Why physicists refuse to measure the one-way speed of light directly?
in [Physics]
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From: glird on 2 Feb 2010 13:43 On Feb 1, 3:18 pm, PD wrote: > On Feb 1, 2:00 pm, glird wrote: > > > gl: You are right, Ken, but you are also wrong. Although the clocks WILL have different readings, they ARE "synchronized" as per Einstein's rather silly definition. > pd: What makes you think they will have different readings? > > gl: > > Did you read Einstein's demo re "the relativity of simultaneity"? If so, you know that he let the clocks of the moving system keep the same time per clock as those of a stationary one. > > > PD: Ah, so what you are saying is that clocks can be synchronized in frame A, and clocks can be synchronized in frame B, but frame B's clocks will not be synchronized in frame A and frame A's clocks will not be synchronized in frame B. And that's right. > > > > > That's why, as he wrote, the moving system will NOT measure c as a constant. > > > pd: Oh, yes it will. It is not necessary for frame B's clocks to be synchronized in frame A for frame B to measure the speed of light to still be c. > > > > gl: That's right; but its irrelevant wrt this discussion, in which -- as Einstein said, clocks of frame B ARE set identically to those of frame A. > > That means the same procedure is used. It does not mean they are set to read the same readings. > You either didn't read or didn't understand Einstein's paper. Or perhaps you don't understand simple English, as in "clocks of frame B ARE set identically to those of frame A"? (If you do, then how come you don't know the difference between "are set identically TO those of frame A" and (are set identically AS those of frame A), as in "the same procedure is used"? It's a shame that such an elementary misunderstanding has gotten in your way. > > gl: Why didn't you do what I suggested next? (Here is that suggestion, which you omitted) Now reset those clocks, via Einstein's defined method, so they DO measure c as a constant and you will see that they WILL have different readings than each other, as per Lorentz's Voigtian "local time" equation in which x, t and t' are co-ordinates of the same system and v is its velocity in Einstein's "empty space". glird
From: PD on 2 Feb 2010 14:02 On Feb 2, 12:43 pm, glird <gl...(a)aol.com> wrote: > On Feb 1, 3:18 pm, PD wrote:> On Feb 1, 2:00 pm, glird wrote: > > > > gl: You are right, Ken, but you are also wrong. > > Although the clocks WILL have different readings, they ARE > "synchronized" as per Einstein's rather silly definition. > > pd: What makes you think they will have different readings? > > > gl: > > Did you read Einstein's demo re "the relativity of > simultaneity"? If so, you know that he let the clocks of the moving > system keep the same time per clock as those of a stationary one. > > > > PD: Ah, so what you are saying is that clocks can be synchronized in > > frame A, and clocks can be synchronized in frame B, but frame B's > clocks will not be synchronized in frame A and frame A's clocks will > not be synchronized in frame B. And that's right. > > > > > > That's why, as he wrote, the moving system will NOT measure c as a constant. > > > > pd: Oh, yes it will. It is not necessary for frame B's clocks to be > > synchronized in frame A for frame B to measure the speed of light to > still be c. > > > > > gl: That's right; but its irrelevant wrt this discussion, in which -- as Einstein said, clocks of frame B ARE set identically to those of frame A. > > > That means the same procedure is used. It does not mean they are set > > to read the same readings. > > > You either didn't read or didn't understand Einstein's paper. Or > perhaps you don't understand simple English, as in "clocks of frame B > ARE set identically to those of frame A"? > (If you do, then how come you don't know the difference between > "are set identically TO those of frame A" and (are set identically AS > those of frame A), as in "the same procedure is used"? > It's a shame that such an elementary misunderstanding has gotten in > your way. He wrote in German. What are the prepositions used in the original? > > > > gl: Why didn't you do what I suggested next? (Here is that suggestion, which you omitted) > > Now reset those clocks, via Einstein's defined method, so they DO > measure c as a constant and you will see that they WILL have different > readings than each other, as per Lorentz's Voigtian "local time" > equation in which x, t and t' are co-ordinates of the same system and > v is its velocity in Einstein's "empty space". > > glird
