# Loss of simultaneity in special relativity

I was recently reading Introduction to Classical Mechanics by David Morin and while reading special relativity I came up with a paradox I am not able to resolve. It is basically related to the loss in simultaneity of events.

Let there be two observers A and B where A is in a train of length L moving at a constant speed with respect to the platform standing at its center while B is on the platform. A shoots 2 photons towards the 2 ends of the trains and raises his right hand when the first photon hits the front of the train and his left hand when the second photon hits the back of the train. There are two detectors situated at the front and the back which record whether a photon has hit the end or not. In his frame both the photons hit the ends at the same time so both his hands are raised together.

Meanwhile in B's frame both the photon hitting events are non simultaneous so in his frame he sees the back photon hit the end before the front photon. So there exists a time period in B's frame when one photon has already hit the end while the other has not. During this particular time interval in B's frame he should expect to see A to have raised his left hand and not his right hand (please see footnote below). But because A raises both his hands together this scenario is never possible.

How do I solve this paradox?

Footnote: I believe (please correct me if I'm wrong) that if a photon has been detected to hit the end in one frame then it will be detected to have hit that end in all frames because if any event has occurred in one frame then it should have occurred in all frames.

PS: I came up with this paradox myself so it may have some framing difficulties. However, I have tried to keep it as clear as possible. Please inform me if this requires editing.

• In A's frame, first one photon hits, then both hands go up, then the other photon hits. Jun 17, 2019 at 14:26
• Why is this a paradox? The detectors register photons simultaneously in A's frame, but not in B's frame. Why should B expect A's hand signals to indicate anything in particular about events measured in B's frame? Bear in mind that A and both of A's detectors are all spacelike separated. Jun 17, 2019 at 14:40

During this particular time interval in B's frame he should expect to see A to have raised his left hand

B wouldn't expect that. B knows that A is in a different frame of reference.

This is the relativity of simultaneity

the relativity of simultaneity is the concept that distant simultaneity – whether two spatially separated events occur at the same time – is not absolute, but depends on the observer's reference frame.

What B expects obviously depends on how much B knows about special relativity. If B only knows about Newtonian physics, the difference between expectation (i.e. prediction from theory) and observation would be a clue that either B's knowledge is not up to date or that a Nobel prize awaits if she can figure out a new mathematical solution that fits the new observation and all prior observations.

• But during the time gap in between the 2 photon hitting events in B's frame, what would A be observing. Will he see both photns to have hit or none of them? Jun 18, 2019 at 4:27
• @DHYEY: A and B will see different things and each will be able to predict that the other sees things that are different and which would be contradictory (or at least inconsistent) if A and B were in the same frame of reference. The existence or not of a time between events depends very much on your frame of reference. You are in effect asking why special relativity is different to Newtonian physics. Jun 18, 2019 at 10:01

A raises both hands together when the two distant events happen simultaneously in train time.

B sees the event at the rear of the train.

Later, B sees A raise both hands together.

Later again, by the same amount, B sees the event at the front of the train.

Simultaneous events separated in space can become separated in time in a different inertial frame.