Has anyone come across the following argument somewhere?

Can anyone suggest a relativistic solution to the paradox?

I've done my best for the diagrams.

Regards.

The standard imaginary model of special relativity (without the mirror) is that of a beam of light (AB) seen as perpendicular to the motion inside the reference frame. While the observer outside the moving reference frame sees the beam of light (AC) pointing ahead of the perpendicular.

____A .

v->

____B .---. C

If c is the speed of light and v is the speed of the moving reference frame and if t is the time inside the reference frame and t' the time outside, then :

AB = ct, AC = ct' and BC = vt'

If ABC is a right angle triangle, then :

(ct)Â² = (ct')Â² - (vt')Â²

t = t sqrt(1 - vÂ²/cÂ²) ... (1)

Now imagine that the beam of light, as seen inside the reference frame (AB), is pointing slightly ahead of the perpendicular.

____A .

v->

____O .---B .---. C

If c is the speed of light and v is the speed of the moving reference frame and if t is the time inside the reference frame and t' the time outside, then :

AB = ct, AC = ct' and BC = vt'

If AOB and AOC are right angle triangles and if OA = y and OB = x, then :

(ct)Â² = xÂ² + yÂ² and (ct')Â² = (x + vt')Â² + yÂ²

Therefore:

(ct)Â² - xÂ² = (ct')Â² - xÂ² - 2xvt' - (vt')Â²

(ct)Â² = t'Â²(cÂ² - vÂ² - 2xv/t')

t = t' sqrt(1 - vÂ²/cÂ² - 2xv/t'cÂ²) ... (2)

Then imagine that the beam of light, as seen in the reference frame (AB), is pointing slightly behind the perpendicular.

______A .

v->

____B .---. C

If c is the speed of light and v is the speed of the moving reference frame and if t is the time inside the reference frame and t' the time outside, then :

AB = ct, AC = ct' and BC = vt'

If ACB is a right angle triangle, then :

(ct)Â² = (ct')Â² + (vt')Â²

t = t' sqrt(1 + vÂ²/cÂ²) ... (3)

This means that the proportion t : t' varies according to the direction of the beam of light, with respect to that of the moving reference frame, and that the relativity of time depends on what is being shown.