From: glird on 3 Feb 2010 13:57 On Feb 2, 2:02 pm, PD <thedraperfam...(a)gmail.com> wrote: > On Feb 2, 12:43 pm, glird <gl...(a)aol.com> wrote: > > On Feb 1, 3:18 pm, PD wrote: On Feb 1, 2:00 pm, glird wrote: gl: You are right, Ken, but you are also wrong. Although the clocks WILL have different readings, they ARE "synchronized" as per Einstein's rather silly definition. pd: What makes you think they will have different readings? gl: Did you read Einstein's demo re "the relativity of simultaneity"? If so, you know that he let the clocks of the moving system keep the same time per clock as those of a stationary one. PD: Ah, so what you are saying is that clocks can be synchronized in frame A, and clocks can be synchronized in frame B, but frame B's clocks will not be synchronized in frame A and frame A's clocks will not be synchronized in frame B. And that's right. > > > > > > That's why, as he wrote, the moving system will NOT measure c as a constant. pd: Oh, yes it will. It is not necessary for frame B's clocks to be synchronized in frame A for frame B to measure the speed of light to still be c. gl: That's right; but its irrelevant wrt this discussion, in which -- as Einstein said, clocks of frame B are set identically to those of frame A. pd: That means the same procedure is used. It does not mean they are set to read the same readings. gl: You either didn't read or didn't understand Einstein's paper. Or perhaps you don't understand simple English, as in "clocks of frame B are set identically TO those of frame A"? (If you do, then how come you don't know the difference between "are set identically TO those of frame A" and (are set identically AS those of frame A), as in your "the same procedure is used"? It's a shame that such an elementary misunderstanding has gotten in your way. > He wrote in German. What are the prepositions used in the original? The original doesn't exist. (Eintein said he lost it.) Here is the relevant portion in English. < We imagine further that at the two ends A and B of the rod, clocks are placed which synchronize with the clocks of the stationary system, that is to say that their indications correspond at any instant to the ``time of the stationary system'' at the places where they happen to be. These clocks are therefore ``synchronous in the stationary system.'' > Note that "as Einstein said, clocks of frame B are set identically to those of frame A." Note further that "their indications correspond at any instant to the ``time of the stationary system'' at the places where they happen to be" does NOT mean that "the same procedure is used". It means that "they are set to read the same readings". In his next paragraph Einstein wrote: < We imagine further that with each clock there is a moving observer, and that these observers apply to both clocks the criterion established in § 1 for the synchronization of two clocks. Let a ray of light depart from A at the time tA , let it be reflected at B at the time tB, and reach A again at the time t'A. Taking into consideration the principle of the constancy of the velocity of light we find that tb - tA = rAB/(c-v) and t'A - tb = rAB/(c+v) where rAB denotes the length of the moving rod--measured in the stationary system. Observers moving with the moving rod would thus find that the two clocks were not synchronous, while observers in the stationary system would declare the clocks to be synchronous. > Instead of being bamboozled by the next sentence in E's paper, as most physicists are, why don't you do what I suggested next? (Here is that suggestion, which you repeatedly omitted): "Now reset those clocks via Einstein's defined method, so they DO measure c as a constant". If you do, you will see that clocks A and B of the moving system WILL have different readings than each other, as per Lorentz's Voigtian "local time" equation t' = t - vx/c^2 in which x, t and t' are co-ordinates of the same system and v is its velocity in Einstein's "empty space". In case anyone didn't understand what I mean by "x, t and t' are co- ordinates of the same system", here is an explanation: The time of clock A of frame A is t, and the time of clock B of frame A is t', and x is the distance between A and B; and all three are as plotted by frame A, thus are co-ordinates of the same system. glird
