# Reciprocal Lorentz Contraction Mismatch

1. Nov 3, 2009

### nutgeb

Consider two groups of 3 spaceships each. The groups are numbered 1 and 2. The ships in group 1 are numbered 1A, 1B and 1C. The ships in group 2 are numbered 2A, 2B and 2C. In each group, ship A leads, followed directly in line, at fixed and equal proper intervals (as measured in each line's own rest frame), by ships B and C respectively.

The two lines of ships pass very close by each other in opposite directions on parallel paths. The speed of each line relative to the other is constant and relativistic, at gamma = 2. They are deep in space and it is dark out.

Each time a ship in one line passes directly abreast of a ship in the other line, a flash goes off and both of those passing ships take a snapshot of each other. After both lines have completely passed by each other, they circle around, meet up and compare photos.

The ships within each line considered their own line to be at rest while the other line of ships passed by them at relativistic velocity. During the passing maneuver each line of ships figured the ships in the opposing line to be Lorentz contracted (not relevant to the problem), and the interval between those ships to be Lorentz contracted as well (which is relevant), in each case to half its proper length (gamma = 2). Naturally, the Lorentz contraction of the opposing group is reciprocal from the perspective of each group.

The time interval between successive photos is constant, and is the same for each group, so we will assign it an incremental value of t=1. As the ships passed by each other, photos were snapped at a total of 7 discrete points in time which were separated by 6 equal time increments. Here are the photos that were taken at each point in time:

t=1: Ship A1 photo of ship A2; and ship A2 photo of ship A1.

t=2: A1 photo of B2; A2 photo of B1.

t=3: A1 photo of C2, B1 photo of A2; A2 photo of C1, and B2 photo of A1.

t=4: B1 photo of B2; B2 photo of B1.

t=5: B1 photo of C2, C1 photo of A2; B2 photo of C1, C2 photo of A1.

t=6: C1 photo of B2; C2 photo of B1.

t=7: C1 photo of C2; C2 photo of C1.

The crews realize that their photo collections are mismatched: the photos taken by each group at each interval ought to include the same ship(s) in the opposite group that photo'd them at the same instant of passing. Sometimes they do, but sometimes they don't.

For example, at t=2, when A1 photo'd B2 (because the two ships were abreast from A1's perspective) why didn't B2 also photo A1 in a reciprocal way? Well, because A1 did not arrive abreast of B2 (from B2's perspective) until t=3. At which point A1 no longer saw B2 to be abreast.

Reciprocal pairs of photos do not seem to be offset by any consistent time differential function. For example, the photos taken at t=1, t=4 and t=7 are directly reciprocal, i.e., no time lag. At the other extreme, C2's photo of A1 lags two time intervals behind A1's photo of C2. {EDIT: There is a pattern here: the time lag is 0 for ships with the same "letter", 1 for ships 1 letter apart (e.g A and B), and 2 for ships 2 letters apart (A and C).}

A failure of simultaneity seems ruled out because whenever pairs of ships passed abreast and took a photo of each other, they were essentially at the same spacetime event.

How should the crews apply Special Relativity to reconcile their mismatched photo collections?

Last edited: Nov 3, 2009
2. Nov 3, 2009

### ZikZak

You are incorrect; you are conflating events that happen at, say t=2, with events that happen at t'=2 and writing them all down as "t=2." But events at t=2 are completely different events from those that happen at t'=2 because of relativity of simultaneity. I suggest drawing a spacetime diagram of the events.

3. Nov 3, 2009

### nutgeb

I get your point, but how can each event of a symmetrical pair of events be different events if each pair occurs as a single spacetime event? The only spacetime events that are photo'd in this scenario are those which are shared simultaneously at the same spatial location in t and t'. Relativity of simultaneity cannot be the issue here.

4. Nov 3, 2009

### ZikZak

They don't. Your sequence of events isn't even wrong. You are not allowed to construct an absolute sequence of events, as you have tried to do. At best, you can write down the sequence of events in each of the two reference frames separately.

When A1 takes a picture of A2, then A2 takes a picture of A1, because those two pictures happen at the same point in spacetime. If your analysis leads to to a different conclusion, then your analysis is wrong. Draw a spacetime diagram of the events. "A1 passes A2" may happen at a different t than t', but that makes no difference. It is one event labeled with two different time coordinates.

5. Nov 4, 2009

### JesseM

Both frames agree on which pairs of ships meet, and what the clocks on each ship read as they meet, but the clocks may show different times, indicating each frame may assign a different time to a giving event of two ships passing. In group 1's frame, the events would have these time coordinates:

t=1: A1 passes A2
t=2: A1 passes B2
t=3: B1 passes A2, A1 passes C2
t=4: B1 passes B2
t=5: C1 passes A2, B1 passes C2
t=6: C1 passes B2
t=7: C1 passes C2

In group 2's frame, we simply reverse the 1's and 2's, giving:

t'=1: A2 passes A1
t'=2: A2 passes B1
t'=3: B2 passes A1, A2 passes C1
t'=4: B2 passes B1
t'=5: C2 passes A1, B2 passes C1
t'=6: C2 passes B1
t'=7: C2 passes C1

So, you can see each's view of the other is totally symmetrical. And again, for any event of two ships passing next to each other, they'll both agree on what the different ships' clocks read at the moment of the passing; for example, they both agree that as A1 passes C2, A1's clock reads t=3 while C2's clock reads t'=5 (and to illustrate the symmetry, they both also agree that as A2 passes C1, A2's clock reads t'=3 while C1's clock reads t=5). So, what's the problem? What inconsistency are you referring to?

6. Nov 4, 2009

### nutgeb

Jesse, your list of the photos taken by the two groups of passing ships and the time intervals is the same as mine, and shows the same symmetry. It's helpful to separate the t photos from the t' photos as you've done.

The underlying problem I was worried about is that group 1 and group 2 do not agree on the time sequence in which some of the photos were taken. But on further reflection, I see that there is nothing wrong with two frames observing events in a different sequence, as long as the two events are spacelike separated.

The sequence mismatch arises as follows: In your example, A1 and C2 passed and photo'd each other when group 1's clock read t=3 and group 2's clock read t'=5. They also agree that B1 and B2 photo'd each other at t=4 and t'=4. That means that they disagree about the sequence of photos. Group 1 says B1 photo'd B2 after A1 and C2 photo'd each other. Group 2 on the other hand says B1 and B2 photo'd each other before A1 and C2 photo'd each other. Both groups agree that A1 and C2 photo'd each other simultaneously, and B1 and B2 also photo'd each other simultaneously.

Here are the two groups' respective sequences of ship pair passing photos, with the problem highlighted:

Group 1: A/A, A/B, B/A, A/C, B/B, C/A, B/C, C/B, C/C.
Group 2: A/A, A/B, B/A, A/C, B/B, C/A, B/C, C/B, C/C.

Likewise, Group 1 says A1 and B2 photo'd each other before B1 and A2 photo'd each other, while Group 2 says the opposite sequence occurred. And so on.

These events are clearly spacelike separated, since in a single time interval the Lorentz-contracted ships travel a distance which is just half the interval between one's own group of ships, and the Lorentz-contracted ships are moving at almost the speed of light. So there is no way in this scenario to reverse cause and effect, e.g. the light from the flash when C2 passes A1 won't arrive at B2 until after B2 passes B1.

So, I now think there is no SR problem with this scenario.

Last edited: Nov 5, 2009