Horace,
I can't find a good citation on this but there seems to be A LOT of
conflicting views regarding the way mass "sums" or as you mentioned " shell
mass outside that radius sums to zero", specifically "when" and "where" the
summing occurs (in the medium or in the matter). If the summing occurs in the
matter one would assume it is being pulled equally in all directions by a
larger quantity of gravity and therefore feel weightless but have higher
dilation than on the surface. Wikipedia seems to support this view but the
majority of the forums lean toward your position.
Best Regards
Fran
>From Wikipedia Gravitational time dilation is the effect of time passing at
>different rates in regions of different gravitational potential; the lower the
>gravitational potential (closer to the center of a massive object), the more
>slowly clocks run. Albert Einstein originally predicted this effect in his
>theory of relativity and it has since been confirmed by tests of general
>relativity. http://en.wikipedia.org/wiki/Gravitational_time_dilation
-----Original Message-----
From: Horace Heffner [mailto:[email protected]]
Sent: Friday, October 23, 2009 10:27 PM
To: [email protected]
Subject: Re: [Vo]:Gravity role in fusion
On Oct 23, 2009, at 10:11 AM, Roarty, Francis X wrote:
> What about the acceleration at the very center of the earth, I know
> It cancels but is time dilated even slower than on the surface or
> does dilation cancel too?
> Regards
> Fran
If an elevator existed to the center of the earth and you took it to
the center, time dilation, (1 - (2 G M)/(r c^2))^0.5, would reduce
to zero when you reached the center. This is due to the fact that the
*effective* M is given by M = k * r^3, because the effect of the
spherical shell mass outside that radius sums to zero. The time
dilation factor f is thus:
f = (1 - (2 G (k * r^3))/(r c^2))^0.5
f = (1 - (2 G (k * r^2))/(c^2))^0.5
and:
(limit r ->0) f = (limit r ->0) (1 - (2 G (k * r^2))/(c^2))^0.5
(limit r ->0) f = (1 - 0)^0.5 = 1
and there is no time dilation. Not that you would notice.
Given the mass of the earth is 5.9736 × 10^24 kg and radius r is
6,378.1 km, and G = 6.67259x10^-11 m^3/(kg s^2), we have:
(2 G M)/(r c^2) = ( 2 (6.67259x10^-11 m^3/(kg s^2)) (5.9736×10^24
kg)) / ( (6,378.1 km) / (2.9979x10^8 m/s)^2)
(2 G M)/(r c^2) = 1.3878x10^-9
and
(1 - (2 G M)/(r c^2))^0.5 = 0.999999999326
so time is slowed by about 1.4828 parts in billion, or about 0.02128
seconds per year at the earth's surface, if I did all that right
(which is doubtful given my track record!)
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
