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From: paparios on 4 Jan 2010 09:49 On 4 ene, 11:27, "papar...(a)gmail.com" <papar...(a)gmail.com> wrote: > On 4 ene, 10:48, mpc755 <mpc...(a)gmail.com> wrote: Let us now introduce Special Relativity. Let frame K be defined by coordinates (x, ct) and frame K be defined by coordinates (x, ct). Coordinates (x, ct) of frame K and (x, ct) of frame K are related among them through the Lorentz Equations in the following way: x = gamma (x+ vt) ; t = gamma (t+ vx/c^2) x' = gamma (x - vt) ; t = gamma (t - vx/c^2) where gamma = 1 / sqrt(1-v^2/c^2) = 1.25 How can we draw these relationships? Well we already did the drawing of coordinates (x, ct), which we repeat here: ct x=ct | * | * | * ___________|*_________ x *| * | * | * | Now, coordinates (x, ct) look as a rotation of the coordinates (x, ct), as follows: ct' x'=ct' x' / * . / * . / * . /*. * .*/ . * / . * / * / Note how the constant relationship between these two graphs is the 45 degree line which denotes the relation x = ct (on frame K) and equation x = ct (on frame K). What happens, now in the train frame K, with the same lightning strike of location A, at time t=0? Coordinates of location A in frame K, are given by Lorentz as: x_A = x' = gamma (x - vt) = gamma (x_A - vt) So with t = 0, x_A = 100000km, gamma = 1.25, we have x_A = 125000km . This result also shows that, as observed in the embankment frame K, the train is seen contracted in length (distance AM on the train frame is 125000 km, but distance AM on the embankment frame is 100000km). t_A = t = gamma (t - vx/c^2) = -gamma (vx_A/ c^2) = 0.25sec Light signal moves, in frame K, according to equation: (x_M x_A) / (t_M t_A) = c Where x_M = 0 at the train observer location. Therefore we have, as measured on frame K: t_M = ( x_A + ct_A)/c = (125000 + 75000)/300000 = 0.667sec (1) What happens, now in the train frame K, with the same lightning strike of location B, at time t=0? Coordinates of location B in frame K, are given by Lorentz as: x_B = x' = gamma (x - vt) = gamma (x_B - vt) So with t = 0, x_B = +100000km, gamma = 1.25, we have x_B = +125000km t_B = t = gamma (t - vx/c^2) = -gamma (vx_B/ c^2) = 0.25sec Light signal moves, in frame K, from point B towards M, according to equation: (x_M x_B) / (t_M t_B) = c Where x_M = 0 at the train observer location. Therefore we have, measured on frame K: t_M = (x_B ct_B)/c = (125000 75000)/300000 = 0.167sec (2) Therefore, for the train observer M, as shown in equations (1) and (2) the strikes ARE NOT simultaneous. Miguel Rios
From: mpc755 on 4 Jan 2010 09:49 On Jan 4, 9:27 am, "papar...(a)gmail.com" <papar...(a)gmail.com> wrote: > On 4 ene, 10:48, mpc755 <mpc...(a)gmail.com> wrote: > > > > > On Dec 30 2009, 4:09 pm, moro...(a)world.std.spaamtrap.com (Michael > > > The Observer knows the water is at rest with respect to the > > embankment. The Observer knows the light is traveling at 0.75c > > relative to the water at rest with respect to the embankment. The > > Observer knows the train is moving at 0.25c relative to the water at > > rest with respect to the embankment. The Observer at M' notes the time > > on the clock at M' when the light from the lightning strike at B/B' > > arrives at M'. Based on the light propagating at 0.8421c relative to > > the train from B' towards M' and the mark made by the lightning strike > > at B' one year from M', the Observer at M' concludes the lightning > > strike at B/B' occurred 0.75c from where M' is relative to the water > > when the light from the lightning strike at B/B' arrived at M'. Since > > light propagates at 0.75c in stationary water, the Observer at M' > > concludes the lightning strike at B/B' occurred one year prior to the > > light arriving at M'. The light from the lightning strike at A/A' > > arrives at M'. Based on the light propagating at 0.6154c relative to > > the train from A' towards M' and the mark made by the lightning strike > > at A' one year from M', the Observer at M' concludes the lightning > > strike at A/A' occurred 1.5c from where M' is relative to the water > > when the light from the lightning strike at A/A' arrived at M'. Since > > light propagates at 0.75c in stationary water, the Observer at M' > > concludes the lightning strike at A/A' occurred two years prior to the > > light arriving at M'. Since the clock at M' notes one year has passed > > between the light from B/B' arriving at M' and the light from A/A' > > arriving at M', the Observer at M' concludes the lightning strikes > > were simultaneous, in nature. > > What matters here is what a theory says it will be observed (and > measured). In this respect Special Relativity has once and again > proved