Darren New <dne### [at] sanrrcom> wrote:
> Warp wrote:
> >   I think that the vernacular interpretation is that the single photon
> > traverses as a wavefront and interferes with itself when it passes through
> > both slits at the same time.

> Well, not really. The basic problem is people calling it an "interference 
> pattern" to start with. It's not really "interference" as such, since 
> there's only one particle. Calling it "interference" just encourages people 
> to go looking for what wave is causing the interference.

  I see a pattern here (pun semi-intended). Shooting individual photons one
at a time results in a pattern which is identical to the interference pattern
which results from shooting huge bunches of photons at the same time, a
pattern which is what one expects from a wave. But since a photon is not a
wave, it's not an interference pattern. It's just a pattern which happens
to look exactly like an interference pattern, but without being one.

  And if I follow your earlier logic, the pattern looks like an interference
pattern *because* it's not an interference pattern.

-- 
                                                          - Warp

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Darren New <dne### [at] sanrrcom> wrote:
> Warp wrote:
> >   It looks like a wave, it behaves like a wave, it produces all the effects
> > that a wave would produce, but it's not a wave.

> No, it doesn't. It has the same math as a wave, *if* there's only one. If 
> there's more than one, the behavior isn't like waves.

> What function of waves produces lasing? Polarization?

  Your logic seems to be that since photons do *more* than waves do, in
other words, photons seems to be a superset of pure waves, then they are
not waves at all.

  Why not apply the same logic in reverse? Since photons seem to do more
than pure particles would be expected to do, then they are not particles
either.

  It feels a bit like saying that a seaplane is not a plane because normal
planes can't land on water but a seaplane can.

> >>> "OK, so why do I still get the exact same patterns if there's only one 
> >>> photon there?"
> > 
> >> Because there are no waves.
> > 
> >   You get a wave interference pattern *because* (not "even though") there's
> > no wave. That makes sense.

> No, you get the exact same patterns when there's only one photon there, 
> because it's not caused by interference between photons.

  I find it rather amusing that QM doesn't seem to have any problem in
accepting phenomena which feel completely supernatural and counterintuitive,
such as the state of one particle instantly determining the state of another
particle which is light-years away, when the state of the particle is
"observed" (whatever that might mean), but the idea of a photon passing
through two slits at the same time and interfering with itself seems too
hard to swallow.

  (I'm not saying that's what's happening. I'm saying that the way you write
makes it sound like that.)

-- 
                                                          - Warp

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Warp wrote:
> It's just a pattern which happens
> to look exactly like an interference pattern, but without being one.

Right!  The math works out to be the same, in some limited cases.

Did you watch the youtube video I linked to? Or are you trying to apply 
common sense to a field which even nobel prize winners in that field agree 
it doesn't make common sense?

>   And if I follow your earlier logic, the pattern looks like an interference
> pattern *because* it's not an interference pattern.

No, you misread that. I said you know it's not an interference pattern 
because it happens with only one photon at a time.

Look, don't argue with me. Argue with Feynman if you want.

-- 
Darren New, San Diego CA, USA (PST)
   Human nature dictates that toothpaste tubes spend
   much longer being almost empty than almost full.

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Warp wrote:
>   Your logic seems to be that since photons do *more* than waves do, in
> other words, photons seems to be a superset of pure waves, then they are
> not waves at all.

No, that isn't my logic. Did you watch the video I linked to, where Feynman 
talks about a few of the things that show why photons aren't waves? Do you 
realize that Einstein got his nobel prize for showing that photons can't be 
waves?

The point is that they aren't waves when you look at more than one at a 
time, because they don't even cause interference patterns like you would 
expect. (Otherwise, lasers wouldn't work.) They aren't waves when you use 
electrons, even though you get "interference" patterns as long as you *only* 
shoot one at a time. One photon traveling through vacuum (or otherwise not 
interacting) follows the same math as a wave. Two photons don't. Two 
electrons don't. A photon interacting with an electron 100 feet down the 
road doesn't act like a wave *here*.

I'll agree it isn't intuitive that the motion of particles obeys some 
unobvious rules. That doesn't make them waves.

>   (I'm not saying that's what's happening. I'm saying that the way you write
> makes it sound like that.)

It's not "too hard to swallow". It's simply counter to experimental 
evidence. The interference has nothing to do with how many slits the photon 
passes through. Indeed, that's the *point* of the two-slit experiment. You 
*only* get "interference" patterns if you *don't look*.  If the 
"interference" patterns were caused by the photon passing through both 
slits, then they would persist if you did something to the photon *after* it 
passed through both slits. But it doesn't.

Indeed, if it was caused by the photon passing through two slits at once, 
you'd expect to get the interference pattern if you closed the slits only 
after you'd measured where the photon landed. But you don't. If you can 
explain that in terms of waves, there's a nobel prize waiting for you.

Is it really that hard to believe that there are multiple ways to come up 
with the same numbers that waves give you?

-- 
Darren New, San Diego CA, USA (PST)
   Human nature dictates that toothpaste tubes spend
   much longer being almost empty than almost full.

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Darren New <dne### [at] sanrrcom> wrote:
> Or are you trying to apply 
> common sense to a field which even nobel prize winners in that field agree 
> it doesn't make common sense?

  That is contradictory with:

> I said you know it's not an interference pattern 
> because it happens with only one photon at a time.

-- 
                                                          - Warp

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Warp wrote:
> Darren New <dne### [at] sanrrcom> wrote:
>> Or are you trying to apply 
>> common sense to a field which even nobel prize winners in that field agree 
>> it doesn't make common sense?
> 
>   That is contradictory with:
> 
>> I said you know it's not an interference pattern 
>> because it happens with only one photon at a time.

I don't see that as contradictory, unless you mean that if it doesn't make 
common sense, you can't talk about it with common words at all. I'm down 
with that. The math doesn't work out the same either.  :-)

But seriously, for there to be an interference pattern, something has to 
interfere with something else, right?  As soon as you interfere with the 
photon, the pattern goes away.  Indeed, the pattern is *only* there exactly 
to the extent that there is *no* interference.

-- 
Darren New, San Diego CA, USA (PST)
   Human nature dictates that toothpaste tubes spend
   much longer being almost empty than almost full.

