I can understand the mathematical explanation for the reason why there should be a length contraction, but I fail to understand it intuitively. That is why I tried to explain it using spacetime diagrams, but for some reason, I was unable to do so.

Let us use the following procedure to measure the length of the rod from S's reference frame: S moves with $\vec{v} = v \hat x$ and sets it's watch to $t = 0$ when it is in one end of the rod, and looks at it's watch when it is in the other end of the rod and sets $t = t_2$. Let us also put one end of the rod, the one which $S$ visits first, to the origin of $S'$. We have two events $$e_1: \quad (t_1', x_1') = (t_1', 0) \quad and \quad (t_1, x_1) = (0,0),$$ $$e_2. \quad (t_2', x_2') = (t_2', L_0)) \quad and \quad (t_2, x_2 = (t_2, 0))$$ Note that, the rod is stationary wrt S' and both events happen at the origin of S wrt S.

Since we are relying on $t_2$ to calculate the length of the rod, graphically (see the above figure), $$t_2 = \sqrt{L_0^2 + (t_2')^2}.$$

If we just cheat a bit (to see whether we are on the right track) and use Lorentz transformations, we can see that $t_2' = \frac{t_2}{\sqrt{1-v^2}}$, which means that the above equation implies $$t_2 = \sqrt{v^2 + 1}t_2' \quad \Rightarrow \quad x_2 = \sqrt{v^2 + 1}L_o,$$ which is clearly wrong.


What am I doing wrong?


4 Answers 4


As @Umaxo says, you are using the wrong metric. Spacetime-intervals on the diagram are not measured with a ruler that you rotate (in a Euclidean way).

Here is a diagram on rotated graph paper (to help us visualize the tickmarks better) that show the elapsed time along worldlines.

UPDATE: The key geometrical idea here is that the ticks are marked off by what I call "light-clock diamonds" (the causal diamond* between successive ticks) which are traced out by light signals in a light clock. The area of the light-clock diamond is an invariant. Boosting the first light-clock diamond with corner at e1 traces out the unit-hyperbola. (In the fact, the square-interval along the diagonal of a causal diamond is equal to the number of light-clock diamond areas contained in the causal diamond.)

UPDATE2: The "causal diamond of the segment from e1 to B" is the intersection of the future [light-cone and its interior] of e1 and the past of B. In the diagram shown, the area is 25 light-clock diamond areas, which is the square-interval of the segment from e1 to B.

UPDATE3: Since the OP provided an answer to the original question with the labels switched and a new strategy, I modified my original diagram to show how length contraction can be shown on a spacetime diagram (on rotated graph paper) for each observer measuring the other observer's ladder. The relative velocity is (3/5)c. Although each observer carries a ladder 5 units long, the other observer measures the length of the ladder (the spatial distance in the measurer's frame between parallel worldlines of the endpoints of the moving ladder) as 4 units long.

(This spacetime diagram shows length contraction and time dilation and relativity of simultaneity--and the symmetry between the inertial observers.)


UPDATE4: Here's a cleaned-up version of the diagram.
For v=(3/5)c, the Doppler factor k=2.
The boosted light-clock diamond will be stretched by k in the frontward-null direction and shrunk by k in the rearward-null direction (to preserve area... the boost in light-cone coordinates with k and 1/k as eigenvalues and the lightcone directions as eigenvectors). Subdivided grids help you draw the diamonds. (Relative speeds with rational Doppler factors work best since these lead to Pythagorean triples which can be drawn exactly on this rotated graph paper and lead to calculations with simple fractions... )

https://doi.org/10.1119/1.4943251 "Relativity on rotated graph paper" American Journal of Physics 84, 344 (2016)



You are using wrong metric. The relation $t_2=\sqrt{L_0'^2+(t_2')^2}$ is correct on your sheet of paper (I would advice using prime for all quantities measured in primed frame). This is not correct in spacetime. In spacetime, the relation is $$t_2=\sqrt{-L_0'^2+(t_2')^2}=\sqrt{-v^2+1}t_2'$$ which is consistent with your time dilation.

The problem is that you are assuming your picture correctly depicts all relations in the sense of Euclidean geometry. But this is something you need to show first and the result is that you need to interpret the picture in the sense of Minkowski geometry.


For an intuitive explanation you need a better diagram. The spacecraft measures length L at equal time in the spacecraft frame.

enter image description here

The spacecraft’s clock is in the bow. The spacecraft and Earth set their clocks to zero when the bow passes the Earth clock. Earth uses radar to measure the distance, $l$, from bow to stern, by sending a signal at time $-l$, which returns at time on the Earth clock. The same signal is used to determine the proper length, $L$, as measured on the spaceship. Using the Doppler shift, the outgoing signal passes the bow at time on the spacecraft’s clock. The returning signal reaches the bow at time $-l/k$. You should already have $$k^2 = \frac{1+v}{1-v}$$ $$\gamma = \frac{1}{\sqrt{1-v^2}}$$

So the spacecraft’s proper length is

$$L=\frac{kl+l/k}{2} = \frac{k^2+1}{2k}l=\frac{(1+v) + (1-v)}{2(1-v)}\sqrt{\frac{1-v}{1+v}l} = \gamma l$$

  • $\begingroup$ But, I am specifically giving you another measurement method which does not do "[...]measures length L at equal time in the spacecraft frame" $\endgroup$
    – Our
    Commented Apr 22, 2020 at 8:29
  • $\begingroup$ Thanks a lot for the effort and the answer, but if I wanted a textbook explanation, I would look at it, I am trying to understand what is wrong with MY argument $\endgroup$
    – Our
    Commented Apr 22, 2020 at 8:30
  • $\begingroup$ You just said. If you do not measure length at equal time, you are not doing Lorentz transformation. You will need the maths of gtr to work in other coordinates. $\endgroup$ Commented Apr 22, 2020 at 9:18

Let $S'$ be a reference frame fixed on the rod and let us measure the positions of both ends of the rod (at the same time) when one end is at $x = 0$ wrt (with respect to) $S$. We have two events $$e_1: \quad (t, x_1) = (0, 0)$$ $$e_2. \quad (t, x_2) = (0, L) \quad and \quad (t_2', x_2') = (t_2, L_0))$$

Using the Lorentz transformations, we have $$t_1' = \gamma (0 - v*0) = 0$$ $$x_1' = \gamma (0 - v*0) = 0$$ $$t_2' = \gamma (t - v*0)$$ $$L_0 = x_2' = \gamma (L - v*0) = \gamma L \quad \Rightarrow L = \frac{L_0}{\gamma},$$ as desired.

Moreover, it can be described graphically as,

enter image description here

  • 1
    $\begingroup$ Compared to your original diagram, you now have switched prime and unprimed?...Not just boosted into the unprimed frame. $\endgroup$
    – robphy
    Commented Apr 22, 2020 at 14:38
  • $\begingroup$ I updated the diagram in my answer to handle that situation (independent of the labeling). $\endgroup$
    – robphy
    Commented Apr 22, 2020 at 15:12

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