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I ran into a serious problem with the Lorentz transformation and time dilation. In the standard configuration you have one observer S and another one S' with their x-axis aligned. I assume S to be at rest and S' to be moving in direction of the x-axis at a speed v = 0.5*c (the relativistic gamma is then g = 1.15). Each observer has a clock and they meet when both clocks show 0 s.

Now suppose a firecracker goes off at the origin of S. This happens at a time when the clock of S reads 5 s. So he assigns the event the coordinates x = 0 m and t = 5 s. I want to know what coordinates observer S' assigns the same event. According to the Lorentz transformation, these are x' = -862,500,000 m and t' = 5.75 s.

(Just to check, I inserted x' = -862,500,000 m and t' = 5.75 s into the inverse Lorentz transformation and got x = 0 m and t = 5 s as expected)

So far, so good. But I'm having trouble interpreting this. So observer S says 5 s passed between their meeting and the explosion of the firecracker, observer S' says 5.75 s passed between their meeting and the explosion of the firecracker. That's also fine. But this means that S says the clock of S' is ticking faster while S' says the clock of S is running slower. Shouldn't both say that the clock of the other is ticking at a slower rate? Where is the problem in my logic, what do I misunderstand?

I initially expected that when insert x = 0 m and t = 5 s into the Lorentz transformation, I get t' < t (clock runs slower), but this doesn't happen!

Would be fantastic if somebody could help me here. It's a very exciting topic, but I feel like I have hit a dead end. No matter how I try to resolve it, I always get the same problem. One observer sees time dilation, the other time "acceleration", but I know that it should always be time dilation (moving clocks run slower, the mantra of SR).

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  • $\begingroup$ In Special Relativity the S' frame is assumed to be the one "in which the clock is moving, which means it is the one you assume to be stationary. $\endgroup$ – bright magus Nov 5 '14 at 15:29
  • $\begingroup$ Thanks for the reply, but I don't fully understand it. So in the standard configuration we have to think of S and his clock moving and S' and his clock being stationary? So then the moving clock of S measures 5 s and the stationary clock 5.75 s, meaning the moving clock is ticking at a slower rate. That still leaves a few problems though for me and my thinking, one of which is: I've seen numerical examples in which t' < t, which would again lead to the moving clock (always S, if I understood you correctly) going faster. Mhhhhh ... $\endgroup$ – MeBe Nov 5 '14 at 15:45
  • $\begingroup$ I would have to see the examples. Perhaps they marked they ref. frames differently. Still, "the rule of thumb" here is that the moving clock is slower, which means the period between the ticks is slower. Fullstop :) Remember, the S clock is local time, where 1s=1s. You don't assume the moving person is comparing anything. The S' is where the observer is and he compares - his own clock to the moving clock. If S begins to observe, he sees S' as moving (and assumes he is stationary himself) and then he calculates time dilatation for the other frame. But then S and S' are switched for calculations $\endgroup$ – bright magus Nov 5 '14 at 16:21
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    $\begingroup$ @bright magus--Not sure what you mean by "you don't assume the moving person is comparing anything"--when you are analyzing a problem in relativity using two frames A and B, you are free to first analyze things from frame A's perspective where B is the "moving" one, and then analyze things from frame B's perspective where A is the "moving" one, and in both cases you must get the same answers to questions about local physical events (like which clock in a given frame was right next to the event when it happened, and what reading that clock was showing at the moment of the event). $\endgroup$ – Hypnosifl Nov 5 '14 at 16:30
  • $\begingroup$ So to sum up, both observers see each other's clocks running slower, because they see each other moving. So A assumes B is moving and B assumes A is moving. In this case each of them assumes he is S' and the other is S. So they both calculate the the time according to the same equation, although each of them puts his own time as S'. $\endgroup$ – bright magus Nov 5 '14 at 16:33
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"So far, so good. But I'm having trouble interpreting this. So observer S says 5 s passed between their meeting and the explosion of the firecracker, observer S' says 5.75 s passed between their meeting and the explosion of the firecracker. That's also fine. But this means that S says the clock of S' is ticking faster while S' says the clock of S is running slower. Shouldn't both say that the clock of the other is ticking at a slower rate? Where is the problem in my logic, what do I misunderstand?"

