Re: Is light a wave or a particle?

From: Bjoern Feuerbacher (feuerbac_at_thphys.uni-heidelberg.de)
Date: 09/21/04 

Date: Tue, 21 Sep 2004 12:00:05 +0200

TomGee wrote:
> Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<cim9c3$445$1@news.urz.uni-heidelberg.de>...
>
>>TomGee wrote:
>>
>>>Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<cijuq0$jbi$2@news.urz.uni-heidelberg.de>...
>>>
>>>
>>>>mitch perkins wrote:
>>>>
>>>>
>>>>>Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in message news:<cih93n$3do$1@news.urz.uni-heidelberg.de>...
>>>>>
>>>>>
>>>>>
>>>>>>TomGee wrote:
>>>>>
>>>>>
>>>No, but I'm familiar with his "sums-over-histories" work
>>
>>That's another name for the same thing. But "sum over paths" is used
>>more often.
>>
>>
>>
>>>which has to
>>>do with electron standing waves in atoms and the countless
>>>probabilities where an electron may be found within it.
>>
>>Actually, it has to do with describing the propagation of particles.
>>
>>
>
> No, it doesn't.

Yes, it does. If you think otherwise, the burden of proof is on
*you*, since *you* claimed above that it has to do with electron
standing waves in atoms.

Why do *you* think it is commonly called "sum over paths"?
Why do *you* think that the object which is calculated using the
"sum over paths" method is called the "propagator"?
<http://galileo.phys.virginia.edu/classes/752.mf1i.spring03/PathIntegrals04.htm>

> It simply counts the many possible paths to somewhere
> and then it claims that it is probable an object will take one of
> those to get there.

*sigh* You contradict yourself. What you described here *is* the
propagation of particles!!!

And, no, that's *not* what it actually says. It says there is a
probability *amplitude* for the particle to go on *all* these
paths; essentially it says that the particles does indeed go on
all of these paths at once!

Again: look at Feynman's popular science book on QED! Or at the first
few chapters of his lecture notes, volume 3!

Or look at this quote from this page (the first which comes up when
one does a Google search for "sum over paths"!):
"According to Feynman's sum over paths method photons will travel
different ways, but when we add the amplitudes from all possible paths,
most of them will cancel out."

> Hardly a mind-blowing revelation.

Well, what the method *actually* says *is* mind-blowing. Unfortunately,
you did not understand it.

> But it is just
> another math construct designed to get you what you want to get out of
> reality.

Nonsense. You can't get all you want out of reality by using math.

> The number of paths one can take to get somewhere are for
> all practical purposes infinite, since one can get there by way of any
> place in the universe. Another Mandrakian sleight-of-hand which saves
> some scientists from having to face reality.

And yet again, you show that you haven't the faintest clue what you
are talking abouz.

>>>Are you applying his work to photons?
>>
>>Yes. Hint: he did do that himself. The resulting theory is called
>>"Quantum Electrodynamics".
>>Hey, he even described that in his popular science book on QED!
>>
>>
>
> Hey, aren't you the one who made fun of pop sci books?

I did not make fun of them. I simply pointed out that they are often
not very reliable. However, there are some exceptions. Hint: Feynman
presents the same argumentation, only with a bit more math, in this
lecture notes, volume 3, which obviosly is not a pop science book.

> Anyway, here
> you are again applying an entire branch of physics

Feynman's method is in no way "an entire branch of physics".

> to try to win a
> discussion about a single part of it!

One uses the methods appropriate to the problem when one wants to
solve a problem. And Feynman's sum over paths is entirely appropriate here.

> Cheap postgrad TA tricks which
> show the fact that you don't have an answer.

I notice that you simply chose to ignore my point that Feynman himself
did apply his method to photons, although you accused me that I misuse
this idea when I applied it to photons. Nice try of obfuscation. Shows
nicely *who* has no answer here.

>>What I am actually saying is that photons, just like electrons,
>>are described by a wave function, and can be *pictured* as "clouds".
>>
>>
>
> But isn't that true of all visible particles?

Yes. Your point?

>>As I already said: they can be found at all places where the wave is
>>non-zero, since the amplitude of an electromagnetic wave gives
>>the probability to find a photon. Just like the wave functions for
>>electrons.
>>
>>
>
> The concept provides that _it is possible_ to find them anywhere the
> wave is non-zero, but you say not that it is possible, but you instead
> assert that "they _can_ be found at all places where the wave is
> non-zero." > Isn't that stretching the concept beyond reasonable
> thought?

