0
$\begingroup$

I tried to understand what is the reason of different clock's measurements in different frames.

For this I looked at easy example with distance S traveled by man with moving clock (velocity V1=2m/s) and distance traveled by man with another moving clock(velocity V2=1m/s). If clock is moving with velocity V1 then measurement T1=S/V1 and for V2 we have measurement T2=S/V2. First clock is more slowly than second clock.

It was bad example but something like example with time dilation effect where relation between velocity V and time T is more complicated (example with light clock, Pythagorean theorem).

Am I right that reason of different clock's measurements is velocity or am I missing something?

$\endgroup$
6
  • $\begingroup$ If the reason for different clock measurements were velocity we could have lived on with Newtonian mechanics and Galilean transformations forever. The ultimate reason is explained in short in this comment. $\endgroup$
    – Kurt G.
    Commented Aug 15, 2023 at 10:10
  • $\begingroup$ @KurtG. Ok. I mean that if velocity of moving clock is zero we have same time in different frames. But if we have non-zero velocity we can use formula with complicated relation between velocity V and times in different frames. $\endgroup$
    – Mike_bb
    Commented Aug 15, 2023 at 10:19
  • 1
    $\begingroup$ If you mean the relationship between $t$ and $t'$ in the well-known Lorentz transformation this should be obvious. $\endgroup$
    – Kurt G.
    Commented Aug 15, 2023 at 10:22
  • 1
    $\begingroup$ If it takes you T seconds to cover the distance S at 2 m/s, then it takes you 2T seconds to cover the distance S at 1 m/s. That's just due to the definition of speed. It has nothing to do with time dilation. $\endgroup$
    – PM 2Ring
    Commented Aug 15, 2023 at 11:03
  • 1
    $\begingroup$ @Mike_bb Let's say we have two (inertial) frames F and F', and F' is moving with constant velocity v relative to F. And we have two events (points in spacetime), A and B. In frame F, the time between A & B equals t and the distance between A & B equals s. In frame F', the time between the same A & B equals t' and the distance between A & B equals s'. The Lorentz transformation tells you how to calculate t' and s' from t, s, and v. $\endgroup$
    – PM 2Ring
    Commented Aug 15, 2023 at 11:34

1 Answer 1

2
$\begingroup$

In physics there are two different concepts of time. “Proper time” is the physical time, denoted $\tau$, read by a good clock. Proper time is invariant, meaning that all reference frames agree on what a given clock reads. “Coordinate time” is the mathematical time, denoted $t$, used in some coordinate system. Different frames use different coordinates, so they disagree on coordinate time.

If the coordinate system is a standard inertial frame then the relationship between the two is $$c^2 d\tau^2=c^2 dt^2-dx^2-dy^2-dz^2$$$$\frac{d\tau^2}{dt^2}=1-\frac{dx^2}{c^2 dt^2}-\frac{dy^2}{c^2 dt^2}-\frac{dz^2}{c^2 dt^2}$$$$\frac{d\tau}{dt}=\sqrt{1-\frac{v^2}{c^2}}=\frac{1}{\gamma}$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.