itself as correct. > > Let us consider Einstein gedanken as set inwww.bartleby.com/173/9.html. > Frame K, which is the embankment, is defined by coordinates (x,ct) and > there is an observer M initially is located at coordinates (0,0). > There are also two points A and B (where the striking events occur), > which initially have the following coordinates: > > Point A: (x_A, ct_A) = (-100000, 0) > Point B: (x_B, ct_B) = (+100000, 0) > > Where distances are expressed in kilometers and c = 300000 km/sec. > This is shown in the following figure: > > ct > | > | > ___________|__________x > A M B > | > | > > Let us first consider only the embankment frame. We know observer M > does not move in this frame, so his coordinate world line is defined > by (0, ct) , which is the vertical line shown in the figure. > > Now we consider the train. This train is very long and moving at a > very high speed v = 0.6c = 180000 km/sec towards the positive > direction of the x axis. The following figure shows this train, as > viewed from the embankment frame: > > t > | > | > | > A'_________M'________B'--> v > ___________|__________ x > A M B > > The figure shows the initial location, as Einstein explains just > when the flashes of lightning occur, this point M naturally coincides > with the point M . We also see from the figure that points A and B > coincide with points A and B, on frame K. > > We will now consider the propagation of light signals on this > embankment scenario. If observer M flashes a light at t=0, this light > signal will propagate towards the right of the figure (the x axis) at > a speed c. This is shown in the following figure, as a 45 degree line, > denoting the equation of movement of that light signal x = ct. > > ct x=ct > | * > | * > | * > ___________|*_________ x > M > > So, in one year time (t = 1 year), we will have that the light signal > has advanced to x = ct = 1 light year distance. > > Let us now consider the lightning strikes occurring at points A/A and > B/B. As shown in the following figure, light signals from points A > and B propagate towards observer M at a speed c and, consequently, > they can be drawn as the diagonal lines shown: > > ct > | > *|* > * | * > * | * > * | * > * | * > *__________|__________*__ x > A |M B > | > > The lines describe the equations of movement of the light signals (for > point A it is x = ct 100000. For point B it is x = -ct + 100000). > Both signals will arrive to the location of observer M at ct = 100000 > km, that is at t = 1/3 sec, as measured by observer M clock. > > Now, let us consider the train (again viewed from the embankment frame > K). Suppose we have previously located several helper observers along > the embankment (every km there is one of these helpers), and each of > these helpers have a synchronized clock, with respect to the one of > observer M. > > Just when M observes the light signals (at t = 1/3 sec), we can affirm > the following will also be accounted: > > a)Helper H1, located at x = 60000 km, will record that he saw observer > M passing through his location at t=1/3 sec. This is clearly correct, > since as the train is moving at v = 180000 km/sec, observer M will > have move from coordinate x = 0 to coordinate x = vt = 180000 (1/3) = > 60000 km. > b)Helper H1 will also point out that he previously saw a light signal > coming from point B at t = 40000/300000 = 0.1333 sec. > c)Helper H1 will later record that he saw a light signal coming from > point A at t = 160000/300000 = 0.5333 sec. > > These are shown in the following figure: > > * 0.5333c > * | > * | > ct * | > |* | > *|* | > * | * | > * | * M'0.3333c > * | * | > * | * 0.1333c > *__________|________|_*__ x > A |M H1 B > | > > The logical conclusion of helper H1, is that the light signal from > point B must have encountered observer M on a point located at the > left of H1, and that the light signal from point A will encounter > observer M at a point located at the right of H1. It is quite clear > that, based on these observation made on the embankment frame, the > observer M' will not receive these light signals at the same time. > Later using Lorentz Transformation we will find exactly when and where > M' receives these light signals. > > These conclusions have been obtained by using just basic geometric > relationships. > > Miguel Rios Geometric relationships are not nature. The Observer at M' on the train needs to determine where the light propagates from relative to the medium in which the light propagates in order to determine simultaneity. When the Observer at M' on the train does this, as I have described above, the Observer at M' determines the lightning strikes to be simultaneous, in nature.
