- #1
babbar.ankit
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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.
What do they read respectively?
Tough the solution seem easy that we can calculate the time dilation
(a) ∆t'= ∆t/√(1-(v/c)^2)=30/√(1-(2/3)^2)=40.24min, so A shall see B's clock showing 1:40
(b) Since all the inertial frames are equivalent (and time dilation is a reciprocal effect) then B shall also see 1:40 in A's clock
But I have some fundamental doubts:
i) does the question is framed properly, if we consider {A's watch reads 1:30 and he looks into the telescope} as an event, then he cannot instantaneously now the exact time in B's watch ( since the speed of light is finite)
ii) if we consider the twin paradox similarly, then A must find that B is younger, and B must find that A is younger (since both are inertial frames)
(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.
What do they read respectively?
Tough the solution seem easy that we can calculate the time dilation
(a) ∆t'= ∆t/√(1-(v/c)^2)=30/√(1-(2/3)^2)=40.24min, so A shall see B's clock showing 1:40
(b) Since all the inertial frames are equivalent (and time dilation is a reciprocal effect) then B shall also see 1:40 in A's clock
But I have some fundamental doubts:
i) does the question is framed properly, if we consider {A's watch reads 1:30 and he looks into the telescope} as an event, then he cannot instantaneously now the exact time in B's watch ( since the speed of light is finite)
ii) if we consider the twin paradox similarly, then A must find that B is younger, and B must find that A is younger (since both are inertial frames)