Moving Mirror Proves Einstein Wrong
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From: glird on 30 Dec 2009 10:00 On Dec 29, 3:14 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. > That is so easy that even Einstein showed how. If you really want to know how he did it, don't limit your request to his jock straps.
From: glird on 30 Dec 2009 10:48 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
From: YBM on 30 Dec 2009 11:20 waldofj a �crit : >>> p xor q => p or q, dumby? Explain us what this is not true. >> Easy as 1,2,3, 4. >> >> 1) (FALSE xor FALSE) = TRUE Ahem... From http://mathworld.wolfram.com/XOR.html The binary XOR operator has the following truth table. A B A xor B T T F T F T F T T F F F Look at the last line: FALSE xor FALSE = FALSE >> 2) (FALSE or FALSE) = FALSE >> 3) (TRUE(from 1) => FALSE(from 2) ) = NOT TRUE 3) (FALSE(from 1) => FALSE(from 2) ) = TRUE > just for the record: > 1) (TRUE xor TRUE) = FALSE > 2) (TRUE or TRUE) = TRUE > 3) (FALSE(from 1) => TRUE(from 2) ) = NOT TRUE Ahem... 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 Look at the third line: 3) FALSE (from 1) => TRUE (from 2) = TRUE
From: Androcles on 30 Dec 2009 12:22 "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. 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. Taking my umbrella when it doesn't rain is valid. The contrapositive (NOT B => NOT A) is: IF I do not take my umbrella THEN it will not rain. I can neither cause rain nor not cause rain by carrying an umbrella as Carnap and Mendelson seem to imagine.
From: waldofj on 30 Dec 2009 16:19
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? |