sysred (01/18/83)
I submitted this to net.physics with no response. Is the answer too obvious to bother answering, has net.physics disappeared or is everyone afraid to flame about a question that can be investigated experimentally (thanks to D. Wall for the last observation!)? Come on, gang, turn up those burners to full broil and tell me the answer!! ==================================================================== Here's the scenario: I have a digital clock in my office. I neglected to reset it for Eastern Standard Time last fall, so it's still an hour ahead. Someone walks into my office and gives me a hard time about the fact that my clock's an hour off (actually, I'm just a few months ahead of fixing it next spring...). So, we start making funny conjectures about how to set the right time. With a subconscious memory of previous net.physics discussions running around inside my ahead, I propose to speed up the clock (a LOT!!) until it relativistically slows down and loses the hour. How to do this, you ask? Well, suppose the line cord is REAL strong, and I swing it around my head REAL fast to make the clock lose the hour. It's a digital clock and counts line frequency signals to keep time. Where do the extra pulses on the line go? X pulses go in, and X come out, but the clock sees X-(however many it needs to lose an hour). Right? - Ralph Droms Computer Science Dept. Penn State (what does physics have to do with football, anyway?)
CAD:moore (01/19/83)
#R:psuvax:-115300:ucbcad:26000003:000:590
ucbcad!moore Jan 18 21:41:00 1983
The answer to this paradox is that the clock does NOT lose
any time. As you pointed out, your clock is not really a clock, but
rather a line pulse counter. So no matter what the actual time rate
for the clock is, the displayed time will be only a function of the
pulse rate from the power plant. Since the power plant is not being
spun around, the pulse rate will stay constant and thus the clock
will not lose any time.
Remember, clocks never directly measure time, they only measure
other quantities that we assume have a particular functional dependance
on time.
Peter Mooreneiman (01/19/83)
Someone just made the claim that the clock does not lose time because it always receives the same number of pulses. Remember that these pulses are crossing a energy potential barrier. In the improbable case that the clock runs at all, the signal coming over the line would appear to be (I think) very high frequency, probably beyond the capacity of the pulse counter to respond. Actually, I wonder how the innards of a digital clock would respond to x-rays coming over the line cord. But anyways, my guess is that the clock would not detect any pulses so that after an hour of vigorous swinging it would show the correct time. It would be radiation damaged, melted by friction, and torn apart by tidal forces, but it would show the correct time. Not afraid to admit I'm just guessing, dann
thomson (01/20/83)
The answer is simple. When you whirl the clock around your head, it does indeed count 60 Hz pulses and doesn't miss any. Therefore, the clock does not run slow as SR predicts. So, the clock is wrong. It happens because you aren't whirling the "clock" around your head, you're only whirling its display. The timebase is at your local generating station, which is not whirling over your head.
mark (01/21/83)
net.physics hasn't gone away here, at least. i suspect the original posting simply got lost, which seems a common enough occurance. the question was: can i make a digital clock, which simply counts waves in the power line, lose an hour by whirling it over my head very fast? what happens to the extra line pulses? this much-abused instrument is not in fact a clock; it is just a display. the clock is whatever is keeping the line frequency at a nice even 60 hertz. if you put a self-contained clock on the other "clock" while spinning it, the new clock would lose an hour. there are actually things in the display which are local-time-dependent, ie the electronics of it. while it is being whirled, it is slowed, so the line frequency looks very high to it. but the actual number of cycles is, of course, not changed by this. mARK bLOORE univ. of toronto
dag (01/24/83)
Disclaimer: I don't know what I'm talking about... Comment: It is my understanding that the time compression (1-(v**2/c**2)) uses a vector v, ie. if you were to spin a clock around at a given SPEED, very fast, so long as it was constant during each revolution, and ignoring the fractional turns during acceleration/deceleration, the velocity is zero since it is measured relative to the observer (In the center). So even if the clock was a crystal or mechanical clock, not being supplied with its time base from a source stationary with respects to the observer, no relatavistic time compression could be observed since none would take place. -EOC-