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Darren New <dne### [at] sanrrcom> wrote:
> But seriously, for there to be an interference pattern, something has to 
> interfere with something else, right?

  If the photon is a wavefront which traverses from the emitter to the
detector, passing throug the slits splits it into two wavefronts, which
interfere with each other. When the wavefront collides with the detector,
it collapses back into a particle.

  I'm not saying that is what happens. I'm just saying it's exactly as
plausible as eg. a particle being in many places at the same time or
affecting another particle instantly.

>  As soon as you interfere with the 
> photon, the pattern goes away.  Indeed, the pattern is *only* there exactly 
> to the extent that there is *no* interference.

  So why does the interference pattern appear? Is nature trying to confuse
us to make it look exactly like it was a wave, without it being so?

-- 
                                                          - Warp

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Warp wrote:
> Darren New <dne### [at] sanrrcom> wrote:
>> But seriously, for there to be an interference pattern, something has to 
>> interfere with something else, right?
> 
>   If the photon is a wavefront which traverses from the emitter to the
> detector, passing throug the slits splits it into two wavefronts, which
> interfere with each other. When the wavefront collides with the detector,
> it collapses back into a particle.

But it doesn't work that way. If you look to see which slit the particle 
passed through, you find out that it only went through one. You never, ever 
see it go through both slits. You basically never see a wave or measure a 
wave. You always measure a particle, even as the whatever goes through the 
slits, even *after* the whatever goes through the slits.

>   I'm not saying that is what happens. I'm just saying it's exactly as
> plausible as eg. a particle being in many places at the same time or
> affecting another particle instantly.

Plausible? Yes. Confirmed by experimental evidence? No. Contradicted by 
experimental evidence? Yes.

I'm just stating what the guys who study this say. You're trying to tell me 
it sounds absurd. Sure, maybe it does. Maybe the world *might* work a 
different way than it does. Maybe if the only experiment you did was the 
two-slit experiment, you'd think photons are sometimes waves. But they're not.

>   So why does the interference pattern appear? 

You know when you do something with macro-sized objects, you get 
one-dimensional probabilities, right?  If you say "flip a coin, twice", 
you'll have the probability of getting two heads being the probability of 
getting one head *times* the probability of getting one head, right? (I'll 
leave it as an exercise for you to decide *why* that is the case, beyond 
"common sense.")  The probability of getting heads either time is the 
probability of getting heads the first time *plus* the probability of 
getting heads the second time *minus* the probability of getting both (just 
so you don't count the same event twice).

It's the same way with quantum particles, except the probabilities are 
complex numbers, hence "amplitudes". The probability of it hitting the 
screen at a given position is the amplitude of it going from the emitter to 
the left slit to that point on the screen *plus* the probability of it going 
from the emitter to the right slit to that point on the screen. But those 
probabilities change based on the distance the particle travels, and when 
you do all the adding and multiplying, the imaginary components wind up 
canceling out some of the "real" components in a way that looks like waves.
Because the probability of a quantum particle going in any given direction 
is a function of its amplitude.  "Amplitude" is a technical term meaning 
"two-dimensional probability." You get the probability of finding a particle 
at a particular place by adding together all the amplitudes of it getting 
there, then taking the absolute value (i.e., the length of the result).

In other words, the probability of something going from A to B is the sum of 
all the amplitudes of the ways it can get from A to B, then you take the 
resulting size. Just like if you want to get a heads on the first flip or a 
heads on the second flip, you add the probabilities.

The probability of two things happening is the multiplication of the two 
*probabilities*, because you know the things happened.

If you measure the likelihood of the particle going from the emitter to the 
screen without checking which slit it went through, the probability is the 
absolute value of the sum of all the ways it could get there.

If you check which slit it went through, it's the probability that it got 
there by going through the first slit plus the probability that it got there 
by going thru the second slit. (Since it never goes through both, you don't 
have to subtract out that possibility like you do with flipping coins.) The 
probability of it going from the emitter to either one of the slits, or from 
the slit to any point on the screen, you still measure by adding up complex 
amplitudes for each possible path, and not just using probabilities.

Note that when you don't check which slit, you add up a bunch of complex 
numbers, then take the absolute value. When you check which slit, you add up 
a bunch of *real* numbers (one for each possible measurement), each of which 
is the result of adding up a bunch of complex numbers.

That's why doing things in the *future* can affect what you measure *now*, 
which isn't how waves work.

> Is nature trying to confuse
> us to make it look exactly like it was a wave, without it being so?

Not every addition of complex numbers is a wave. Just so, not every 
collection of converging lines is caused by perspective. Having converging 
lines in nature that don't converge to the horizon isn't any more 
"confusion" than complex numbers not caused by waves.  In much the same way, 
I can show you that *those* lines are *not* perspective, because if you 
shift your point of view, they don't converge at the horizon. But if you 
only look at exactly one experiment, namely the two slits with no other 
measurements made, it *look* like perspective/waves, but as soon as you do 
another experiment, it stops looking like that.

-- 
Darren New, San Diego CA, USA (PST)
   Human nature dictates that toothpaste tubes spend
   much longer being almost empty than almost full.

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Darren New <dne### [at] sanrrcom> wrote:
> Warp wrote:
> > Darren New <dne### [at] sanrrcom> wrote:
> >> But seriously, for there to be an interference pattern, something has to 
> >> interfere with something else, right?
> > 
> >   If the photon is a wavefront which traverses from the emitter to the
> > detector, passing throug the slits splits it into two wavefronts, which
> > interfere with each other. When the wavefront collides with the detector,
> > it collapses back into a particle.

> But it doesn't work that way. If you look to see which slit the particle 
> passed through, you find out that it only went through one. You never, ever 
> see it go through both slits.

  But if you observe it going only through one of the slits, the interference
pattern doesn't appear, IIRC.

  Wouldn't that be kind of evidence that when the interference pattern appears,
it did to through both slits? When someone forces it to go through only one of
the slits (by observing it) the interference pattern disappears.

> You basically never see a wave or measure a 
> wave. You always measure a particle, even as the whatever goes through the 
> slits, even *after* the whatever goes through the slits.

