I define a reference frame $S$ and a reference frame $S^\prime$ moving with velocity $v$ in the positive $x$-direction relative to $S$. There is a meter stick at rest in $S$ with left and right ends $L^\mu$ and $R^\mu$ respectively, which can be represented as spacetime points:



With $c=1$, the Lorentz transformation from $S$ to $S^\prime$ in matrix form is

$ \Lambda^{\mu^\prime}_{\;\;\mu} = \begin{pmatrix} \gamma & -\gamma v & 0 & 0 \\ -\gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} . $

Using this transformation on a general spacetime point in $S$, $x^\mu=(t,x,y,z)$, produces the coordinate relationships:

$t^\prime=\gamma t -\gamma v x$

$x^\prime=\gamma x -\gamma v t$.

Using these relationships, $L$ and $R$ can be written as points in $S^\prime$:

$L^{\mu^\prime}=(\gamma t -\gamma v x_L, \gamma x_L -\gamma v t, 0, 0)$

$R^{\mu^\prime}=(\gamma t -\gamma v x_R, \gamma x_R -\gamma v t, 0, 0)$.

(I believe my logical flaw is here) If our goal is to measure the distance of the meter stick with respect to $S^\prime$, we need to find the difference in the $x^\prime$ components of $L^{\mu^\prime}$ and $R^{\mu^\prime}$ when their $t^\prime$ components are equal, so we equate their time components:

$\gamma t -\gamma v x_L = \gamma t -\gamma v x_R$.

With $x_L=0$ and $x_R=1$, we get

$0=\gamma v x_R$,

which is only true when $v=0$ (which it isn't) or $x_R=0$ (which it isn't).

My question is: What is the logical flaw in equating the time components of the left and right spacetime points of the meter stick in order to find their $x^\prime$ components at a synchronized time $t^\prime$ so that the meter stick's contracted length can be measured as $\Delta x^\prime=R^{x^\prime}-L^{x^\prime}$?

edit: I want to clarify why I thought to equate their time components in the first place. My understanding is that the length of an object in a reference frame $F$ is the difference in their spatial coordinates when measured at the same time within that reference frame.

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    $\begingroup$ Try to measure them at different times then and check what times they should be measured to be simultaneous in the other reference system? (since you know it doesnt move in your reference system this could be an approach perhaps ?) $\endgroup$
    – Emil
    Aug 13, 2016 at 9:20

1 Answer 1


Using the Lorentz transformations the position of the left end of the rod at some time $t_L$ in $S^\prime$ is:

$$ (t_L, 0) \rightarrow (\gamma t_L, -\gamma vt_L) \tag{1} $$

Likewise the position of the right end of the rod at some time $t_R$ in $S^\prime$ is (taking the rod length to be $\ell$:

$$ (t_R, \ell) \rightarrow \left( \gamma(t_R-v\ell), \gamma(\ell-vt_R) \right) \tag{2} $$

We want to compare the $x^\prime$ position of the ends at the same $t^\prime$ but this will be at different values of $t$ i.e. $t_L \ne t_R$. If we require that $t^\prime$ be the same for the left and right ends then we get:

$$ \gamma t_L = \gamma(t_R-v\ell) $$

giving us:

$$ t_R = t_L + v\ell $$

Now we take the equation for $x^\prime_R$ from equation (2) above:

$$ x^\prime_R = \gamma(\ell-vt_R) $$

and substitute $t_R = t_L + v\ell$ to get:

$$\begin{align} x^\prime_R &= \gamma(\ell-v(t_L + v\ell)) \\ &= \gamma\ell(1 - v^2) -\gamma vt_L \\ &= \frac{\ell}{\gamma} + x^\prime_L \end{align}$$

which is the correct result.

Looking at your approach I think you are assuming that $t^\prime_L = t^\prime_R$ and $t_L = t_R$ and this can't be correct because both times can't be equal in both frames.

  • $\begingroup$ "I think you are assuming that $t^\prime_L = t^\prime_R$ and $t_L = t_R$" is completely correct. Defining $L^\mu=(t_L,0,0,0)$ and $R^\mu=(t_R,1,0,0)$ with separate time components made the math fall right into place. Thanks for the insight. $\endgroup$
    – xish
    Aug 13, 2016 at 10:42
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    $\begingroup$ And your clarification also makes it clear why I end up with the implication that $t^\prime_L=t^\prime_R$ only when $v=0$ or the meter stick has zero proper length, since those are the only situations when the times can be equal in both frames simultaneously. $\endgroup$
    – xish
    Aug 13, 2016 at 11:04

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