Special Relativity Clocks

JM

Right on George!

DaleSpam

I would call them "clocks".

DaleSpam

What is the question and what are m, u, and v?

harrylin

It looks as if my next remark didn't reach:
"
- the straight same constant speed trajectory BC is another line of the polygon. The Lorentz transformation relating "time" in a co-moving frame along AB is identical to that along BC: x is by definition the direction of motion. And obviously from point B the clock has to continue its counting from where it was the moment before - that's just common sense.
"
I'll try again. For the first leg, the X-axis of K and K' is by definition chosen along the line AB. It is you who draws the lines and defines the frames for the calculation. Thus you give A and B the same Y and Z coordinate (in this case you can keep them both 0), and v along x is simply v. That's how the Lorentz transformations are defined. And how the math between the polygon lines is connected I explained next. So, I'm afraid that you could not follow me.
To elaborate: you choose for the calculation for BC new reference frames with X and X' oriented along BC.

ghwellsjr

Then what would you do to JesseM's quote to make it satisfactory to you?

JM

m is the coefficient in the LT often refered to as gamma. v is the speed of the frame moving in the x direction of the stationary frame K. u is the speed of a single clock moving in the + x direction of K. The question is 'what is the time on the single moving clock'?

JM

I think you didn't understand my question.
Section 4 of 1905 envisions a single clock moving along a polygon path wrt a stationary frame K. The clock starts at a point of K and returns to the same point of K. What you have described is the time of the clock wrt the polygon path. What you have not described is the time of the clock wrt the original frame K. This is the time that is required in order to make a valid comparison with the K time at the end of the path.

GAsahi

If the time on the clock at rest wrt the polygon is [itex]T[/itex] and the speed of the second observer is [itex]v[/itex], then the clock for the observer moving wrt the polygon will show [itex]t=T \sqrt{1-(v/c)^2}[/itex] when the observers are reunited so they can compare clocks.
Does this answer your question?

DaleSpam

For approximately the 100th time I refer you to the formula I posted back in post 36. The time displayed on the clock is:

[tex]\tau = \int \sqrt{1-v(t)^2/c^2} dt[/tex]
So in frame K
[tex]\tau = t \sqrt{1-u^2/c^2}+C_1[/tex]
And in the other frame
[tex]\tau = t \sqrt{1-\frac{(u+v)^2}{c^2(1+uv/c^2)^2}}+C_2[/tex]