  But does the interference pattern remain if the particle is measured?
If it does, then *that* would be indicative that the pattern is not
appearing because the photon behaved like a wave.

> >   I'm not saying that is what happens. I'm just saying it's exactly as
> > plausible as eg. a particle being in many places at the same time or
> > affecting another particle instantly.

> Plausible? Yes. Confirmed by experimental evidence? No. Contradicted by 
> experimental evidence? Yes.

  Contradicted how? "We forced the photon to pass through only one of the
slits and what do you know, the interference pattern disappeared." That
would be confirming evidence, not contradicting one.

> I'm just stating what the guys who study this say. You're trying to tell me 
> it sounds absurd.

  I'm not saying it sounds absurd. I'm saying that the argument of "it's
only one photon, it cannot interfere with itself" all by itself doesn't
convince *me*. I need more.

  Explaining the reason why the interference pattern appears even though
the photon does not behave like a wave would help.

> The probability of it hitting the 
> screen at a given position is the amplitude of it going from the emitter to 
> the left slit to that point on the screen *plus* the probability of it going 
> from the emitter to the right slit to that point on the screen.

  I don't think you can talk about amplitudes after claiming so firmly that
photons are not waves.

-- 
                                                          - Warp

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Warp wrote:
>   But if you observe it going only through one of the slits, the interference
> pattern doesn't appear, IIRC.

Correct.

>   Wouldn't that be kind of evidence that when the interference pattern appears,
> it did to through both slits? 

Not really, no.

> When someone forces it to go through only one of
> the slits (by observing it) the interference pattern disappears.

That would only be true if the observation occurs before the particle goes 
through the slits. If it *really* goes through both slits, wouldn't it have 
to do so *before* it gets to the detectors on the screen side of the slits?

The whole "quantum eraser" thing is designed to prove you can measure which 
slit it went through after it has already hit the screen, and you still 
erase the interference pattern.  If the interference pattern was based on 
the particle turning into a wave and going through both slits, that wouldn't 
work.

>> You basically never see a wave or measure a 
>> wave. You always measure a particle, even as the whatever goes through the 
>> slits, even *after* the whatever goes through the slits.
> 
>   But does the interference pattern remain if the particle is measured?

No. The "interference" is caused not by the particle, but by the slits.

> If it does, then *that* would be indicative that the pattern is not
> appearing because the photon behaved like a wave.

Correct. But the pattern goes away even if you measure the path of some 
*other* particle *after* the particle that went thru the slits has already 
been measured. Hence, it's not a property of the particle/wave by itself.

>   Contradicted how? "We forced the photon to pass through only one of the
> slits and what do you know, the interference pattern disappeared." That
> would be confirming evidence, not contradicting one.

Contradicted by the fact that the pattern also disappears if you measure 
which slit some other particle went through *after* the original particle 
has already been measured. In other words, it also disappears if you decide 
whether or not to look which slit it went through *after* it has already 
been detected.

You have your choice of the particle looking at your experimental setup and 
telling itself backwards in time whether to go thru one slit or both, or 
just accepting that the particle never goes through both slits.

What you're doing is saying that the medical screening procedure *causes* 
cancer, because every time you do the blood test and it comes up positive, 
the person has a higher chance of being sick.

>   Explaining the reason why the interference pattern appears even though
> the photon does not behave like a wave would help.

http://vega.org.uk/video/subseries/8

http://www.amazon.com/QED-Strange-Theory-Light-Matter/dp/0691024170

I'm trying to. You get an interference pattern because the probability for a 
particle to go somewhere is based on 2D numbers that when you multiply them 
wind up "interfering" with each other. But there's no wave there.

Watch this, which is Feynman answering exactly your question:

http://www.youtube.com/watch?v=_7OEzyEfzgg

>   I don't think you can talk about amplitudes after claiming so firmly that
> photons are not waves.

As I said, "amplitude" is a technical term that means a two-dimensional 
probability. It's not talking about the distance from the trough to the 
crest of a wave. You can replace it with any word you want, but you won't 
understand what quantum mechanics are talking about if you don't know what 
the word means.

-- 
Darren New, San Diego CA, USA (PST)
   Human nature dictates that toothpaste tubes spend
   much longer being almost empty than almost full.

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Warp wrote:
> when the state of the particle is "observed" (whatever that might mean)

Merely that you have done something to it which changes its unknown 
state, usually causing it to make contact with another particle(s), such 
as a detector. In effect, if the particle is on its own, and hasn't hit 
anything, it can be in a lot of places at once. In the case of light, 
"hitting" requires that the thing it hits have optical properties that 
have a high chance of making such a change, which things like air don't. 
But, don't ask me why in that case, other than it maybe involves optical 
properties and perhaps how often it "will" hit the other particles in 
the air itself. lol

-- 
void main () {
   If Schrödingers_cat is alive or version > 98 {
     if version = "Vista" {
       call slow_by_half();
       call DRM_everything();
     }
     call functional_code();
   }
   else
     call crash_windows();
}

<A HREF='http://www.daz3d.com/index.php?refid=16130551'>Get 3D Models, 
3D Content, and 3D Software at DAZ3D!</A>

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Jim Henderson wrote:

> You could probably see a fair bit from Stonehenge....

With Salisbury 11 miles to the south, and Amesbury 4 miles to the east?

Unrelated but... I gather they don't let you go near Stonehenge any more.

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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Warp wrote:
>   I don't think you can talk about amplitudes after claiming so firmly that
> photons are not waves.

Here's another analogy that might help.

You're flipping coins. You notice that coins come up 50% heads and 50% 
tails.  You say "How can that be, if every coin has to come up either heads 
or tails? How do you get a coin coming up only 50% heads?"

Well, in spite of that, every *other* measurement tells you a coin comes up 
heads or tails, never 50% of each.

Same thing with photons. There doesn't have to be waves to get an 
interference pattern, just interference. But what's interfering is the 
possibilities, not the electrons. Just like the thing that's 1/2 in the 
coins is that each possibility is only half likely, when any given 
individual coin flip always comes up 100% heads or 100% tails.

-- 
Darren New, San Diego CA, USA (PST)
   Human nature dictates that toothpaste tubes spend
   much longer being almost empty than almost full.

