Re: alice 'n bob again.. those two..

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• I must admit I thought it would be longer. I was under the impression that to get really decent time dilation (e.g. a visit to the galactic core or - for an
Message 1 of 21 , Jan 31, 2002
I must admit I thought it would be longer. I was under the impression that to
get really decent time dilation (e.g. a visit to the galactic core or - for an
extra couple of years, the furthest visible galaxy - in a human lifetime) you
needed to accelerate at slightly more than 1g. (Also, of course this is a purely
centre would burn up in the dust clouds between us and there.)

Charles

----- Original Message -----
From: "Doug Donaghue at 054" <DDonaghue@...>
To: <Fabric-of-Reality@yahoogroups.com>
Sent: Friday, February 01, 2002 2:48 AM
Subject: RE: alice 'n bob again.. those two..

>
>
> > -----Original Message-----
> > From: Charles Goodwin [mailto:charles@...]
> > Sent: Tuesday, January 29, 2002 7:49 PM
> > To: Fabric-of-Reality@yahoogroups.com
> > Subject: Re: alice 'n bob again.. those two..
> >
> >
> > So what's the answer?! :-)
> >
> > Charles
> >
>
> I got 23.4 years of aging for Bob (that's each way, for a total of 46.8
> years) using:
>
>
> where T is the proper time (as measured by Bob in his reference frame), a is
> acceleration, d is
> distance and c is 3*10^8. (It also helps to convert 30,000 light years to
> 2.84*10^20 meters <g>)
>
> There is a derivation of this formula in "Gravitation" by Misner, Thorne and
> Wheeler.
>
>
> Doug
>
>
>
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
• In the old days people could calculate these things using paper and pencil. Here we are in the year 2002 with computers that can perform more computations in a
Message 2 of 21 , Feb 2, 2002
In the old days people could calculate these things using paper and pencil.
Here we are in the year 2002 with computers that can perform more
computations in a second than one person can perform during his entire life,
and guess what? We can't compute a simple hyperbolic cosine. I wonder what
the future will look like once quantum computers have replaced pc's.

Saibal

Charles wrote:
----- Original Message -----
From: Charles Goodwin
To: Fabric-of-Reality@yahoogroups.com
Sent: Wednesday, January 30, 2002 3:48 AM
Subject: Re: alice 'n bob again.. those two..

Charles

----- Original Message -----
From: "Doug Donaghue at 054" <DDonaghue@...>
To: <Fabric-of-Reality@yahoogroups.com>
Sent: Wednesday, January 30, 2002 7:48 AM
Subject: RE: alice 'n bob again.. those two..

>
> Hi again,
>
> > calculator on my PC doesn't do hyperbolic cosines so I can't
> > tell you the
> >
> > Charles
>
>
> I just noticed that the calculator on my PC *does* do hyperbolic
functions.
> Put it in scientific mode (click on 'View') and then click the 'Hyp' box
> just to the right of the 'Inv' box. We're running NT Workstation 4.0
here.
>
>
> Doug
>
>
>
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>

• Hi, ... In the old days (up until about the 1900 s) the word computer had a very different meaning than it does today. But I still have the Pickett Log
Message 3 of 21 , Feb 4, 2002
Hi,

> -----Original Message-----
> From: Saibal Mitra [mailto:smitra@...]
> Sent: Saturday, February 02, 2002 3:24 PM
> To: Fabric-of-Reality@yahoogroups.com
> Subject: Re: alice 'n bob again.. those two..
>
>
> In the old days people could calculate these things using
> paper and pencil.
> Here we are in the year 2002 with computers that can perform more
> computations in a second than one person can perform during
> his entire life,
> and guess what? We can't compute a simple hyperbolic cosine.
> I wonder what
> the future will look like once quantum computers have replaced pc's.

In the 'old days' (up until about the 1900's) the word 'computer' had a very
different meaning than it does today. But I still have the Pickett Log Log
Duplex Decitrig slide rule (12") that I bought in 1960 (for the, then
princely, sum of \$25.00 <g>) and it does hyperbolics just fine.

OTOH, if you're a *real* purist, use e^u =
1+u+(u^2/2!)+(u^3/3!)+.....(u^n/n!) and calculate the hyperbolic functions
as I pointed out earlier.

It converges fairly quickly (the first 6 or 8 terms are sufficient for 3 or
4 digits of precision) And, of course, it also works for imaginary or
complex arguments <g>

Doug
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