From: Rock Brentwood on 4 Feb 2010 21:30 On Jan 11, 9:23 am, Igor <thoov...(a)excite.com> wrote: > No, you've got it backwards. The speed of light in vacuum is a > universal constant and the meter is defined based on it and the time > standard. The correct statement is: the speed of light in a "vacuum" (i.e. a boost-invariant, rotation-invariant, translation-invariant medium) in a flat Minkowski space is a constant. Correct though this may be, it is not physically relevant for two main reasons: (1) there is no such thing as a vacuum, not even outer space, and (2) the cosmos is not flat, as a space-time geometry. The metric describing the universe is: ds^2 = dt^2 - 1/c(t)^2 (dx^2 + dy^2 + dz^2) where the speed of light c(t) is a function of time given (in a matter- dominant era) as c(t) = c (T/t)^{2/3} where T represents the current time and c represents the speed of light at time T. In the radiation-dominant era, it goes as t^{-1/2} and in a deSitter era (i.e. now) it goes EXPONENTIAL -- here exponentially decaying such that a light ray in comoving coordinates never gets beyond a distance of sqrt(Lambda/3) from its starting point. But in all cases: it's not constant. Now, it is true that by a change of coordinates, one can rewrite the metric in a conformally Minkowski form. In particular, taking tau = 3 c(T) T (t/T)^{1/3}, one has the following metric (for the dust- dominant era): ds^2 = (tau/(3T))^2 (dt^2 - 1/c^2 (dx^2 + dy^2 + dz^2)) but this assumption is flawed: it runs afoul precisely where the "King's Thumb" problem occurs -- where c -> infinity (i.e. where t -> 0 or tau -> 0). Plus, this is STILL not Minkowski geometry, but only conformally flat geometry. You still have the fact that the metric is time-dependent, which is just a fancy way of restating the very point made with the first form of the metric. Plus it turns a non-problem into a pseudo-problem. What was originally simply the fact that light speed reaches infinity at time 0 now becomes a seemingly intractible "singularity". In fact, the Weyl curvature is 0, it's only the field sources that approach infinity and the "curvature" that one refers to in the term "curvature singularity" actually has absolutely nothing to do with what one normally visualizes as geometric curvature. The "curvature" (i.e. the Ricci scalar) is just the c''/c + a multiple of (c'/c)^2 -- i.e., it's the "acceleration" of c. It goes to infinity because the characteristic surfaces at time 0 (i.e. the light cone) flatten out and become the simultaneity surfaces of non-relativistic spacetime. (For the radiation dominant era, it might even be a good exercise to integrate the geodesic equation ds^2 = 0, assuming c goes like t^{-1/2} all the way to time 0. You'll get a very interesting picture, which puts into proper perspective the so-called "horizon problem", showing that how this is actually a pseudo-problem). That you can have a metric with a variable c and with c -> infinity also shows the problem behind directly incorporating the theoretical axiom of spacetime signature into the DEFINITION of units: the moment you incorporate a theoretical axiom into the very system of units you use for carrying out the measurements use to test theory, you've just made it impossible to even TALK about the question of whether that axiom is true (at all points in space and time), never mind actually examining the question, itself, of the axiom's universal veracity. An empirical science is supposed to be falsifiable. That means, first and foremost, you must ALWAYS keep the language intact for counter- factuals, regardless of whether they are true or not! Because when you don't you've just introduced a hole in the rug where oversights can slip in unseen that you can't even address or see (because the conventions in place linguistically filter them out), much less address. One of the places oversights crop in with the flawed notational conventions (e.g. "set c = 1", or as the ISO says, "set c-second/meter = 299792458") is you lose meaningful and HIGHLY physically relevant discussion about such issues as the variation of the permittivity (plus, you get the apparent discrepancy above that 1 meter -> 0 as time t -> 0). In fact, the variability of both permeability and permittivity are precisely what the metric above imply -- not just for electromagnetism, but