Need some education for the super mentors?
selfAdjoint said:
Since the speed of the plane is constant, this value will never increase, so it will never reach the limit of 1 second you are looking for.
Exactly what are the qualifications for a "Super Mentor"? The man asked, "How much time must pass before a clock in the airplane and one on the ground differ by 1.0 s?" And selfAdjoint gave the answer above!
Oh, he is ok as far as he goes but he makes a very major error: he confuses apparent rate with reading on the clock, two very different issues. Yes, the rate difference is given by the formula Adjoint quotes but the readings on the clock change with time and the man asked how long it would take for the difference to be one second. If the answer were "never" then the clocks would have to go at the same rate! Plus that, out "super mentor" has failed to take the opportunity to point out some very interesting things.
The speed of light is roughly 3 x 10^8 meters/sec which is about 666 x 10^6miles/hour (a number easy to remember). So, for the fun of it why don’t we make life simple and let the jet be going 666 miles/hour (closer to what jets run too). Then v/c is simple, it's just 10^{-6}; square it and we have 10^{-12}. That is a very small number! Subtract it from one and the result is 0.999999999999, awfully close to "one". We now have to take the square root of that. Clearly the answer is going to very close to one (if you try and do it on a calculator, you will probably get one but that is the wrong answer).
The trick is to look at the fact that the answer will differ from one by a very small amount, x = 1-\delta. We have to find out what \delta is. Well x^2\,=\,1-2\delta+\delta^2. Since \delta is very small, \delta^2 is much much smaller and can be (at least initially) be ignored. It is then simple to conclude that \delta is very close to .5 x 10^{-12} with an error of about .25 x10^{24} (the \delta^2 term).
In fact, it is the \delta term which is of interest here, not the \gamma Ajoint concludes with. Yes, t'\,=\, t/\gamma (or \gamma t' \,=\,t) but it is (t-t') which the Rocker is asking about. That's just \delta t and it become quite clear that, in order to get an answer of one second, we have to wait 2 x 10^{12} seconds or about 11.12 x 10^8 hours. That is something close to one hundred and thirty thousand years (about 126,800 years actually)!
Now remember, this is only true for uniformly moving coordinate systems (special relativity, not general relativity), which means the plane (maybe we should change it to a rocket ship) has to keep that speed in a straight line for close to one hundred and thirty thousand years. At that rate it will be some 740 billion miles away. That's a long way away; however, light could make the same trip in just over 46 days so it is only a fraction of a light year (about an eighth). So what's big and what's little depends on your point of view.
And, speaking of "point of view", if you are going to look at that clock on the plane/rocket ship with a telescope, the answer above is dead wrong! In that case, you have to include the fact that every second the plane is about 977 feet further away (I am presuming here that the clocks agreed as the plane passed you). That means the light you are using to look at that clock had to travel 977 feet more than what you saw a second ago. It follows that, even if there were no relativistic effect (and at 666 mph there practically isn't one), the clocks would "appear" to differ by one second after the ship has traveled for a mere 277 hours (11.5 days, considerably smaller than 130,000 years). This is called the Doppler effect; the frequency appears to be higher for things approaching you and lower for things going away.
If I have made a numerical error in the above, I apologize; I will stand behind the ideas but I often make little math errors.
Take a look at
http://home.jam.rr.com/dicksfiles/StarCurv.htm
you might find it interesting -- Dick
