the ends of the train are at L*sqrt(1-v^2/c^2)/2 in each direction from the observers. Yes. From the _frame_ of reference of the train, the ends of the train are L/2 in each direction, where L is the length of the train at rest. Fine. Frank,  Ok, let's take it from where we agree.    We agreed that the two observers could be opposite one another.  We agreed where the Lorentz equations put the front and rear of the train at that particular moment from each _frame_ of reference.    Now suppose that we emit a light from the position of the observers in each _frame_ of reference.  What will this light do in each _frame_ of reference?    In the _frame_ of reference of the track the light will go in each direction from the observer by the track at a rate of c.  In the _frame_ of reference of the train, the light will go in each direction from the position of the observer on the train at a rate of c.     So, if we have two marks on the track a distance of L/2 in each direction from the observer by the track, the light will reach the marks in a time of t=L/2c.  How far has the light gone at this time in the _frame_ of reference of the train?     t'=(L/2c) (c-v)/c    d=[(L/2c)(c-v)/c]*c=(L/2c)(c-v) This means that the light emitted at the position of the observer on the train has not yet reached the ends of the train when t has expired in the _frame_ of reference of the track. Robert B. Winn

has not yet reached the ends of the train when t has expired in the _frame_ of reference of the track. No, it has, in the backward/towards_rear case + ; and yes, it hasn't, in the forwards/towards_front case - . Btw., using both directions, as prescribed in the distance definition, would result in measurement of length contraction : what the track claim to be L/2 forwards + L/2 backwards == L is being measured by the train as   L/2 (1 - (v/c)) / sqrt( 1 - (v/c)^2 ) +         L/2 (1 + (v/c)) / sqrt( 1 - (v/c)^2 ) == L/2 sqrt( 1 - (v/c)^2 ) . Length contraction is mutual, as expected; or rather: as one requires when deriving the Lorentz transforms in the first place. Regards,   Frank  W ~@) R

ahead and show it. Robert B. Winn This may be so if light has infinite speed. But for scientist who out of train A whit time becomes closer and D farther. There for when light reaches A, D will be still in dark. For scientist in the train A and D move with him and light reaches them at the same time. This is because the speed of light is finite and does not depend on reference _frame_. Aleks Kleyn.

send him by the section of the accelerator with the two marks.   When astronaut reaches the first mark, a scientist by the acceleratro starts his stopwatch, and the astronaut starts his stopwatch.  When the astronaut reaches the second mark, both stopwatches are stopped.  The scientist's stopwatch reads t, the astronaut's stopwatch reads t', where t'=t(c-v)/c.  This means that from the _frame_ of reference of the astronaut, the sides of the accelerator are going by him at a rate of vt/t', which could be faster than c.  My belief is that if a particle is accelerated to a velocity faster than c it will emit light. Robert B. Winn Hi I like how you follow. But in SR definition of velocity is not 3-dimentional. We have space-time, so 4 dimentions. In general, we draw the history line. It goes from past into future. Then we find vector that is tangent to this line and has length 1. This is 4-dimentional velocity. Its derivative is 4-dimentional aceleration. For regular body acceleration is orthogonal to velocity. Aleks Kleyn. Aleks,     Sorry to take so long to answer this.  I was in Montana where computers are scarce.   I am only a welder in a steel fabrication shop, and only have a high school education, so I cannot really address the considerations of four dimensions, since I cannot even visualize the concept.    What I can do is describe what I have described here, which seems to me to conform to reality.  For instance, an astronaut in a sattelite will get a different value if he divides the distance of his orbit by the time registered on his clock than a scientist on earth would get if he divides the distance of the orbit by the time on his clock.    My belief is that the two times in question will conform to the equation t'=t(c-v)/c, where t' is the time on the astronaut's clock and t is the time on the scientist's clock, and v is velocity as calculated by the scientist on earth.    My belief is that the astronaut and the scientist will get the same value for the distance of the orbit, which can be proven fairly simply, a radar on the sattelite will give the same distance to earth as a radar on earth will give to the sattelite, making the radius of orbit the same from either place.    Therefore, we have the relationship, d/t'd/t, where d is the distance of the orbit. There is no distance contraction indicated by this, but velocity has to be calculated from earth because the sattelite is moving relative to earth, whereas, earth is not moving relative to the sattelite.    The Lorentz equations do not show this, but indicate that velocity is the same from either _frame_ of reference.  The fact remains that the earth is not moving relative to the sattelite, except perhaps rotating on its axis, which will not affect the orbit of the sattelite.   My belief is that the Lorentz equations do not show time as it actually exists, but are only valid for comparisons of one atom to another in the same field of gravitation, such scientists have used these equations in the past.   Robert B. Winn

The  observer on the train reaches the position of observer o on the ground, the front of the train is at C, and the rear of the train is a B, lightning strikes the front and rear of the train simultaneously in both _frame_s of reference.  The distance from B to C is the rest length of the train.    This is what I believe happens.  If you can find an error in what I say, go ahead and show it. Robert B. Winn This may be so if light has infinite speed. But for scientist who out of train A whit time becomes closer and D farther. There for when light reaches A, D will be still in dark. For scientist in the train A and D move with him and light reaches them at the same time. This is because the speed of light is finite and does not depend on reference _frame_. Aleks Kleyn. Aleks,      Once again, I am not pretending to know exactly what happens.  However, I believe that scientists are reading too much into the result of the Michaelson-Morley experiment when they say that they know exactly how light gets from one place to another and how fast it is going.  I say that what the experiment shows is that light reacts with the elements in that particular time and place at a rate of c, and that is about all you can say about it. Robert B. Winn