# Tricky relativity problem

So how is possible the moving object to have a higher reading that the stationary. Since time slows down then we should have a higher reading that her. Right?

jambaugh
Gold Member
So how is possible the moving object to have a higher reading that the stationary. Since time slows down then we should have a higher reading that her. Right?

It is not good conceptually to say "time slows down". Time is how far we move through space-time. We each see a time axis in the direction we're moving (we each see ourselves as stationary through space and "moving" through our time at a rate of 1 second per second.). If you and I are not moving in parallel (i.e. we're not spatially stationary w.r.t. each other) then my time direction and yours don't agree. When you compare what my watch reads to what yours does at a given event we see distinct numbers because we are measuring temporal distances along distinct paths.

It is not different in principle than the fact that you walking North will see me, walking North-East, take more steps to cross a given latitude line. (Just looking at footprints and do not invoking time out of context.) Likewise I defining N' latitude' lines so I'm moving North' and you're moving North'-West' will see you take more footsteps than I to cross my latitude' lines.

The distinction, obvious here, is the analogous principle to time dilation and relativity of simultaneity but the actual effect is reversed with time due to the peculiar hyperbolic geometry of space-time. Relatively moving observers will each see the other taking less clock-steps to pass our own temporal "latitude" lines.

If you understand this then you can correctly parse the problems you are given and read in them the implicit assumptions (e.g. one observer O is said to be stationary and so it is relative to his frame that statements of simultaneity are intended even though they are stated in absolute terms.)

BTW I find it much clearer to express Lorentz transformations in terms of hyperbolic trig:
For some boost parameter b we have:

$$\beta = \tanh(b),\quad \gamma = \cosh(b),\quad \beta\cdot\gamma = \sinh(b)$$
thence
$$\Delta x' = \cosh(b) \Delta x + \sinh(b) c\Delta t$$
$$c\Delta t' = \sinh(b)\Delta x + \cosh(b) c\Delta t$$
where you solve for the boost parameter b using known quantities. Note here the sign of beta and b matter. Be sure beta refers to the velocity of the unprimed (pre-transform) observer relative to the primed (post-transform) observer.

Note also the problem you were given makes use of the Pythagorean triple:
$$3^2 + 4^2 = 5^2$$
but in hyperbolic form:
$$5^2 - 4^2 = 3^2$$
Thence $$\cosh(b) = 5/3,\quad \sinh(b) = 4/3,\quad \tanh(b) = 4/5 = 0.8$$