# Special relativity - loss of simultaneity - Is that real?

Below is normal example that is generally given for loss of simultaneity in relativity

A person (A) is on platform. Another person (B) is travelling in train (moving left to right). When person A is right at middle of moving train (outside), lightning strikes at both the ends.

• A sees lighting strike at both ends as simultaneous

• B doesn’t see both the events to be simultaneous. Strike on the right is seen to be before left

There are couple of questions I am struggling with:

• Typical explanation given is since B is moving towards right & light from right side is coming towards person, it will meet person B first and hence lighting on right will appear to B to happen before. So is this loss of simultaneity real or just because light takes more time to travel from left side. You can potentially have events which are simultaneously but light from them need not reach observe in same time.

• If light didn’t have same velocity in all frame of reference and behaved like other velocities what will B see? both the events simultaneous.? If yes why.?

• Let it also be noted that if when lightning strikes both ends of the train it also illuminates clocks that are located at each opposite ends of the train, the two flash images of the clock readings, when reaching B, will also confirm to B that the lightning struck the front of the train first. Understand how this occurs, and you will be on your way to understanding SR and simultaneity. – Sean Dec 6 '17 at 17:27
• This animation provides a decent visual depiction of not only the relativity of simultaneity but also of the physical consequences of time dilation and length contraction associated with it -- youtube.com/watch?v=C2VMO7pcWhg – JM1 Dec 17 '17 at 4:47
• – Frobenius Dec 31 '19 at 18:50

First, as a Special Relativity example it has absolutely nothing to do with how long it takes the light signal to reach the observer's eye. The observer in the middle of the train sees light taking the same time to arrive from either end.

Special Relativity observers calculate the time at which events happen by allowing for the time in their inertial frame it takes for light to reach them from the clock they are observing.

Two simultaneous events in one inertial frame (observer A) are not simultaneous in the moving inertial frame (observer B)

In fact, if observer A looks carefully at the watches the passengers on the train are wearing, the passengers' watches in the front of the train show an earlier time than those at the back when the lightning strikes.

This is the relativity of simultaneity. It is crucial to understanding Special Relativity. Google it if it is not explained in your relativity text book.

[Here you'll get a very clear, and descriptive approach to your answer, so that you can have a strong foundation on the Concept of Relativity, which is important]

First you need to have a clear conception that, $$Simultaneity$$ $$is$$ $$Not$$ $$Absolute$$, and this $$Simultaneity$$ is defined on a certain $$Inertial$$ $$Frame$$, which might differ from other $$Inertial$$ $$Frame$$, and $$loss$$ $$of$$ $$Simultaneity$$ depends on observers, in such a way, that one might not agree the event of Simultaneity with other, but both being true. So the Simultaneity is real, but on certain $$Inertial$$ $$Frame$$.

Now, for such this example, that you're asking:

• For the $$Inertial$$ $$Frame$$ $$of$$ $$Reference$$ on the train, w.r.t. the person B, the event of the Lightning, striking both the ends of the carriage, is $$NOT$$ $$Simultaneous$$, and $$loss$$ $$of$$ $$Simultaneity$$ is real here. (The other part of this statement of yours is unclear, but i'm telling you something, which might clear your conception) When we say two events to be $$simultaneous$$, we of course say that from a certain $$Inertial$$ $$Frame$$, and clearly in there, the light from two events, reach the observer's eye at the same time. But being in a restricted $$Inertial$$ $$Frame$$, one CANNOT say, any two events to be $$simultaneous$$, if light from both the events come to the observer at the different times.

• (This is a good question clearly) If light didn't have the property that it has same velocity in all $$Inertial$$ $$Frame$$, then there would be a great o ambiguity of time, upon which any two events are primarily defined. Precisely,

$$Considering$$ $$that$$, light travels like all other velocities, and is not same in all $$Inertial$$ $$Frame$$, we have, for the $$photon$$, from the lightning at the $$rear$$ of the carriage, will have velocity $$(c+v)$$ $$Towards$$ $$the$$ $$Passenger$$ $$B$$ (who is at the middle of the train), as the train is moving in the forward direction. And for the $$photon$$, from the lightning at the $$front$$ $$end$$ of the carriage, will have velocity $$(c-v)$$ $$Towards$$ $$the$$ $$Passenger$$ $$B$$.

Taking $$t_1$$ to be the time for the $$photon$$ from $$rear$$ $$end$$, to reach $$B$$, and $$t_2$$ for the $$photon$$ from $$front$$ $$end$$ and the carriage distance to be $$s$$.

