Freixas said:
Let's remove the infinite line of closely spaced B's and replace it with a line of B's that are spaced evenly 4 LY apart as measured by A.
Ok so far. (Note, though, that this is a
different spacing than the one we have been assuming, implicitly, thus far. See below.)
Freixas said:
Traveling at 0.8c, A will expect to see a B pass by every 5 years.
Ok so far.
Freixas said:
Each time a B passes by and sees A, it should see that A's time is a multiple of 5 (if the first conjunction occurred at 0).
Yes,
if the first conjunction occurred at 0. But,
if the B clocks are all synchronized with each other--meaning that all of them are synchronized using the natural notion of simultaneity in the B rest frame--then only
one of the B clocks will have a first conjunction with an A clock that reads zero at the same instant that that B clock reads zero. That clock is the clock we have been calling B1.
So here is the first logical error you are making: you are assuming that all of the B clocks have a conjunction with an A clock at which the B's clock reads zero
and the A clock it is passing reads zero. That is only true of
one B clock.
Freixas said:
You tell me that the first trailing B (what you call BN) will see A's clock as reading 1.8 years at conjunction. I can't see how. Help!
You were the one who defined what clock BN sees, not me. I was just working out consequences of your definition that you did not realize.
First, let's be clear about clock B1. It meets A when its clock reads zero and A's clock reads zero. It meets M when its clock reads 3 years and M's clock reads five years. We agree on those things, at least.
Now here is how you defined clock BN:
Freixas said:
BN's clock reads 3 years when it reaches A
Let's analyze this in the A frame, since you said that's all you're concerned about. In this frame, the B clocks are moving to the right (positive x direction) at speed 0.8. Clock M is at rest in this frame 4 light years to the right of clock A (i.e., at x = 4). Clock B1 therefore passes through the events ##(x, t) = (0, 0)## and ##(x, t) = (4, 5)##. And the spacetime interval between these two events is ##\sqrt{5^2 - 4^2} = 3##, which is consistent with clock B1 reading 3 years when it passes M (because we have specified that it reads zero when it passes A).
Now consider clock BN. Its clock reads 3 years when it passes A, which means that this event has coordinates ##(0, T)## for some ##T## that we have to calculate (the ##x## coordinate of this event must be zero because A is always at ##x = 0##). You have been implicitly assuming that all of those events have the same value of ##t## in the A frame--in other words, that since clock B1 reads 3 years at ##t = 5## (when it passes M), then any other B clock must also read 3 years at ##t = 5## (but at some other ##x## value)--which would mean clock BN would read three years at the event ##(x, t) = (0, 5)##. But that is obviously false, because of relativity of simultaneity--or, to put it another way, because the B clocks must be offset in the A frame.
How much are they offset? There are a number of ways to calculate this, but the simplest is to use the Lorentz transformation. We know that clock B1 reads 3 years at event ##(4, 5)##. We know that clock BN reads 3 years at event ##(0, T)##. And we know that, for both of these events, the equation ##t' = 3 = \gamma \left( t - v x \right)## must hold. So we have ##3 = (5/3) \left( 5 - 4 * 0.8 \right) = (5/3) T##. The first equality is obviously correct and gives us a sanity check; the second equality easily gives us ##T = 3 (3/5) = 9/5 = 1.8##. So, as I have been saying, when clock BN, defined as you defined it, meets clock A, clock BN reads 3 years (by your definition) and clock A reads 1.8 years (by the calculation I have just done).
So to summarize: each B clock passes an A clock every 3 years by its own time, and each A clock reads 5 years later than the last A clock it passed. But if we compare adjacent B clocks, the readings they see on the same A clock when they pass it are not separated by 5 years; they are separated by only 1.8 years. And, as should be evident, these B clocks are
not spaced 4 LY apart in the A frame. They are spaced only 1.44 LY apart (as I calculated before). And the B clocks do not all see the A clock readings be exact multiples of 5 years when they pass; only clock B1 does. Clock BN, for example, sees the A clock readings as multiples of 5 years plus a fixed offset of 1.8 years.
Now, let's ask a different question: suppose we define "adjacent" B clocks by the requirement I quoted at the start of this post--that they are spaced 4 LY apart in the A frame. Under this requirement, yes, each B clock will see each A clock read some multiple of 5 years when it passes it, and will also see each A clock read 5 years later than the previous B clock did when it passed the same A clock. But now the
B clock readings will not all be multiples of 3 years; they will still be 3 years apart as they pass each successive A clock, but only clock B1 will see its own readings be multiples of 3 years (0, 3, ...).
For example, call the next clock in the B series under this definition clock BA--it is the clock that passes clock A when clock A reads 5 years. What will clock BA read at this passing? We didn't quite calculate this above, but it's easy enough to do. We calculated above that clock BN reads 3 years at ##(x, t) = (0, 1.8)## in the A frame. And we know clock B1 reads zero years at ##(x, t) = (0, 0)##. So if we move along the ##t## axis of the A frame (i.e., along clock A's worldline, at ##x = 0##), we find the readings of B clocks passing us increasing at a rate of ##3 / 1.8 = 5/3##, i.e., ##\gamma##. So at ##t = 5##, we should see the B clock passing us read ##5 * 5/3 = 8 + 1/3## years. But this is just clock BA, so we have shown that clock BA reads 8 1/3 years when it passes clock A at A's clock reading of 5 years. And since 3 years will elapse on clock BA until it passes clock M, it will pass clock M at 11 1/3 years. In other words, clock BA's readings when it passes A clocks will be multiples of 3 plus a fixed offset of 2 1/3 years (whereas clock B1's readings are exact multiples of 3 years).
And since B1 is at rest in B's frame, B1's clock time is the same as time in B's frame.
