In article <jofjd3$f4l$1@blue-new.rahul.net>, spope33@speedymail.org
says...
>
> josephkk <joseph_barrett@sbcglobal.net> wrote:
>
> >On Sat, 5 May 2012 05:53:41 +0000 (UTC), spope33@speedymail.org (Steve
>
> >>Actually I think there is a size lump of silicon, that is sufficiently
> >>big that at a given temperature, the number of silicon atoms in that
> >>lump varies based on the uncertainty principle. Virtual
> >>silicon atoms appear/disappear out of the vacuum all the time.
> >>(Trust me, as a EE I know this.) There is no counting the exact number
> >>of silicon nucleii in a large lump.
>
> >This is the first time i have seen virtual particles extended to complete
> >nuclei with associated mass rather than the usual boson and lepton virtual
> >particles.
>
> Cool. You heard it from me first. If the lump of silicon is large
> enough, this should happen.
>
> Someone who is actually a physicist can tell you how large. :--)
>
>
> Steve
Wow. And given a whale big enough, do whole cells pop in and out of
existence?
--
Saludos.
Ignacio G.T.
Reply by Steve Pope●May 10, 20122012-05-10
josephkk <joseph_barrett@sbcglobal.net> wrote:
>On Sat, 5 May 2012 05:53:41 +0000 (UTC), spope33@speedymail.org (Steve
>>Actually I think there is a size lump of silicon, that is sufficiently
>>big that at a given temperature, the number of silicon atoms in that
>>lump varies based on the uncertainty principle. Virtual
>>silicon atoms appear/disappear out of the vacuum all the time.
>>(Trust me, as a EE I know this.) There is no counting the exact number
>>of silicon nucleii in a large lump.
>This is the first time i have seen virtual particles extended to complete
>nuclei with associated mass rather than the usual boson and lepton virtual
>particles.
Cool. You heard it from me first. If the lump of silicon is large
enough, this should happen.
Someone who is actually a physicist can tell you how large. :--)
Steve
Reply by josephkk●May 10, 20122012-05-10
On Sat, 5 May 2012 05:53:41 +0000 (UTC), spope33@speedymail.org (Steve
Pope) wrote:
>Jerry Avins <jya@ieee.org> wrote:
>
>>On 5/3/2012 5:49 PM, Robert Wessel wrote:
>
>>> A lump of pure silicon must, by definition, contain an integral number
>>> of silicon atoms.
>
>>Certainly. But its weight need not be a rational fraction of any
>>particular weight standard.
>
>Actually I think there is a size lump of silicon, that is sufficiently
>big that at a given temperature, the number of silicon atoms in that
>lump varies based on the uncertainty principle. Virtual
>silicon atoms appear/disappear out of the vacuum all the time.
>(Trust me, as a EE I know this.) There is no counting the exact number
>of silicon nucleii in a large lump.
>
>
>Steve
This is the first time i have seen virtual particles extended to complete
nuclei with associated mass rather than the usual boson and lepton virtual
particles.
?-)
Reply by Mac Decman●May 8, 20122012-05-08
On Tue, 01 May 2012 18:16:25 -0500, Tim Wescott <tim@seemywebsite.com>
wrote:
>Instead of doing productive work, I just spent a few enjoyable minutes
>with Scilab finding approximations to pi of the form m/n.
>
>Because I'm posting to a couple of nerd groups, I can be confident that
>most of you probably know 22/7 off the tops of your heads.
>
>What interested me is how spotty things are -- after 22/7, the error
>drops for a bit until you get down to 355/113 (which, if you're at an
>equal level of nerdiness to me will ring a bell, but not have been
>swimming around in your brain to be found).
>
>But what's _really_ interesting, is that the next better fit isn't found
>until you get up to 52163/16604. Then things get steadily better until
>you hit 104348/33215 -- at which point the next lowest ratio which
>improves anything is 208341/66317, then 312689/99532. At this point I
>decided that I would post my answers for your amusement, and get back to
>being productive.
>
>Discrete math is so fun. And these newfangled chips are just destroying
>the joy, by making floating point efficient and cheap enough that you
>don't need to know little tricks like pi = (almost) 355/113.
Thanks Tim, (I've cross posted just for fun) for bombing out my usenet
with this thread. I can't believe this was the one which was better
than any troll post!
Mark DeArman
Reply by David Brown●May 6, 20122012-05-06
On 06/05/12 23:02, Jerry Avins wrote:
> On 5/4/2012 5:59 PM, David Brown wrote:
>
> ...
>
>> You are /still/ missing the point - just because you divide two numbers,
>> does not make the ratio a /rational/ number. It is only /rational/ if
>> the two numbers are integers (or at least rationals). If the numbers are
>> unrelated "random" numbers, such as different physical measurements,
>> then their ratio will be irrational because there is no fundamental
>> common unit of measurement.
>>
>> It's simple probability. (Actually, it's quite hard probability to do
>> this stuff rigorously - but the layman's "it works for finite cases, so
>> we extend it to infinite cases" is good enough for now.) There are
>> /many/ more irrational numbers than rational ones. In any given range,
>> there are only countably infinite rational numbers. But there is 2 to
>> the power that many /real/ numbers, so they vastly outweigh the
>> rationals. Thus if you pick any two random numbers and divide them, the
>> result will be irrational.
