Course 3 on Lorentz-Four-Contractions Session 13, Part 3

To find the velocities of these three lines, here. The lines are to be considered so extremely thin that the only impetus derived from these is in their endpoints. The lines connecting these points are merely a ghost of delineation. So a theoretical line in the middle of the directoral line that was orthogonal with the directoral line would bear no reality, and thus no motion while the horizontal (directoral) line moves. So, here, motion of interconnected lines is minimized at 90 degrees, and maximized at 0 degrees, when calling the horizontal line 0 degrees and the orthogonal position 90 degrees. So, if the "X lines" were at 3 degrees from the flat line each, and neither line fidgeted at all, then as the flat line traveled at .8c, then the two lines would travel at cosine of 3 degrees * .8c each. If there were 3 billion of such lines interconnected in the same fashion, then, to find out their velocity, multiply the cosine of the angle of each by .8c to find their velocity as a unit in the given direction. All three lines, however, would be traveling at .8c relative to the direction of the directoral line. You see, each line would be traveling as a unit in a sense quicker if all of their lengths were the same in a given direction, yet, take the endpoints into consideration. Take two pencils of equal length. Lay one flat. Angle the other one. Yes, the flat pencil goes the furthest along the surface. So, again, faster than light we shall discuss in future courses.