Now clearly, as the train is moving in front direction, for the $$photon$$ $$from$$ $$rear$$ $$end$$ we have:

$$\frac{s}{t_1} = V_{photon} + V_{train} \implies \frac{s}{t_1} = (c-v)+v \implies t_1 = \frac{s}{c}$$ $$Since$$ when the photon reaches the position, of B, from the rear end, the train have gone away $$(V_{train}*t_1)$$ amount of distance, and we need to compensate that.

Similarly for the $$photon$$ $$from$$ $$the$$ $$front$$ $$end$$ we have:

$$\frac{s}{t_2} = V_{photon} - V_{train} \implies \frac{s}{t_1} = (c+v)-v \implies t_2 = \frac{s}{c}$$ $$Since$$ when the photon reaches the position, of B, from the from end, the train have came near $$(V_{train}*t_2)$$ amount of distance, and we need to subtract that.

$$Therefore$$, $$t_1 = t_2 \implies$$ that, $$Both$$ $$the$$ $$Events$$ $$are$$ $$Simultaneous$$ $$If$$ $$we$$ $$don't$$ $$consider$$ $$light$$ $$to$$ $$have$$ $$same$$ $$velocity$$ $$in$$ $$all$$ $$Inertial$$ $$Frames.$$

$$That$$ $$means$$ $$both$$ $$Maxwell$$ $$and$$ $$Newton$$ $$cannot$$ $$be$$ $$true$$ $$simultaneously$$ $$and$$ $$it$$ $$turned$$ $$out$$ $$that$$ $$Newton$$ $$fails$$ $$to$$ $$explain$$ $$These$$ $$events.$$ And $$Maxwell$$ $$Is$$ $$Correct$$, that LIGHT NEEDS TO HAVE SAME VELOCITY IN ALL INERTIAL FRAMES.

• Please don't mind that I gave all algebraic expressions, but in order to understand every bit of it, i think it is necessary. – Pritam Sarkar May 4 '20 at 11:20
• This argument is hard to follow, and whether or not it provides a strong foundation is not for you to decide.. But aside from that, you start with the answer that the OP is asking: is lack of simultaneity real, or just an artifact of light travel time? You can improve this and remove the uncertainty concerning light travel time by presenting an argument that does not use it. – garyp May 4 '20 at 11:38
• No, please don't take it in that context, I said that, because these thought experiments gave rise to the whole relativistic kinematics, and, we can only conclude whether or not something in Relativistic mechanics, real or certain artifact, when we choose a fixed Inertial Frame. And it can be real or an artifact both at the same time, by the choice of our Inertial Frames, therefore all I mentioned, was that depending on the choice of Inertial frame, it comes to be real. Otherwise, how does that happen, physically! – Pritam Sarkar May 4 '20 at 11:44

So is this loss of simultaneity real or just because light takes more time to travel from left side?

Yes, it is real, and yes, it is because light takes more time to travel from the left side. To understand why that is the case, we need to dive into what we actually mean when we say that two events are simultaneous.

A common definition for two events to be "simultaneous", is if they happen at the same time as indicated on two clocks, each clock local to one of the events. But in order for those two clocks to be able to indicate simultaneous events, they need to be synchronised. That is, they need to show the same time at the same time. The common definition for when two clocks are properly synchronised, is if an observer who is located exactly in the middle between two clocks sees that they always indicate the same time. In order for the observer to see the clocks, light needs to travel from them to the observer. That is, the clocks are synchronised if the light of one clock reaching 12:00 arrives at the middle observer at the same time as light from the other clock reaching 12:00.

In that sense, whether two events are simultaneous is defined by the time it takes for light to reach a middle observer. But it is also very real. If the train car had two synchronised clocks, one in the front and one in the back, and these clocks would be stopped by the lightning strikes, then they would indicate different times. That is, as considered from the train car, the two lightning strikes happened at different times.

If light didn’t have same velocity in all frame of reference and behaved like other velocities what will B see?

Light does behave like other velocities in Special Relativity. For example, suppose we have two synchronised clocks, and an observer in the middle as described above. And suppose that, at exactly 12:00, each of the clocks triggers a gun to fire a bullet at identical velocities towards the middle observer. Then those two bullets would arrive simultaneously at the observer as well, independent of whether the train car - containing the clocks, guns, and observer - is moving or not.

There is a common mathematical formula in Special Relativity that describes how to add two velocities together:

$$w' = \frac{v+w}{1 + vw/c^{2}}$$

where v and w are two velocities, and w' the resulting velocity. That formula is valid for all velocities, it just happens to always produce c when any velocity is added to c.

If we reformulate your question as:

If light behaves according to a different model from Special Relativity, what will B see?

As shown above, our definition of "simultaneous" and "synchronised clocks", depends on the travel time of light to reach a middle observer. If we keep that definition in any alternative physics model, then B would still only consider two events simultaneous, if the light arrives at the same time.