4Newton said:
Object A takes the path C-A’-A and object B has the path of C-B’-B. At any time a light pulse may be sent from either object to the other. This is shown by light lines A’ to B and B’ to A. The length of each light path must be equal. Therefore from the point of any simultaneous events on either objects the light must arrive at the same time on the other object. The start of the simultaneous event may be all the way back to the BB. The point of the BB must be simultaneous for all events in the universe. By this method it is shown that at all times all events in the universe are simultaneous and that all objects in the universe have the same normal, or proper time.
This is an interesting argument, but is it correct? Let's find out.
"Object A takes the path C-A’-A and object B has the path of C-B’-B"
You should have explained what this drawing represents before you started talking about the path of objects. I will do it for you. We started out by attempting to find solutions of Einstein's equation that describe spacetimes that can be "sliced" into one-parameter families of spacelike hypersurfaces that are homogeneous and isotropic. The drawing is a spacetime diagram that represents
one such solution. In this particular solution, the spacelike hypersurfaces are 3-spheres.
The easiest way to construct a coordinate system on a spacetime that is a one-parameter family of 3-spheres is to take the parameter that labels the 3-spheres to be the 0th coordinate (i.e. the "time" coordinate) and use a coordinate system that's appropriate for a 3-sphere to define the other three coordinates. (This is not the
only way however).
With this choice of coordinates, any two points on the same 3-sphere (from this one-parameter set of 3-spheres) are simultaneous in the sense that they have the same time coordinate.
We can
think of these 3-spheres as "space" at a different times, but it's important to realize that this "time" is just a parameter that labels the different 3-spheres. We can't just
assume that e.g. A and B are simultaneous in any kind of "absolute" way. We can however be certain that any two clocks that are stationary in this frame will tick at the same rate. This is obvious because of the symmetries we have imposed on these solutions (homogeneity and isotropy). Because of this, it is natural to choose the value of the time parameter to be the time displayed by a clock (any clock) that's stationary in this frame.
"At any time a light pulse may be sent from either object to the other. This is shown by light lines A’ to B and B’ to A."
There are two things you could mean by "at any time":
1. At any value of the parameter that labels the 3-spheres (i.e. "on anyone of the 3-spheres")
2. At any "real" time, where you have
assumed that the parameter represents some kind of "real" time
The second alternative is pretty much what you're trying to prove, so you wouldn't want that to be one of your assumptions, would you? You probably meant 2, but I'm going to try to find a way to make sense of your proof, so I'll pretend that I believe that you meant 1.
The light lines are drawn incorrectly. See my earlier post about this.
"The length of each light path must be equal."
OK. (You should have explained why, but since I already know why, it doesn't really matter to me).
"Therefore from the point of any simultaneous events on either objects the light must arrive at the same time on the other object."
Do you think anyone understands what you're trying to say here? I have no idea what "from the point of any simultaneous events on either objects" is supposed to mean. You're not making sense.
"The start of the simultaneous event may be all the way back to the BB. The point of the BB must be simultaneous for all events in the universe."
You're still not really making sense. When you say stuff like "simultaneous for all events ..." it really sounds like you have absolutely no idea what you're talking about.
"By this method it is shown that at all times all events in the universe are simultaneous and that all objects in the universe have the same normal, or proper time."
This is a bizarre claim. "By this method it is shown..." No, it isn't.
It's hard to tell what you're thinking, but I can make a guess, based on what I've read here, and in your previous posts. This is what I believe you're trying to say:
"If the big bang happened N seconds ago to the observer at A whose world line is GA'A, it must also have happened N seconds ago to the observer at B whose world line is GB'B. Since A and B are just two arbitrary points in space, this means that every stationary observer at a point on the same 3-sphere (i.e. on the blue circle) will agree that the big bang happened N seconds ago. This proves that all points in space are simultaneous, in an absolute sense, and that anyone who says that it makes just as much sense to think of two events that are not on the same 3-sphere as simultaneous is just wrong".
The last sentence is OK until the first comma, but not after that. I would agree that it makes sense to think of two points on the same 3-sphere as simultaneous, but this is not a very profound statement. It's kind of like saying that "bald guys have no hair".
There is nothing in your analysis that justifies the conclusion that simultaneity is absolute.