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The most energetic proton ever detected in the cosmic rays coming to Earth from space had an astounding kinetic energy of 3.0xx10^(20)eV (enough energy to warm a teaspoon of water by a few degrees). How long does the trip take as measured in the reference frame of the proton? |
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Answer» Solution :KEY IDEAS 1. This problem involves measurements made from two (inertial) reference frames: one is the Earth-Milky Way frame and the other is attached to the proton. 2. This problem also involves two events: the first is when the proton passes one end of the diameter along the Galaxy, and the second is when it passes the opposite end. 3. The time interval between those two events as measured in teh proton.s reference frame is the proper time interval `Deltat_(0)` because the events occur at the same location in that frame-namely, at the proton itself. 4. We can find the proper time interval `Deltat_(0)` from the time interval `Deltat` measured in the Earth-Milky Way frame by using Eq.`(Deltat=gammaDeltat_(0))` for time dilation. (Note that we can use that EQUATION because one of the time MEASURES is a proper time. However, we get the same RELATION if we use a Lorentz transformation.) Calculation: Solving Eq. for `Deltat_(0)` and substituting `gamma` from (a) and `Deltat` from (b), we find `Delta t_(0)=(Delta t)/(gamma)=(9.8xx10^(4)y)/(3.198xx10^(11))` `=3.06xx10^(-7)y=9.7s.` In our frame, the trip takes 98 000 y. In the proton.s frame, it takes 9.7 s! As PROMISED at the start of this CHAPTER, relative motion can alter the rate at which time passes, and we have here an extreme example. |
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