chinglu:
You're still making the same mistake.
I have made this simple. There is a clock at k in the context of the primed frame with k < 0.
The question is from the time the origins are the same to the time k and the unprimed origin is the same, what will be the time on the clock at k and what will be the time on the unprimed origin clock.
You mean there's a clock at (x',t')=(-k,0). Is that the clock you're talking about?
Apply the Lorentz transformation to that event:
$$x = \gamma (x' + vt') = -\gamma k$$
$$t = \gamma (t' + vx'/c^2) = \gamma (0 -vk/c^2) = -\gamma vk/c^2$$
So, in the unprimed frame, the spacetime coordinates of that event are:
$$(x,t) = (-\gamma k, -\gamma vk/c^2)$$
Do you agree? Yes or no?
You then introduce some clock at k/ γ in the unprimed frame when that clock is not even in the problem.
Well, let's ignore that clock then, since it's not in the problem.
I assume you understand when the origins are the same, t'=t=0.
Yes. That's how you set up the problem.
Note that the event (x',t')=(-k,0) occurs at the same time the origins are "the same" in the primed frame. However, in the unprimed frame the two events are not simultaneous.
Do you agree? Yes or no?
Next, from the view of the primed frame, the unprimed origin is a distance k from the clock at the unprimed origin when the two origins are the same.
Hopefully, you can understand, the clock at k will elapse -k/v for the unprimed origin to reach k. Note all measurements are in the context of the primed frame so this should be simple with not disagreements.
So the end event in the primed frame occurs at:
$$(x',t')=(-k,k/v)$$
Do you agree? Yes or no?
Next, the final question is what is the elapsed time on the unprimed origin clock?
1) From the view of the primed frame, this is standard time dilation so, the primed frame concludes the unprimed origin clock elapses -k/(γv)
2) From the view of the unprimed frame, since the clock at k is measured in primed frame coordinates, then when the origins are the same, the unprimed frame concludes primed clock at k is a distance k/γ from the unprimed origin. Since the distance is k/γ to the unprimed origin and the speed is v, then the elapsed time from the view of the unprimed origin on the unprimed origin clock is -k/(γv).
In the unprimed frame, the end event is found by applying the Lorentz tranformation:
$$(x',t')=(-k,k/v)$$
$$x=\gamma (-k + vk/v) = 0$$
$$t=\gamma (k/v - vk/c^2) = \gamma (k/v)(1-(v/c)^2) = k/\gamma v$$
The elapsed time in the unprimed frame is the difference between the final event and the initial event:
$$\Delta t = \gamma (k/v)(1-(v/c)^2) - (-\gamma vk/c^2) = \gamma k/v$$
Do you agree? Yes or no?
Thus, the time elapsed in the unprimed frame is
longer than in the primed frame.
Compare:
$$\Delta t_{Chinglu} = k/\gamma v$$
$$\Delta t_{Correct} = \gamma k/v$$
The rest of your post is based on your mistake, so there's no need to reply to it.
To summarize your error, you consider a clock in the unprimed frame at the same location as k when the origins were the same and placed some time on it. That clock at that location is not in any way involved in this problem and has no affect on the conclusions. We are only concerned with two clocks, the clock at the unprimed origin and the clock at k in the primed frame.
In this post I have only used the clock that YOU specified.
Do you agree? Yes or no?
Can you see your mistake, now, finally?
If not, please review posts #117, #118, $165 and #177, where I explained your error in more detail.
If you can find a mistake in any of those posts, please post the mistake. I note again that you have still NEVER gone through those posts line-by-line to point out any mistake.
You do not because you cannot.
You're a troll.