Naked Science Forum

Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: DoctorBeaver on 29/11/2005 19:03:24

Title: Uncertain about Heisenberg
Post by: DoctorBeaver on 29/11/2005 19:03:24
This uncertainty principle is bugging me. We know the weights of the different particles. So, if we get a detector that they collide with we can measure their speed by the force with which they hit it; and we know where they are as they've just hit the damned thing!
Isn't that measuring both position & speed simultaneaously or am I missing something?
Title: Re: Uncertain about Heisenberg
Post by: Solvay_1927 on 29/11/2005 21:11:29
Eth, the Uncertainty Principle was discussed in an earlier thread ("Uncertainty about speed cameras"), where I tried explaining to someone called DoctorBeaver what it was all about.  If you contact this DoctorBeaver person I'm sure he can explain it all to you. [:D]

(But I would warn you first - he does have memory problems.)
Title: Re: Uncertain about Heisenberg
Post by: DoctorBeaver on 29/11/2005 21:26:52
Paul - I'm aware of that but this is a different aspect of it that's bugging me
Title: Re: Uncertain about Heisenberg
Post by: neilep on 29/11/2005 22:28:48
Are you certain about that Eth ?

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Title: Re: Uncertain about Heisenberg
Post by: Solvay_1927 on 29/11/2005 23:22:05
OK, OK, just kidding! [;)]

In simple terms (which are all I understand anyway), the HUP states that:

dp.dx >= h

i.e. the uncertainty in momentum (delta p) times the uncertainty in position (delta x) must always be (at least) as big as Planck's constant (h).

(Incidentally, in simple terms, this can also be written as:  dv.dx >= h/m  -  i.e. divide both sides of the previous equation by the mass, m, so the momentum turns into velocity, giving: error in velocity times error in position is at least h divided by the mass).

But h is VERY small: 6 x 10^-34 Joules seconds.

When you say you know the position of a particle hitting a detector, you don't really mean that you know with ABSOLUTE PRECISE accuracy where it is.  What you mean is that you know it to within an accuracy of, maybe, 0.000000000001 metres.  And you know the momentum to an accuracy of, maybe, 0.000000001 kg-metres-per-second.  So then dp.dx is the product of these two numbers, i.e. 10^-12 * 10^-9 = 10^-21, which is greater than h, so the HUP still holds.

The HUP is more useful in theoretical considerations.

For example, another version of the equation is:  dE.dt >= h  -  i.e. uncertainty in energy times uncertainty in time is at least h.  If you apply this to a vacuum, it means (apparently, in very simplistic terms) that for any small region of space, you can't be sure that there's absolutely never any energy in that space.  (For a small period of time dt, you can't have E=0, there must be some fluctuation in E, dE, to ensure that dE.dt >= h.)  Hence virtual pairs of particles and antiparticles popping into and out of existence for (incredibly small) periods of time.

I don't know if this is helping or not (you probably know all this already). Have I missed the point in your question?

P.S. One link I've found re: the HUP which I think is moderately interesting is as follow:
http://www.physlink.com/Education/AskExperts/ae689.cfm
(It's from that other physics forum I don't like much - but occasionally it's useful.)
Title: Re: Uncertain about Heisenberg
Post by: Solvay_1927 on 29/11/2005 23:36:56
Just done a quick search.  Here's another basic description (which is probably better than my explanation above):
http://www.bbc.co.uk/dna/h2g2/A408638
Title: Re: Uncertain about Heisenberg
Post by: DoctorBeaver on 30/11/2005 10:15:58
quote:
When you say you know the position of a particle hitting a detector, you don't really mean that you know with ABSOLUTE PRECISE accuracy where it is. What you mean is that you know it to within an accuracy of, maybe, 0.000000000001 metres. And you know the momentum to an accuracy of, maybe, 0.000000001 kg-metres-per-second. So then dp.dx is the product of these two numbers, i.e. 10^-12 * 10^-9 = 10^-21, which is greater than h, so the HUP still holds.


Point taken. Thank you for the clarification. I suppose the actual particle in the receptor that the incoming particle hits is also positionally uncertain (is that a real term?) so it cannot be ascertained exactly where the particle has hit.

And thanks for the links. I had trouble getting my head around that 1st 1, though. Is it saying that a vertically polarized beam of light isn't polarized vertically? [xx(]
Title: Re: Uncertain about Heisenberg
Post by: extrasense on 30/11/2005 15:13:39
quote:
Originally posted by DoctorBeaver
I suppose the actual particle...



To confuse it a bit more, the "particle" is a wave too.

e[8D]s
Title: Re: Uncertain about Heisenberg
Post by: DoctorBeaver on 30/11/2005 17:20:42
OK - the incoming particle waves to the receptor... [xx(]
Title: Re: Uncertain about Heisenberg
Post by: Solvay_1927 on 30/11/2005 21:28:23
And sometimes particles wink at receptors and flutter their eyelids at them too![:o)]
 
quote:
I had trouble getting my head around that 1st 1, though. Is it saying that a vertically polarized beam of light isn't polarized vertically?

Er, yes ... I think.  I read more about this some time ago, but I've forgotten it all now and I can't find a decent link to it anymore.
(Anyone out there who can give a simple summary of how plane polarised light is actually made up of appropriate circularly polarised photons?)
Title: Re: Uncertain about Heisenberg
Post by: DoctorBeaver on 02/12/2005 13:58:48
quote:
And sometimes particles wink at receptors and flutter their eyelids at them too


They must be charms
Title: Re: Uncertain about Heisenberg
Post by: Solvay_1927 on 09/12/2005 23:37:10
Eth, found a reference related to your question "Is it saying that a vertically polarized beam of light isn't polarized vertically?", in case it's of interest:
http://www.thenakedscientists.com/forum/topic.asp?TOPIC_ID=2391
Title: Re: Uncertain about Heisenberg
Post by: DoctorBeaver on 09/12/2005 23:40:01
Thanks, Paul

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