: From: Dale Jenkins <100575.143@CompuServe.COM> : Message-ID: <47ggsh$a6k$2@mhadf.production.compuserve.com> : We zoom past the Earth in a rocket at 240,000 kps. [...] : According to SR, each side judges every minute to be worth only 0.6 of : the other's minute. Now, suppose a dust cloud moves across our line : of sight. Further suppose it was agreed beforehand that, exactly ten : Earth days after the fly- by, we would make a radio broadcast (which : the dust cloud will not affect because it is transparent to radio : waves), while the Earthlings would turn on their receivers 18 days : after that fly-by, the 8 extra days being the time that light will : require to cover the intervening distance. Since we can no longer see : the Earthlings' signals, we have to resort to calculation to estimate : their time. But how do we do this? Ten shipboard days are equal to : only six Earth days, according to SR. But, if we wait a further six : and two thirds days (ie 4 Earth days according to our clocks), by then : the Earthlings will have given up, and switched off their receivers. : However, trying to work out how the other side calculates the time : difference is scarcely more useful because, if both sides think that : one of their own minutes equals 0.6 of the other's, then each side, : trying to work out what the other's arithmetic is, can calculate : backwards indefinitely, ending up in an infinite regression: 1 rocket : minute = 3/5 Earth minutes = 9/25 rocket minutes = 27/125 Earth : minutes, etc. The only way out is to say that Earth is moving slower : than the rocket.
No, that's not the only way out.
The situation is that the earthlings turn on their receivers after 18 days, and the question is, at what time on their shipboard clock should the astronauts send the signal the earthlings will listen for at 18 days past flyby?
If we presume (as the earthlings in the problem might) that the earth is motionless and the spacecraft is receding, the problem is simple. The .8 lightspeed trip away from earth must equal the distance of the lightspeed signal back, so
0.8*t(wait) = 1.0*t(signal)The earthlings conclude the spaceship should send the signal at t=10*(1-0.8^2)^(1/2)=6 by the shipboard clock.
0.8*t(wait) = 1.0*(18-t(wait))
t(wait) = 10
If we presume (as the astronauts might) that the spaceship is motionless, and the earth is receding, the problem is a bit more complicated because we have to account for earth being in motion while the signal travels, but it's not too bad. We simply equate the sum of how far earth recedes during the wait and how far it recedes during the signal to the total signal transit time. Note that in this case we must presume the earth clock ticks slower than the shipboard clock, as would happen if the spaceship were motionless and the earth were moving. We have
0.8*t(wait) + 0.8*t(signal) = 1.0*t(signal)The astronauts also conclude that they need to send the signal when the shipboard clock reads 6 days. And they came to this conclusion without ever needing to suppose that their own clocks were ticking slowly because of motion. From their perspective, they had to send that signal so early, just so it'd catch up to the receding earth, not because their own clocks tick slow. Indeed, from the astronauts' perspective, earth clocks had better be ticking slow, or they'd have to send the signal even earlier.
0.8*t(wait) = 1.0*((18/(1-0.8^2)^(1/2))-t(wait))
t(wait) = 6
Note that the astronauts are sending the signal when (from their perspective) the earth clock reads 3.6, not 10. This is because the statement in the problem as posed that the signal is sent when the earth clock reads 10 is giving a time to an event occuring at a distance. This is fundamentally ambiguous in relativity, and in fact it turns out to be false according to the astronauts. If they intend to hit the earth with a radio message at earth's t=18, they must send at (according to themselves) earth t=3.6.
Thus we see that the only reason one might claim that "the only way out is to say that earth is moving slower than the rocket" is that the problem statement was overspecified with that viewpoint, not that the physics of special relativity in this case force such a viewpoint.
Here is a pair of space-time diagrams summarizing the above calculations in graphical form.
Time is plotted horizontally, space vertically. The rocketship's path in spacetime is shown in red, and its space axes shown in thinner red lines, and its clock values annotated in red. The earth's path in spacetime is shown in blue, and its space axes shown in thinner blue lines, and its clock values annotated in blue. The radio signal is shown as a black line at 45 degrees in both diagrams.
The left-hand diagram shows the situation from the earth's perspective (the first calculation). The right-hand diagram shows the situation from the spacecraft's perspective (the second calculation). Note that these two diagrams show all the same events with all the same relationships between them. There are no contradictions, and no "infinite regress". In fact, either calculation can be seen represented in either diagram; the two diagrams are actually identical (except for scale and whether the blue or red coordinates are drawn orthogonally).
For more on this subject, there is also a comparison of time dilation to rotation effects in normal analytic geometry.