| Topic: |
Science > Physics |
| User: |
"Barrow" |
| Date: |
20 May 2007 01:51:51 AM |
| Object: |
geometrized units and numerical calculation |
Dear All,
For simplicity in the formula, we often drop the light speed c and
gravitational constant G in equations, i.e., let c = G = 1.
I was studying the dynamics of a relativistic system, so I
calculated some field equation and worked out the numerical result.
But I am confused that what my t = 1 stands for?
At first glance, I think t = 1 is equivalent to 1/(3*10^8) sec in SI
units. But I also let G = 1, doesn't this cause anything?
I came up with that Plank time = \sqrt{hG/c^5} ~ 10^-44 sec, if I
just used geometrized units, that is, c = G = 1, then my unit of time
is equal to 10^-44/sqrt{10^-34} ~ 10^-28 sec?? (where 10^-34 is the
value of Planck's constant)
I really got confused about the units to my numerical result, any
help will be appreciated! what my t = 1 stands for...
Sincerely Barrow
.
|
|
| User: "Ben Rudiak-Gould" |
|
| Title: Re: geometrized units and numerical calculation |
20 May 2007 11:50:27 PM |
|
|
Barrow wrote:
For simplicity in the formula, we often drop the light speed c and
gravitational constant G in equations, i.e., let c = G = 1.
I was studying the dynamics of a relativistic system, so I
calculated some field equation and worked out the numerical result.
But I am confused that what my t = 1 stands for?
It means you made a mistake somewhere. Your times should have units of distance.
-- Ben
.
|
|
|
|
| User: "The Ghost In The Machine" |
|
| Title: Re: geometrized units and numerical calculation |
20 May 2007 10:33:31 PM |
|
|
In sci.physics, Barrow
<GRseminar@gmail.com>
wrote
on 19 May 2007 23:51:51 -0700
<1179643911.427324.76430@k79g2000hse.googlegroups.com>:
Dear All,
For simplicity in the formula, we often drop the light speed c and
gravitational constant G in equations, i.e., let c = G = 1.
I was studying the dynamics of a relativistic system, so I
calculated some field equation and worked out the numerical result.
But I am confused that what my t = 1 stands for?
At first glance, I think t = 1 is equivalent to 1/(3*10^8) sec in SI
units. But I also let G = 1, doesn't this cause anything?
I came up with that Plank time = \sqrt{hG/c^5} ~ 10^-44 sec, if I
just used geometrized units, that is, c = G = 1, then my unit of time
is equal to 10^-44/sqrt{10^-34} ~ 10^-28 sec?? (where 10^-34 is the
value of Planck's constant)
I really got confused about the units to my numerical result, any
help will be appreciated! what my t = 1 stands for...
Sincerely Barrow
You are no doubt aware of such unit conversions as
1 in = 2.54 * 10^-2 m; a measurement of 10 inches
can be converted by simply multiplying:
10 in = 10 * (2.54 * 10^-2 m/in) = 25.4 cm.
Time units such as minutes and hours are similarly
handled:
1.25 hours * (60 min/hour) = 75 minutes
(There are some considerations regarding Kelvin,
Centigrade, Rankine, and Fahrenheit which are more
complicated than shown here, unless one is converting
Rankine to Kelvin and vice versa. Such conversions
are rather rare nowadays; most people in the US use
Fahrenheit, and the EU Celsius. The scientific
community uses Kelvin almost exclusively, at least
for low temperatures.)
c = 1 by a judicious choice of space and time units.
For example, one could measure things using seconds and
light-seconds (a light-second is, of course, 299792458 m).
If one prefers, one can keep the meter and define a time
unit, which would logically be called "light-meter" but
that's a bit unwieldly (not to mention already in use
amongst shutterbugs).
G is commonly expressed 6.674215 * 10^-11 m^3 / (kg-s^2);
a slightly more logical set of units might be
6.674215 * 10^-11 N/kg^2, but the two are entirely
equivalent (since 1 Newton is 1 kg accelerated 1 m/s/s).
Since we've already defined time and space, we are left
with a new mass unit such that G = 1.
If, as you suggest, we use meters, ticks, and nuggets [*] such that
c = G = 1, then a tick is 1/299792458th of a second, and a
nugget will be such that G = 1 m^3 / (nugget-tick^2).
