Simultaneity of Relativity
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From: PD on 13 Oct 2009 19:10 On Oct 13, 6:00 pm, mpc755 <mpc...(a)gmail.com> wrote: > On Oct 13, 6:52 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > Neither one. Did you want me to explain the Einstein gedanken to you? > > > > Just have to ask. > > > > How is it the Observers who adhere to Relativity of Simultaneity > > > cannot determine the lightning strikes in my thought experiment were > > > simultaneous but in Simultaneity of Relativity they can? > > > Observers in nature CAN determine if lightning strikes are > > simultaneous. It's just that this determination is frame-dependent. > > No, it is not frame-dependent. Yes, it is, experimentally so. Trying to tell nature that it really isn't what it is, is a futile exercise. It is called denial of reality and some people go so far as to call it psychosis. Just like it is *experimentally determined* that some physical properties are frame-dependent -- such as velocity, momentum, kinetic energy, electric field, magnetic field -- it is also *experimentally verified* that simultaneity is frame-dependent. Two events that are *experimentally* simultaneous in one frame are *experimentally* not simultaneous in another frame. Denial of experimental observations is a bad idea in science. > That is what is incorrect with > Relativity of Simultaneity. The Observers at M and M' in my thought > experiment, using Relativity of Simultaneity incorrectly conclude all > four flashes were not simultaneous because the Observer at M > arbitrarily and incorrectly concludes the light from the lightning > strike at A' traveled from where A' *was*. This is incorrect. Just > like the Observer at M' determines the light traveled from where A' > *is* when the light reaches M', the Observer at M, in Simultaneity of > Relativity, determines the light traveled from where A' *is* to where > M *is*, does the same for the distance the light travels from A, B, > and B' and correctly concludes all four lightning strikes occurred > simultaneously. > > > > > It's similar to being at rest. It's easy to determine if something is > > at rest. However, that determination is frame-dependent, because in > > another reference frame the same object is not at rest. This is > > something that Galileo understood well over 300 years ago, and school > > children ever since have too. Do you see this? > > > > Why does the light travel from where A' *is* to where M' *is* but the > > > light travels from where A' *was* to where M *is*? > > > It doesn't. > > > You haven't asked me to explain the Einstein gedanken yet. > And you still haven't. Are you interested in having it explained to you? Why is it emotionally difficult for you to ask for this? You seem to have emotional difficulties about a number of things.
From: glird on 13 Oct 2009 19:13 On Oct 13, 4:56 pm, mpc755 wrote: > An Observer exists on a line equi-distant between A and A'. He pushes > a button that allows flashes to occur at A and A' when A and A' are as > close to him as possible. The light from A and A' reaches the Observer > between A and A' simultaneously. "Simultaneously" is the key word. (See below.) > The same is true for an Observer between B and B'. Presumably, these two observers are at rest wrt each other. If not, then their esynched clocks at B and B' have different settings than each other. The following bit from A Flower for Einstein is relevant here: << In short, for any deformations whatsoever in the perpendicular axes, ø is a function of the velocity ±v only if both systems move at identical absolute velocities in opposite directions and only if plotted either by identically deformed each other or by a third system, stationary Euclidean cs S whose velocity determinations are absolute. Hence, and the second point of this temporarily trivial exercise, it is {and should always have been} self evident that the degree of physical deformation of moving systems is not a function of the relative velocities. Not even as found by a moving Euclidean system. Once the physical meaning of ø(v) and ø( v) are defined, it is clear that the inverse transformation, y = ø( v)y', has the same form as the direct transformation, y' = ø(v)y, simply because ø(v) is equal to 1/ø( v) even if neither of them is equal to 1. It is equally clear that if y' = ly then y = yâ/l, just as Lorentz accurately provided in some of his "wrong" 1904 equations. To gain reciprocity for two systems plotting each other's perpendicular co-ordinates, all we need do is recognize that in Lorentz, compared to Einstein, l = ø(v) and l = ø( v) and ø( v) = 1/ø(v) wherefore y' = ly = ø(v)y and y = ly' = y'/ø (v) = ø( v)y'. It is even easier to just define ø(v) as the inverse of the deformation value in the axes perpendicular to the direction of motion, as determined by the viewing system. (It is its own numerical inverse when the other system is the viewer.) Given that, and letting the primes always denote the viewed system, then all by itself and for all values of ø, y' = ø(v)y is both the direct and inverse transformation. What could be simpler? With one exception (case A, below) a change in unit rate is also required by relativistic equations. The easiest way to find the needed rate change is to check events on a perpendicular axis. We will therefore let rod AB be on the Y axis. Z results would be identical. (Because c' = q both ways on a perpendicular plane, no local time offsets have to be inserted perpendicularly to X or X'.) Again setting ø = q, 1/q, 1 in above cases a, b and c, and letting v = .6c upon X, we will now examine the consequent degree of rate change cs Kâ must insert in obeying the definition of sinchronism: tâB - tâA = t'Aâ - tâB = AB/c, in which tâA is the time the ray leaves A (as marked by clock A), tâB is the time it reaches B (as marked by clock B) and t'Aâ is the time it returns to A. If we set AB equal to one unit length as marked by Kâ, this becomes tâB - tâA = t'Aâ - tâB = 1. In the three cases, Y" remains a Euclidean cs but this time is attached to deformed relativistic cs k' just as in Lorentz's 1904 paper. The two attached systems thus co-move at .6c on X of cs S. (a) Y' unit rod AB is contracted to ø = q = .8 units long compared to those of Y" and S. At c' = q, it takes one second for a ray to traverse such a q shrunken unit length. Therefore, no rate change is needed since the ray travels "one unit" of cs kâ in one Y" and S second in the Y', Z' directions.. (b) Y' units are expanded to 1/q = 1.25 units long. At c' = q it takes 1.5625 seconds for the ray to reach B. The ratio of rates has to be set so that t'/t * 1.5625 = 1, hence t'/t = .64 = Q. Therefore, if ø = 1/q then t'/t = Q. (c) Y' units arenât deformed. It thus takes 1.25 seconds for a ray to reach B. Accordingly, t'/t * 1.25 = 1; so t'/t = q. Therefore, if ø = 1 then t'/t = q. This represents the LTE case. Note that this q rate change, imposed by setting ø(v) equal to unity as in the LTE, is totally independent of any differently moving system. Even co-moving Euclidean cs Y" will find a .8 slowdown in rates of cs kâ. Clearly, the degree of rate change is a function of the deformation in the perpendicular planes, and vice versa. Mathematically, neither of them can be derived by itself! To check sinchrony for cs kâ clocks situated in the direction of motion, we will use the degrees of length and rate deformation set forth in a,b,c, above; noting that the rate changes in the perpendicular axes remain valid with respect to the axis of motion. (Any k' clock in an X' direction also exists in the Y' and Z' directions, and can have only one rate. We will therefore assume that the rates were set in accord with those obtained above using the Y' rods. and that all clocks of k' initially have identical settings.) (a) Rates: X' unit rod AB is contracted to ï³ï = qø = Q = .64 S units long. No rate change was needed since the ray travels "two units" of cs kâ in two S seconds. Local time settings: At c' = c v = .4c a ray takes 1.6 seconds to traverse Q contracted AB. Hence t'B = 1.6. At c' = c+v = 1.6c it takes .4 seconds for the return leg, so t'A' = 2. In order for the clock at B to sinchronize with that at A, clock B has to be turned back by .6 seconds; which is equal to vx'/c2. Similarly, for clocks of cs kâ a local time lag ("offset" hereafter) of vx'/c2 seconds has to be inserted, where x' is the distance between any two of them as plotted by cs kâ itself. In terms of its own origin clock, successive clocks in the direction of absolute motion have readings such that t'o = t'x' vx'/c2; in which t'x' is the time of the clock at any x', t'o is the time of the origin clock and x' is the k' distance between the origin and the target clock, all exclusively in terms of cs kâ alone. >> > If the light from A and B reaches M simultaneously, does the light > from A' and B' reach M' simultaneously? Define "simultaneously". > When the four flashes of light reach the Observers at M and M', are > they able to determine the flashes occurred simultaneously? Define "simultaneously"!! > > In Simultaneity of Relativity, the light from A' reaches M and the > light from B reaches M' simultaneously and the Observers at M and M' > note the time and the distance A' and B are from them, respectively, > at the time they see the flashes of light and determine when the > flashes