You are forgetting to take into account the relativity of simultaneity, which is relevant because each observer is supposed to assign time to events using readings on a network of synchronized clocks at rest at different locations along their own rulers (the clocks synchronized using the Einstein synchronization convention), so that each event's time can be judged by a clock that was right next to the event when it happened. That way you don't have to worry about light delays--for example, if I see light from an explosion 5 light-years away in 2005, and light from an explosion 10 light-years away in 2010, I can look at the clock in my system that was next to each event when it happened, and see that each explosion happened next to a clock that read t=2000 at the moment it happened, so I judge that these events happened simultaneously in my frame despite the fact that I saw them at different times. This page has an illustration of such a lattice of rulers and clocks, which give physical meaning to the position and time coordinates assigned by a given inertial frame:

enter image description here

In your example, there are two events that the observer in S' is assigning coordinates to--the first is the event of the clock belonging to frame S reading t=0, the second is the event of the firecracker going off. S' has a clock #1 at position x'=0 that was next to the first event when it happened, and that clock read t'=0 at that moment; and S' has a separate clock #2 at position x'=-865,426,282 m that was next to the second event when it happened, and that clock read t'=5.77 s (my numbers are slightly different than yours, it looks like you had some roundoff error). The key to understand why this does not contradict the fact that S should see all of the clocks in S' running slow is that from the perspective of frame S, these two clocks in S' are not synchronized, due to the relativity of simultaneity. If the two clocks are a distance L apart in S' and synchronized in that frame, then from the perspective of frame S which judges S' to have velocity v, the back clock's time will be ahead of the front clock's time by an amount Lv/c^2 at any given moment. So in this case, with L=865,426,282 m and v = 0.5c = 149,896,229 m/s, frame S will say that at any given moment, clock #2's time is ahead of clock #1's time by the amount (865,426,282)*(149,896,229)/(299,792,458)^2 = 1.4385 seconds. So at t=0 in frame S, when clock #1 reads t'= 0 s, clock #2 reads t'= 1.4434 s. Then 5 seconds later in frame S, each of the clocks in S' has only advanced forward by 4.3301 s, because time dilation says that the clocks in S' must be running slow in frame S. But since clock #2 already had a head start of 1.4434 s, at t = 5 in frame S, clock #2 will read 1.4434 + 4.3301 = 5.7735 s, and that's the moment at which the firecracker goes off next to it.

