BurtJordaan
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I'm trying to understand Lincoln's values for frame C. So I added some values from GrayGhost's Minkowski diagram for essentially the same (ABC) scenario:

With $\large v= 0.866c$, $\large L/v =4$ and $\large \gamma = 2$:

Lincoln's transformations with Gray's values calculated:
$\large (x,t)_{1A} = (0, 0)$
$\large (x,t)_{2A} = (L, \frac{L}{v})=(3.464,4)$
$\large (x,t)_{3A} = (0, \frac{2L}{v})=(0,8)$

$\large (x,t)_{1B} = (0, 0)$
$\large (x,t)_{2B} = (0, \frac{1}{\gamma} \frac{L}{v})=(0,2)$
$\large (x,t)_{3B} = (-2 \gamma L, 2 \gamma \frac{L}{v})=(-13.856,16)$

$\large (x,t)_{1C} = (0, 0)$
$\large (x,t)_{2C} = (2 \gamma L, \gamma \frac{L}{v} (1 + \frac{v^2}{c^2}))=(13.856,14)$
$\large (x,t)_{3C} = (2 \gamma L, 2 \gamma \frac{L}{v})=(13.856,16)$

Now to get the time passage of C from 2 to 3, we need to subtract their times:

$\large \Delta t_{3,2 C} = t_{3C} - t_{2C} = \frac{1}{\gamma} \frac{L}{v} = 2$

So the total time experienced by two, inertial observers (as compared to A) is

$\large t_{2B} + \Delta t_{3,2 C} = \frac{2}{\gamma} \frac{L}{v} = 4$

Lincoln used the reference frame of observer C in 'standard configuration', i.e. frame C's origin is also where A and B's are (Event 1). Gray and me 'resynchronized' the clock of observer C so that it read 2 yrs at event 2, which we called the 'time-handoff', i.e. C sets her clock to momentarily read the same as B's clock, i.e. 2 years.

To show how the C's values on the Minkowski correlate with Lincoln's equations, one can do a Lorentz transform for Event 0 from reference frame A to frame C. Event 0 is where observer C crosses the t=0 line of frame A, so in frame A, Event O sits at (6.928,0) and v= -0.866, i.e. gamma = 2.

$\large (x,t)_{0C} = (2(6.928 + 0), 2(0 + 0.866 \times 2\times 6.928)) = (13.856, 12)$

This means Gray's clock C is exactly 12 years behind what Lincoln's 'standard configuration' clock would have been, explaining the values for tC shown on the diagram.

Last edited by BurtJordaan on December 23rd, 2014, 12:45 am, edited 4 times in total.
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