occurred. In STR, contrary to Wheeler, the length of a moving rod is different in the Y and Z directions than in the direction of motion, taken as X; and the rate at which clocks of a given system beat is a function of the amount of that deformation and of the velocity of the given system. The "time" per successive clock of any esynched system is different from that of any neighbors in the direction of motion by -vx/ c^2 seconds, where v is the speed of the system wrt the space in which light moves at c, and x is the distance between two such neighbors as measured by the given system itself. > The flashes of light from A and B reach M and the > flashes of light from A' and B' reach M'. The Observers at M and M' > note the time and determine how far the light has traveled from where > the source of the flash *is*. The Observers determine when the flashes > occurred. .. If you are discussing relativity, "when" the flashes occur and "how far" it travels is an abstraction, being a mathematical function of how fast the system is moving and how far apart the clocks that mark the time at A, B, M etc are. If you are discussing classical physics, the time and length is constant in all systems. > The light from B' reaches the Observer at M and the light > from A reaches the Observer at M' simultaneously. The Observers at M > and M' note the time and how far the light traveled from B' and A, > respectively, and determine when the flashes of light occurred. > The Observers at M and M' both conclude correctly the four flashes > occurred simultaneously. > > What is incorrect in the above description of Simultaneity of > Relativity? I think you are trying to compare two entirely different scenarios as though they were identical. > Explain what is incorrect and what the sequence of flashes > as determined by the Observers at M and M' is. Clocks set via Einstein's method MUST mark the time it takes a ray to get from any point A to any point B as equal to the time it takes for the return trip. If you really want to understand the latter portion of your sentence, do this: Let cs M be at rest in the luminiferous material that Einstein called "empty space". Let system M' be moving in the X direction at v = .8c. Let A and A' be at the coinciding origins at t = t' = 0 and M, M', B and B' on Y, and let AM = A'M' = MB = M'B'. Now plot how long it will take a ray to get from A to M or B and from B to M or A and then do it wrt the moving system. (Remember to let the path of the ray be on the hypotenuse of a right-angled triangle wrt system M'!) Let the "time" of system M' now run slower than that of the stationary system by exactly enough so that t' is identical with t for each leg of the light's trips. (If - as in RT - we hold vertical lengths unchanged, then clocks of M' must run slow by q = sqrt(1-v^2/c^2) Letting M' clocks run slow by q compared to those of M, now let lengths in M' contract by q, as in RT. At t' = 0 let a ray emit from A' and travel to M' and then on to B', where M' and B' are on X of cs M' thus on or parallel to X of M. Solve for the value of t' in both cases. Then let the ray reflect from B' and plot the time t it takes to get to M' and back to A'. Now find out how much you have to change the times t' in order for t' to be identicle for the trip from A' to B' as from B' to A'; i.e for the clocks to be set in accord with Einstein's (defective) definition of "synchronous" clocks. glird
From: mpc755 on 13 Oct 2009 19:26 On Oct 13, 7:10 pm, PD <thedraperfam...(a)gmail.com> wrote: > On Oct 13, 6:00 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > On Oct 13, 6:52 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > Neither one. Did you want me to explain the Einstein gedanken to you? > > > > > Just have to ask. > > > > > How is it the Observers who adhere to Relativity of Simultaneity > > > > cannot determine the lightning strikes in my thought experiment were > > > > simultaneous but in Simultaneity of Relativity they can? > > > > Observers in nature CAN determine if lightning strikes are > > > simultaneous. It's just that this determination is frame-dependent. > > > No, it is not frame-dependent. > > Yes, it is, experimentally so. Trying to tell nature that it really > isn't what it is, is a futile exercise. It is called denial of reality > and some people go so far as to call it psychosis. > > Just like it is *experimentally determined* that some physical > properties are frame-dependent -- such as velocity, momentum, kinetic > energy, electric field, magnetic field -- it is also *experimentally > verified* that simultaneity is frame-dependent. Two events that are > *experimentally* simultaneous in one frame are *experimentally* not > simultaneous in another frame. > Yes, velocity of objects is frame dependent, except for photons. It's not like throwing a softball on a train where you can be on the embankment and see the softball's origination point on the train and add the softballs velocity to the trains velocity and when you catch the softball on the embankment it is traveling at the sum of the velocities. Photons do not work that way. You do not 'see' the photon's origination point on the train in three dimensional space relative to where you are on the embankment. You 'see' the photon when it hits your eye and then it has traveled from where the source *is* to where you are on the embankment when the photon hits your eye and the photon has propagated as a wave at 'c' from the source to your eye. I know you will never understand this concept and I know you do not care to understand this concept because it is different than you were taught. There is no difference in the distance the photon travels if it winds up hitting an observer on the train in the eye, or the observer on the embankment leans into the train and pushes the observer on the train out of the way and has the photon hit the observer on the embankment in the eye. The photon simply travels from where the source *is* to where the destination *is* when the photon reaches its destination. > Denial of experimental observations is a bad idea in science. > > > > > That is what is incorrect with > > Relativity of Simultaneity. The Observers at M and M' in my thought > > experiment, using Relativity of Simultaneity incorrectly conclude all > > four flashes were not simultaneous because the Observer at M > > arbitrarily and incorrectly concludes the light from the lightning > > strike at A' traveled from where A' *was*. This is incorrect. Just > > like the Observer at M' determines the light traveled from where A' > > *is* when the light reaches M', the Observer at M, in Simultaneity of > > Relativity, determines the light traveled from where A' *is* to where > > M *is*, does the same for the distance the light travels from A, B, > > and B' and correctly concludes all four lightning strikes occurred > > simultaneously. > > > > It's similar to being at rest. It's easy to determine if something is > > > at rest. However, that determination is frame-dependent, because in > > > another reference frame the same object is not at rest. This is > > > something that Galileo understood well over 300 years ago, and school > > > children ever since have too. > > Do you see this? > > > > > > > Why does the light travel from where A' *is* to where M' *is* but the > > > > light travels from where A' *was* to where M *is*? > > > > It doesn't. > > > > You haven't asked me to explain the Einstein gedanken yet. > > And you still haven't. Are you interested in having it explained to > you? Why is it emotionally difficult for you to ask for this? You seem > to have emotional difficulties about a number of things. Because I understand Relativity of Simultaneity.
From: mpc755 on 13 Oct 2009 19:33 On Oct 13, 7:13 pm, glird <gl...(a)aol.com> wrote: > On Oct 13, 4:56 pm, mpc755 wrote: > > > An Observer exists on a line equi-distant between A and A'. He pushes > > a button that allows flashes to occur at A and A' when A and A' are as > > close to him as possible. The light from A and A' reaches the Observer > > between A and A' simultaneously. > > "Simultaneously" is the key word.  (See below.) > > > The same is true for an Observer between B and B'. > > Presumably, these two observers are at rest wrt each other. If not, > then their esynched clocks at B and B' have different settings than > each other. The following bit from A Flower for Einstein is relevant > here: > << In short, for any deformations whatsoever in the perpendicular > axes, ø is a function of the velocity ±v only if both systems move at > identical absolute velocities in opposite directions and only if > plotted either by identically deformed each other or by a third > system, stationary Euclidean cs S whose velocity determinations are > absolute.  Hence, and the second point of this temporarily trivial > exercise, it is {and should always have been} self evident that the > degree of physical deformation of moving systems is not a function of > the relative velocities. Not even as found by a moving Euclidean > system. > Once the physical meaning of ø(v) and ø( v) are defined, it is clear > that the inverse transformation,  y = ø( v)y', has the same form as > the direct transformation, y' = ø(v)y, simply because ø(v) is equal to > 1/ø( v) even if neither of them is equal to 1. It is equally clear > that if y' = ly then y = yâ/l, just as Lorentz accurately provided in > some of his "wrong" 1904 equations. > To gain reciprocity for two systems plotting each other's > perpendicular co-ordinates, all we need do is recognize that in > Lorentz, compared to Einstein, > l = ø(v) and l = ø( v) and ø( v) = 1/ø(v) > wherefore              y' = ly = ø(v)y and y = ly' = y'/ø > (v) = ø( v)y'. > It is even easier to just define ø(v) as the inverse of the > deformation value in the axes perpendicular to the direction of > motion, as determined by the viewing system. (It is its own numerical > inverse when the other system is the viewer.)  