If a person A is stationary, let's say, with respect to a universal inertial frame, and another person B starts to move near light speed with respect to the same frame.

When both encounter each other again, both will have a clock measuring different time; they are not synchronized. Person B will have a clock that si delayed with respect to the person stopped.

Now, let a pipe that is stationary with respect to A be in the background. Person A will claim that person B should measure a pipe contracted.

That's the thing: Person A can say that the time interval passed of B will be $\Delta T_B = \Delta T_A / \gamma$

And he wil say also that the length of the B's tube is $L_B = L_A / \gamma$

That's my point, why is one called time dilatation, and the other length contraction, if both are lesser = contracted wrt to a person stopped?


3 Answers 3


If you draw the standard Minkowski diagram, with the (arbitrary) "at rest" frame having 90 degree $(x,t)$ axes, then the boosted frame has both $x'$ ($t'$) axis tilted inward by and angle:

$$ \alpha = \tan^{-1}\beta $$

with respect to $x$ ($t$). Moreover, both scales are dilated by a factor:

$$ U'/U = \sqrt{\frac{1+\beta^2}{1-\beta^2}}$$

as per wikiedpia:

enter image description here

Time and space are on equal footing in relativity, so this is mandatory. The primed frame has dilated clocks and dilated rulers.

To correct for that scaling, one can make a Loedel diagram in which each frame moves in opposite direction at equal speeds. Time dilation in the (green) primed frame then looks like: enter image description here

While length contraction arises via: enter image description here

So for time dilation, we're looking at the time interval $\Delta t'$ measured at fixed $x'$, from a position of fixed $x$, and comparing that with $\Delta t$.

In Length contraction, we're looking at the space interval $\Delta x$ measured at fixed $t$, and comparing that with $\Delta x'$ at fixed $t'$.

Which is a longwinded way of saying which reference frame gets the hypotenuse of the right triangle and which gets the base in the Loedel diagram.

  • $\begingroup$ Interesting. Note that Loedel diagram uses right angles for Euclidean-trig-calculation purposes only to relate these two frames when drawn in their “center frame”. The right angle in the Loedel diagram is between one line (say, worldline) of one observer and the other type (x-axis) of line of the other observer. This is different from the lorentz-invariant Minkowski right-angle which between the worldline and x-axis of each observer, which has direct physical meaning. (Analyzed in a third frame, the corresponding Loedel diagram would not show the corresponding two lines with a right angle.) $\endgroup$
    – robphy
    Oct 2, 2021 at 23:35

Time dilation is indeed a "dilation". Person A will find that clocks in the frame of reference co-moving with person B run slower. The correct relation is $$ \Delta t^\prime = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} $$ $\Delta t^\prime$ is the duration of a tick in B's frame as measured by A, while $\Delta t$ is the duration of a tick in the frame of A as measured by A himself. The former interval is larger, hence the term dilation. $v$ is of course the relative speed between the two frames A and B.

  • $\begingroup$ So when they talk about time dilatation, in not about time itself, but about the interval of the clicks? (Of course, if B' click take longer, its total time passed will be lesser than A, as we know) $\endgroup$
    – LSS
    Oct 2, 2021 at 13:28
  • $\begingroup$ It is about time, as you mentioned in the parenthesis. A will see that while his heart beats 60 times in 1 minute, B's heart will beat 40 times in that same minute. The 1 minute mentioned in the last sentence was measured using A's own clock. $\endgroup$ Oct 2, 2021 at 13:32

This spacetime diagram might help.

It's drawn on rotated graph paper to help us accurately visualize the tickmarks, which are based on the light-signals in a ticking light clock.


In this diagram, Alice (in red) is [inertially] at rest and Bob (in Blue) is in inertial motion, with velocity (6/10)c.

Alice measures Bob's time-interval OQ (an elapsed time on Bob's watch) and a space-interval OL (say, the length of a meterstick carried by Bob, marked as the distance between two worldlines parallel to Bob's worldline using a segment "perpendicular" to the parallel lines).

Alice measures OQ using OT, the time-component of the vector OQ.
Alice says event T on her worldline is simultaneous with distant event Q.
The ratio of what Alice measures to what Bob measures is $$\frac{OT}{OQ}=(10/8)=(5/4)>1,$$ hence this is called "time dilation".
Alice says that Bob's 8-second time-interval OQ takes 10 seconds to elapse according to her clock.

Alice measures the distance between the meterstick worldlines using OD.
OD is the segment on Alice's x-axis cut by the parallel lines.
The ratio of what Alice measures to what Bob measures is $$\frac{OD}{OL}=(4/5)<1,$$ hence this is called "length contraction".
Alice says that Bob's meterstick (which Bob says is 1 meter long according to the space markings along Bob's x-axis) is only (4/5) meter long according to the markings of Alice's x-axis.

That the above ratios are reciprocal is not accidental.

We work in the Minkowskian-trigonometric viewpoint--the natural trigonometry of special relativity (in the Cayley-Klein classification of geometry).

OTQ is a Minkowski-right-triangle,
with Minkowski-right-angle at T since $\vec{OT}\cdot\vec{TQ}=0$ with the Minkowski-dot-product.
The green "angle" between Alice's and Bob's worldlines ("timelines") is called the rapidity $\theta$, whose hyperbolic tangent is equal to the relative-velocity $v_{B,wrtA}=\tanh\theta$.
OQ is the hypotenuse [opposite the Minkowski-right-angle], and OT [what Alice measures] is the adjacent side. Call the ratio of adjacent-to-hypotenuse (think "[hyperbolic-]cosine") $$\gamma_T=\frac{adj}{hyp}=\frac{OT}{OQ}.$$ So, $\frac{OT}{OQ}=\gamma_T.$ In the usual notation, this is $$\frac{OT}{OQ}=\gamma \geq 1.$$

Similarly, OLD is a Minkowski-right-triangle,
with Minkowski-right-angle at L since $\vec{OL}\cdot\vec{LD}=0$ with the Minkowski-dot-product.

The green "angle" between Alice's and Bob's x-axes ("spacelines") is numerically equal to the rapidity.
OD [what Alice measures] is the hypotenuse and OL is the adjacent side. Call the ratio of adjacent-to-hypotenuse
$$\gamma_X=\frac{adj}{hyp}=\frac{OL}{OD}.$$ Since the green "angles" are numerically equal, the ratio $\gamma_X$ is numerically equal to $\gamma_T$.
So, $\frac{OD}{OL}=\frac{1}{\gamma_X}=\frac{1}{\gamma_T}.$ In the usual notation, this is $$\frac{OD}{OL}=\frac{1}{\gamma} \leq 1.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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