# Confusing time dilation - proper time is higher?

The problem states that 2 rockets of proper length 100m are going in opposite directions. From the system of rocket A, the tip of B took 5 microseconds to pass the rocket A. If a clock on the tip of B marked t=0 when their tips met, what does the clock says when rocket B reaches the end of A. (I assume that all of this is measured from rocket A)

First, I computed the relative velocity (dividing the length travelled by the time it took), $$v= 2 \times 10^7$$ m/s. So $$\gamma=1.00223$$.

Then I used the Lorentz transform of times: $$t' = \gamma(t-(v \times 100)/c^2)$$, then $$t' = 4.989 \times 10^{-6}$$ seconds.

I understand the math but this doesn't match with the statement "proper time is always the lowest" because this proper time $$5 > 4.989$$ microseconds.

## 1 Answer

Your calculation is correct but you muddled which one is the proper time. Proper time is the time elapsed between events as observed in the frame in which those events are at the same spatial location. In other words, it is the time registered by a clock that is carried from one event to the other. This is the clock on the tip of B in this example. It registers the time $$t'$$.