Given that, and letting > the primes always denote the viewed system, then all by itself and for > all values of ø, y' = ø(v)y is both the direct and inverse > transformation.  What could be simpler? > With one exception (case A, below) a change in unit rate is also > required by relativistic equations. The easiest way to find the needed > rate change is to check events on a perpendicular axis. We will > therefore let rod AB be on the Y axis. Z results would be identical. > (Because c' = q both ways on a perpendicular plane, no local time > offsets have to be inserted perpendicularly to X or X'.) > Again setting ø = q, 1/q, 1 in above cases a, b and c, and letting v > = .6c upon X, we will now examine the consequent degree of rate change > cs Kâ must insert in obeying the definition of sinchronism: > tâB - tâA = t'Aâ - tâB = AB/c, > in which tâA is the time the ray leaves A (as marked by clock A), tâB > is the time it reaches B (as marked by clock B) and t'Aâ is the time > it returns to A. If we set AB equal to one unit length as marked by > Kâ, this becomes tâB - tâA = t'Aâ - tâB = 1.  In the three cases, Y" > remains a Euclidean cs but this time is attached to deformed > relativistic cs k' just as in Lorentz's 1904 paper. The two attached > systems thus co-move at .6c on X of cs S. > (a)  Y' unit rod AB is contracted to ø = q = .8 units long compared > to those of Y" and S. > At c' = q, it takes one second for a ray to traverse such a q shrunken > unit length. Therefore, no rate change is needed since the ray travels > "one unit" of cs kâ in one Y" and S second in the Y', Z' directions. > (b)  Y' units are expanded to 1/q = 1.25 units long.  At c' = q it > takes 1.5625 seconds for the ray to reach B. The ratio of rates has to > be set so that t'/t * 1.5625 = 1, hence t'/t = .64 = Q. Therefore, if > ø = 1/q then t'/t = Q. > (c)  Y' units arenât deformed.  It thus takes 1.25 seconds for a ray > to reach B.  Accordingly, t'/t * 1.25 = 1; so t'/t = q. Therefore, if > ø = 1 then t'/t = q.  This represents the LTE case. > Note that this q rate change, imposed by setting ø(v) equal to unity > as in the LTE, is totally independent of any differently moving > system. Even co-moving Euclidean cs Y" will find a .8 slowdown in > rates of cs kâ. > Clearly, the degree of rate change is a function of the deformation in > the perpendicular planes, and vice versa. Mathematically, neither of > them can be derived by itself! > To check sinchrony for cs kâ clocks situated in the direction of > motion, we will use the degrees of length and rate deformation set > forth in a,b,c, above; noting that the rate changes in the > perpendicular axes remain valid with respect to the axis of motion. > (Any k' clock in an X' direction also exists in the Y' and Z' > directions, and can have only one rate. We will therefore assume that > the rates were set in accord with those obtained above using the Y' > rods. and that all clocks of k' initially have identical settings.) > (a)  Rates: X' unit rod AB is contracted to ï³ï = qø = Q = .64 S units > long. No rate change was needed since the ray travels "two units" of > cs kâ in two S seconds. > Local time settings: At c' = c v = .4c a ray takes 1.6 seconds to > traverse Q contracted AB.  Hence t'B = 1.6.  At c' = c+v = 1.6c it > takes .4 seconds for the return leg, so t'A' = 2. In order for the > clock at B to sinchronize with that at A, clock B has to be turned > back by .6 seconds; which is equal to  vx'/c2. Similarly, for clocks > of cs kâ a local time lag ("offset" hereafter) of  vx'/c2 seconds has > to be inserted, where x' is the distance between any two of them as > plotted by cs kâ itself.  