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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, $$x_1=0,t_1=0$$
$$x_1'=0,t_1'=0$$

Event-2 $B$ reaches at point $P$ which is at $x=3\;light\;years$
$$x_2=3ly,t_2=\frac{3ly}{0.6c}=5\;years$$
In $S'$ frame, $B$ always is at rest and see the point $P$ to coming towards it by velocity $0.6c$
$$x_2'=0,t_2'=?$$

By Lorentz transformation,
$t_2'=\gamma\Big(t_2-\frac{vx_2}{c^2}\Big)$
$t_2'=1.25(5yr-(0.6\times3)yr)$
$\implies t_2'=1.25(3.2)=4yr$
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 $S'$=$\frac{2.4ly}{0.6c}=4yr$

Event-3
Now suppose if I am interested in finding out the position of origin of $S$ in $S'$ when $B$ reaches $P$
$$x_3=0,t_3=5yr$$
$$x_3'=?$$
$x_3'=\gamma(x-vt)=1.25(-0.6c\times5yr)=-3.75ly$

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.

Addendum
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?

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    $\begingroup$ 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 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 at 13:28
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    $\begingroup$ In S', when O is at -2.4 ly , the event of P reaching B occurs $\endgroup$ May 17 at 13:44
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Additional comment regarding the use of the Lorentz transformation for this problem.
SEE BELOW.


UPDATE:

(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.
    Instead,
    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.
    ADDITIONAL COMMENT:
    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.

robphy-RRGP-lengthContraction

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!

robphy-RRGP-lengthContraction-boosted


ORIGINAL:

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).

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  • $\begingroup$ 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 at 12:38
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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.

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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:

$$x_3'=\gamma(x-vt)=1.25(-0.6c\times5yr)=-3.75ly$$

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

$$t_3'=\gamma(t-\frac{vx}{c^2})=6,25(year)$$

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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