4Newton said:
In order to remain in sync with BD any transition in the spatial dimension must exceed the speed of light in that space-time vector BE as indicated by the length of the lines.
I think I see what you mean now. I thought you were wrong about the definition of four-velocity, but you were talking about ordinary velocity. What you are wrong about is what the world line would look like. (It starts out almost vertical, but a little to the right, and then changes to almost vertical, but a little to the left. It can be almost vertical through the whole trip).
4Newton said:
Minkowski space-time is not an inertial frame.
That is correct. It's not a frame at all. It's a manifold. (If you really want to understand relativity you should study differential geometry).
4Newton said:
Even if you assume that it is a relative viewpoint there is nothing that prohibits a viewpoint that has absolute zero spatial transition.
Yes, there is. The idea of "absolute zero spatial transition" contradicts the speed of light postulate, time dilation, and any other relativistic effect you've ever heard of. We can talk more about this when you've done the excercise I suggest that you do at the end of this post.
4Newton said:
In Minkowski space-time you are only able to show space as a single dimension.
This is of course wrong. We can all visualize three dimensions, so you can draw
two spatial dimensions ("the x-y plane") and the t axis. (I don't understand how you can say that we can't).
4Newton said:
It is not possible to have all three spatial dimensions with 45-degree light lines.
I hope that what you
really mean is that we can't
visualize all four dimensions in our minds (because there
are three spatial dimensions in Minkowski space, and the world line of any light ray through the origin makes a 45° angle with all the spatial axes).
4Newton said:
The speed of light limit is the vector sum of the three spatial dimensions.
Wrong. Have you never heard of the "light cone"? Imagine a spacetime diagram with
two spatial dimensions (because you won't be able to visualize one with three). The set {(t,x,y)|-t²+x²+y²=0} is the union of all lightlike worldlines through the origin. This is a
cone, not a line. Hence the name "light cone".
4Newton said:
This is also true because you can only move in one direction at a time, in the spatial dimension, so although you have an unlimited degree of freedom to move in any direction any selected direction is perpendicular to the time dimension.
I can't tell if this is just wrong or "not even wrong".
4Newton said:
You still insist on referring to the drawing as a moving frame. It is not. This is an absolute coordinate system. All moving frames would lie on BE like lines. All vertical lines are at spatial positions without motion.
In that case, your spacetime diagram doesn't represent Minkowski space and has nothing to do with relativity, or the universe as we know it.
Do you understand that you can draw another spacetime diagram that represents the coordinates used by the observer whose world line is BE, and that in that diagram the line AD has a negative slope. In
that frame AD is moving!
I also think you should make up your mind. There are no "absolute" coordinate systems in SR. In a previous post you said that you have no objections to relativity. Then why are you contradicting it?
4Newton said:
The exercise of simultaneous actions is to show proof as you have stated to validate the absolute Minkowski space-time drawing. In the simultaneous exercise equal distance is just that. Equal distance in all directions to form a sphere from a center point.
There are a few things you should realize:
1. In your spacetime diagram (with one spatial dimension), the "sphere" is just the two points (0,-a) and (0,a).
2. The events on this "sphere" are simultaneous
by definition (in
this frame), so it makes no sense at all to try to
prove that they are simultaneous.
3. You're doing all of this from one observer's point of view, and that means that if you're able to prove a statement such as "these events are simultaneous" you have only proved that it's true in
that observer's frame.
4Newton said:
I start with an undisputed simultaneous action, two flashing lights at the same point in space, I then move them apart with the observer at the center still having verification the two lights are simultaneous. I then state the obvious, that the change of the observers position changes his view of the lights flashing simultaneously but obviously his change of position does not change the action of the flashing lights.
This is correct (but pointless).
4Newton said:
I have thus separated the reality of observation from the reality of action. This is something SR failed to do.
You haven't separated anything that wasn't separated already.
That second sentence is so wrong I don't even know where to begin.
4Newton said:
It is also obvious that the flashing lights may be moved in any direction of any equal distance, line, circle or spherical, with the same results. This may then be done a second time. You now have an infinite number of simultaneous spheres intersecting another set of infinite number of simultaneous spheres resulting in the conclusion that all action is simultaneous through out all points of the universe.
So? You're just saying that (0,-a) and (0,a) are simultaneous regardless of what a is. This is nothing new.
If you're going to prove something, you can't begin with the assumption that what you're trying to prove is true, and than use that in your proof. No logical fallacy can be worse than that.
4Newton said:
...again this is absolute space-time with inertial frames moving on BE or ED like lines.
It seems pointless to continue this discussion unless you first do this excercise:
The observer whose world line is AE (extended to infinity in both directions) shoots a laser beam in the positive x direction at time -t (according to his own clock). It is reflected off a mirror that is perpendicular to the laser beam. At time t (according to the same clock) the reflected laser beam has returned, and hits the laser. Your task is to draw the path of the laser beam in your spacetime diagram. At what point in the spacetime diagram is the beam reflected? Can you figure out what time the AE observer's clock is displaying when the reflection event happens?