I've been away for a while because I've had a devil of a time with the math. This was from a few posts back to prove if Alice chenges her velocity from .6c to c away from Bob, she will age 6 yrs from 4 to 10 while Bob will only age 4 yrs from 4 to 8 during the time of relative velocity imbalance. It turns out my initial equation for v_{h }was about .5 c while it's the slope of the line of perspective simultaneity of slope 1/v_{h }that is the line which intersects the Alice's velocity line c. So let's start over:

**Correction: my new formulas are ( x-3 ) /t = v/c and x / ( t - 3 ) = c/v**_{h}

I don't quite understand why each velocity is divided by c but it doesn't affect the numbers and makes the math really crisp. The time results end up without units though.

x = (t-3) / (v_{h}/c)

plugging in x = vt/c + 3

we get (t-3) / (v_{h}/c) = vt/c + 3

t(1-vv_{h}/c^{2}) = 3(v_{h}/c + 1)

so **t = 3/c (v**_{h} + c)/ ( 1 - vv_{h}/c^{2})

knowing t'=t/Y and plugging in **Y = (c**^{2} + v_{h}^{2}) / (c^{2} - v_{h}^{2}) and **v = 2 c**^{2} v_{h}/ (c^{2} + v_{h}^{2})

t' = 3/c (v_{h} + c)(c^{2}-v_{h}^{2})/ ((( 1 - 2v_{h}^{2}c^{2} / (c^{2}(c^{2}+v_{h}^{2}))(c^{2}+v_{h}^{2}))

= 3/c (v_{h} + c)(c^{2}-v_{h}^{2})/ (c^{2}+v_{h}^{2 }-2v_{h}^{2}))

The term (c^{2}-v_{h}^{2 }) that was causing infinity gets cancelled out so

we get **t' = 3(v**_{h} + c) **/c**

Bob's age is 8 and Alice's age is 4 + t'. So as v_{h }-> c, t' =6

QED

Here are some more numbers to play around with in the last equation

v c v_{h }c Age diff (Alice ends up older by)

1 1 2

40/41 4/5 1.4

15/17 3/5 .8

4/5 1/2 .5

3/5 1/3 0

Now we can continue the math opening up new velocities of relativity's landscape that are forbidden by relativity to determine age difference.

**Edited by ralfcis, Today, 06:34 AM.**