Oh my goodness. Where on earth is the difference between "they can
be found" and "it is possible to find them"?

>>>Huygens' principle has to do with any point on wave crests, from where
>>>it may be thought that secondary waves can emanate, but it was a way
>>>to explain what happens when waves encounter barriers or other media.
>>>It has nothing to do with our discussion about wave gaps nor does it
>>>explain how photons "spread out", as you say.
>>
>>Well, then I have no clue what you mean by the "smaller circles"
>>you mentioned above.
>>
>>
>
> No clue indeed.

Well, why don't you tell me?

>>>>>>>If the
>>>>>>>larger circle is the wave and the smaller circles are particles,
>>>>>>
>>>>>>They aren't. For the 10th time: the photons are "spread out" over the
>>>>>>sphere, not located at distinct points on it.
>>>>>
>>>>>Reading the above, I wonder if when you say, 'the photons are
>>>>>"spread out"', Tom is thinking that you mean many point like particles
>>>>>are grouped less closely together,
>>>>
>>>>
>>>Yes, that is precisely what I thought. Thank you.
>>
>>Sorry for the misunderstanding. I hope you understand now better what
>>I actually mean?
>>
>>
>
> Yes. You mean you don't know what you're talking about and your
> knee-jerk reactions to new ideas define you as one of those people
> whose life preservation defense mechanisms overreact much more than
> normal and so they take much longer to get back to where most people
> accept new ideas.
>
> Your ideas about "spread out photons" are childish and bankrupt, and
> it is time for you as an adult to discharge your responsibility to
> think in reasonable terms wrt reality.

Thanks for showing yet again that you haven't got the faintest clue what
you are talking about, and your method to deal with a clear explanation,
based on sound physics, is to throw insults around and ignore everything
your are told.

Idiot.

>>>>>whereas you actually mean that each
>>>>>point like particle is "stretched out", elongated as it were, to cover
>>>>>the entire distance of the wave?
>>>>
>>>>
>>>If that were to occur, it would solve the "gap" problem, alright, but
>>>in order for that to happen, Bjoern must show us how that could be
>>>possible.
>>
>>As I said about 10 times now: the amplitude of the electromagnetic
>>wave describes the probability to find a photon.
>>
>>
>
> Well, let's say that is so, and we observe two waves of differing
> amplitudes, which means that each wave describes different
> probabilities of finding a particle at a particlular place, what does
> that have to do with your assertion that the particles are "spread
> out"?

I notice you ignore my explanation that this "spread out" is merely
a way to visualize this.

Since a spherical wave is non-zero on an entire sphere, and one can
find photons anywhere on it, and the probability to find a photon at any
given place on the sphere decreases with 1/r^2, i.e. inversely
proportional to the surface of the sphere (I hope you learned in the
meantime what this means...), it is entirely reasonable to describe this
by saying that the photons are spread out over the wave front. Just as
one often pictures electrons in atoms as "probability clouds".

> Absolutely nothing since probabilities are just about where a
> single particle may be found in any one place at any given instant,

Exactly the same applies to electrons in atoms. And nevertheless, when
one visualizes them, one draws "clouds" - i.e. one pretends that they
are spread out.

> and since that has nothing to do with the many particles which must be
> created to fill in the gaps as the light wave expands.

*sigh* For the 20th time: there are no gaps.

>>So the probability
>>to find a photon is the same everywhere on the wave front,
>>
>>
>
> Okay, let's say the probability for a wave of a certain amplitude is
> 30000:1

You probably mean 1:30000? Probabilities greater than one make no sense.

> that a photon will be found at a particular place on the wave
> front, doesn't that tell you that it will not be found anywhere else
> when it is found there?

After it had been found, yes, obviously, you can't find it anywhere
else. But before it has been found, it can be anywhere. That's what
"collapse of the wave function" means, if you didn't notice.

> You may want to argue that as a photon will be found there, there is
> another probability that one will be found at a point immediately next
> to it,

Yes.

> and of course, the probability exists, but in order to have
> particles all along the front of the wave crest, it must be a
> certainty and not a probability.

It is a certainty that it is possible to find photons anywhere you look.

> Here's another shocker to your brain lobes: How do you explain wave
> crests and troughs in terms of a photon particle other than the way my
> model does it?

