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Now from the point of view of the observer on ground what will happen? If the signals don’t arrive simultaneously the device must not explode. Constructive interference occurs at a given event when the phase, $\phi$, of the two waves differs by an even multiple of $\pi$ at that event. Destructive interference occurs when they differ by an odd multiple of $\... 2 I really wish more books used spacetime diagrams to teach relativity, because 90% of confusion in relativity problems can be resolved by drawing a careful spacetime diagram. The idea is to superimpose the$t$and$x$axes for one observer (call them observer B) and the$t'$and$x'$axes for a moving observer (observer A) on the same diagram. All events ... 2 'How old is B from his own (B's) frame of reference at the moment A made the above observation?' Your problem lies in the very moment you are talking about. In SR there isn't an unique moment for universe. A moment will be meaningful only if you specify your refrence frame of that moment. i.e you should have asked something like this: 1.'How old is B ... 2 As is often the case, the "culprit" is the relativity of simultaneity. In the$S$frame, the events of A being at location$.75ct$and B being at$.25ct$are simultaneous, but that's not the case in the$A$frame or the$B$frame. You can't just use the$\gamma$separately on distance & time and combine them, you need to use the full Lorentz ... 1 Suppose we choose some coordinates. These can be any coordinates that are convenient such as Schwartzschild, Gullstrand-Painleve, Kruskal-Szekeres or whatever. If we denote your position in these coordinates$(x^0, x^1, x^2, x^3)$then we can differentiate twice to get your acceleration in these coordinates: $$\frac{d^2x^\alpha}{d\tau^2}$$ This is called ... 1 The outcome depends on the method, how A (or B) measures and which clock he compares to which clock. The problem is that A and B can only directly compare readings of their clocks when they are at the same point. If they are at some distance from each other, to say what "another clock" shows they must make some assumptions about the one-way speed of light ... 1 The way to think about problems in relativity is to avoid reference frames and think in terms of invariants, in this case proper time. Proper time is just path length in Minkowski space. So if the twins travel equal path lengths their clocks will measure equal durations. It's not acceleration per se that produces the paradox, it's the fact that a straight ... 1 It's all in the Lorentz Transform:$t'=\gamma (t-v x/c^2)$In your case$\gamma=2$,$t=60$and as$A$is presumably at the origin,$t'=120 $y, as @WillO says. However different values of$x$would give different$t'$. So this is not 'a particular moment in the first inertial frame' which is what your question asks about. At the$t=60$moment there are ... 1 It's a little hard to interpret your quoted question: "How old is B from his own frame of reference at the moment A made the above observation?" I'm interpereting this to mean "According to B's frame of reference, how old is B at the moment when A makes his observation?" If you meant something else, then the following might not apply: I'm assuming that ... 1 Both the distance traveled and the time elapsed are multiplied by$\gamma$, so if you were to just divided the distance between the two ships by the time elapsed, you would get .5c. However, Lorentz boosts don't simply scale space and time, they convert some time to space and some space to time. Imagine at time$t_0$you were to measure the positions of$A\$...