 #51
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Hi mysearch,
I have attached a slightly modified version of your first diagram to the aid the discussion. (hope you don't mind).
From the attached diagram:
e1 and e(0,0) are causally connected and said to be "time like separated". Ther is no reference frame in which they can be considered to be simultaneous. Same goes for events e(0,0) and e2 and e(0,0) and e3''.
e3 and e(0,0) are "space like separated" events as they can not be casually connected. e3 and e(0,0) are example of where delta(x) > delta(t)*c and in this case delta(t) =0. It is also true that delta(x)/c < delta(t) in this case. A reference frame in which two space like separated events are considered to be simultaneous can always be found. e4 and e(0,0) are also space like separated.
e(0,0) and e3' are "light like separated" and the two events can always be connected by a light signal according to any observer.
Ref: http://en.wikipedia.org/wiki/Spacetime
It is also relevant to the Bell's spaceship paradox where the observers comoving with the rockets measure the separation distance of the rockets to greater than the constant separation that the non accelerated observer measures. (The comoving observers are using length contracted rulers).
I have attached a slightly modified version of your first diagram to the aid the discussion. (hope you don't mind).
From the attached diagram:
e1 and e(0,0) are causally connected and said to be "time like separated". Ther is no reference frame in which they can be considered to be simultaneous. Same goes for events e(0,0) and e2 and e(0,0) and e3''.
e3 and e(0,0) are "space like separated" events as they can not be casually connected. e3 and e(0,0) are example of where delta(x) > delta(t)*c and in this case delta(t) =0. It is also true that delta(x)/c < delta(t) in this case. A reference frame in which two space like separated events are considered to be simultaneous can always be found. e4 and e(0,0) are also space like separated.
e(0,0) and e3' are "light like separated" and the two events can always be connected by a light signal according to any observer.
Ref: http://en.wikipedia.org/wiki/Spacetime
There is always an ambiguity of which observer is moving relative to the other in realtivity :P...
Let’s take this line of logic one step further and assume that there are 2 identical spaceships. One used by the moving observer and one left behind with the stationary observer. Now, even though there is no ambiguity about which observer is actually moving relative to the another, ...
Correct. Each observer will consider their own spaceship to be shorter than the other....
the definition of proper length seems to imply that the moving observer perceives his spaceship to be 3 units in length, while perceiving the length of the spaceship on the ground to be only 2.4 units. While, the stationary observer on the ground perceives the lengths to be the other way round.
The comparison is a little tricky because the accelerating observer has velocity as well as acceleration. In a comparison of the proper times of two observers leaving location A simultaneously and arriving at location B simultaneously, with one observer moving with constant velocity and the other accelerating continuously then the the accelerating observer experiences more time dilation and less proper time. The accelerating observer takes a longer curved path "through" spacetime. So yes, I would tend to agree with your conclusion....
Proper Time Caveat?
If the moving observer’s was accelerating between events (A) and (B), would the proper time, i.e. wristwatch time, be shorter than that measured by a nonaccelerated (inertial) wristwatch?
If yes, does the equivalence principle between acceleration and gravity, allow this additional time dilation to be interpreted in terms of an equivalent gravitational effect in conjunction with velocity?
I would tend to say yes....
Proper Length Caveat?
If yes, is there an equal implication of there being an extra effect on space curvature due to this acceleration as per gravity?
Depends who defines the distance the light has to travel. Local observers in the gravity well would measure the radial distance as greater than the coordinateradius if they tried to measure it directly because they would effectively be using length contracted rulers. For the same reason an observer on the perimeter of a rotating disk would measure a greater radius than a non rotating observer....
If curvature increases with gravity, does this mean that light has to travel a greater distance than implied by the coordinateradius?
It is also relevant to the Bell's spaceship paradox where the observers comoving with the rockets measure the separation distance of the rockets to greater than the constant separation that the non accelerated observer measures. (The comoving observers are using length contracted rulers).
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