>
> You are almost right. The possibility exists that a pair of irrational
> numbers have a rational ratio. Sqrt(8)/sqtrt(2) is an example.
>
If the numbers are unrelated, then the probability is /really/ small -
it is infinitesimal.
Reply by Jerry Avins●May 6, 20122012-05-06
On 5/4/2012 6:45 PM, Jasen Betts wrote:
> On 2012-05-04, Les Cargill<lcargill99@comcast.com> wrote:
>
>>> Of course, it doesn't make much sense to talk about the irrationality of
>>> a measured number, since measurements are by definition limited in
>>> precision, and irrationality is a property of the pure number. It may
>>> also turn out that it is not constant - perhaps it varies gradually with
>>> the expansion of the universe or the strength of surrounding fields.
>>>
>>> Without any explanation or definition otherwise, however, Avogadro's
>>> number is like any other arbitrary number - irrational.
>>
>> But it's *defined* as a number of atoms. That's the point. If it's an
>> indeterminate quantity, it makes no sense to describe it as either
>> real nor irrational.
>
> real numbers are the set of numbers containing all the rational and
> irrational numbers, most of real numbers are irrational by a factor of
> some infinite amount more than the rational numbers.
>
> it makes sense to describe any real number that is not known to
> be rational as irrational.
Numbers obtained by measurement rather than counting necessarily
describe a range. There is an infinite count of numbers within that
range. If you must describe measurements in terms of "rational" and
"irrational", the correct description is "both".
Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Reply by Steve Pope●May 6, 20122012-05-06
Jerry Avins <jya@ieee.org> wrote:
>On 5/4/2012 5:59 PM, David Brown wrote:
>> You are /still/ missing the point - just because you divide two numbers,
>> does not make the ratio a /rational/ number. It is only /rational/ if
>> the two numbers are integers (or at least rationals). If the numbers are
>> unrelated "random" numbers, such as different physical measurements,
>> then their ratio will be irrational because there is no fundamental
>> common unit of measurement.
>>
>> It's simple probability. (Actually, it's quite hard probability to do
>> this stuff rigorously - but the layman's "it works for finite cases, so
>> we extend it to infinite cases" is good enough for now.) There are
>> /many/ more irrational numbers than rational ones. In any given range,
>> there are only countably infinite rational numbers. But there is 2 to
>> the power that many /real/ numbers, so they vastly outweigh the
>> rationals. Thus if you pick any two random numbers and divide them, the
>> result will be irrational.
>You are almost right. The possibility exists that a pair of irrational
>numbers have a rational ratio. Sqrt(8)/sqtrt(2) is an example.
Hence the qualifying term "unrelated". These two number are related.
Steve
Reply by Jerry Avins●May 6, 20122012-05-06
On 5/4/2012 5:59 PM, David Brown wrote:
...
> You are /still/ missing the point - just because you divide two numbers,
> does not make the ratio a /rational/ number. It is only /rational/ if
> the two numbers are integers (or at least rationals). If the numbers are
> unrelated "random" numbers, such as different physical measurements,
> then their ratio will be irrational because there is no fundamental
> common unit of measurement.
>
> It's simple probability. (Actually, it's quite hard probability to do
> this stuff rigorously - but the layman's "it works for finite cases, so
> we extend it to infinite cases" is good enough for now.) There are
> /many/ more irrational numbers than rational ones. In any given range,
> there are only countably infinite rational numbers. But there is 2 to
> the power that many /real/ numbers, so they vastly outweigh the
> rationals. Thus if you pick any two random numbers and divide them, the
> result will be irrational.
You are almost right. The possibility exists that a pair of irrational
numbers have a rational ratio. Sqrt(8)/sqtrt(2) is an example.
...
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by David Brown●May 6, 20122012-05-06
On 05/05/12 00:45, glen herrmannsfeldt wrote:
> In comp.dsp David Brown<david.brown@removethis.hesbynett.no> wrote:
>
> (snip, someone wrote)
>>> Then it's a *fraction*. It's not irrational... I thought I
>>> had adequately corrected "integer" to "rational"... in truth,
>>> there's a bijective map between them, so...
>
>> You are /still/ missing the point - just because you divide two numbers,
>> does not make the ratio a /rational/ number. It is only /rational/ if
>> the two numbers are integers (or at least rationals). If the numbers
>> are unrelated "random" numbers, such as different physical measurements,
>> then their ratio will be irrational because there is no fundamental
>> common unit of measurement.
>
> In the case of measured quantities, there is no point in saying that
> one is irrational. That is, quantities that have an uncertainty.
True.
>
> Now, that doesn't mean that physical constants can't be irrational.
>
Absolutely.
As has been noted already, this discussion is pretty pedantic. All I am
really saying is that the theoretical "true" values - that you can never
measure, even if they can be said to exist at all - are irrational. In
particular, I am saying that you can't call them "rational" nor
"integer", unless of course your units are defined to make them rational.
Reply by Fred Abse●May 5, 20122012-05-05
On Wed, 02 May 2012 06:23:45 +0100, John Devereux wrote:
> We had a teacher that insisted it was exactly equal!
What did he/she teach?
History? drama? civics? languages? phys.ed?
--
"For a successful technology, reality must take precedence
over public relations, for nature cannot be fooled."
(Richard Feynman)