Since G = 6.674215 * 10^-11 m^3/(kg-s^2) * (1/299792458 s/tick)^2
= 7.426066 * 10^-28 m^3/(kg-ticks^2),
one gets 7.426066 * 10^-28 nuggets/kg, or 1.3466081 * 10^27 kg/nugget,
or about 673.3 microSols (1 Sol is just under 2 * 10^30 kg -- the
mass of our sun; Earth is therefore just shy of 3 microSols in mass).
(Note that the multiplication above is treating the units as more or less
cancellable variables. This is more or less typical in physics.)
Since the second is commonly defined as 9192631770 cycles
of a certain Cs-133 transition, one might be better off
using seconds, light-seconds, and granules. In this case,
G = 6.674215 * 10^-11 m^3/(kg-s^2) * (1/299792458 lsec/m)^3
= 2.477069 * 10^-36 lsec^3/(kg-s^2)
and therefore a kg is about 2.477069 * 10^-36 of a granule -- and
therefore a granule is about 200,000 times as massive as Sol, or
even bigger than a nugget.
A third possibility is cesium-ticks (cticks), zits, and globs.
A cesium-tick is exactly 1/9192631770th of a second. A zit
is exactly 299792458/9192631770 m (about 3.261 cm) in length. So:
G = 6.674215 * 10^-11 m^3/(kg-s^2) * (9192631770/299792458 zits/m)^3
* (9192631770 cticks/s)^(-2)
= 2.27707821 * 10^-26 zits/(kg-cticks^2)
and therefore a glob is about 22 microSols.
The units for Planck's Constant are joule-seconds, or kg-m^2/s.
Therefore, one can apply straightforward conversions:
h = 6.6260693 * 10^-34 kg-m^2/s
= 6.6260693 * 10^-34 kg-m^2/s * 7.426066 * 10^-28 nuggets/kg
* (1/299792458 s/tick) = 1.64132308 * 10^-69 nugget-m^2/tick
= 6.6260693 * 10^-34 kg-m^2/s * (299792458 m/lsec)^2
* (2.477069 * 10^-36 granule/kg)
= 1.4751476 * 10^-52 * granule-lsec^2/sec
= 6.6260693 * 10^-34 kg-m^2/s * (2.27707821 * 10^-26 globs/kg)
* (9192631770/299792458 zits/m)^2 * (9192631770 cticks/s)^(-1)
= 1.54323612 * 10^-66 globs-zits^2/cticks
HTH
[*] the names are, of course, my own invention.
--
#191,
Insert random misquote here.
--
Posted via a free Usenet account from http://www.teranews.com
.
|
|
|
|
| User: "H. Wabnig .... .-- .- -... -. .. --. @ .- --- -. DOT .- -" |
|
| Title: Re: geometrized units and numerical calculation |
20 May 2007 02:11:53 AM |
|
|
On 19 May 2007 23:51:51 -0700, Barrow <GRseminar@gmail.com> wrote:
Dear All,
For simplicity in the formula, we often drop the light speed c and
gravitational constant G in equations, i.e., let c = G = 1.
I was studying the dynamics of a relativistic system, so I
calculated some field equation and worked out the numerical result.
But I am confused that what my t = 1 stands for?
At first glance, I think t = 1 is equivalent to 1/(3*10^8) sec in SI
units. But I also let G = 1, doesn't this cause anything?
I came up with that Plank time = \sqrt{hG/c^5} ~ 10^-44 sec, if I
just used geometrized units, that is, c = G = 1, then my unit of time
is equal to 10^-44/sqrt{10^-34} ~ 10^-28 sec?? (where 10^-34 is the
value of Planck's constant)
I really got confused about the units to my numerical result, any
help will be appreciated! what my t = 1 stands for...
Sincerely Barrow
t stands for "Tea Time"
The t unit is 3 minutes soaking duration.
Then lift -dip -lift the tea bag 5 times, squeeze it dry,
add some teaspoons of Scotch Whiskey, get some muffins,
and enjoy.
Remember, t equals 3 minutes.
w.
.
|
|
|
|
|
Related Articles |
|
|
