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## What does a proton look like?

[QUOTE=SpaceTiger]
 Theory without experimental support is just philosophy
No, It is just a theory. Supergravity is not philosophy. Any way, I meant to say that experimental physics is not a business of mine. And I do not find it interesting to talk about the various lepton-nucleon DIS experiments.
 Color is indeed based upon experimental evidence -- for example, the lack of qq hadrons mentioned in another thread.
The introduction of the colour degree of freedom "was" based purely on theoretical ground, without it the $(3/2)^{+}$ decuplet baryon wavefunctions would violate Pauli principle. This is why it was called the colour hypothesis. The story here is similar to that of the neutrino hypothesis. Do you see what I meant by theoretical "facts"; we know "it" is there even though experiment does not show "it".
At that time, there was no experiment that led us to the introduction of colours.
The absence of qq hadron does not mean or imply that quarks carry colour. However, if the quark does carry colour, then the absence of qq means that colour is a hidden degree of freedom, i.e. coloured hadrons are not observable. This is why (after the introduction of colour) it was "assumed":
"only colour-singlets are observable"
But what is it about colour that makes it a hidden degree of freedom? I wish, I know the answer. It seems that the colour-singlet conjecture is not derivable from the mathematical tools of the theory.
We still do not know why the quarks can not confine themselves in a bound state of coloured hadrons like qq. But we know why we do not know: the exact form of q-q potential (which we do not know) should be able to rule out such bound-states. That is if, it is meaningful to talk about such thing as the "exact q-q potential".
We now know several direct pieces of experimental evidence which support the colour hypothesis. The first comes from the analysis of the $\pi^{0}$ lifetime (as discussed by many textbooks), this is "wrong" by roughly a factor $N^{2}=9$ without the inclusion of colour. Further support comes from measurements of:

$$R_h = \sigma ( e^{+}e^{-} \rightarrow hadrons) / \sigma (e^{+}e^{-} \rightarrow \mu^{+} \mu^{-})$$

with the inclusion of the colour factor 3, the calculations shows R = 11/3 in reasonable agreement with the data.

 I was having a little trouble understanding how this fit in with the conventional picture of hadron structure that jtbell described.
I thought my statement was clear and simple

 do you simply mean what you said in your last post, that there is a non-negligible probability of finding the proton in the five-quark configuration due to the sea of virtual pairs?
O Yes, this is what I meant, and this is (it seems) exactly how jtbell understood my statement.

regards

sam

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Quote by Imop45
 Hi, sorry for posting this, but I was wanting to know if there is an actual picture of an atom. Picture as in a scan or something like that. Has a Microscope even been made that can see that small a structure?
It is not possible to see the atom. Heisenberg uncertainty principle does not let you take a picture.

regards

sam

 Quote by Hurkyl I wanted to object to this. The electron doesn't look like a point particle revolving around the nucleous: it's smeared out across its entire orbital! It really is stationary in the sense that it doesn't change over time. (Sort of like a stationary current)
Actually, electrons are not smeared out at all. This is a misconception. They are actual particles. The smearing out refers to the probability of the location.

 What does a proton look like? Jerry Friedman, Henry Kendall, and Richard Taylor participated in experiments (under Robert Hofstadter) at the Mark III electron accelerator (~ 300 MeV) at Stanford to measure the form factor of the proton in the late 1950's. The electron accelerator at SLAC (~ 20 GeV, Stanford) was built in part to make better proton form factor measurements, because the momentum transfer would be much higher. But when they did the experiment, the proton "broke" apart ~1970). Here is Jerry Friedman's 1990 Nobel Prize Lecture: http://nobelprize.virtual.museum/nob...an-lecture.pdf Bob S

 Quote by samalkhaiat It is not possible to see the atom. Heisenberg uncertainty principle does not let you take a picture. regards sam
Now I'm a chemist, and not a physicist -- but I believe this isn't right. Not too far off, but not right.

 Quote by samalkhaiat It is not possible to see the atom. Heisenberg uncertainty principle does not let you take a picture.
Δp Δx ≥ h-bar/2

If Δp > ~300 MeV/c

then Δx < ~1 fermi

So momentum transfers over ~300 MeV/c can "see" structure inside the proton (radius ~ 0.8 fermi).

Bob S

 What ANYTHING "looks like" is really a function of its Compton wavelength.I mean,the ratio of "h-bar/momentum".That's as good as it gets.Start by calculating,say,the Compton wavelength of the Earth for example,then,work yer way "down" to smaller and smaller critters.The attempts to give a "familiar structure" to leptons/hadrons tormented the minds of some very bright people and it led nowhere.Read-up on Heisenberg's "gamma ray microscope",read Feynman Chapter 37-38 Lectures Vol.1.Get a feel for the limits of how far "mental pictures" can go before they are,...,worthless.
 I just read Bob S about the Mark III and the Stanford experiments of Hofstadter.They were beutiful experiments,and Hofstedter richly deserved the 1961 Nobel.Annual Reviews contains his take,as well as an early volume from the old Benjamin-Cummings "Frontiers In Physics" series.These experiments proved beyond a reasonable doubt that nucleons are NOT point-particles,as leptons continue to be.Nope.Hadrons have a "structure",they are ALL "resonances",as far as I am concerned(though not with the same intent as Chew and the "bootstrap" gang,...,)Keep studying,....
 If single particles cannot emit light , then where does the light come from in a neutron star. I know a neutron star has electrons and other particles in it but i don't think it has atoms of molecules in it .

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