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>> It's true that in very low light conditions, vision takes on a 
>> "speckley" character, presumably due to some combination of small 
>> numbers of photons or small numbers of individual nerve impulses 
>> generating a fairly noisey signal. I'm not sure whether one single 
>> photon is enough to generate a nerve action potential though; maybe it 
>> takes 10 or so?
> 
> Not a bad guess.
> 
> http://math.ucr.edu/home/baez/physics/Quantum/see_a_photon.html

Well, I have a book - a truly awful book written by a guy who thinks 
that molecular biology *proves* that Darwin is wrong. The book spends 
most of its time with broken analogies and flawed logic, but inbetween 
is some really quite interesting stuff about various chemical processes 
of the cell - including the cascade for either retinol or rhodopsin, I 
forget which.

Essentially, when a single molecule of this protein absorbs a single 
photon, it undergoes a structural change which causes it to react with 
another protein, which causes one of the ion pumps on the cell wall to 
shut off, thereby macroscopically altering the ion concentration inside 
the cell, generating a minute but measurable net charge for the entire 
cell, which leads to a nerve impulse being fired... or something like 
that. It's *complicated*!

But anyway, I don't recall the exact pathway off the top of my head, but 
it would seem that you'd need a few photons for one of them to actually 
hit a rod (rather than a cone or some other random cell), in the right 
place to collide with a rhodopsin molecule (instead of some other 
protein floating around in the cytol), and it seems reasonable that 
you'd need to switch off a few ion pumps at once to make an action 
potential big enough to "fire" the neuron... so it seems reasonable that 
some smallish number of photons is required. Biological systems can 
become ludicrously efficient where there's some advantage to it, so it's 
not going to be thousands of photons either...

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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On Tue, 01 Dec 2009 21:42:56 +0000, Orchid XP v8 wrote:

> Jim Henderson wrote:
> 
>> You could probably see a fair bit from Stonehenge....
> 
> With Salisbury 11 miles to the south, and Amesbury 4 miles to the east?

That's not that far, really.

> Unrelated but... I gather they don't let you go near Stonehenge any
> more.

Depends on your definition of "near" - when we visited, they had the 
stones themselves roped off, but you could still walk around them.  
Problems with vandalism apparently led them to cordon them off except for 
during special events (like the Solstice).

Jim

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Warp wrote:
> Darren New <dne### [at] sanrrcom> wrote:
>> That's mistaken. There are no waves.
> 
>   It looks like a wave, it behaves like a wave, it produces all the
>   effects
> that a wave would produce, but it's not a wave.

Conclusion: quantum physics doesn't have duck typing :)

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Nicolas Alvarez wrote:
> Warp wrote:
>> Darren New <dne### [at] sanrrcom> wrote:
>>> That's mistaken. There are no waves.
>>   It looks like a wave, it behaves like a wave, it produces all the
>>   effects
>> that a wave would produce, but it's not a wave.
> 
> Conclusion: quantum physics doesn't have duck typing :)

Heh.  More precisely, there's one situation in which it behaves like a wave 
in one aspect.  So it's kind of like a java interface.

-- 
Darren New, San Diego CA, USA (PST)
   Human nature dictates that toothpaste tubes spend
   much longer being almost empty than almost full.

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Warp wrote:
>   It looks like a wave, it behaves like a wave, it produces all the effects
> that a wave would produce, but it's not a wave.

By the way, just so you know, this sort of statement comes across to me as 
sarcastic, as if you're ridiculing my statements. I don't know if that's 
your intention here, but that's why after a number of exchanges I sometimes 
get ruder than I need to be. I don't know if it's a communication issue or 
you getting frustrated at my inability to explain quantum mechanics in a 
2-paragraph post ;-) but sometimes when you don't agree, your expressions of 
that lack of agreement (not necessarily disagreement, mind) sound sarcastic.

-- 
Darren New, San Diego CA, USA (PST)
   Human nature dictates that toothpaste tubes spend
   much longer being almost empty than almost full.

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> Essentially, when a single molecule of this protein absorbs a single 
> photon, it undergoes a structural change which causes it to react with 
> another protein, which causes one of the ion pumps on the cell wall to 
> shut off, thereby macroscopically altering the ion concentration inside 
> the cell, generating a minute but measurable net charge for the entire 
> cell, which leads to a nerve impulse being fired... or something like 
> that. It's *complicated*!

With some quick calculations and rather dubious assumptions I estimate that 
from a star like Betelgeuse approximately 10^6 photons are going into your 
eye per second.  This is about the same as a 100 W light bulb 10 km away.

Post a reply to this message

>> Essentially, when a single molecule of this protein absorbs a single 
>> photon, it undergoes a structural change which causes it to react with 
>> another protein, which causes one of the ion pumps on the cell wall to 
>> shut off, thereby macroscopically altering the ion concentration 
>> inside the cell, generating a minute but measurable net charge for the 
>> entire cell, which leads to a nerve impulse being fired... or 
>> something like that. It's *complicated*!
> 
> With some quick calculations and rather dubious assumptions I estimate 
> that from a star like Betelgeuse approximately 10^6 photons are going 
> into your eye per second.  This is about the same as a 100 W light bulb 
> 10 km away.

Dubious assumptions FTW! :-D

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"scott" <sco### [at] scottcom> wrote:
> With some quick calculations and rather dubious assumptions I estimate that
> from a star like Betelgeuse approximately 10^6 photons are going into your
> eye per second.  This is about the same as a 100 W light bulb 10 km away.

Interesting. Perhaps bizarrely, I know exactly what Betelgeuse looks like to the
naked eye, but not a lightbulb 10km distant!

What are your dubious assumptions? This is just the sort of 'sensible
guesstimation' calculation I used to enjoy at university. :)

Post a reply to this message

> What are your dubious assumptions?

That all the energy from the star/light bulb comes out equally in all 
directions in the form of 550 nm photons, and your eye has an opening of 2mm 
x 2mm.

Obviously all the energy doesn't come out as visible photons, and you could 
probably do something clever with the spectrum to get a more accurate 
figure.

