Simplyh said:
Any two events with coordinates (x,t) equal to (0,0) and (x,0), with x <>0 are not simultaneous in the same reference frame.
What do you mean "in the same reference frame"? They are certainly simultaneous in the frame where the first event had coordinates (0,0) and the second had coordinates (x,0), because all "simultaneous" means is "same time coordinate", and both have a time coordinate of 0 in this frame. In a second inertial frame in motion relative to this one, the events would be assigned different coordinates and their time coordinates might no longer be equal, in which case they'd be non-simultaneous in the second frame.
Simplyh said:
If we have two events at distance, then “... it is not possible without further assumption to compare, in respect of time, an event at A with an event at B. We have so far defined only an ``A time'' and a ``B time.'' We have not defined a common ``time'' for A and B, for the latter cannot be defined at all unless we establish by definition that the ``time'' required by light to travel from A to B equals the ``time'' it requires to travel from B to A. Let a ray of light start at the ``A time'' from A towards B, let it at the ``B time'' be reflected at B in the direction of A, and arrive again at A at the ``A time'' .
In accordance with definition the two clocks synchronize if TB – TA = T'A – TB “(Einstein – 1905 paper).
So, according to this definition from Einstein any events (x,t) are only simultaneous in the same reference frame if tx = x/c.
How do you get that from Einstein's definitions? The units in your equation tx = x/c don't make sense since the left side has units of time*distance (unless you mean to write t
x, the x just being a subscript) while the right has units of time. And your equation doesn't seem to make sense as a definition of simultaneity, because it seems to deal only with a
single set of x and t coordinates, while simultaneity always deals with a
pair of events with different coordinates, we might call them (x1, t1) and (x2, t2). And Einsteins' definition above was really dealing with
three different events which could be labeled (XA, TA), (XB, TB) and (X'A, T'A). The first event is light being sent from clock A, the second event is the light beam being reflected from clock B, and the third event is the reflected light being received by clock A. Since these coordinates are supposed to in the rest frame of the clocks, XA = X'A, the distance between clocks A and B is (XB - XA), and if the clocks are synchronized in this frame then TB - TA = (XB - XA)/c. Of course, this assumes the coordinate system we are using to describe the events is the same as the one where the clocks are at rest and we want to synchronize them, if we want to apply Einstein's definition using coordinates of a frame that is not the rest frame of the two clocks, translating the definition into equations involving the frame's coordinates would be more complicated (see below for an example).
Simplyh said:
This means that your events would only be simultaneous if one had coordinates (0,0) and the other (8,8/c).
You mean, they would be simultaneous
in the train frame if those were the coordinates
in the platform frame? Again you can only talk about events being "simultaneous" relative to a particular choice of frame, never in an absolute sense!
But even if that is what you meant, your math is wrong. If the train is moving at 0.6c relative to the platform frame, and its length is 8 light-seconds in the platform frame, then the event with coordinates in the platform frame x=12.5, t=7.5 will occur at the front end of the train and will be simultaneous in the train frame with the event at the back end of the train at coordinates x=0, t=0 in the platform frame. In other words, if a clock A at the back of the train reads 0 at x=0, t=0, then a clock B at the front of the train must read 0 at x=12.5, t=7.5 in order for it to be "synchronized" with clock A in the train's own rest frame.
With a little analysis we can show that this does make sense in terms of Einstein's definition. Suppose we have a clock A at the back of the train and a clock B at the front, and their worldlines in the coordinates of the platform frame are giving by x=0.6c*t for clock A, and x=0.6c*t + 8 for clock B (these equations ensure that they are both moving at 0.6c, and that at t=0 clock A is at x=0 while clock B is at x=8). Suppose at t=-12.5 seconds in the platform frame, a light signal is sent from clock A; by the equation x=0.6c*t, clock A must be at position x=-7.5 light-seconds when the signal is sent. We assume that at x=0, t=0 clock A read a time of zero, but clock A is running slow by a factor of \sqrt{1 - 0.6^2} = 0.8 in the platform frame, so 12.5 seconds earlier at t=-12.5 seconds, clock A read a time of 0.8*-12.5 = -10 seconds, and we call this reading TA in Einstein's notation.
