# How to interpret negative time in Lorentz transformation?

I am somewhat confused about how to interpret negative time in Lorentz transformation. In the usual case of two reference systems S and S' where the distance X (the one that measures S) to an event, is very large with respect to the distance between S and S' (also measured by S) the time t' of S' gives a negative result.

I don't understand why if at $$t = t '= 0$$, S and S' were together, the event has been observed by S' before this synchronization..., so it could not be in front of S, and also, in this synchronization, S' could warn S about the event that is future for S ... which seems to me paradoxical.

Just in case my general interpretation of the Lorentz transformation is incorrect, I clarify that I assume that S and S' are in positive X, they are together at $$t = t' = 0$$ and after a certain time t, S observes S' at a certain distance and an event X; with X and t I can calculate that for S' the event has occurred at X' and at t '.

Then, using natural units for simplicity, if $$V = 0.8$$, and at $$t = 7.5$$, S observes an event at $$X = 11$$, then S' which at that time is $$V * t$$ away from S, observes ,applying Lorentz, $$X’= 8.3333$$ and $$T’ = - 2.16666$$

How should we interpret this result? Did S' really saw the event before synchronization?

According to Marco's answer and comments, I tried some numbers playing with http://www.trell.org/div/minkowski.html, and now I can answer my own question.

We have two events, event $$A$$ in $$x=x'=0$$ ; $$t=t'=0$$ and event $$B:$$ $$x=11$$ and $$t=7.5$$ with $$v=0.8$$ ($$t' = -2.1666$$, $$x'=8.333$$)

Lorentz transformation gives you the distances regarding each observer, and what this negative time $$t' = -2.1666$$ means is that regarding $$S'$$, the event $$B$$ occurs before event $$A$$, event $$B$$ occurs before the synchronization... BUT, and this is the important point, S' is not going to be aware of that event until $$t' = 6.1167$$ ($$x'=0$$, $$x=8.22$$, $$t=10.277$$) so there is no way S' can warn S in the $$A$$ event moment of the synchronization.

No, the difference in time simply reflects the fact that the clocks of the two observers are out of synch with each other.

A good way to understand this is to imagine that the speed of light is, say, one foot per second. Now imagine the classic set-up of a 100ft long railway carriage on a platform. All along the platform are people waiting to see the train off, and all along the carriage are people looking out of the window. Initially everyone's watch is synchronised with everyone else's.

As the train pulls off, at say one inch per second, someone standing by the back of the train flashes a light along the platform parallel to the train. As the light moves along the platform, the people on the platform and the people on the train detect it reaching them. When the light reaches the people halfway down the carriage (ie fifty feet along it) their watches will tell them that 50 seconds have passed. However, by that time the train will have moved fifty inches down the platform, so for the people outside, the light is now fifty feet and fifty inches from where it started, which is roughly 54 feet, so for them 54 seconds seem to have passed. There is a 4 second difference between the reading on the watch of the person on the train and the reading on the watch of the person right alongside them on the platform, even though they both see the same light at the same instant. One doesn't see the light 4 seconds earlier than the other, they just have different readings on their watches.

• Ok, then, would it be more appropriate to think that the problem statement is incorrect? I wonder that If we obtain such a pronounced time difference, perhaps the result suggests that the situation presenting the data is not possible ... perhaps they had to synchronize clocks much earlier in order to observe the event at the suggested distances... – Pedro Oct 8 '19 at 9:11
• What I mean is that if they really have seen the event at the same instant, there is no possible S' has a negative time on his watch if they had sincr – Pedro Oct 8 '19 at 9:54
• Hi Pedro. Your problem statement is a little ambiguous. It would help if you could clarify it with a diagram, perhaps. If they had synchronised their watches at t=t'=0, and if S' subsequently sees the event, he cannot have a negative time on his watch. – Marco Ocram Oct 8 '19 at 10:04
• I'm going to find a way to draw the diagram, but it's simple. on the X axis S and S' are together at $t = t' = 0$, S' goes at $V=0.8$. At $T= 7.5$ seconds , S' is at the right of S at a distance of 6 units $(V*T)$. At that moment S observes an event X at 11 units. If you calculate X 'and T' you get a negative time.... In fact always that X is much larger that V*T you are gonna get a negative time – Pedro Oct 8 '19 at 10:35
• When you say at T=7.5 S observes an event, do you mean that S actually receives the light from that event (ie that the event happened at T= -3.5s). Or do you mean that the event happens at X=11 and T=7.5? – Marco Ocram Oct 8 '19 at 10:44

Good question. The relativity of simultaneity is the key to understanding this.

If you check the (perfectly synchronised) watches the passengers on Einstein's train are wearing, at an instant when a passenger in the middle of the train is passing you, you will see (after making allowance for the time it takes the light to reach you) the watches at the front of the train show an earlier time than those at the back. You are sharing 'now' with the passenger passing you. You are sharing what that passenger thinks is history with his fellow passenger at the front of the train, and you are sharing what the central passenger thinks of as the future with his co-passenger at the rear of the train. Your 'now' is smeared in train time down the length of the train.

In terms of train time and train distances, you see a change of train time $$\Delta t=\frac{v}{c^2} \Delta l$$ on their watches for every unit of train length separating them.

So yes, events behind the passenger passing you happen in your time frame before they happen for him. In your time frame they happen when he passes you, in his they happen later.

And that is why, even though at rest the train would be the same length as the station you are standing in, you see his moving train shorter than the station, and he sees it longer.

And yes, the same is true from his point of view if he looks at the watches of the passengers waiting on the station for a stopping train.

And yes, it is mind blowing that as soon as you start walking in flat spacetime, somewhere a long way behind you is suddenly back in the Jurassic age. If you were able to 'watch' a clock that far behind you, and compensate for the time it took for the light to reach you, you would see the distant clock races backwards as you accelerate to walking speed. Everybody back to the dinosaurs walks backwards too.

Of course, once you are walking at a steady space, the distant clock runs forwards again, and you see it all start again, forwards from the age of the dinosaurs.

Or you might prefer to look forwards to a distant lottery draw on a distant planet and try to buy a ticket while you walk.

Enjoy.