Re: question

Liste des GroupesRevenir à s math 
Sujet : Re: question
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.math
Date : 18. May 2024, 18:30:01
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <3b43df19-9011-49f3-a7ee-ea7040af7092@att.net>
References : 1 2 3
User-Agent : Mozilla Thunderbird
On 5/17/2024 11:14 PM, Peter Fairbrother wrote:
On 17/05/2024 13:56, Jim Burns wrote:
On 5/16/2024 7:41 PM, Peter Fairbrother wrote:

Is lim (cos pi/2n)^n = 1 as n -> infinity?
>
Yes, 1.
>
a) thanks
My pleasure.

b) yikes!
I remember doing some of the below math
at school and university, and thinking
I'd probably never need it again.
This is now the second time
I have needed something like that,
in 50 years,
so maybe I should learn it to immediately usable status,
or maybe not:
but in any case
I will have to look at the details
which you have so kindly provided.
Sorry if my answer was a little too.much.information.ish.
It might have been shaped by my recent discussions,
in which every last #$%^ing.obvious detail
gets disputed.
That, of course, has nothing to do with you.
Expanding functions into series
and throwing away higher order terms
is a more physicsy presentation:
cos(x)  ≅  1 - x²/2 + ...
(1 - x)^n  ≅  1 - n⋅x + ...
cos(π/2n)^n  ≅
(1 - (π/2n)²/2)^n  ≅
1 - n⋅(π/2n)²/2  ≅
1 - (π²/8)/n  ⟶  1
Whew! Same answer! Yay me.

It's about a physics thing,
passing light through polarising filters:
eg when n=1 the expression =0,
no light passes through a pair of polarisers which are crossed at 90 degrees.
>
Hmmm,
when n=0 does the expression (cos pi/2n)^n = 1/2?
(cos π/0)^0 is undefined.
(cos π/2⋅2)^2 = 1/2

Insert another suitably orientated filter
in between those filters (n=2),
and some light passes.
Oooh, quantum weirdness, recent Nobel prizes,
Bell's theorem, spooky-action-at-a-distance,
and so on.
That effect would make a nice exhibit at
a science museum
https://cosi.org/
https://www.exploratorium.edu/
etc
I'm thinking of
two fixed polarizers at 90°, backlit,
and shaped polarizers: letters animals puppets
which can be inserted between them.
Where the shapes are, more light, not less.
Weirdness.
Budding Nobel.prize.winners galore!

Insert lots and lots of filters
in a rotating pattern,
and all (1/2) the light passes through.
>
Except -
suppose it's all really simple instead,
and light which passes through a filter
just has its polarisation changed a little.
Then all the Bell hidden variables are irrelevant,
as they change depending on their particles' history.
I can do the math for all that,
but I struggle with the concepts.
What I read tells me I have illustrious company.
I imagine some future generation,
maybe not very far off in time,
being as comfortable with 21st cen. quantum weirdness
as we are with 19th cen. electromagnetic weirdness.
It seems to me that the de.weirding will be brought by
the widespread use of quantum technologies --
not very far off at all.
I imagine us today pecking out "Chopsticks" on
the grand piano of quantum physics, with
our quantum computing and our quantum key distribution.
There will come a Mozart, a Van Cliburn,
and, until they come,
we won't be able to imagine what they'll play.

Note,
there is necessarily a measurement of the photons
in a polarising filter -
but this does not necessarily involve
a complete wavefunction collapse.
My probably.out.of.date understanding of
quantum wave collapse
is that
pre.collapse, there are interference patterns
post.collapse, there aren't interference patterns.
I'd be surprised if the interference patterns
of multiply.polarized light goes away,
even partially.
My intuition gives it a thumbs.down,
but I haven't looked into this carefully.

On to entanglement... -
but that's another story, for later.
I hope I'll get to see someone tell that story.

Date Sujet#  Auteur
17 May 24 * question4Peter Fairbrother
17 May 24 `* Re: question3Jim Burns
18 May 24  `* Re: question2Peter Fairbrother
18 May 24   `- Re: question1Jim Burns

Haut de la page

Les messages affichés proviennent d'usenet.

NewsPortal