If a light beam is emitted at x=-7.5, t=-12.5 in the platform frame, the light beam's position as a function of time in this frame must be x = c*t + 5 (this ensures that the light moves at a coordinate speed of c, and that at t=-12.5 it is at position x=-12.5 + 5 = -7.5). So if the front of the train has position as a function of time given by x=0.6c*t + 8, then to figure out when the light beam will catch up with the front, we set these equal, giving c*t + 5 = 0.6c*t + 8, and solving for t gives t = (8 - 5)/(1c - 0.6c) = 3/0.4c = 7.5 seconds. If you plug this time into either the light beam equation or the equation for the front of the train, you find that the x-coordinate of the light catching up to the front of the train must be x=12.5 light-seconds. And remember, I said above that the clock B should read 0 at exactly these coordinates in the platform frame in order for it to be "synchronized" with clock A in the train frame. So, in Einstein's notation we can say that TB = 0, since TB is supposed to be the time on clock B when the light hits it and is reflected back to A.
If the light beam is reflected at these coordinates, then its speed remains constant but its direction changes, so its new position as a function of time is given by x = -c*t + 20 (this ensures that the light is now moving at -c along the x-axis, and that at t=7.5 it has position x=-7.5 + 20 = 12.5 light-seconds). And the position as a function of time for clock A was x=0.6c*t, so to find when the reflected light gets back to clock A we set them equal, giving -c*t + 20 = 0.6c*t, solving for t gives t = 20/(0.6c + 1c) = 12.5 seconds. Since clock A read 0 at t=0, and it's running slow by a factor of 0.8, at t=12.5 seconds clock A reads a time of 12.5*0.8 = 10 seconds. So in Einstein's notation, we have T'A = 10 seconds.
Putting it all together, with TA = -10 seconds, TB = 0 seconds, and T'A = 10 seconds, you can see that Einstein's equation TB – TA = T'A – TB is indeed satisfied here. Thus, in order for clocks A and B to be synchronized in their own frame, if clock A reads 0 at x=0, t=0 in the platform frame, clock B must read 0 at x=12.5 light-seconds, t=7.5 seconds.
Now to check that this matches the Lorentz transformation. Transforming from the platform frame to the train frame, x=0, t=0 in the platform frame corresponds to x'=0, t'=0 in the train frame. As for x=12.5, t=7.5, plugging into the Lorentz transformation gives:
x' = gamma*(x - vt) = 1.25 * (12.5 - 0.6*7.5) = 10
t' = gamma*(t - vx/c^2) = 1.25 * (7.5 - 0.6*12.5) = 0
So, both events have a time coordinate of t'=0 in the train frame, so they are indeed simultaneous in this frame according to the Lorentz transformation.
Simplyh said:
Let's please go back to the 1905 experiment which is easier to understand because everything happens in the same reference frame.
We have a solid arm with two clocks and two observers at extremities A and B. At A there is a light source; at B a mirror. Let a light beam come from A, be reflected at B and go back to A. Time for light to travel from A to B should always be equal to time for light to travel from B to A because all happens in the same reference frame.
"time" in what frame? If the device is moving in the direction of B in your frame, then in your frame the time from A to B is
not equal to the time from B to A, because B is moving
away from the light and A is moving
towards the light, therefore the light has a longer distance to travel to go from A to B then it does to go from B to A, and since it travels at the same speed in both directions in your frame, the time to go from A to B must be larger. But if there are clocks attached to A and B, and these clocks are "synchronized" in the device's own frame, then if TA is the reading of the clock at A when the light leaves it, TB is the reading of the clock at B when the light hits it, and T'A is the reading of the clock at A when the light returns to it, the equation (TB - TA) = (T'A - TB) will be satisfied. In
your frame this is because the clocks are out-of-sync, so even though the time in
your frame between the first pair of events is different than the time in
your frame between the second pair, the actual readings on the clocks at each event will satisfy (TB - TA) = (T'A - TB). In the device's own rest frame the clocks are "really" synchronized", and the reason (TB - TA) = (T'A - TB) is that the coordinate time between each pair of events really is the same in the device's frame, since the device is at rest and the light travels the same distance in both directions in this frame.
SimplyH said:
But it is not so: if the frame is stationary relatively to Earth this is correct; if the frame displaces relatively to Earth than this is not correct. This obviously means that light does not travel at the same speed in all directions in that inertial system whenever that system travels relatively to Earth.
Wrong, see above. The light
does travel at the same speed in all directions in the device's rest frame, since that frame defines distance/time in terms of distance as measured by something at rest in the frame (like the device itself), and assigns time coordinates to events using clocks which are considered to be "synchronized" in that frame (like the pair of clocks at either end of the device). So the distance between the two pairs of events above will be equal, and the time between each pair will also be equal since it is true of the clock readings TA, TB and T'A that (TB - TA) = (T'A - TB).