Moving Mirror Proves Einstein Wrong
in [Relativity]
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From: waldofj on 30 Dec 2009 16:33 > We have a situation whereby the travel times for a light pulse to go from > points A to B and from B to A are UNEQUAL in the ground frame yet EQUAL in the > mirror frame. correct. And like I said this a consequence of the principle of the constancy of the speed of light. In other words, this is what must be for the speed of light to be the same constant value in both frames. > This directly violates Einstein's first postulate, which says that the laws of > physics are the same in ALL FRAMES. No. travel time is not a physical law. Not only is there no physical law that says travel time has to be the same in all frames there is a specific physical law (the principle of the constancy of the speed of light) that says they must be different in different frames. > P1 insists that two entities which are the same in one frame must be the same > in ALL frames whether or not they have different magnitudes in the different > frames. no it doesn't. Again, "entities" are not physical laws. And how can two entities be the same and have different magnitudes? What do you mean by entity?
From: Androcles on 30 Dec 2009 16:59 "waldofj" <waldofj(a)verizon.net> wrote in message news:ea47baf0-a8f9-4f58-935d-812452ab275d(a)o19g2000vbj.googlegroups.com... On Dec 30, 7:36 am, "Androcles" <Headmas...(a)Hogwarts.physics_q> wrote: > "waldofj" <wald...(a)verizon.net> wrote in message > > news:ee0312b4-f009-44c3-b06f-4a051a24d103(a)d20g2000yqh.googlegroups.com... > > > > > > >> The task for relativists is to show how the pulse travel times between > >> the > >> clocks can be the same in the mirror frame yet different in the source > >> frame . > > > you come close to answering your question with this paragraph. > > >> In the source frame, the pulse's travel time between C1 and C2 on the > >> forward > >> trip is d/(c+v). The clocks NUMERICALLY read t0 and t1 as the light > >> passes. > >> After reflection, its travel time between C2 and C1 is d/(c-v) and the > >> clocks > >> record the numbers t1 and t2. > > Although that should be d / gamma rather then just d. > > > what makes this work is according to Lorentz the clocks C1 and C2 are > > NOT synchronized in the source frame. Specifically C1 will lag C2 by > > an offset determined by both d and v. > > So what happens is this: > > As seen from the source frame light takes a time d / gamma / (c + v) > > to go from C1 to C2. This time is less than d / c but because C2 is > > more advanced than C1 it records a later (larger) value for t1 by just > > the right amount such that d / (t1 - t0) = c. > > As seen from the source frame light takes d / gamma / (c - v) to go > > from C2 to C1. This time is greater than d / c but because C1 lags C2 > > it records an earlier (smaller) value for t2 by just the right amount > > such that d / (t2 - t1) = c. > > Don't bother looking for a physical explanation of this effect (clocks > > synchronized in one frame are not synchronized in other frames). There > > is no explanation, this is just what has to be if the principle of the > > constancy of the speed of light is correct. > > Remember this is not proof of anything, it's just a description of > > what SR says about this scenario. > > If you want proof, well: > >http://www.cottonexpressions.com/ccp0-prodshow/YouWantProof.html > > 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] > > Chess boards have 63 squares and squares have 6 sides in relativity. > http://www.androcles01.pwp.blueyonder.co.uk/MC2.htm > > Don't bother looking for a physical explanation of this effect, the board > area is (8-1) * (8+1) = 63. 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. All triangles are isosceles. http://www.jimloy.com/geometry/every.htm 1892 is just the right time period for a boy like Einstein without a Playstation or TV to get interested in mathematical curiosities in magazines. So he said all rectangles are squares, I'll prove it. And the dumbos bought it, the greatest huckster of all time is still going strong even after he's been dead 54 years.
From: PD on 30 Dec 2009 18:20 On Dec 29, 2:14 pm, ..@..(Henry Wilson DSc) wrote: > This experiment involves a light source and a mirror. > > The source emits pulses of light towards the mirror, which is moving at v > towards it, as measured in the source frame. > > According to SR, both the incoming and reflected pulses move at c in the source > frame. They 'close on' the mirror at c+v and leave it at c-v. > > Can any Einstein supporter show the world how, under these circumstances, the > incident and reflected speeds of the pulses can be equal when measured in the > mirror frame. > Yes. In the frame of the source, the speed of light is c as it approaches the mirror. If you want to know what the speed of this same signal is in the mirror frame, you do the appropriate velocity transform: (c+v)/(1+cv/c^2) = c(1+v/c)/(1+v/c) = c. In the frame of the source, the speed of light is c as it departs form the mirror. If you want to know what the speed of this same signal is in the mirror frame, you do the appropriate velocity transform: (c-v)/(1-cv/c^2) = c(1-v/c)/(1-v/c) = c. Now, I don't know why you expected that if the closing speed is c+v in one frame, then the closing speed should be c+v in any other frame.
