Deriving the Lorentz Factor: Exploring SR & GR with Einstein

[asymmetric]

I Recently Went Through The Derivation Of The Lorentz Transformations Again...
I'm Reading A Book By Einstein On Sr And Gr.
It Came To Me That Instead Of Using Only K And K1 Frames When Explaining The Vertical Mental Experiment Of K1 With A Mirror Placed Vertically Above It At A Distance L And Moving On X Axis At Uniform S@v And K Being Observer We Could Use A Third Frame K2 And Leave K2 At Origin K2 On Its X=0.
For K And K1 We Have The T Dilation.
Now K2 Would Register, I Believe, Exactly What K Would Register After Vt If The Mirror Had Certain Characteristics That Could Take Part Of The Same Ray To X=vt And The Other Part Back To X=0.
Could This Special Mirror Be Found In The Electromagnetic Wave Itself?
Has Any Work Been Done On This Question?

 

[Doc Al]

Sorry, but I don't understand what you're saying. I think you're talking about the "light clock" thought experiment, where light is bounced back and forth between two vertical placed mirrors. The "clock" (in frame K1 say) moves with speed v with respect to a "stationary" observer (in frame K). From this (and the assumptions of relativity) one can deduce time dilation.

You introduced another frame K2. How does this frame fit in? How does it move with respect to K and K1?

 

[asymmetric]

yes we are referring to this thought experiment. thank you. and i apologise.
we are also talking about 2 uniformly moving frames. ie K AND K1.
IT TOOK ME SOME TIME TO FIGURE WHAT THIS WAS ALL ABOUT.
the observer K is never stationary.
if this were not so, that is if K DID NOT MOVE AT V then we would find in the horizontal thought experiment where K is measuring the stick that is moving, that the information received by K at the beginning is registered with gamma factor involved but if K DOES NOT MOVE THEN the information or same ray does not get back to K to complete the registering OR MEASURING of same.
in the vertical experiment this doesn't have to be so ...at least i think not in one special circumstance, ie with a special mirror.
simply because the rays diagonal outwards towards mirror and then down to the new x would take the same t as the diagonal outwards and back to our k2 who hasn't moved.
so i added another frame K2 THAT WAS GENUINELY at standstill. THOUGHT EXPERIMENT.
So along side K and k1...please PLACE K2. BUT WITHOUT K2 MOVING.
now repeat the vertical mirror experiment.
when k finishes observing or registering k1, where k1 is the frame that's got a mirror placed at an L distance above him, K2 THAT HASN'T MOVED WILL , I STRONGLY BELIEVE, DEPENDING ON THE ''MAKE UP'' OF THE MIRROR, REGISTER THE INFO in same t AS K.
For k the ray is along x-axis and upwards on the y axis, then down on y-axis and further along on x-axis creating a triangle when the registering is finished.
for k2 that hasn't moved, depending on the ''make up'' of the mirror...imagine that the mirror can return part of same ray back to k2, maybe split ray, then k2 will register things as does k.
is this an improvement in my explanation?...
so i looked at nature and asked if this type of mirror could exist within say the em wave itself...that is why i asked for help.
if the answer is a positive one then we could maybe suggest a ''riding back or return feature'' existing in all em waves...
thank you again.

 

[Doc Al]

so i added another frame K2 THAT WAS GENUINELY at standstill. THOUGHT EXPERIMENT.
So along side K and k1...please PLACE K2. BUT WITHOUT K2 MOVING.
now repeat the vertical mirror experiment.

Sorry, but I don't understand what you mean by a frame that is "genuinely at a standstill". Such a concept seems meaningless to me. Movement is relative, so all frames are moving with respect to some other frame.

when k finishes observing or registering k1, where k1 is the frame that's got a mirror placed at an L distance above him, K2 THAT HASN'T MOVED WILL , I STRONGLY BELIEVE, DEPENDING ON THE ''MAKE UP'' OF THE MIRROR, REGISTER THE INFO in same t AS K.

If the K and K2 frames are both moving together (or both "at rest", if you prefer) then measurements made in both frames will match. They are really the same frame, of course.

 

[asymmetric]

q on lorentz

what I'm envisaging is k2 not running parallel to k and with no apparent v.
lets place k2 on the other side of k1.
in a diferent frame altogether. that is, k on one side of k1 and k2 on the other side.
if the k2 frame is on its own axis at x=0 y=0 and z=o when we begin with the flash or ray, and k2 somehow loops there at its zero coordinates and / or remains there it will receive information the same time as k's time.
remembering always what i mention about the special mirror to split , divide or return part of the light back to x=0 for k2.
when i was working the transformations out i saw that it was necessary for k to be moving also for the transformations to be valid..specially in the horizontal stick thought experiment otherwise we could not, i seriously supposed, be able to apply t dilation found in vertical thought experiment...
but until i read the book twice over i didn't know that this was a condition in the first place, ie that both frames were on the move...infact the transformation x" = x - vt / gamma...really means that the distance d equal to vt has to be deducted from new x because k has moved on also.
the tricky point I'm trying to attend to is the posible lagging behind of one of the three frames, that is, a posible natural occurence or characteristic to be found in the emwave itself that maybe is producing this type of lagging behind so that at origin for k2 info gets to him at same t as for k...
you see sir,
if this was the case we would maybe seeing some sort of return or ride back in the em wave that drags or pulls on the k2 frame...or at least effects it!...anyway I'm going on a bit so again i apologise.
thank you kindly again.