Consider two persons $A$ and $B$ and $B$ is moving with velocity $+0.6c$ in $+x$ direction.
Take the frame $S$ in which A is at rest but $B$ appears to move in the $+x$ direction with velocity $0.6c$. $S'$ is the frame in which $B$ always is at rest and is at origin.
So, $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{1}{\sqrt{1-0.36}}=\frac{1}{0.8}=1.25$
Event-1 Origin of $S$ and $S'$ are coinciding at $t=t'=0$.
So, $$x_1=0,t_1=0$$

Event-2 $B$ reaches at point $P$ which is at $x=3\;light\;years$
In $S'$ frame, $B$ always is at rest and see the point $P$ to coming towards it by velocity $0.6c$

By Lorentz transformation,
$\implies t_2'=1.25(3.2)=4yr$
So, in $S'$ frame after $4yr$, P reaches the origin of $S'$.

Now we can get the same result using Lorentz contraction,
$Length\;OP$ in $S'$ frame=$\frac{Length\;OP\;in\;S\;frame}{\gamma}=\frac{3ly}{1.25}=2.4ly$
So, Time taken by far end to reach the origin in $S'$=$\frac{2.4ly}{0.6c}=4yr$

Now suppose if I am interested in finding out the position of origin of $S$ in $S'$ when $B$ reaches $P$

But $Length\;OP$ in $S'$ frame=$2.4ly$. This means when the far end reaches $x'=0$ this means the other end reaches $x'=-2.4ly$.

The question is using Lorentz transformation, $x_3'=-3.75ly$. But by considering $OP$ as a rod and using Lorentz contraction, I get $x_3'=-2.4ly$.

Why this is so? And which one is correct? I am very confused.

Basically the question is how the two events "position of $O$ in $S'$ frame when $P$ is at the origin of $S'$" and "position of $O$ in $S'$ frame when an event occurs at the origin of $S$ at $t=5ly$" are different?

  • 1
    $\begingroup$ In frame S, the event (0,5) is simultaneous with B reaching P. In frame S', these 2 events would not be simultaneous. So, in frame S', the event of B reaching P would be simultaneous with O at -2.4 ly , not with O at -3.75 ly $\endgroup$ May 17, 2021 at 13:20
  • $\begingroup$ So, in $S'$ when $O$ is at $-3.75ly$ then $P$ is not $0$ but at some $-ve$ value (such that the separation between them is $2.4ly$). $\endgroup$
    – Iti
    May 17, 2021 at 13:28
  • 1
    $\begingroup$ In S', when O is at -2.4 ly , the event of P reaching B occurs $\endgroup$ May 17, 2021 at 13:44

3 Answers 3


Additional comment regarding the use of the Lorentz transformation for this problem.


(A note on notation, you use capital letters to refer to points in "space", which trace out worldlines. In my original answer, I referred to "P" as an event (akin to a point in the diagram). In your notation, I should have referred to it as "event-2", when worldline-P meets worldline-B.)

A spacetime diagram will help interpret your numbers.
Drawing it on rotated graph paper will help me draw in the ticks ("light clock diamonds") as traced out by light-rays in various light-clocks.

In your "Event-3" section,

  • Your first attempt uses the event (that I call) ev5
    which has coordinates $(t_5, x_5)=(5,0)$ and $(t'_5, x'_5)=(6.25, -3.75)$.

    Although the (Alice) S-frame regards ev5 to be simultaneous with ev2,
    the (Bob) S'-frame does NOT regard them as simultaneous.
    That is, the (Bob) S'-frame does NOT regard the spacetime-segment ev2-ev5 as purely-spatial.
    it is the spacetime-segment ev4-ev5 that is purely spatial according to the (Bob) S'-frame.
    That displacement is $-3.75$, as you computed.

    Your method using the Lorentz transformation is correct... but you used the wrong event.
    Use the other part of the Lorentz transformation:
    $$t_3'=\gamma(t_3-vx_3)$$ With the ev5 , one gets $t_3' =\frac{5}{4}(5-\frac{3}{5}0)=6.25$.
    According to the (Bob) S'-frame, this is NOT simultaneous with ev2 $(t'_2,x'_2)=(4,0)$.
    With the ev3 [below] , one gets $t_3' =\frac{5}{4}(3.2-\frac{3}{5}0)=4$.
    According to the (Bob) S'-frame, this IS simultaneous with ev2 $(t'_2,x'_2)=(4,0)$.

