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

2 Answers 2


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

Actually, it's $$L^{μ\prime}=(γt_L - γv x_L, γx_L - γvt_L, 0, 0), \hspace 1em R^{μ\prime}=(γt_R - γv x_R, γx_R - γvt_R, 0, 0)$$ and this is actually what you set: $$γt_R - γv x_R = γt_L - γv x_L \hspace 1em⇒\hspace 1em t_R - t_L = v\left(x_R - x_L\right),$$ and this is what you actually get: $$\begin{align} \left(γx_R - γvt_R\right) - \left(γx_L - γvt_L\right) &= γ \left(x_R - x_L - v\left(t_R - t_L\right)\right) \\ &= γ (1 - v^2) \left(x_R - x_L\right) \\ &= \sqrt{1 - v^2} \left(x_R - x_L\right), \end{align}$$ after noting that $γ = 1/\sqrt{1 - v^2}$.

So, $t_R ≠ t_L$: simultaneity is relative. Another way of saying the same thing is that infinite speed - the speed of being simultaneous - is relative, and the "simultaneous" of the left and right ends now become a speed: $$\frac{x_R - x_L}{t_R - t_L} = \frac{1}{v}.$$ That's $c^2/v$ in your units; which is faster than light. In Relativity, "faster than light" means "simultaneous in someone's frame". Unlike the world of Newtonian physics, infinity is not the absolute speed, but relative. Instead, the absolute speed is finite and non-zero.

This can be generalized to cover all cases for absolute speed being zero, finite/non-zero or infinite. Write $c = \sqrt{β/α}$ and generalize from that. Write the transforms as: $$(t',x',y',z') = (γ(t - αvx), γ(x - βvt), y, z), \hspace 1em γ = \frac{1}{\sqrt{1 - αβv^2}}.$$ This is suitable for a geometry in which the invariants are: $$ β dt^2 - α\left(dx^2 + dy^2 + dz^2\right), \\ dt \frac{∂}{∂t} + dx \frac{∂}{∂x} + dy \frac{∂}{∂y} + dz \frac{∂}{∂z}, \\ β\left(\left(\frac{∂}{∂x}\right)^2 + \left(\frac{∂}{∂y}\right)^2 + \left(\frac{∂}{∂z}\right)^2\right) - α\left(\frac{∂}{∂t}\right)^2. $$

Now, repeat the exercise: $$γ\left(t_R - αvx_R\right) = γ\left(t_L - αvx_L\right) \hspace 1em⇒\hspace 1em t_R - t_L = αv\left(x_R - x_L\right).$$ Then, $$\begin{align} γ\left(x_R - βvt_R\right) - γ\left(x_L - βvt_L\right) &= γ\left(x_R - x_L - βv\left(t_R - t_L\right)\right) \\ &= γ\left(1 - αβv^2\right) \left(x_R - x_L\right) \\ &= \sqrt{1 - αβv^2} \left(x_R - x_L\right). \end{align}$$

An absolute zero speed is the case $α ≠ 0$, $β = 0$; an absolute infinite speed is the case $α = 0$, $β ≠ 0$. The case where all speeds are absolute is where $α = 0$, $β = 0$. In each of those cases, there is no length contraction, since $αβ = 0$ and $\sqrt{1 - αβv^2} = 1$, and also $γ = 1$.

In the "static" case $α = 0$, $β = 0$, in place of those listed above, the invariants are now: $$ dt, \hspace 1em dx^2 + dy^2 + dz^2, \\ \hspace 1em dx \frac{∂}{∂x} + dy \frac{∂}{∂y} + dz \frac{∂}{∂z}, \\ \left(\frac{∂}{∂x}\right)^2 + \left(\frac{∂}{∂y}\right)^2 + \left(\frac{∂}{∂z}\right)^2, \hspace 1em \frac{∂}{∂t}. $$

The case of finite, non-zero absolute speed is $αβ$, with the speed being $c = \sqrt{β/α}$. In that case, $βv$ is what actually has the units of speed ($αv$ has units of "slowness", or inverse speed, which is more suitable for the case of zero absolute speed), so the length contraction factor is $\sqrt{1 - (βv)^2/c^2}$.

Finally ... there is the as-yet-unmentioned case of $αβ < 0$. What's that? That's the one where the coordinate invariant is $$|β|dt^2 + |α|\left(dx^2 + dy^2 + dz^2\right).$$ That's the case, not of a (3+1)D space-time, but of a 4D timeless space, where $t$ is now a space coordinate. Then, the "contraction" factor is now a "stretching" factor $\sqrt{1 + |αβ|v^2}$.

That's how you tell the difference between $t$ being a temporal dimension versus $t$ being a spatial dimension. The transform above also applies to the Euclidean 4D case, except that $(t',x',y',z') = (-t,-x,y,z)$ can now also be reached, since the transform is just ordinary rotation in the $tx$ plane, but are not listed above, while in the other cases they cannot, but have to be separately stipulated - if at all.

What you tried to do, in setting $t_R = t = t_L$, is force simultaneity back to being absolute, i.e. to make infinity an absolute speed, again; and you tried forcing that, in conjunction with the transforms which already assume that the absolute speed $c$ is finite and non-zero.

But, in the general setting, for transforms in a direction collinear to a trajectory $x = ut$, the speed $u$ transforms as $$u' = \frac{u - βv}{1 - αvu}.$$ The only way to get two speeds $u_0$ and $u_1$ to be absolute, where $|u_0| ≠ |u_1|$ is where (for any transform speed $v ≠ 0$): $$u_0 = \frac{u_0 - βv}{1 - αvu_0}, \hspace 1em u_1 = \frac{u_1 - βv}{1 - αvu_1}.$$ That is: $$u_0 - αv\left(u_0\right)^2 = u_0 - βv, \hspace 1em u_1 - αv\left(u_1\right)^2 = u_1 - βv,$$ or $$α\left(u_0\right)^2 = β = α\left(u_1\right)^2.$$ Thus $$α\left(|u_0|^2 - |u_1|^2\right) = 0,$$ from which it follows that $α = 0$ and, by virtue of the above relations, that $β = 0$. That's the case where all speeds are absolute.

Setting even two speeds absolute makes them all absolute; and by setting $t_R = t_L$ - in conjunction with the Lorentz transforms - you just made them all absolute.


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