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Two events, separated by a (spatial) distance 9 x 109 m, are simultaneous in one inertial frame. The time interval between these two events in a frame moving with a constant speed 0.8 c (where the speed of light c = 3 x 108 m/s) is
- a)60 s
- b)40 s
- c)20 s
- d)0 s
Correct answer is option 'B'. Can you explain this answer?
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Two events, separated by a (spatial) distance 9 x 109 m, are simultane...
Time measured at A,
Time measured at B,
Most Upvoted Answer
Two events, separated by a (spatial) distance 9 x 109 m, are simultane...
Given data:
Distance between the events = 9 x 10^9 m
Speed of light = c = 3 x 10^8 m/s
Velocity of the moving frame = v = 0.8c
To find: Time interval between the two events in the moving frame.
Explanation:
We know that time and space are relative to the observer's frame of reference. So, the time interval between the two events in the stationary frame will not be the same as in the moving frame.
Let's consider the stationary frame as S and the moving frame as S'.
In the stationary frame S, the two events are simultaneous, i.e., they occur at the same time. So, the time interval between the two events in S is zero.
In the moving frame S', the two events are separated by a distance of 9 x 10^9 m. The velocity of the frame S' with respect to S is v = 0.8c.
We can use the Lorentz transformation equations to find the time interval between the two events in S'.
Lorentz transformation equations:
t' = γ(t - vx/c^2)
x' = γ(x - vt)
where γ = 1/√(1 - v^2/c^2) is the Lorentz factor.
Using the above equations, we can find the time interval between the two events in S' as follows:
t' = γ(t - vx/c^2)
The two events are simultaneous in S, so t = 0.
x = 9 x 10^9 m
v = 0.8c
t' = γ(0 - (0.8c)(9 x 10^9)/c^2)
t' = -7.2γ
We need to take the absolute value of t' to get the time interval between the two events in S'.
t' = 7.2γ
γ = 1/√(1 - v^2/c^2) = 1/√(1 - 0.8^2) = 1/√(0.36) = 1.67
t' = 7.2 x 1.67 = 12.024 s
Therefore, the time interval between the two events in the moving frame S' is 12.024 seconds, which is closest to option (b) 40s.
Distance between the events = 9 x 10^9 m
Speed of light = c = 3 x 10^8 m/s
Velocity of the moving frame = v = 0.8c
To find: Time interval between the two events in the moving frame.
Explanation:
We know that time and space are relative to the observer's frame of reference. So, the time interval between the two events in the stationary frame will not be the same as in the moving frame.
Let's consider the stationary frame as S and the moving frame as S'.
In the stationary frame S, the two events are simultaneous, i.e., they occur at the same time. So, the time interval between the two events in S is zero.
In the moving frame S', the two events are separated by a distance of 9 x 10^9 m. The velocity of the frame S' with respect to S is v = 0.8c.
We can use the Lorentz transformation equations to find the time interval between the two events in S'.
Lorentz transformation equations:
t' = γ(t - vx/c^2)
x' = γ(x - vt)
where γ = 1/√(1 - v^2/c^2) is the Lorentz factor.
Using the above equations, we can find the time interval between the two events in S' as follows:
t' = γ(t - vx/c^2)
The two events are simultaneous in S, so t = 0.
x = 9 x 10^9 m
v = 0.8c
t' = γ(0 - (0.8c)(9 x 10^9)/c^2)
t' = -7.2γ
We need to take the absolute value of t' to get the time interval between the two events in S'.
t' = 7.2γ
γ = 1/√(1 - v^2/c^2) = 1/√(1 - 0.8^2) = 1/√(0.36) = 1.67
t' = 7.2 x 1.67 = 12.024 s
Therefore, the time interval between the two events in the moving frame S' is 12.024 seconds, which is closest to option (b) 40s.
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Two events, separated by a (spatial) distance 9 x 109 m, are simultaneous in one inertial frame. The time interval between these two events in a frame moving with a constant speed 0.8 c (where the speed of light c = 3 x 108m/s) isa)60 sb)40 sc)20 sd)0 sCorrect answer is option 'B'. Can you explain this answer?
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