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 vh was about .5 c while it's the slope of the line of perspective simultaneity of slope 1/vh 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/vh
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) / (vh/c)
plugging in x = vt/c + 3
we get (t-3) / (vh/c) = vt/c + 3
t(1-vvh/c2) = 3(vh/c + 1)
so t = 3/c (vh + c)/ ( 1 - vvh/c2)
knowing t'=t/Y and plugging in Y = (c2 + vh2) / (c2 - vh2) and v = 2 c2 vh/ (c2 + vh2)
t' = 3/c (vh + c)(c2-vh2)/ ((( 1 - 2vh2c2 / (c2(c2+vh2))(c2+vh2))
= 3/c (vh + c)(c2-vh2)/ (c2+vh2 -2vh2))
The term (c2-vh2 ) that was causing infinity gets cancelled out so
we get t' = 3(vh + c) /c
Bob's age is 8 and Alice's age is 4 + t'. So as vh -> c, t' =6
Here are some more numbers to play around with in the last equation
v c vh 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.