Quite simple: by pointing out that there is indeed a wave. The wave
gives the wave function for the photons.

>>inversely proportional to the distance to the source squared (the usual
>>inverse square law!). The "spreading out" is just a way to picture this.
>>Just as electrons in atoms are often pictured as "charge clouds" or
>>"probability clouds".
>>
>>
>
> Finally we see your argument, which you have until now refused to
> support, about my use of the term "inversely proportional"!

Liar. I gave three links which explained in detail what "inversely
proportional" means. You snipped them and pretended they never were there.

> It seems
> you are arguing that no one can use the term unless they add the
> suffix "to the distance to the source squared",

Absolute utter bullshitting NONSENSE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

I SAID NOTHING AT ALL LIKE THAT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

For the third time, at least:
"A is inversely proportional to B" means "there is a constant C so that
A = C/B".

"x is inversely proportional to y" means "there is a constant z so that
x = z/y".

"energy density is inversely proportional to distance squared" means
"there is a constant E0 so that energy density = E0/(distance squared)"

etc.

It is *really* incredible that you are so *totally* unable to understand
what such a basic term like "inversely proportional" means!!!!!

> and my using it in
> "inversely proportional to the state of motion of discrete
> objects/systems" is a wrong usage of it! What crap!

*sigh* No, I said nothing like that. What was wrong with your usage
is that "time rate is inversely proportional to the state of motion"
*means* (because of the *definition* of the term "inversely
proportional"!!!!!) that "there is a constant (let's call it k) so
that time rate = k/(state of motion)". And that latter statement is
obviously false.

PLEASE get a ***BASIC*** education!!!!!

It is *really* incredible that you are so *totally* unable to understand
what such a basic term like "inversely proportional" means!!!!!

>>>>>Not saying that that *is* what you mean; just wondering if the word
>>>>>"spread" is being correctly interpreted.
>>>>
>>>>Perhaps. I gave the analogy of electrons in atoms in order to make
>>>>clear what I mean - but probably he managed anyway to misunderstand me.
>>>>
>>>>
>
> As you can see above, I did not misunderstand you at all, I just
> disagree with your argument, and so I have destroyed it, as it so
> deserves.

You have destroyed nothing here. You merely showed once again that
you are totally unable to understand my arguments.

PLEASE get a ***BASIC*** education!!!!!

It is *really* incredible that you are so *totally* unable to understand
what such a basic term like "inversely proportional" means!!!!!

>>>but as you can see by my response, it doesn't hold water
>>
>>Oh, you think by merely asserting that I am confused about Feynman's
>>work (and thereby showing nicely that you yourself know next to nothing
>>about it), you have shown that my explanation holds no water?
>>
>>
>
> I said, clearly, "but as you can see by my response,....", but your
> closed and drowning mind understood only, "...it doesn't hold water".
> I did not merely assert, but I supported my assertion with my
> response.

Err, my point was that your response consisted only of assertions.

>>Try looking into his popular science book on QED, and try understanding
>>that what he explains there is applying his "sum over paths" approach
>>to photons.
>>
>>
>
> Why should I waste my time?

You might learn something.

> You tried it and failed to understand it even in a pop sci book.

You have no evidence that I failed to understand it.

>>>Just how do you mean that photons are
>>>spread out over the sphere of expanding light wave.
>>
>>
>>For the 11th time: the amplitude of the electromagnetic
>>wave describes the probability to find a photon. So the probability
>>to find a photon is the same everywhere on the wave front, and decreases
>>inversely proportional to the distance to the source squared (the usual
>>inverse square law!). The "spreading out" is just a way to picture this.
>>Just as electrons in atoms are often pictured as "charge clouds" or
>>"probability clouds".
>>
>>
>
> I truly wish I could cut you some slack here, but honestly, it would
> not be fair to those who may hold you in high esteem, or to those who
> may have been convinced of your propaganda, therefore, I must call
> your argument stupid and groundless.

What I wrote is in no way propaganda.

And that you call the argument above "stupid and groundless" reveals
only your own stupidiy, yet again.

>>>Neither Feynman
>>>nor Huygens helped you in this particular case, so give us another
>>>explanation, won't you?
>>
>>Feynman indeed helped me. Again, read his popular science book on QED.
>>
>>
>
> So you have no other explanations, do you?

Since the explanation I already gave is perfectly appropriate,
no other is needed.

> You lose.

Only in your wet dreams.

Bye,
Bjoern

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