Post a reply to this message

scott wrote:
>> What are your dubious assumptions?
> 
> That all the energy from the star/light bulb comes out equally in all 
> directions in the form of 550 nm photons, and your eye has an opening of 
> 2mm x 2mm.
> 
> Obviously all the energy doesn't come out as visible photons, and you 
> could probably do something clever with the spectrum to get a more 
> accurate figure.

To say nothing of the absorption and scattering of the Earth's atmosphere...

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Invisible <voi### [at] devnull> wrote:
> scott wrote:
> >> What are your dubious assumptions?
> >
> > That all the energy from the star/light bulb comes out equally in all
> > directions in the form of 550 nm photons, and your eye has an opening of
> > 2mm x 2mm.
> >
> > Obviously all the energy doesn't come out as visible photons, and you
> > could probably do something clever with the spectrum to get a more
> > accurate figure.
>
> To say nothing of the absorption and scattering of the Earth's atmosphere...

Probably not nearly as important as the error introduced by assuming both star
and bulb emit only in the visible.

Otherwise, not such dubious assumptions at all!

Post a reply to this message

>> To say nothing of the absorption and scattering of the Earth's atmosphere...
> 
> Probably not nearly as important as the error introduced by assuming both star
> and bulb emit only in the visible.
> 
> Otherwise, not such dubious assumptions at all!

Well, at least he didn't say an LED or a flourescent tube or something 
with a complex emission system. A normal light bulb and a star are 
(AFAIK) both simple black body radiators, so you just need to know their 
colour temperature...

Post a reply to this message

Invisible <voi### [at] devnull> wrote:
> >> To say nothing of the absorption and scattering of the Earth's atmosphere...
> >
> > Probably not nearly as important as the error introduced by assuming both star
> > and bulb emit only in the visible.
> >
> > Otherwise, not such dubious assumptions at all!
>
> Well, at least he didn't say an LED or a flourescent tube or something
> with a complex emission system. A normal light bulb and a star are
> (AFAIK) both simple black body radiators, so you just need to know their
> colour temperature...

That just tells you the peak power output. The total power output might not be
as much as an order of magnitude different, but it will be different.

In any case, for a lightbulb, less than 10% of that 100W will be visible. An LED
or fluorescent tube will emit mostly in the visible, so the power will be much
more accurate for this comparison.

Post a reply to this message

>> Well, at least he didn't say an LED or a flourescent tube or something
>> with a complex emission system. A normal light bulb and a star are
>> (AFAIK) both simple black body radiators, so you just need to know their
>> colour temperature...
> 
> That just tells you the peak power output. The total power output might not be
> as much as an order of magnitude different, but it will be different.

I was under the impression that the colour temperature tells you roughly 
how the signal power is distributed over the whole spectrum. (Initially 
just IR, then red as well, and gradually more high frequencies, while 
the power at lower frequences decreases.)

> In any case, for a lightbulb, less than 10% of that 100W will be visible. An LED
> or fluorescent tube will emit mostly in the visible, so the power will be much
> more accurate for this comparison.

Yeah, both stars and lightbulbs presumably put out far more IR and radio 
waves than visible light...

Post a reply to this message

Invisible <voi### [at] devnull> wrote:
> I was under the impression that the colour temperature tells you roughly
> how the signal power is distributed over the whole spectrum.

Roughly, in that it tells you where the peak is and the general shape of the
rest of the spectrum.

> Yeah, both stars and lightbulbs presumably put out far more IR and radio
> waves than visible light...

Actually, many stars peak in the visible - Betelgeuse is red. Ours is yellow. I
think maybe lightbulbs peak in the IR (lower temperature), so all we see is a
section of the tail of the spectrum.

http://en.wikipedia.org/wiki/Black_body

Post a reply to this message

> Invisible <voi### [at] devnull> wrote:
>> I was under the impression that the colour temperature tells you roughly
>> how the signal power is distributed over the whole spectrum.
> 
> Roughly, in that it tells you where the peak is and the general shape of the
> rest of the spectrum.

Sure. So if you know how hot the star is, you know roughly the shape of 
the entire emission spectrum. :-)

>> Yeah, both stars and lightbulbs presumably put out far more IR and radio
>> waves than visible light...
> 
> Actually, many stars peak in the visible - Betelgeuse is red. Ours is yellow. I
> think maybe lightbulbs peak in the IR (lower temperature), so all we see is a
> section of the tail of the spectrum.

The peak is just the single wavelength at which there is the most 
energy. I think if you look at *bands* of the spectrum, it's possible 
for a wide band with low energy at any specific single wavelength to 
have a greater total power and a band containing only a single 
high-power (but narrow) peak.

Post a reply to this message

> for a wide band with low energy at any specific single wavelength to

How can you have a "wide band" at a specific single wavelength?  Surely the 
definition of a "band" is a range of wavelengths?

Sure, if the peak is at 600 nm, then the energy in the band from 599.99 to 
600.01 will most likely be less than the band from 100 to 599.

Post a reply to this message

scott wrote:
>> for a wide band with low energy at any specific single wavelength to
> 
> How can you have a "wide band" at a specific single wavelength?  Surely 
> the definition of a "band" is a range of wavelengths?
> 
> Sure, if the peak is at 600 nm, then the energy in the band from 599.99 
> to 600.01 will most likely be less than the band from 100 to 599.

What I actually meant was that a "peak" is just one wavelength with a 
lot of energy. The band containing the peak isn't necessarily the band 
with the highest total energy.

Post a reply to this message

Invisible <voi### [at] devnull> wrote:
> Sure. So if you know how hot the star is, you know roughly the shape of
> the entire emission spectrum. :-)

Yes. Except of course it works the other way around in practice.

> > think maybe lightbulbs peak in the IR (lower temperature), so all we see is a
> > section of the tail of the spectrum.
>
> The peak is just the single wavelength at which there is the most
> energy. I think if you look at *bands* of the spectrum, it's possible
> for a wide band with low energy at any specific single wavelength to
> have a greater total power and a band containing only a single
> high-power (but narrow) peak.

What Scott said. But I know what you mean - and yes, of course. I was just
explaining why most of an incandescent bulb's energy output is not in the
visible. Look at the shape of a blackbody spectrum for 3000K. Clearly, most of
the power output is IR and beyond.

