No, it is possible to conclude t'=t and in SR t' and t mean dt' and dt. So, the only way to make that happen is time expansion + time dilation.
But nothing can be measured and reconciled until both frames are brought together in the same inertial framework. So it is not possible to conclude anything without taking into consideration the accelerated motion that at least one, or both frames must experience, to synchronize their relative velocities.
Nothing was "added". That result is a simple application of the Lorentz transformation to an event that YOU specified. It's hard to make any sense of this bizarre statement. If you have the same object at two different events, then the proper time is always the shortest time interval that can be measured between those two events. Every frame other than the rest frame of the object will measure a longer time for the same two events. Do you agree?
Why do you think it is invalid? It comes straight out of the Lorentz transform: t' = 0, x' = -k \(t = \gamma(t' + \frac{vx'}{c^2})\) \(t=-\frac{\gamma k v}{c^2}\) This just tells that the primed clock at x'=-k doesn't show t'=0 until it is passing an unprimed clock reading \(t=-\frac{\gamma k v}{c^2}\) What do you not understand about this?
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No, this is not the case that this was the problem I set up. We were in the context of the unprimed frame and you added a time to the clock in the unprimed frame. Next, I will go ahead and ignore that and assume you wanted to add some time on the clock in the primed frame as I see Pete has done in the post following yours. First, it does not matter the start time of any clock. LT is about elapsed times. So, any start time you claim is not part of the answer. For example, assume Pete's initial condition start time on the clock at k in the negative x direction. Now, when the frames are the same, assume a light pulse is emiitted. If it is you contention that this "start time" is part of the problem, then try it out with the time of LT with the light sphere + "your start time" and you will find the light like space time interval is not invariant. That proves your method is wrong.
chinglu, you said: Note that you asked for the time interval in each frame, which is exactly what James did in post #117. In the primed frame, the time interval begins when t'=0, right? And at x'=-k and t'=0, the unprimed clock passing by reads \(t=-\frac{\gamma k v}{c^2}\) What do you not understand?
Obviously you're simply mistaken about this. Please review posts #117, #118, #165 and #177 of the current thread. If you find any errors in those, please point them out by quoting the relevant parts of those posts and explaining exactly where the mistake was made. Right now you're just making false claims about what I wrote. My entire explanation of where you went wrong is in those four posts. If you want to discuss this further, you'll have to step through those posts line by line and show an error. I note that I have invited you to do this explicitly about 10 times or more by now, and you have NEVER even started to try to do so. I regard you as a troll with nothing interesting to say. Your entire opening post and the entire point of this thread is about the time intervals between various events. To specify a time interval you need a start time and an end time - two events in spacetime. The Lorentz transformation first and foremost transforms spacetime events, not intervals. What do you not understand about this? There's no assumption. YOU, chinglu, specified which events would be used, not me. What do you mean by "when the frames are the same"? The frames are never the same. Why don't YOU "try it out"? Show me your maths. Show me your calculations. I'm sick of doing all your thinking for you. I'm sick of pointing out the same mistakes of yours over and over again. Get up off your backside, switch on your brain, and start thinking and actually doing something, rather than continuing to make idiotic claims.
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 then introduce some clock at k/ γ in the unprimed frame when that clock is not even in the problem. I assume you understand when the origins are the same, t'=t=0. 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. 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). Note how both frames agree on the time of the unprimed origin clock. There is no disagreement. So far, we have both frame agree the unprimed origin clock will elapse -k/(γv) from the time the origins are the same to the time the clock at k meets the unprimed origin. We have the primed frame claiming the clock at k will elapse -k/v. Last, we need the conclusion of the unprimed frame for the clock at k. This is standard LT. k us measured in light seconds and v is measured in terms of c. t = -k/(γv) x = 0 t' = ( t - vx/c² )γ =( -k/(γv) - 0)γ = -k/v. Let's makes sure we have the correct x'. x' = ( x - vt )γ = ( 0 - v(-k/(γv))γ = k. Yes, we have the correct x'. Therefore, both frames agree the clock at k will elapse -k/v. Both frames agree the clock at the unprimed origin will elapse -k/(γv). Finally, we take the view of the unprimed origin and the moving clock at k coming toward the origin beats time expanded as agreed on both frames. And frame agreement is necessary because of the invertibility of LT. 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.
chinglu: You're still making the same mistake. 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? Well, let's ignore that clock then, since it's not in the problem. 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? So the end event in the primed frame occurs at: \((x',t')=(-k,k/v)\) Do you agree? Yes or no? 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. 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.
