From: tmyoung@rodan.acs.syr.edu (Thomas M. Young)
Subject: Re: general relativity
Organization: Syracuse University, Syracuse, NY
Date: Thu, 13 Aug 92 09:40:14 EDT

In article <drenze.713555212@icaen.uiowa.edu> drenze@icaen.uiowa.edu 
(Douglass J. Renze) writes:
>I've been re-reading Hawking's "A Brief History of Time" (yes, I know--pop-
>
>Now, I've a question:  Out of curiosity, how would one calculate this time
>differential?  Assuming that we could take "instantaneous" readings of
>a clock in Earth's orbit around the sun, and a clock, say, in Saturn's
>orbit (yes, I know...instantaeous, relative to what?  Howzabout relative
>to the frame of reference which you and I inhabit?  Or, just the frame of
>reference which I inhabit, since your frame of reference is entirely a
>figment of my imagination ;) ... what would the time differential be?
>(eg, supposing a reading were taken at t(0), at which the clocks both read
>00:00:00 UTC...when the clocks were again read at t(1), assuming that the
>clock in Earth's orbit read 01:00:00 UTC, what should the clock in Saturn's
>orbit theoretically say?

OK - well my (non-expert) understanding of how you would figure it 
is basically this:

The clock that was lifted up away from the Earth, (ex: your clock at Saturn)
will run faster by a fraction (Lorentz contraction a.k.a Fitzgerald ratio?).
This is also the ratio for how much the mass of the clock is increased to 
store the potential energy you add by lifting it.

So to figure the rate for the clock,
1. you figure the amount of potential energy you added by lifting the clock.  
2. you figure the energy-equivalent of the clock's rest-mass.
3. Add #1 and #2, and divide the sum by #2.

Ex: A 1kg clock is lifted to the top of a 100m tower.  The potential energy
added in order to do this is given by the formula P.E.= m*g*h, where m=1kg,
g=9.8m/s^2, and h=100m. (Note this formula only works for smallish heights,
you have to use a more detailed one for orbital distances & such).  So
P.E.= 980 J.  OK, now the energy-equivalent of the 1kg clock is given by
E = mc^2, so E = (1kg)(3e8)^2 = 9e16 J.  We now take our clock's rate as
(P.E. + E)/E = ((980J)+(9e16J))/(9e16J) = 1.000000000000098 = 1 + 9.8e-14.
The clock on the top of the tower will gain about 3 microseconds a year.

Now I have been told that potential energy is a pseudotensor in general
relativity.  My understanding is that it is perfectly valid to think of the
Earth as sitting in the clock's gravitational field, and to calculate the
increase in P.E. of the Earth as it is pushed away from the clock.  The
P.E. turns out to be the same, but of course E above is much larger, so
there is no noticable effect on the rate of passage of time for Earth. 
(mass of earth =approx 6e24 kg, the act of lifting that clock causes all
the other clocks on Earth to gain about .5e-29 nanoseconds/year)

If anybody out there can give me a better way to figure this,
please do.  Particularly, I'm confused about why adding potential energy
would speed up a clock, while adding kinetic would slow it down relative to
a clock that you don't add any kind of energy to.

--
Tom Young                                         tmyoung@rodan.acs.syr.edu
Newsgroups: rec.arts.sf.science