Re: Variation of Twins Problem

seppala51_at_sbcglobal.net
Date: 08/26/04 

Date: Thu, 26 Aug 2004 23:22:48 GMT

On 25 Aug 2004 23:07:19 -0700, suzysewnshow@yahoo.com.au
(suzysewnshow) wrote:

>seppala51@sbcglobal.net wrote in message news:<412d105e.31469663@news.houston.sbcglobal.net>...
>> <snip>
>> >Actually, A observes B's clock to switch from ticking slower than his own
>> >clock to ticking faster than his own clock while A is _still decelerating_.
>> >This ''switch point'' occurs where R = 1 (R being the ratio of B's clock
>> >rate to A's clock rate as observed by A). According to the formula we have
>> >
>> >R = Sqrt[1 - u^2/c^2] { 1+ |g| D/c^2}
>> >
>> >I replaced -g by +|g| since g is negative in your example.
>> >
>> >u is the velocity of B as observed by A
>> >D is the distance of B as observed by A
>> >g is the acceleration of A as measured by an accelerometer carried by A
>> >
>> >Just after A and B initially pass each other, R < 1 because u is still
>> >relatively large while D is very small. R<1 means that A observes B ticking
>> >slower than A.
>> >
>> >As they continue to decelerate u becomes smaller while D becomes larger. At
>> >some point *before* u reaches zero, R will become greater than 1 and A will
>> >start to observe B's clock to be ticking at a faster rate than his own
>> >clock.
>> >
>> >JJ
>> Thanks for the post, and the link to Tom Snyder's site (who helped
>> explain things to me). But after looking at things, there's a few
>> questions I have.
>> 1. I don't understand the following regarding Tom's equation (above).
>> In that equation g doesn't necessarily go to zero when u goes to zero.
>> Clocks A and B can be accelerating (or decelerating in my problem) and
>> reach a relative velocity of zero at some point. Let's say there are
>> four clocks, A1 and A2 at the same point in space, and clocks B1 and
>> B2 at the same point a distance D away from A1 and B2. Let A1 and B1
>> have zero relative velocity. Let A2 and B2 decelerate from some
>> velocity V to zero relative to clocks A1 and B1 and wrt to each other
>> and then continuing their deceleration. At this point (when all four
>> clocks have zero relative velocity) the ratio of the clock rate of A1
>> to the clock rate of A2 equals some constant C1 (they have zero
>> relative velocity at this point in time). The ratio of the clock rate
>> of B1 to the clock rate of B2 equals the same constant. There is
>> zero relative velocity between all four clocks. The ratio of the
>> clock rate of A1 to the clock rate of B1 equals one (the clocks have
>> zero relative velocity). Doesn't that make C1 = 1? Tom's equation
>> says no. C1 does not equal 1 in this case. Can you explain what is
>> going on here? If A1/A2 = B1/B2 and A1=B1 doesn't that mean that
>> A2=B2. I don't see how Tom's equation shows that.
>> 2. Is there any experiment observers in craft A and craft B of my
>> problem can do to measure how their respective clocks tick relative to
>> each other?
>> 3. Responses to various length contraction problems I've posted
>> resulted in responders saying the length contraction is a physical
>> phenomena that occurs at the speed of sound. When an accelerated
>> clock stops accelerating is there some time response necessary for the
>> clock to start ticking at the same rate as other non-accelerated
>> clocks in the same rest frame?
>> David
>> >
>
>If the twins agree on the number of times a satellite orbits a body,
>which they must if they have a continuous view of the satellite, why
>in heaven and earth would you think that clocks should behave so
>strangely?
If one twin is near the orbiting satellite, and the other twin is many
light years away, and say the satellite makes an orbit once a day, the
two twins won't have the same count of the number of orbits.
Einstein's theory is why one thinks the clocks act so strangely,
because he hypothesized that all observers always measure the speed of
light in a vacuum to be the same independent of how fast they are
moving relative to each other, and furthermore he hypothesized
observers moving at different velocities can do no experiment which
reveals a frame's constant velocity with respect to space.
> When my watch behaves that way I know it is time to change
>the battery. Have you condidered that?
>Kind regards, Sue...

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