#physics #relativity #special-relativity

idea

Time is relativistic. In inertial reference frames, it is perceived to pass slower in objects that move faster.

Thought experiment

If a space ship at a speed $v$ compared to Earth has an experiment where a beam of light is reflected on a screen situated at a distance $h$, a single photon will take $t_0$ to move to the screen and then back.

Given that a photon moves at the speed of light:

$$ t_0 = 2 \times \frac{h}{c} $$

For an observer on Earth, given that the space ship is moving, the path followed by the photon is not a straight line, but forms a /\.

For the observer it travels a the distance $d$. Using Pythagora:

$$ d = 2 \times \sqrt{h^2 + L^2 } $$

Where $L$ is 1/2 the distance traveled by the spaceship.

The speed of light is a constant that does not change based on the inertial frame of reference. This means that for the observer, the photon takes $t$ to travel $d$:

$$ t = \frac{d}{c} \implies t = \frac{2 \times \sqrt{h^2 + L^2 }}{c} $$

We can then express $t$ as a function^[1] of $t_0$:

$$ t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}} $$

Comparing $t$ and $t_0$, for the observer on earth, time seems longer due to as $v$ increases. Time is dilated and seems to pass slower for the observer.

Application

The speed of light is extremely high, and for speeds encountered in Newtonian models, time dilation is mostly irrelevant ($v \ll c$, $\sqrt{1-v^2/c^2} \approx 0$ and $t \approx t_0$).

It applies to GPS satellites for example, where time-sync is crucial, and which travel at high speeds compared to the inertial reference frame of Earth, which leads to micro-seconds of drift. Their clock is re-adjusted using the special relativity formula.

links

references

Professor Dave / Special relativity part II

proof

[1]: We simplify by considering only the trip up.

$$ t_0 = \frac{h}{c} ; t = \frac{\sqrt{h^2 + L^2 }}{c} $$

Using $L = v \times t$:

$$ h = c \times t_0 \implies t = \frac{ \sqrt{c^2 t_0^2 + L^2 } }{ c } \newline \implies t^2 = \frac{c^2 t_0^2 + v^2 t^2}{c^2} \newline \implies c^2 t_0^2 + v^2 t^2 - t^2 c^2 = 0 \newline \implies c^2 t_0^2 + t^2 (v^2 - c^2) = 0 \newline \implies t^2 = \frac{t_0^2 c^2}{c^2 - v^2} \newline \implies t^2 = \frac{t_0^2 c^2}{c^2(1 - \frac{v^2}{c^2})} \newline \implies t = \frac { t_0 } { \sqrt{1 - \frac{ v^2}{c^2}} } $$