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The spatial distance between 2 points supposes being taken at the same time. In this case, the moving observer must measure the distance $0a$ at the same time t'. If we take t = 0 and t'= 0 for x = 0 …

I don't know if it helps, but one problem that I see with distances in relativity is how to measure it in typical examples: A trip from earth to Alpha Centauri, that is 4 ly distance, at a speed of 0, …

Suppose a ship with a velocity U to an inertial frame of reference. The force term relates to the acceleration as measured by the crew of the ship. If someone drops an object inside the ship, it falls …

For the platform frame, any position in the train is: $x' = \gamma(x - vt)$ If $C$ is middle point of $A$ and $B$ in the train's frame: $x'_A + x'_B = 2x'_C$ So, for the same time $t$ measured in …

In the first part, we can assume that two distant stars at opposing distances were colonized by humans in previous trips, so clocks and calenders there are synchronized to the earth time. When the twi …

Alice point of view: Bob is moving toward her so his clock is counting time two times slower and when it stops it displays half-hour. Their clocks can be compared that way when both are momentarily …

The answer is relativity of simultaneity. When $t = 0$ for B frame, A has at this very instant (for A frame) $t_A = vX_B$, where $X_B$ is negative if A is approaching with positive velocity $v$.(I use …

The factor for time dilation can be understood from the Minkowskian metric with one spatial dimension for inertial frames: $$c^2d\tau^2 = c^2dt^2 - dx^2 \implies \frac{d\tau^2}{dt^2} = 1 - \frac{v^2}{ …

The experience of person 1 and 2 about the simultaneity is not falsified by the observer on the station. What they say is that the events were simultaneous in the train's frame where they were. The ob …

It is useful to draw a diagram $t \times x$. Then calculate the equations for the straight lines of the beam and of the rocket trajectory. The interception is the event $t,x$ where the light reach the …

Suppose a rocket going from Earth to Alpha Centauri. From the rocket frame, the distance between these two locations is shorter than the 4 light years that an Earth observer measures. That is length c …

Even for the reference frame of the hose, the fact that water is passing through the tap B doesn't mean that the tap A is open at the same instant. The 2 tap operators could arrange to open them at in …

We can think of 2 events in space-time. A pulse being emitted by the source and the spot receiving the pulse. Let's take the angle $\theta = 90^{\circ}$ to simplify the analysis, and the z axis being …

If so then the star should have moved close to me at faster than c. Note that if you invert the situation, and the star were accelerating to an inertial ship, it would not be coming faster than c. S …

What is the distance separating Alice and Bob at the U-turn? Instead of call Bob's frame as "immobile", it is more precise to take its frame as the reference for coordinate time. That means: there a …