also the gauge-theoretic analogue of epsilon and mu for gauge fields. This can be seen easily and directly. Write down the potential 1-form A = *A*.dr - phi dt, *A* = (A_x, A_y, A_z), dr = (dx, dy, dz) and field 2 form F = B.dS + E.dr ^ dt, B = (B^x, B^y, B^z), dS = (dy^dz, dz^dx, dx^dy), E = (E_x, E_y, E_z). The Maxwell-Lorentz Lagrangian is given by L = -1/4 k root(|g|) g^{mr} g^{ns} F_{mn} F_{rs} (summation convention used) k = coupling coefficient with the component forms F = 1/2 F_{mn} dx^m ^ dx^n given by E = (F_{10}, F_{20}, F_{30}), B = (F_{23}, F_{31}, F_{12}) and g_{00} = 1, g_{i0} = 0 = g_{0j}, i, j = 1, 2, 3 g_{ij} = -1/c^2 delta_{ij}, i, j = 1, 2, 3 (delta_{ij} = 1 if i = j, 0 else). Then g = det(g) = -1/c^6, root(|g|) = 1/c^3 g^{00} = 1, g^{i0} = 0 = g^{0j}, g^{ij} = -c^2 delta^{ij} and L = 1/2 k/c (E^2 - c^2 B^2). The electric induction D = (D^x, D^y, D^z) = dL/dE and magnetic field strength H = (H_x, H_y, H_z) = -dL/dB are then determined to be D = k/c E, H = kc B. Hence epsilon(t) = k/c(t) = k/c (t/T)^{2/3}, mu(t) = 1/(kc(t)) = 1/(kc) (t/T)^{2/3}. This is more directly seen by simply writing out the Hodge-deRham operator for the above metric: Delta = -1/root(|g|) @_m (g^{mn} root(|g|) @_n (_)) @_m = d/dx^m. This expands out to Delta = -c^3 d_t (c^{-3} d_t (_)) + 1/c^2 del^2. Apply this operator to the potentials and you get the same equation as would be obtained by using the Lorenz gauge 1/root(|g|) @_m (g^{mn} root(|g|) A_n) = 0 for the potentials. All the above considerations apply generally to gauge fields, with the only modification being that the E and B fields now have an extra Lie index (E^a, B^a: a = 1, 2, .... dimension of underlying Lie group), and D and H have the indices in the lower position D_a, H_a, and epsilon and mu are now matrices (epsilon_{ab}, mu^{ab}), while k generalizes to the gauge group metric k_{ab} and 1/k to the inverse of the gauge group metric k^{ab}. Since one normally assumes the gauge theory is given with an adjoint- invariant metric, then for simple gauge groups, k reduces to a multiple k_{ab} = (1/g^2) kappa_{ab} of the Killing metric. So, the variability of epsilon translates into a variability of g -- which is now proportional to t^{-2/3}.
From: Jerry on 4 Feb 2010 23:45
On Feb 4, 8:30 pm, Rock Brentwood <markw...(a)yahoo.com> wrote: > An empirical science is supposed to be falsifiable. That means, first > and foremost, you must ALWAYS keep the language intact for counter- > factuals, regardless of whether they are true or not! Because when you > don't you've just introduced a hole in the rug where oversights can > slip in unseen that you can't even address or see (because the > conventions in place linguistically filter them out), much less > address. > > One of the places oversights crop in with the flawed notational > conventions (e.g. "set c = 1", or as the ISO says, "set c-second/meter > = 299792458") is you lose meaningful and HIGHLY physically relevant > discussion about such issues as the variation of the permittivity > (plus, you get the apparent discrepancy above that 1 meter -> 0 as > time t -> 0). In fact, the variability of both permeability and > permittivity are precisely what the metric above imply -- not just for > electromagnetism, but also the gauge-theoretic analogue of epsilon and > mu for gauge fields. I find the above line of argument to be unconvincing, perhaps even somewhat specious. Every choice of notational convention has at its basis a set of theoretical and/or practical assumptions, but the mere fact that one is necessarily forced to adopt one or another convention does -not- preclude investigation into the validity of one's choice. For example, the definition of the kilogram in terms of a material standard, the International Prototype Kilogram (IPK), has -not- precluded investigations on the stability of the IPK, which exhibits rapid, short term mass swings on the order of 30 ug when compared against the worldwide ensemble of prototype standards, and which been estimated to have lost on the order of 50 ug over the last century. Likewise, modern-day repetitions of the MMX are -not- interpreted as checks on the dimensional stability of the cryogenic optical resonators employed in the most precise of these recent tests, even though the ISO definition of the meter would seemingly preclude any possibility of measuring anisotropies in the speed of light. Jerry |