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  • $\begingroup$ Though I can not follow everything you said at the moment, the idea that the network of clocks a certain observer S has (and that is synchronized in S) are seen as out of sync for an observer S' that moves relative to observer S seems very helpful. If I understood you correctly, I should not think of S' carrying one clock with him, but rather see him in the center of his network of clocks. $\endgroup$ – MeBe Nov 5 '14 at 19:04
  • $\begingroup$ Yes, exactly--and S' used one of his own clocks to mark the time of the clock carried by S reading t=0, then S' had to use a different one of his own clocks to mark the time of the firecracker explosion, since these two events happened at different positions in the rest frame of S', and S' assigns time to each event using whatever clock in his network happened to be next to the event when it happened. And in the frame of S, these two clocks belonging to S' are perpetually out-of-sync by 1.4434 seconds. Let me know if there's any part of my answer you can't follow even after rereading. $\endgroup$ – Hypnosifl Nov 5 '14 at 19:11
  • $\begingroup$ OK, I think it's slowly getting through, your help is greatly appreciated! if I'm not mistaken, the problem was that I used two different clocks in the frame S', the one in his origin that I used to determine the time of the meeting of S and S' (it read t' = 0) and the "different" clock x'=-865,426,282 m that I used to determine the time of the firecracker event (it read t' = 4.3301). These clock are in sync in frame S', but not in frame S. To find out how much time actually passed for S' between the events, I have to compare the readings on his "origin clock". Wow, that's pretty complicated. $\endgroup$ – MeBe Nov 5 '14 at 19:26
  • $\begingroup$ I think I'll have to go back to the Lorentz transformation to try to determine the 1.4434 s you mentioned. At this point the situation is still a bit foggy, it seems there's much I missed, but your input is very helpful. $\endgroup$ – MeBe Nov 5 '14 at 19:29
  • $\begingroup$ If by "how much time actually passed for S' between the events" you mean the time between the events in the S' frame, you don't have to worry about which clock measured which event because all the clocks belonging to S' are synchronized in the S' frame. So in this frame, if the first event is assigned time t'=0 and the second event is assigned t'=5.77, then the time between these events was indeed 5.77 seconds in the S' frame. But if you want to know how this is compatible with the fact that these clocks only ticked 4.33 s between the same events in the S frame, you must consider simultaneity. $\endgroup$ – Hypnosifl Nov 5 '14 at 19:31
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In the frame $S$ the two ticks of the clock, at $t = 0$ and $t = 5$, happen at the same place. But in frame $S'$ they happen in different places. So you cannot just take the two points in $S'$, treat them as ticks of a clock and transform back to $S$ to calculate the time dilation. If you want to calculate the time dilation going from $S'$ to $S$ you need to take a clock that is stationary in $S'$.

To illustrate this consider the following:

Take your clock stationary at the origin of $S$ as before, so the two ticks are at the spacetime points $(0, 0)$ and $(5, 0)$. But now have a second firecraker at $t = 5$ but at a distance $4c$ from the origin in $S$ i.e. the spacetime point for the second firecracker is $(5, 4c)$. In $S$ both firecrackers go off when the clock read $5$ seconds. But if we Lorentz transform the two points $(5, 0)$ and $(5, 4c)$ into $S'$ we get:

$$ (5, 0) \rightarrow (\gamma 5, \gamma -5v) $$

$$ (5, 4c) \rightarrow (\gamma (5 - \frac{4vc}{c^2}), \gamma (4c - 5v)) $$

So though the two firecrackers go off simultaneously in $S$ they do not go off simultaneously in $S'$. So should you calculate the time dilation using the firecracker at the origin or the one at $x = 4c$? Obviously you use the one at the origin because you need the clock to be stationary.

The point of this is that when you take the two transformed points in $S'$ and transform them back to $S$ you are effectively trying to calculate the time dilation of a clock that is moving in $S'$. the initial tick and the final tick don't happen at the same place.

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  • $\begingroup$ I think with the help of your and Hypnosifl's answer I'm starting to understand my problem, but it's still a bit unclear. In stupid words: a clock can only be a clock if it ticks in the same place. For S this wasn't problematic, his clock was stationary at x = 0, it ticked once at t = 0 and then again at t = 5 when the firecracker went kaboom. But for S' I used two different clocks. First I read the time from the clock in his origin, then I read the time from his clock at x'=-865,426,282 m, but they are not in sync as seen from S. Is this the key? Mhhh ... I have to digest that first ... $\endgroup$ – MeBe Nov 5 '14 at 19:17
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Time dilation effect applies when an observer moving with respect to a pair of events (S') measures the time interval with the clock resting in his reference frame, and the events are at the same location in the other rest frame (S). The S' observer then compares his time (5.77 s) with the clock which is at rest with respect to both events (the clock in S). S' will have more time elapsed on the clock with the inference that "the clock in S must be slow," the popularized (but misleading) idea of time dilation.

The clock in S' is moving with respect to the two events, so S will not observe a time dilation effect for the S' clock measuring those events.

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