Try2:-
(a) If A see 30 min in his clock that is coordinate time so proper time in B's frame should be:
∆t= 30 min,
∆t=∆t'/√(1-(v/c)^2)
∆t'=30*(√(1-(2/3)^2))=22.4min
{ And this must be the time when A's coordinate clock is at exactly the position where B is at that time... }
But, how to...
Ques) 2 observers, A on Earth and B in rocket ship whose speed is 2x10(^8) m/s, both set their watches at 1:00 when ship is abreast of the earth.
(a) When A's watch reads 1:30, he looks at B's watch through telescope,
(b) When B's watch reads 1:30, he looks at A's watch through telescope...
A particle has 3 degree of freedom (dof) , so we write 3N dof for N particles...
But I have a fundamental doubt: 3 dof means we consider only transitional motion of particle?
... why can't we say 6 dof?