A rough draft
PART I
Let's consider a thin long rod with proper length L.
Suppose that in two points of the rod there are two synchronously going clocks – a clock C1 and a clock C2 - installed at a distance d away from each other. The distance d can be less then or equal to the length L of the rod, i.e. in the general case L ≥d.
If L> d, then the rod looks approximately so:
----------©--------------------©-----------
The symbols © imitate here the clock C1 (at the left) and the clock C2 (on the right), the line ----imitates the body of the rod.
In specific case, if L = d, the clocks C1 and C2 are located on the ends of the rod.
Let the rod motionlessly laying on axis X of inertial system of reference K, at the same time move with constant speed v' along the axis X' of other system of reference K' (axes X and X' of the systems K and K' slide during their movement relative to each other). ). In this case in the system of reference K' the moving rod has the length L' and the distance between the clocks C1 and C2 is equal d'.
We shall write down Lorentz's reverse transformations for the indications of the clocks of the rod. They look so:
^{2}}})
; ……….(1)
^{2}}})
……….(2)
Let us designate
and
Here
. and
- are coordinates of the clocks C1 and C2, and
- is the length d' of the "section of the rod (between the clock C1 and the clock C2) moving in the frame of reference K' at a rate of v'.
We guess it is understandable for you why the letters t to the left of the equal sign in the lower designation are not supplied with prime symbols, and the letter t to the right is supplied with prime.
Using the designations given above, we obtain from (1) and (2):
^{2}}}})
………..(3)
Designating
^{2}}})
,
we obtain from (3)
,
wherefrom
……….(4)
follows.
Now we can find
via
. What do the deltas Δ here mean, try to guess yourselfs. Thus, from (4) we obtain
...........(5)
If
,
Then from (5)
..........(6)
follows.
If by uncertainty of the coordinate of rod we understand the value
, then (6) takes the form:
...........(7)
Since on the condition of task L≥d, then
. Here
=
.
Therefore we can write (7) as:
………..( 8 )
PART II
Now our little secret. If you are able to decipher it, then do it!
In cases when
,
.
If
, then
...........(9)
Taking into account (9), we obtain from (8)
or, converting,
Since
, then
PART III
..........(10)
PART IV
Obtaining the formula (10) we used simplifications, which make it possible to speak only about an order of magnitude
, but not about its precise value.
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