From: paparios on 4 Jan 2010 09:53 On 4 ene, 11:49, mpc755 <mpc...(a)gmail.com> wrote: > On Jan 4, 9:27 am, "papar...(a)gmail.com" <papar...(a)gmail.com> wrote: > > Geometric relationships are not nature. > Geometry and math are tools to explain to fools like you what is happening in nature. Miguel Rios
From: mpc755 on 4 Jan 2010 09:53 On Jan 4, 9:49 am, "papar...(a)gmail.com" <papar...(a)gmail.com> wrote: > On 4 ene, 11:27, "papar...(a)gmail.com" <papar...(a)gmail.com> wrote: > > > On 4 ene, 10:48, mpc755 <mpc...(a)gmail.com> wrote: > > Let us now introduce Special Relativity. > > Let frame K be defined by coordinates (x, ct) and frame K be defined > by coordinates (x, ct). Coordinates (x, ct) of frame K and (x, ct) > of frame K are related among them through the Lorentz Equations in > the following way: > > x = gamma (x+ vt) ; t = gamma (t+ vx/c^2) > > x' = gamma (x - vt) ; t = gamma (t - vx/c^2) > > where gamma = 1 / sqrt(1-v^2/c^2) = 1.25 > > How can we draw these relationships? > > Well we already did the drawing of coordinates (x, ct), which we > repeat here: > > ct x=ct > | * > | * > | * > ___________|*_________ x > *| > * | > * | > * | > > Now, coordinates (x, ct) look as a rotation of the coordinates (x, > ct), as follows: > > ct' x'=ct' x' > / * . > / * . > / * . > /*. > * > .*/ > . * / > . * / > * / > > Note how the constant relationship between these two graphs is the 45 > degree line which denotes the relation x = ct (on frame K) and > equation x = ct (on frame K). > > What happens, now in the train frame K, with the same lightning > strike of location A, at time t=0? > > Coordinates of location A in frame K, are given by Lorentz as: > > x_A = x' = gamma (x - vt) = gamma (x_A - vt) > > So with t = 0, x_A = 100000km, gamma = 1.25, we have x_A = > 125000km . This result also shows that, as observed in the embankment > frame K, the train is seen contracted in length (distance AM on the > train frame is 125000 km, but distance AM on the embankment frame is > 100000km). > > t_A = t = gamma (t - vx/c^2) = -gamma (vx_A/ c^2) = 0.25sec > > Light signal moves, in frame K, according to equation: > > (x_M x_A) / (t_M t_A) = c > > Where x_M = 0 at the train observer location. Therefore we have, > as measured on frame K: > > t_M = ( x_A + ct_A)/c = (125000 + 75000)/300000 = 0.667sec > (1) > > What happens, now in the train frame K, with the same lightning > strike of location B, at time t=0? > > Coordinates of location B in frame K, are given by Lorentz as: > > x_B = x' = gamma (x - vt) = gamma (x_B - vt) > > So with t = 0, x_B = +100000km, gamma = 1.25, we have x_B = +125000km > > t_B = t = gamma (t - vx/c^2) = -gamma (vx_B/ c^2) = 0..25sec > > Light signal moves, in frame K, from point B towards M, according to > equation: > > (x_M x_B) / (t_M t_B) = c > > Where x_M = 0 at the train observer location. Therefore we have, > measured on frame K: > > t_M = (x_B ct_B)/c = (125000 75000)/300000 = 0.167sec (2) > > Therefore, for the train observer M, as shown in equations (1) and > (2) the strikes ARE NOT simultaneous. > > Miguel Rios Geometric relationships are not nature. The Observer at M' on the train needs to determine where the light propagates from relative to the medium in which the light propagates in order to determine simultaneity. When the Observer at M' on the train does this, as I have described above, the Observer at M' determines the lightning strikes to be simultaneous, in nature.
From: mpc755 on 4 Jan 2010 09:56
On Jan 4, 9:53 am, "papar...(a)gmail.com" <papar...(a)gmail.com> wrote: > On 4 ene, 11:49, mpc755 <mpc...(a)gmail.com> wrote: > > > On Jan 4, 9:27 am, "papar...(a)gmail.com" <papar...(a)gmail.com> wrote: > > > Geometric relationships are not nature. > > Geometry and math are tools to explain to fools like you what is > happening in nature. > > Miguel Rios You are mistaking math for nature. |