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Relativity (Time Dilation)

  • #1
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)
 

Answers and Replies

  • #2
Doc Al
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(a) ∆t'= ∆t/√(1-(v/c)^2)=30/√(1-(2/3)^2)=40.24min, so A shall see B's clock showing 1:40
According to this, A sees B's clock as running faster. (You are applying the time dilation formula backwards.)

(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
You are correct that the effect should be symmetric.

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)
Exactly. You need to take into consideration the time it takes for the light to travel to the observer.
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)
Yes. I think you realize that something's wrong with how you applied time dilation.
 
  • #3
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 indulge the time associated with the delay that comes due to finite speed of light...
(b) Even now, we hold the same argument
{ But then the twin paradox still creates a problem as A will say B is younger and B will say A is younger}
 
  • #4
Doc Al
Mentor
44,940
1,201
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... }
During the time that A's clock shows 30 minutes, A will say that B's clock showed 22.4 minutes elapsing. That's true, but not really relevant to this problem.
But, how to indulge the time associated with the delay that comes due to finite speed of light...
You need to figure it out. Try this. If (according to A) B travels for X minutes then emits a flash of light, when will A see that light? You need to find the value of X so that the total time between the start and the seeing of the flash equals 30 minutes. (Once you find that value of X--which is according to A's clock--then you can apply time dilation.)
(b) Even now, we hold the same argument
{ But then the twin paradox still creates a problem as A will say B is younger and B will say A is younger}
There's really no problem. A measures B's clocks to run slowly and B measures A's clocks to run slowly. The effect is completely symmetric.

With the twin paradox, one twin turns around and rejoins the other. That 'turning around' breaks the symmetry.
 
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