Post a reply to this message

> Probably not nearly as important as the error introduced by assuming both 
> star
> and bulb emit only in the visible.

Wikipedia can tell you the "luminous efficacy" of the sun (14%), a light 
bulb (2%) and LEDs (1.5-22%):

http://en.wikipedia.org/wiki/Luminous_efficacy

Whilst the % values are not exactly the amount of radiation that is visible 
(it's weighted more towards green), it's a better approximation than 100% 
:-)

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> Wikipedia can tell you the "luminous efficacy" of the sun (14%), a light 
> bulb (2%) and LEDs (1.5-22%):

Anybody happen to know the luminous efficacy of a glow worm?

Post a reply to this message

>> Wikipedia can tell you the "luminous efficacy" of the sun (14%), a light 
>> bulb (2%) and LEDs (1.5-22%):
>
> Anybody happen to know the luminous efficacy of a glow worm?

Man Wikipedia knows everything!  We should power our lights with glow worms!

Post a reply to this message

"scott" <sco### [at] scottcom> wrote:
> >> Wikipedia can tell you the "luminous efficacy" of the sun (14%), a light
> >> bulb (2%) and LEDs (1.5-22%):
> >
> > Anybody happen to know the luminous efficacy of a glow worm?
>
> Man Wikipedia knows everything!  We should power our lights with glow worms!

All they need is a reflective cavity and a biochemical population inversion, and
we could have them in CD players too!

Post a reply to this message

>>> Anybody happen to know the luminous efficacy of a glow worm?
>> Man Wikipedia knows everything!  We should power our lights with glow worms!
> 
> All they need is a reflective cavity and a biochemical population inversion, and
> we could have them in CD players too!

...so I'm guessing it's quiet efficient then? (Mother Nature has an 
uncanny way of doing this, it seems.)

Given the number of genetically engineered organisms which glow, I'm 
guessing scientists have a good idea of how this works. Maybe 
electronically-controlled glowing algae or something isn't so 
far-fetched. (I hate to think what the component lifetime would be though!)

Post a reply to this message

Darren New <dne### [at] sanrrcom> wrote:
> No. The "interference" is caused not by the particle, but by the slits.

  How does the photon know there was another slit nearby, and behave
differently if that was so?

-- 
                                                          - Warp

Post a reply to this message

Patrick Elliott <sel### [at] npgcablecom> wrote:
> Warp wrote:
> > when the state of the particle is "observed" (whatever that might mean)

> Merely that you have done something to it which changes its unknown 
> state, usually causing it to make contact with another particle(s), such 
> as a detector.

  I think the Copenhagen interpretation goes beyond that. If the decay of
the radioactive substance causes the flask to be broken, killing the cat,
it has already been "observed" (by whatever detector caused the flask to
be broken) and thus there are no superimposed states, but according to
the Copenhagen interpretation there are, until some external observer opens
the box.

  That's the kind of "observation" I don't understand.

-- 
                                                          - Warp

Post a reply to this message

Darren New <dne### [at] sanrrcom> wrote:
> Warp wrote:
> >   It looks like a wave, it behaves like a wave, it produces all the effects
> > that a wave would produce, but it's not a wave.

> By the way, just so you know, this sort of statement comes across to me as 
> sarcastic, as if you're ridiculing my statements. I don't know if that's 
> your intention here, but that's why after a number of exchanges I sometimes 
> get ruder than I need to be. I don't know if it's a communication issue or 
> you getting frustrated at my inability to explain quantum mechanics in a 
> 2-paragraph post ;-) but sometimes when you don't agree, your expressions of 
> that lack of agreement (not necessarily disagreement, mind) sound sarcastic.

  It was not sarcasm, it was puzzlement.

-- 
                                                          - Warp

Post a reply to this message

>> No. The "interference" is caused not by the particle, but by the slits.
>
>  How does the photon know there was another slit nearby, and behave
> differently if that was so?

If you could answer exactly how a photon "knows" where it can and can't go 
then you'd probably become quite famous.

Post a reply to this message

Warp wrote:
> Darren New <dne### [at] sanrrcom> wrote:
>> No. The "interference" is caused not by the particle, but by the slits.
> 
>   How does the photon know there was another slit nearby, and behave
> differently if that was so?

Nobody knows.

-- 
Darren New, San Diego CA, USA (PST)
   Human nature dictates that toothpaste tubes spend
   much longer being almost empty than almost full.

Post a reply to this message

Warp wrote:
>   That's the kind of "observation" I don't understand.

They're still working on that one.  Someone recently proposed it's actually 
gravity waves causing the "collapse", which is why it doesn't happen until 
you get enough particles that gravity becomes significant. Or some such.

That's another one of those "nobody is sure why yet" kind of things.

-- 
Darren New, San Diego CA, USA (PST)
   Human nature dictates that toothpaste tubes spend
   much longer being almost empty than almost full.

Post a reply to this message

Warp wrote:
>   It was not sarcasm, it was puzzlement.

I will keep that in mind, then. :-)

Generally speaking, as a native speaker of English, when you express 
something like that as a statement rather than a question, it comes across 
with an invisible "Yeah, right, pull the other one" kind of statement. At 
least to me.  Just so ya know.

-- 
Darren New, San Diego CA, USA (PST)
   Human nature dictates that toothpaste tubes spend
   much longer being almost empty than almost full.

Post a reply to this message

Darren New <dne### [at] sanrrcom> wrote:
> Warp wrote:
> > Darren New <dne### [at] sanrrcom> wrote:
> >> No. The "interference" is caused not by the particle, but by the slits.
> > 
> >   How does the photon know there was another slit nearby, and behave
> > differently if that was so?

> Nobody knows.

  But it seems that everybody knows that it does not happen by the photon
going through both slits?

  (After all, it would make sense: If there's no other slit to go through,
then it can't go through two slits, and thus its behavior is different than
if there was another slit and it went through both.)

  Or is this again that kind of almost-supernatural phenomena like the
photon somehow knowing that one year from now it will be measured which
slit it went through and thus it will already know to not to produce the
interference pattern?