In terms of its own origin clock, successive > clocks in the direction of absolute motion have readings such that t'o > = t'x'  vx'/c2; in which t'x' is the time of the clock at any x', t'o > is the time of the origin clock and x' is the k' distance between the > origin and the target clock, all exclusively in terms of cs kâ alone. > > > > > If the light from A and B reaches M simultaneously, does the light > > from A' and B' reach M' simultaneously? > > Define "simultaneously". > > > When the four flashes of light reach the Observers at M and M', are > > they able to determine the flashes occurred simultaneously? > > Define "simultaneously"!! > > > > > In Simultaneity of Relativity, the light from A' reaches M and the > > light from B reaches M' simultaneously and the Observers at M and M' > > note the time and the distance A' and B are from them, respectively, > > at the time they see the flashes of light and determine when the > > flashes occurred. > > In STR, contrary to Wheeler, the length of a moving rod is different > in the Y and Z directions than in the direction of motion, taken as X; > and the rate at which clocks of a given system beat is a function of > the amount of that deformation and of the velocity of the given > system.  The "time" per successive clock of any esynched system is > different from that of any neighbors in the direction of motion by -vx/ > c^2 seconds, where v is the speed of the system wrt the space in which > light moves at c, and x is the distance between two such neighbors as > measured by the given system itself. > > > The flashes of light from A and B reach M and the > > flashes of light from A' and B' reach M'. The Observers at M and M' > > note the time and determine how far the light has traveled from where > > the source of the flash *is*. The Observers determine when the flashes > > occurred. > > .  If you are discussing relativity, "when" the flashes occur and "how > far" it travels is an abstraction, being a mathematical function of > how fast the system is moving and how far apart the clocks that mark > the time at A, B, M etc are.  If you are discussing classical physics, > the time and length is constant in all systems. > > > The light from B' reaches the Observer at M and the light > > from A reaches the Observer at M' simultaneously. The Observers at M > > and M' note the time and how far the light traveled from B' and A, > > respectively, and determine when the flashes of light occurred. > > The Observers at M and M' both conclude correctly the four flashes > > occurred simultaneously. > > > What is incorrect in the above description of Simultaneity of > > Relativity? > > I think you are trying to compare two entirely different scenarios as > though they were identical. > > >  Explain what is incorrect and what the sequence of flashes > > as determined by the Observers at M and M' is. > >   Clocks set via Einstein's method MUST mark the time it takes a ray > to get from any point A to any point B as equal to the time it takes > for the return trip.  If you really want to understand the latter > portion of your sentence, do this: > Let cs M be at rest in the luminiferous material that Einstein called > "empty space". Let system M' be moving in the X direction at v = .8c. > Let A and A' be at the coinciding origins at t = t' = 0 and M, M', B > and B' on Y, and let AM = A'M' = MB = M'B'. >  Now plot how long it will take a ray to get from A to M or B and > from B to M or A and then do it wrt the moving system. (Remember to > let the path of the ray be on the hypotenuse of a right-angled > triangle wrt system M'!)  Let the "time" of system M' now run slower > than that of the stationary system by exactly enough so that t' is > identical with t for each leg of the light's trips. (If - as in RT - > we hold vertical lengths unchanged, then clocks of M' must run slow by > q = sqrt(1-v^2/c^2) >  Letting M' clocks run slow by q compared to those of M, now let > lengths in M' contract by q, as in RT.  At t' = 0 let a ray emit from > A' and travel to M' and then on to B', where M' and B' are on X of cs > M' thus on or parallel to X of M.  Solve for the value of t' in both > cases.  Then let the ray reflect from B' and plot the time t it takes > to get to M' and back to A'.  Now find out how much you have to change > the times t' in order for t' to be identicle for the trip from A' to > B' as from B' to A'; i.e for the clocks to be set in accord with > Einstein's (defective) definition of "synchronous" clocks. > > glird The clocks at M and M' run at the same rate. There is no length contraction. The aether is stationary in the train frame of reference and in the embankment frame of reference. When the light wave from the flash at A' reaches M the flash from the light wave at B reaches M'. When the light from the flashes at A and B reach M the light from the flashes at A' and B' reach M'. When the light from the flash at B' reaches M, the light from the flash at A reaches M'. All of the light waves travel from where the source *is* to where the destination *is* when the light from the flash reaches the destination.
