......How should we see a
linear object (a shaft) in repose and perpendicular to our main sight axis
if we were flying a spaceship and getting close to such object, as shown
in picture 2, at a speed v inferior to c = 300,000 km/s but in terms of
the same greatness?
Picture 2:
explanation in the text
......Apparently we would not
notice any deformation in the object but the one due to its growing as we
approach it. Also, we would not have any conditions to visually realize
its shape on the plan of the figure. If instead of a linear shaft the
object would be like a coin seen in profile, the image would be alike.
......But what should be the
interpretation an observer might give to this image if he knew it actually
was a linear shaft and that the spaceship flies towards him at a constant
speed v close to c, for example, v = 0.5c? As we have seen, the image does
not come instantly to the observer, but the information transmitting it
(light) cover distance b, in between the object and the observer, at a
speed c.
......The central point O on
the shaft, when it sent the image now observed, was at distance longer
than b from the observer, that is, 1.5b. This is easy to be understand for
if light covered distance b = ct at this same time interval, t and the
observer, at a speed v = 0.5c, covered in opposite way distance 0.5b
(picture 2).
......Using arguments from
classical physics and centering the figure on the observer, that is,
assuming the observer's referent is still, we get the impression it is the
shaft that is approaching the observer. Leaving aside the relativist
effects of modern physics and related to this change in referent, let us
examine how the Euclidean geometry would deal with the problem not only in
relationship to point O but also to the other points on the shaft.
......If point O ¾shown
in picture 2 and now draw in this new referent (as if the movement were by
the shaft)¾
occupied in the past (when its present image was sent) position P, 0.5b
from O (picture 3), where ¾in
the same scheme¾
will the other points on the shaft be? Let us think of a point O' above O,
and let us look for its temporal image P'. The image of O' ¾in
order to reach the observer¾
covers a distance which is greater than b; therefore, the image the
observer gets from O' was generated at a previous time, that is, P' did
not occupy the same vertical axis where P is. It is easy to find P'
graphically, if we keep in mind that during time t' in which light covered
distance ct', from O', as far as the observer, the latter covered a
distance vt' = 0.5ct, which at the present referent would represent the
moving of point P' to O' (picture 3a).
Picture
3: explanation in the
text
