- #1

granpa

- 2,268

- 7

obviously if rocket 'C' is considered to be stationary then we just have the twins paradox. but what is the result if we look at it from a different point of view? obviously the result should be the same.

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- Thread starter granpa
- Start date

- #1

granpa

- 2,268

- 7

obviously if rocket 'C' is considered to be stationary then we just have the twins paradox. but what is the result if we look at it from a different point of view? obviously the result should be the same.

- #2

Dale

Mentor

- 33,714

- 11,308

As always I would recommend the spacetime geometric approach. Here is an example, I used v = .6 c (gamma = 1.25) so that plotting it is easier and picked the inertial reference frame where A is at rest. Units of time are years and units of distance are lightyears so that c = 1.

obviously if rocket 'C' is considered to be stationary then we just have the twins paradox. but what is the result if we look at it from a different point of view? obviously the result should be the same.

Recall that the four-vector is (ct,x) and that the spacetime interval along a four-vector dr = (c dt,dx) is ds² = c²dt²-dx² = dr.dr and that ds/c is the proper time experienced by a clock moving along dr.

a0 = b0 = c0 = (0,0)

a1 = b1 = (5,0)

c1 = (5,3)

a2 = (15.625,0)

b2 = c2 = (15.625,9.375)

Let da1=a1-a0 and da2 = a2-a1 then total accumulated time on clock A is:

sqrt(da1.da1)/c + sqrt(da2.da2)/c = sqrt(5²-0²) + sqrt(10.625²-0²) = 15.625

Let db1=b1-b0 and db2 = b2-b1 then total accumulated time on clock B is:

sqrt(db1.db1)/c + sqrt(db2.db2)/c = sqrt(5²-0²) + sqrt(10.625²-9.375²) = 10

Let dc1=c1-c0 and dc2 = c2-c1 then total accumulated time on clock C is:

sqrt(dc1.dc1)/c + sqrt(dc2.dc2)/c = sqrt(5²-3²) + sqrt(10.625²-6.375²) = 12.5

I will leave it as an exercise for you to show that the intervals are the same in the inertial frame where C is at rest.

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