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- Newsgroups: sci.physics
- Subject: Re: Twins Paradox Resolved
- Message-ID: <mcirvin.712337187@husc10>
- From: mcirvin@husc10.harvard.edu (Mcirvin)
- Date: 28 Jul 92 15:26:27 GMT
- References: <BrrztE.J4u@well.sf.ca.us> <712252466@sheol.UUCP>
- Nntp-Posting-Host: husc10.harvard.edu
- Lines: 36
-
- throopw@sheol.UUCP (Wayne Throop) writes:
-
- >> From: metares@well.sf.ca.us (Tom Van Flandern)
- >> Message-ID: <BrrztE.J4u@well.sf.ca.us>
- >> In any frame, the apparent rate of progress of time (and distance) for
- >> a fast-moving traveler is quite different when the traveler is moving
- >> away, from what it is for an approaching traveler.
-
- >This is so clearly wrong that I suspect I'm missing Tom's point
- >altogether. The "rate of progress of time (and distance)" of
- >a fast-moving traveler is identical whether "moving away" or
- >"approaching".
-
- Perhaps the problem here is the meaning of "apparent." Tom
- van Flandern might be talking about what the observer actually
- *sees* looking at emitted light from the traveler, not what
- rate is calculated after taking signal delays into account.
- If I'm traveling toward you at a rate close to c, in your frame
- I'm close on the heels of my emitted light. You see me start
- my trip just before I arrive at your position, and my clock
- *appears* to be sped up. Now, if you know that light has a
- finite speed and take that into account, you'll find that
- my clock is actually slowed down by the usual factor.
-
- Likewise, if I'm traveling away from you, you see my clock
- slowed down to an even more extreme extent than time
- dilation implies, because the signal delay is getting longer
- and longer as I go.
-
- An interesting exercise is to examine how fast the traveling
- and stationary twins' clocks *look* to each other in the
- traditional twin-paradox situation, and to see how the
- elapsed apparent times add up to the correct ratio.
-
- --
- Matt McIrvin mcirvin@husc.harvard.edu
-