL). It was so long ago I dont remember how I did it but I came up with this: X = X VT T = T VX / C^2 So the reverse transform is X = X + VT T = T + VX / C^2 The next step was to see if these equations satisfy the principle of relativity (POR) In this case it means when the reverse transform is applied to the forward transform you should end up with X = X T = T However when I applied the reverse transform to the forward transform I got X = X(1 V^2 / C^2) T = T(1 V^2 / C^2) POR not satisfied. I realized if I could somehow get 1 / (1 V^2 / C^2) into the equations it would divide out. I also realized for the equations to satisfy the POR they had to be symmetrical, the forward and reverse equations have to have the same form. So I tried multiplying everything by 1 / sqrt(1 V^2 / C^2). This worked. The final equations are: X = (X VT) (1 / (1 V^2 / C^2)) T = (T VX / C^2) (1 / (1 V^2 / C^2)) X = (X + VT) (1 / (1 V^2 / C^2)) T = (T + VX / C^2) (1 / (1 V^2 / C^2)) These are of course the LTE. They embody the PCSL and satisfy the POR and gamma is 1 / sqrt(1 V^2 / C^2). From this I realized the role that gamma plays in the LTE is to reconcile the PCSL with the POR. Btw I didnt use c + v or c v here but you are wrong that the use of them (under certain circumstances) is self-contradictory
From: waldofj on 3 Jan 2010 15:53
On Jan 3, 3:08 pm, waldofj <wald...(a)verizon.net> wrote: > >news:ea47baf0-a8f9-4f58-935d-812452ab275d(a)o19g2000vbj.googlegroups.com.... > > On Dec 30, 7:36 am, "Androcles" <Headmas...(a)Hogwarts.physics_q> wrote: > > > > I don't care about physical explanations, there is no mathematical > > > explanation. Anyone handwaving their precious gamma around > > > needs to derive it first. > > > > gamma = sqrt[(c-v)(c+v)/c^2] > > > actually gamma is the reciprocal of that function and deriving it is > > easy. What's the big deal? > > ======================================== > > Go on then, show and tell. Just remember that the speed of light is > > c in all frames of reference and so you can't use c+v or c-v without > > being self-contradictory. > > Ill derive it two ways, the first is the way it was taught to me in > college, the second is a way I came up with myself. > First: compute gamma by analyzing a light clock, two mirrors with a > light beam bouncing between them. The mirrors are oriented so the > light is moving parallel to the Y axis and perpendicular to the X > axis. I will call the time it takes light to travel from one mirror to > the other from the point of view of an observer that is stationary > w.r.t. the clock T and the time from the point of view that is moving > w.r.t. the clock T. Gamma is defined as the ratio between these two > times. > Gamma = T / T > The distance between the mirrors is D and the moving point of view is > moving with velocity V along the X axis. > From the stationary point of view the time for the light to bounce > between the mirrors is just D divided by the speed of light. > T = D / C > From the moving point of view the light traces out a diagonal path and > a right triangle is needed to analyze this motion. The light path is > the hypotenuse of the triangle and its length is the speed of light > times the time it takes for the light to bounce between the mirrors, > CT. The vertical leg of the triangle is the distance between the > mirrors, D. The horizontal leg of the triangle is the distance the > moving point of view moves in the time it takes for the light to > bounce between the mirrors, VT. So from the Pythagorean theorem we > have > (CT)^2 = (VT)^2 + D^2 > (CT)^2 (VT)^2 = D^2 > T^2(C^2 V^2) = D^2 > T^2 = D^2 / (C^2 V^2) > T = D / sqrt(C^2 V^2) > So > Gamma = T / T > Gamma = (D / sqrt(C^2 V^2)) / (D / C) > Gamma = C / sqrt(C^2 V^2) > Gamma = 1 / sqrt((C^2 V^2) / C^2) > Gamma = 1 / sqrt(1 V^2 / C^2) > > The second way I stumbled on to when I tried to derive the LTE from an > idea I had. The first step was to find transformation equations that > embodied the principle of the constancy of the speed of light (PCSL). > It was so long ago I dont remember how I did it but I came up with > this: > X = X VT > T = T VX / C^2 > > So the reverse transform is > X = X + VT > T = T + VX / C^2 > The next step was to see if these equations satisfy the principle of > relativity (POR) > In this case it means when the reverse transform is applied to the > forward transform you should end up with > X = X > T = T > > However when I applied the reverse transform to the forward transform > I got > X = X(1 V^2 / C^2) > T = T(1 V^2 / C^2) > > POR not satisfied. I realized if I could somehow get 1 / (1 V^2 / > C^2) into the equations it would divide out. I also realized for the > equations to satisfy the POR they had to be symmetrical, the forward > and reverse equations have to have the same form. So I tried > multiplying everything by 1 / sqrt(1 V^2 / C^2). This worked. The > final equations are: > X = (X VT) (1 / (1 V^2 / C^2)) > T = (T VX / C^2) (1 / (1 V^2 / C^2)) > X = (X + VT) (1 / (1 V^2 / C^2)) > T = (T + VX / C^2) (1 / (1 V^2 / C^2)) > > These are of course the LTE. They embody the PCSL and satisfy the POR > and > gamma is 1 / sqrt(1 V^2 / C^2). > From this I realized the role that gamma plays in the LTE is to > reconcile the PCSL with the POR. > > Btw I didnt use c + v or c v here but you are wrong that the use of > them (under certain circumstances) is self-contradictory oops! somehow sqrt dropped out of my copy/paste. Those LTE's should of course be X = (X VT) (1 / sqrt(1 V^2 / C^2)) T = (T VX / C^2) sqrt(1 / (1 V^2 / C^2)) X = (X + VT) (1 / sqrt(1 V^2 / C^2)) T = (T + VX / C^2) sqrt(1 / (1 V^2 / C^2)) |