From: Inertial on 30 Dec 2009 19:18 "Henry Wilson DSc" <..@..> wrote in message news:0venj5d0taf7ij9m7ipmkffu10gu6out8k(a)4ax.com... > On Wed, 30 Dec 2009 07:48:04 -0800 (PST), glird <glird(a)aol.com> wrote: > >>On Dec 29, 11:12 pm, ..@..(Henry Wilson DSc) wrote: >> >>>< A light source emits pulses of light towards two clocks and a mirror >>>that are moving towards the source at v. The clocks are rigidly joined >>>and the mirror is connected to C2. In the source frame, they appear as >>>follows: >> S---------------------------------------------------- >> v<--C1___d___ |M,C2 >> The clocks are E-synched in the mirror frame. That is, their readings >>and rates are made the same by sending time signals from each clock >>towards the other and adjusting one accordingly. > >> >> Very good, Henri!! Having gone that far, take the next step and you >>will have answered your own problem. >> >> >>> <According to SR, all light moves at c in the source frame. >> In the experiment, a light pulse is emitted by S towards mirror, M. >>The two clocks register their readings as the pulse passes them. >> C1 reads t0 when the pulse initially passes by. C2 reads t1 when the >>pulse reaches the mirror. C1 reads t2 when the reflected pulse passes >>on the return trip. >> In the source frame, the pulse's travel time between C1 and C2 on the >>forward trip is d/(c+v). The clocks NUMERICALLY read t0 and t1 as the >>light passes. After reflection, its travel time between C2 and C1 is >>d/(c-v) and the clocks record the numbers t1 and t2. >> According to SR, the pulse should also move at c in the mirror frame >>both before and after reflection. In that frame, the source is moving >>towards the mirror at v and the pulse will therefore take the same >>time to travel between the two clocks in both directions (for both SR >>and BaTh) {???}. Therefore, numerically, t1-t0 should equal t2-t1. >>(The value should be d/c according to SR, d/(c+v) for BaTh.) >> The task for relativists is to show how the pulse travel times >>between the clocks can be the same in the mirror frame yet different >>in the source frame. > >> >> You already did that, so you must be a jock strip;;;'' >>(See above, for how "The clocks are E-synched in the mirror frame".) >> >>glird > > You are all missing the point. > > We have a situation whereby the travel times for a light pulse to go from > points A to B and from B to A are UNEQUAL in the ground frame yet EQUAL in > the > mirror frame. Yeup > This directly violates Einstein's first postulate, which says that the > laws of > physics are the same in ALL FRAMES. Nope > P1 insists that two entities which are the same in one frame must be the > same > in ALL frames whether or not they have different magnitudes in the > different > frames. Nope > So P2 leads to a violation of P1....and since P2 is a consequence of P1 we > can > only conclude that the whole theory of Einstein is a bloody big joke.. Nope You bombed out again Henry.
From: Inertial on 30 Dec 2009 19:21
"Androcles" <Headmaster(a)Hogwarts.physics_q> wrote in message news:K%L_m.4731$Ma2.4033(a)newsfe04.ams2... > > "YBM" <ybmess(a)nooos.fr.invalid> wrote in message > news:4b3b7ddf$0$24767$426a74cc(a)news.free.fr... > >> From: http://mathworld.wolfram.com/Implies.html >> A=>B has the following truth table (Carnap 1958, p. 10; Mendelson 1997, >> p. 13). >> A B A=>B >> T T T >> T F F >> F T T >> F F T > > > For the record, "implies" is "if A then B" > or > if A = TRUE then B = TRUE. A naive explanation .. but close > If it rains I'll take my umbrella. > IF "it rains" = TRUE THEN "I'll take my umbrella" = TRUE. > > "It rains" (A) "I'll take my umbrella" (B) A=>B > T T T > T F F > F T undefined. > F F undefined. No .. those 'undefined' values should be 'T's. There is no 'undefined' value in any boolean operators. Anyone with knowledge of boolean logic would know that. |