Video: Determining a Time Interval Change Due to Time Dilation

πœ‹-mesons have a proper lifetime of 2.60 Γ— 10⁻⁸ s. What lifetime do πœ‹-mesons have as measured by an observer moving relative to them at 2.70 Γ— 10⁸ m/s?

03:29

Video Transcript

πœ‹-mesons have a proper lifetime of 2.60 times 10 to the negative eighth seconds. What lifetime do πœ‹-mesons have as measured by an observer moving relative to them at 2.70 times 10 to the eighth meters per second?

As we work through this problem, we’ll assume that the speed of light 𝑐 is exactly 3.00 times 10 to the eighth meters per second. In this statement, we’re told that πœ‹-masons have what’s called a proper lifetime of 2.60 times 10 to the negative eighth seconds. We’ll call that 𝑑 sub 𝑝. We’re then told that there is a reference frame moving relative to the πœ‹-mesons at 2.70 times 10 to the eighth meters per second. We’ll call that speed simply 𝑣. We’re asked to solve for the lifetime of πœ‹-mesons as measured by an observer moving at the speed 𝑣 relative to them. We’ll call that lifetime 𝑑.

This is a problem involving two inertial frames of reference: one in which the πœ‹-mesons are stationary and another which is moving relative to the πœ‹-mesons at the speed that we’ve called 𝑣. There is a correlation or a connection between time experienced in one frame and the other. That correlation in time is that the time passed in one frame, we’ll call it 𝑑 prime, that correlation is that the time passed in one frame, 𝑑 prime, is equal to 𝛾 times the proper time or the time passed in the other frame that we call 𝑑.

Remember that 𝛾 is defined as one divided by the square root of one minus 𝑣 squared over 𝑐 squared. As it relates to our problem, we can use this time correlation relationship to write that 𝑑, the time we want to solve for, is equal to 𝛾 times 𝑑 sub 𝑝, the proper lifetime of πœ‹-mesons. We can also substitute in for 𝛾 the more expanded form of that symbol. So that 𝑑 is equal to one divided by the square root of one minus 𝑣 squared divided by 𝑐 squared all multiplied by 𝑑 sub 𝑝, the proper lifetime.

Since we’re given 𝑣 and 𝑑 sub 𝑝 in the problem, we can insert those values into this equation. Recall that 𝑐, the speed of light, is 3.00 times 10 to the eighth meters per second. So 𝑑 is equal to 2.60 times 10 to the negative eighth seconds divided by the square root of one minus 2.70 times 10 to the eighth meters per second squared divided by 3.00 times 10 to the eighth meters per second squared.

When we calculate this value, we find a time of 5.96 times 10 to the negative eighth seconds. This is the lifetime of πœ‹-mesons as measured by an observer moving relative to them at the speed of 𝑣.

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