-- 
                                                          - Warp

Post a reply to this message

Warp wrote:
> Darren New <dne### [at] sanrrcom> wrote:
>> Warp wrote:
>>> Darren New <dne### [at] sanrrcom> wrote:
>>>> No. The "interference" is caused not by the particle, but by the slits.
>>>   How does the photon know there was another slit nearby, and behave
>>> differently if that was so?
> 
>> Nobody knows.
> 
>   But it seems that everybody knows that it does not happen by the photon
> going through both slits?

Correct.  Or, rather, nobody has ever measured anything that would imply the 
photon goes through both slits as a wave. Nobody has ever measured the 
actual wave-like nature of photons, other than that the probability you 
measure one of those particles at a particular place shares much of the same 
math as waves. (The probability that *two* photons do something does *not* 
share the same math as waves.)

The photon, in some very rare cases, might go through both slits, but not at 
the same time as a wave. Instead, it might actually go in a circle and go 
through both. But it's vanishingly improbable, and maybe even impossible if 
the probabilities actually add up to zero.

Essentially, every single time you find a photon, you find an entire photon. 
There's never a case where you get a half a photon. Hence the "quantum" 
nature of it all. You don't get half a photon just like you never ever find 
an electron with half a charge, and just like you don't find a coin that 
comes up half heads half tails. The only "interference" you get is patterns 
when you measure where lots of photons go.

>   (After all, it would make sense: If there's no other slit to go through,
> then it can't go through two slits, and thus its behavior is different than
> if there was another slit and it went through both.)

But it only goes through one slit if you measure which slit it went thru. 
This is true even *if* you make the measurement *after* the slits. You never 
see the photon going through both slits, like you would if it were actually 
a wave.

>   Or is this again that kind of almost-supernatural phenomena like the
> photon somehow knowing that one year from now it will be measured which
> slit it went through and thus it will already know to not to produce the
> interference pattern?

Again, it's not a question of one photon making an interference pattern. 
It's a bunch of photons making an interference pattern. Any one photon shows 
up all at once and makes a single discrete mark wherever it hits. It's just 
more likely to hit in one area than another.

It's sort of like saying that at a sufficiently small scale, there aren't 
waves in water either. If there's a "high" wave, you're more likely to see a 
molecule of H2O high above the average surface level than if there's a 
trough. You can calculate the probability of H2O molecules being in a 
particular place, but you're never going to see an H2O molecule interfering 
with itself. (At least, not in the "classic" size of water waves.) You can 
ask "how does the beach get more wet near the waterline than higher up if 
water isn't waves?" It's because you're more likely to find an H2O molecule 
close to the waterline than higher up the beach. But that doesn't mean H2O 
is made out of waves.

But in short, yes, it's the same weirdness, which is how people know it's 
not waves per se. Being a wave would imply it's actually the photon itself 
that's controlling where it goes. But we already know you can control where 
the photon goes retroactively, which would obviously make no sense if it was 
deciding based on being a wave. If it's interfering *because* it is going 
through both slits, you wouldn't be able to change whether it went thru both 
slits or not after it already hit the detector. There's something deeply 
unintuitive going on that nobody understands.

BTW, it's also pretty easy to prove it's not something in the photon that we 
can't measure controlling how it works. That's what Bell's Inequality is about.

-- 
Darren New, San Diego CA, USA (PST)
   Human nature dictates that toothpaste tubes spend
   much longer being almost empty than almost full.

Post a reply to this message

Darren New <dne### [at] sanrrcom> wrote:
> >   But it seems that everybody knows that it does not happen by the photon
> > going through both slits?

> Correct.  Or, rather, nobody has ever measured anything that would imply the 
> photon goes through both slits as a wave.

  The interference pattern in the detector is not a measurement?

> Essentially, every single time you find a photon, you find an entire photon. 
> There's never a case where you get a half a photon.

  But I thought that's the whole idea in the Copenhagen interpretation:
Particles are in superpositions until they are measured, in which case
they collapse into a definite state. Thus when you measure a photon, you
will always find a collapsed photon.

> >   (After all, it would make sense: If there's no other slit to go through,
> > then it can't go through two slits, and thus its behavior is different than
> > if there was another slit and it went through both.)

> But it only goes through one slit if you measure which slit it went thru. 
> This is true even *if* you make the measurement *after* the slits. You never 
> see the photon going through both slits, like you would if it were actually 
> a wave.

  You also never see the cat in both live and dead states at the same time.

> >   Or is this again that kind of almost-supernatural phenomena like the
> > photon somehow knowing that one year from now it will be measured which
> > slit it went through and thus it will already know to not to produce the
> > interference pattern?

> Again, it's not a question of one photon making an interference pattern. 
> It's a bunch of photons making an interference pattern. Any one photon shows 
> up all at once and makes a single discrete mark wherever it hits. It's just 
> more likely to hit in one area than another.

  Of course I implied that you perform the test many times and see the overall
pattern that the photons create.

-- 
                                                          - Warp

Post a reply to this message

Warp wrote:
> Darren New <dne### [at] sanrrcom> wrote:
>>>   But it seems that everybody knows that it does not happen by the photon
>>> going through both slits?
> 
>> Correct.  Or, rather, nobody has ever measured anything that would imply the 
>> photon goes through both slits as a wave.
> 
>   The interference pattern in the detector is not a measurement?

No. It's a bunch of individual measurements of different photons. There's no 
way to look at *one* photon and decide whether there was one slit or two. 
Because it's not a wave, it's a particle, so it always makes the same sort 
of single-spot quantum event in a single place.

>   But I thought that's the whole idea in the Copenhagen interpretation:
> Particles are in superpositions until they are measured, in which case
> they collapse into a definite state. Thus when you measure a photon, you
> will always find a collapsed photon.

If every time one measures something, one gets "it isn't a wave", then why 
would one think it's ever a wave?

Particle *probabilities* are in superpositions. That doesn't make them waves.

But again, this is a question of "why does it work that way", and not "how 
does it work", and AFAIK nobody knows the underlying reason why the 
probabilities work out the way they do.

>> But it only goes through one slit if you measure which slit it went thru. 
>> This is true even *if* you make the measurement *after* the slits. You never 
>> see the photon going through both slits, like you would if it were actually 
>> a wave.
> 
>   You also never see the cat in both live and dead states at the same time.