From: PD on 13 Oct 2009 19:36
On Oct 13, 6:26 pm, mpc755 <mpc...(a)gmail.com> wrote: > On Oct 13, 7:10 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > On Oct 13, 6:00 pm, mpc755 <mpc...(a)gmail.com> wrote: > > > > On Oct 13, 6:52 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > Neither one. Did you want me to explain the Einstein gedanken to you? > > > > > > Just have to ask. > > > > > > How is it the Observers who adhere to Relativity of Simultaneity > > > > > cannot determine the lightning strikes in my thought experiment were > > > > > simultaneous but in Simultaneity of Relativity they can? > > > > > Observers in nature CAN determine if lightning strikes are > > > > simultaneous. It's just that this determination is frame-dependent. > > > > No, it is not frame-dependent. > > > Yes, it is, experimentally so. Trying to tell nature that it really > > isn't what it is, is a futile exercise. It is called denial of reality > > and some people go so far as to call it psychosis. > > > Just like it is *experimentally determined* that some physical > > properties are frame-dependent -- such as velocity, momentum, kinetic > > energy, electric field, magnetic field -- it is also *experimentally > > verified* that simultaneity is frame-dependent. Two events that are > > *experimentally* simultaneous in one frame are *experimentally* not > > simultaneous in another frame. > > Yes, velocity of objects is frame dependent, except for photons. It's > not like throwing a softball on a train where you can be on the > embankment and see the softball's origination point on the train and > add the softballs velocity to the trains velocity and when you catch > the softball on the embankment it is traveling at the sum of the > velocities. And actually for softballs, it is not traveling at the sum of the velocities, either, though the sum is a pretty good approximation. > > Photons do not work that way. You do not 'see' the photon's > origination point on the train in three dimensional space relative to > where you are on the embankment. You 'see' the photon when it hits > your eye and then it has traveled from where the source *is* to where > you are on the embankment when the photon hits your eye and the photon > has propagated as a wave at 'c' from the source to your eye. Photons travel from the event where they were generated to the place where they are received. They do not travel from the location where the source eventually is when the photon is received. The source may have been destroyed by that time, or it may have turned around and gone the other direction by that time. The photon has no idea where the source is going to end up by the time the photon is eventually absorbed. This is *also* experimentally confirmed. > > I know you will never understand this concept and I know you do not > care to understand this concept because it is different than you were > taught. I was taught how to find out about the experimental tests that have been done for various ideas. And I was taught how to conduct a fair number of these tests myself. It's awfully hard to argue with what's going on when you're staring at it with your own eyes. > > There is no difference in the distance the photon travels if it winds > up hitting an observer on the train in the eye, or the observer on the > embankment leans into the train and pushes the observer on the train > out of the way and has the photon hit the observer on the embankment > in the eye. The photon simply travels from where the source *is* to > where the destination *is* when the photon reaches its destination. > > > > > > > Denial of experimental observations is a bad idea in science. > > > > That is what is incorrect with > > > Relativity of Simultaneity. The Observers at M and M' in my thought > > > experiment, using Relativity of Simultaneity incorrectly conclude all > > > four flashes were not simultaneous because the Observer at M > > > arbitrarily and incorrectly concludes the light from the lightning > > > strike at A' traveled from where A' *was*. This is incorrect. Just > > > like the Observer at M' determines the light traveled from where A' > > > *is* when the light reaches M', the Observer at M, in Simultaneity of > > > Relativity, determines the light traveled from where A' *is* to where > > > M *is*, does the same for the distance the light travels from A, B, > > > and B' and correctly concludes all four lightning strikes occurred > > > simultaneously. > > > > > It's similar to being at rest. It's easy to determine if something is > > > > at rest. However, that determination is frame-dependent, because in > > > > another reference frame the same object is not at rest. This is > > > > something that Galileo understood well over 300 years ago, and school > > > > children ever since have too. > > > Do you see this? > > > > > > Why does the light travel from where A' *is* to where M' *is* but the > > > > > light travels from where A' *was* to where M *is*? > > > > > It doesn't. > > > > > You haven't asked me to explain the Einstein gedanken yet. > > > And you still haven't. Are you interested in having it explained to > > you? Why is it emotionally difficult for you to ask for this? You seem > > to have emotional difficulties about a number of things. > > Because I understand Relativity of Simultaneity. Apparently not. You've made a number of errors so far in trying to describe it. Do you know the definition of simultaneity for two spatially separated events? |