Oh, I should add a couple more comments on chinglu's error: Here, you are assuming what you should be proving. It is indeed "standard time dilation", but you've got the frames around the wrong way. The way you have set up the problem, the primed frame measures the proper time, so the unprimed frame must measure a LONGER time, not a shorter time as you claim. In other words, as shown in my previous post, the gamma factor should be on top, not on the bottom, of your time. No. The way you have set up this problem has the clock at distance \(\gamma k\) from the unprimed origin when the origins are "the same", as shown in my previous post. No. Since the distance is \(\gamma k\) and the speed is v, the time taken is \(\gamma k/v\).
How many times does this have to be explained to you. You are trying to calculate some time at some clock at the same place as the k clock in the primed frame. This "place" in the unprimed frame is never part of the problem. Perhaps you should sit down and draw a pciture or something. Wharever clock is at the same place as k in the unprimed frame is NOT PART OF THE PROBLEM. Who cares, that event is not under consideration. We are only concerned with the origin of the unprimed frame. This is where you are getting confused. And that event at the unpirmed origin is 0 when (x',t')=(-k,0) since t'=0=t. OK Wrong, you are event mixing. In the unprimed frame, the k clock is a distance k/γ when the origins are the same. How can you possibly claim a clock at a distance k/γ takes kγ/v to reach the origin? You may have to give up on account of problem complexity.
chinglu: If you believe I have used the wrong spacetime events, then you must specify which spacetime events you are using. Please give the spacetime coordinates of all relevant events, in both frames. After you have done that, then we can consider time intervals. You must name the relevant events and give coordinates in the form (x,t)=(...,...), (x',t')=(...,...). That way, there can be no further confusion about which events we are considering. Can you do this? I don't think you are capable. I think you will avoid ever doing this. Because you are a troll.
James R I was very specific. In the unprimed frame, the k clock is a distance k/γ when the origins are the same. How can you possibly claim a clock at a distance k/γ takes kγ/v to reach the origin? You must first confess you were wrong and the correct answer above is k/γv.
I never made the claim in the second sentence, so let's move on and deal with the rest. You say "the k clock is a distance k/γ when the origins are the same." The origins are the same at t=t'=0. Correct? So, you're now talking about the event (x,t)=(-k/γ,0). Is that correct? Once you answer this, we can continue.
No, we are not talking about that event at (x,t)=(-k/γ,0). That clock at that place in the unprimed frame will not play any role in this problem. I have said this over and over. It seems all this is all beyond you. How many times must I explain it to you. We care about the clock at k in the primed frame and the clock at the origin of the unprimed frame. You are never going to get this. Can you find someone to help you ?
Let's consider chinglu's clock at \((x,t)=(-k/\gamma, 0)\). In the primed frame, that event (the "start" event) occurs at: \((x',t')=(-k,kv/c^2)\) In other words, this event occurs after the origins are "the same" in the primed frame. Using the "start" event specified by chinglu, we need the unprimed origin to travel to the previous location of \(x=-k/\gamma\), which takes time \(\Delta t = k/\gamma v\). So chinglu's "end" event is: \((x,t)=(0,k/\gamma v)\) In the primed frame, the coordinates of this event are given by the Lorentz transformations: \(x'=\gamma (0 - vk/\gamma v) = -k\) \(t'=\gamma (k/\gamma v - v(0)/c^2) = k/v\) The time interval in the primed frame is: \(\Delta t' = k/v - kv/c^2 = (k/v)(1-(v/c)^2)= k/\gamma^2 v\) Note that this is SHORTER than the same interval in the unprimed frame, by a factor of \(\gamma\). This makes sense, because again the start and end events occur at the same spatial location in the primed frame, so the primed frame measures the proper time. The proper time is always the SHORTEST time between two events. Times in any other frame between the same events are always longer. This is in accordance with special relativity. --- Note: There is nothing new in this post. This point was already explained in posts #117, #118, and particularly in #165 and #177, above. Those posts remain unexamined by chinglu, who refuses to go through them line by line to find a mistake. Because he knows there are no mistakes.
Please give the spacetime coordinates of the events we are talking about. It's a simple request. Can you do it?