That's actually evidence for it being particles and *not* waves. :-)

>> Again, it's not a question of one photon making an interference pattern. 
>> It's a bunch of photons making an interference pattern. Any one photon shows 
>> up all at once and makes a single discrete mark wherever it hits. It's just 
>> more likely to hit in one area than another.
> 
>   Of course I implied that you perform the test many times and see the overall
> pattern that the photons create.

Yes. But if a photon was a wave, it could interfere with *itself*. It could 
split up and go through two separate slits. One photon doesn't make an 
interference pattern, and that's how you know the photon isn't a wave. You 
don't see half the photon's energy hit here and half hit there with a blank 
spot in the middle.

-- 
Darren New, San Diego CA, USA (PST)
   Human nature dictates that toothpaste tubes spend
   much longer being almost empty than almost full.

Post a reply to this message

Darren New <dne### [at] sanrrcom> wrote:
> Warp wrote:
> > Darren New <dne### [at] sanrrcom> wrote:
> >>>   But it seems that everybody knows that it does not happen by the photon
> >>> going through both slits?
> > 
> >> Correct.  Or, rather, nobody has ever measured anything that would imply the 
> >> photon goes through both slits as a wave.
> > 
> >   The interference pattern in the detector is not a measurement?

> No. It's a bunch of individual measurements of different photons. There's no 
> way to look at *one* photon and decide whether there was one slit or two. 
> Because it's not a wave, it's a particle, so it always makes the same sort 
> of single-spot quantum event in a single place.

  That's like saying that you can't measure the probability of a coin
giving heads because you can only toss one coin at a time.

  I'd say tossing 100 coins one after another is exactly as valid as if you
had tossed one coin 100 times simultaneously (if that was physically possible).

> >   But I thought that's the whole idea in the Copenhagen interpretation:
> > Particles are in superpositions until they are measured, in which case
> > they collapse into a definite state. Thus when you measure a photon, you
> > will always find a collapsed photon.

> If every time one measures something, one gets "it isn't a wave", then why 
> would one think it's ever a wave?

  Because of the interference pattern?

  If it behaves like a wave, what can we deduce from that?

> >> But it only goes through one slit if you measure which slit it went thru. 
> >> This is true even *if* you make the measurement *after* the slits. You never 
> >> see the photon going through both slits, like you would if it were actually 
> >> a wave.
> > 
> >   You also never see the cat in both live and dead states at the same time.

> That's actually evidence for it being particles and *not* waves. :-)

  It's an argument that measurement doesn't necessarily tell what was the
state before the measurement was performed.

> Yes. But if a photon was a wave, it could interfere with *itself*. It could 
> split up and go through two separate slits. One photon doesn't make an 
> interference pattern, and that's how you know the photon isn't a wave.

  That just doesn't convince me. Just because the photon collapses when it
hits the detector doesn't mean that on the way it didn't pass through both
slits at the same time and interfered with itself.

  It sounds like you are arguing that particles cannot be in superimposed
states and that there's no collapse phenomenon when they are measured, but
instead they are always at single places at a time, in particle form. In
other words, that the Copenhagen interpretation is wrong. (Of course, if
I have understood correctly, nobody has proven that it's right, but you
are seemingly claiming it's wrong. Has that been proven?)

  (Not that I have the slightest idea what I'm talking about. The problem
I'm having is that, seemingly, people have no idea why the interference
pattern appears yet are completely sure it's *not* because photons are
in wave state at any point.)

> You 
> don't see half the photon's energy hit here and half hit there with a blank 
> spot in the middle.

  You don't see it because if you try to see it, the photon collapses and
the only thing you see is a particle.

-- 
                                                          - Warp

Post a reply to this message

Warp wrote:
> Patrick Elliott <sel### [at] npgcablecom> wrote:
>> Warp wrote:
>>> when the state of the particle is "observed" (whatever that might mean)
> 
>> Merely that you have done something to it which changes its unknown 
>> state, usually causing it to make contact with another particle(s), such 
>> as a detector.
> 
>   I think the Copenhagen interpretation goes beyond that. If the decay of
> the radioactive substance causes the flask to be broken, killing the cat,
> it has already been "observed" (by whatever detector caused the flask to
> be broken) and thus there are no superimposed states, but according to
> the Copenhagen interpretation there are, until some external observer opens
> the box.
> 
>   That's the kind of "observation" I don't understand.
> 
But, the Copenhagen interpretation is intended as an example, where by 
the "mechanism" isn't part of the system that causes the observation. 
The presumption being that what ever opens the box to look is what 
*causes* the result. It was never intended to be used to imply that the 
state in such a box could/does exist in the real world. In short, you 
are exaggerating the meaning of the example. There may be others doing 
so too, but most understand that the *entire box*, and the cat inside, 
is the particle, and that the observer is basically what ever "looks" 
into the box to see what happened, which could be an X-Ray machine, for 
all that it matters, or a guy with a crowbar, or just some single 
particle, shot at the box, which "quantum tunnels" through the inside, 
and other the other, carrying some information on the state inside. The 
*observer* is never a conscious observer, it is always, "What ever 
happens to interact with the particle, causing it to change."

And, if you think about it, it can't be otherwise. If you observe a 
particle with a detector, the detector "causes" the state change. Turn 
on a flashlight, photons cause the state change. There is no way to 
*observe* an object that doesn't involve us hitting the particle with 
another particle, and measuring what the change in *that* particle is, 
thereby deriving what happened to the other one. All Copenhagen really 
says is, "If nothing is interacting with the particle, the particle's 
state isn't known, but the moment something does (which they confusingly 
termed "observer"), the state becomes fixed." Its sort of like the way 
normal people use "theory" to mean guess, but scientists mean, "already 
supported by evidence". "Observer" doesn't mean what the average person 
would mean by it. Imho, it was a really poor choice, but we are, at this 
point, stuck with it. lol

-- 
void main () {
   If Schrödingers_cat is alive or version > 98 {
     if version = "Vista" {
       call slow_by_half();
       call DRM_everything();
     }
     call functional_code();
   }
   else
     call crash_windows();
}

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