  • Your second attempt (via the length contraction formula) implicitly uses the [correct] event ev3
    which has coordinates $(t_3, x_3)=(3.2,0)$ and $(t'_3, x'_3)=(4, -2.4)$.

    The (Bob) S'-frame DOES regard ev3 to be simultaneous with ev2.
    That is, the (Bob) S'-frame DOES regard the spacetime-segment ev2-ev3 as purely-spatial.

    That displacement is $-2.4$, as you computed.


Below the same situation in the (Bob) S'-frame.
Although it's not shown below, you can fill in the following:
from the separation event ev1, count up 5 ticks for Bob, then count over 3 space-ticks [sticks] to the left. You should meet the worldline of the (Alice) S-frame origin, which occurs 4 ticks along the (Alice) S-frame origin worldline.
SYMMETRY, in accordance with the principle of relativity!



For Event 3,
you used $𝑥_3=0,𝑡_3=5𝑦𝑟$,... which is simultaneous with $P$ in the S-frame.
But, that event is not simultaneous with $P$ in the $S'$-frame,
which would be associated with length-contraction (as a measurement in the S'-frame).

  • $\begingroup$ But when point $P$ reaches the origin of $S'$, then $O$ will reach $-2.4ly$ as the length of rod in $S'$ frame is $2.4ly$. Why can't we use this as $x_3$? May you please explain in a bit more detail. $\endgroup$
    – Iti
    May 17, 2021 at 12:38

The two positions of the origin of S that you have determined, namely -3.75ly and -2,4ly, differ because they relate to the position of the origin of S at two different times.

The first figure, namely -3.75ly, is the position of the origin of S at time t=5yr.

The second figure, namely -2.4ly, is the position it would be at t'=4yr.

The inconsistency arises because you use the word 'when' to imply two different simultaneity conditions without realising it.

In the sentence that reads 'Now suppose if I am interested in finding out the position of origin of S in S' when B reaches P' you use the word 'when' to mean at the same time in S.

In the sentence that reads 'This means when the far end reaches 𝑥′=0 this means the other end reaches 𝑥′=−2.4𝑙𝑦' you use the word 'when' to mean at the same time in S'.

So you are actually comparing two different events.


You assume that, when the origins of the frames coincide, $t=t'$. This isn't rue. The right transformation is:

$$t'=\gamma (t-\frac{vx}{c^2})$$

This means that when the origins of $S$ and $S'$ coincide, the clocks in $S$ will all tell you that $t=0$, while the clocks in $S'$ will tell you something different. If you move along the $x$-axis (while keeping the origins at place) in $S$, the clocks in $S'$ will give you a time $t'\neq 0$. This means that simultaneous events in $S$ are not simultaneous in $S'$.

But I see your problem. You wrote:


You have set $x=0$ and $t=5$. When you fill this in for the $t_3'$ transformation you get:


The first equation seems in contradiction with space contraction, which gives $2,4$. But note that in the case $x_3'=-3,75$ (coinciding with the origin of $S$), $x$ will be $x=3$ (coinciding with the origin of $S'$). So a length of $3,75$ in $S'$ corresponds to a length $3$ in $S$ (as a length $3$ in $S'$ corresponds to a length $2,4$ in $S$). That is, $3,75$ has contracted to $3$. You can't say that a length of $3$ has elongated to $3,75$ though, because you consider $S'$ to be the moving frame and $S$ the stationary one. If $S'$ were the stationary frame then indeed a length of $3,75$ in $S'$ would indeed have elongated in $S$ (to $\gamma\times 3,75$).
A $5(year)$ during kiss, given at the origin of $S$, seems to last $6,25(year)$ in $S'$. The clock at the origin of $S$ points to $5$, while the clock in $S'$ points to $6,25$ at $x_3'=-3,75$. That is, the kiss seems to go slower in $S'$. As it should be. A $4(year)$ during kiss give at the origin of $S'$ seems to last $5(year)$ in $S$. That is time seems to go slower in $S'$, as seen in $S$.
Very confusing! But it all works out fine.
Note one more thing. For $\frac{-2,4}{4}$ you will get the same result as $\frac{-3,75}{6,25}$, namely $-0,6$, the velocity of the frame $S$ wrt to $S'$. Likewise, you get $\frac{3}{5}=0,6$, the velocity of $S'$ wrt to $S$.


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