"I found I could say things with color and shapes that I couldn't say any other way -- things I had no words for." -Georgia O'Keeffe
When it comes to the Universe, it isn't just the stuff that's in it that's important.
It's also how all that stuff interacts with itself and everything else. To the best of our knowledge, there are four fundamental forces in the Universe, and they're all essential to our existence.
Some of them are familiar, like gravitation. On the largest scales in the Universe, gravitation is not only the most important force, but arguably the only important force in play. The amount of mass and energy inherent to objects determines how spacetime itself is curved, and this curvature of spacetime in turn determines how objects move and accelerate.
There's no "anti-mass" or "anti-energy" that causes some objects to be gravitationally repelled while others are gravitationally attracted. Gravity is always attractive, and we can interpret mass/energy as the lone type of gravitational charge, if we want to.
But other forces and interactions can be more complicated than gravity in this regard. Take the electromagnetic force, for example, just by looking at charged particles.
Image credit: http://Maxwells-Equations.com, copyright 2012.
Instead of one type of charge where like-attracts-like, we have two types of electric charge: positive and negative, where like charges repel and unlike charges attract. It's very different than gravity and somewhat more complicated, but there are some things that are familiar to us.
A neutral atom, for instance, is a good example of electromagnetism, where the positively charged nucleus is orbited by a swarm of negatively charged electrons. The electrons repel each other, but they're all attracted to the nucleus even moreso. As long as the total charge of the atom is zero and there is no sufficiently strong external radiation, the atom will remain stable and neutral.
We even understand, at a fundamental quantum level, how this works. The attraction and repulsion between all charged particles is mediated by the same particle: the photon.
It just takes one particle to take care of both attraction and repulsion, because of the relatively simple structure -- two charges, like-repels-like and opposites attract -- of electromagnetism. But things get a lot more complicated if we go inside the nucleus, and ask just how it is, at a fundamental level, that these tiny, charged structures hold themselves together.
An atomic nucleus, of course, is made up of protons and neutrons, except for hydrogen, which is just a proton by itself. But seeing as how protons have a positive electric charge and neutrons have no electric charge at all, there must be some sort of extra force -- a force even stronger than the electromagnetic force -- to hold these nuclei together.
In fact, the creatively-named strong force is required to hold even the individual protons and neutrons themselves together. Because a proton and neutron themselves are not fundamental, but composed of even smaller, fractionally-charged particles known as quarks.
Image credit: Learn EveryWare, © 2009 Alberta Education, error caught by Rich and Mike, edited by me.
The electric forces inside of a proton, for instance, would cause the nucleus itself to fly apart if there weren't another type of charge attached to each of these quarks: in addition to electric charge, they also possess color charge, which comes in not one type (like gravity), nor two (like electromagnetism), but three.
Only unlike gravitation and electromagnetism, you can't just have a color charge off by itself: you need a red, green, and blue together, to add up to "colorless," just like red, green, and blue light together add up to white. (Any Americans who want the three colors to be "red, white, and blue" can leave now, as can any Frenchman who thinks they should be "blue, white, and red." I'm making an analogy, and your flag does not trump physics.)
Just like there's matter and antimatter, there are quarks and anti-quarks, and so there are colors (red, green and blue) and anti-colors: cyan is the anti-red, magenta is the anti-green, and yellow is the anti-blue. So to add up to "colorless," you either need three quarks (or three anti-quarks), or one quark and one anti-quark.
Image credit: McLean County Unit District Number 5, http://www.unit5.org/.
It's a little bit weird: if red + green + blue makes white, but red + anti-red also makes white, does that mean that green+blue is the same as anti-red? Yes, yes it does, at least in terms of color. Which means you can pair a quark with either two other quarks, with an antiquark, or possibly even with three other quarks and one antiquark. As long as the color comes out white (or colorless), you're in business.
And that's why you can have combinations of three quarks, like protons and neutrons, or combinations of one quark and one anti-quark, like mesons. But unlike gravity, which bends spacetime, or electromagnetism, where photons (with no charge) are exchanged, the strong force works by exchanging a new type of particle -- the gluon -- which carries both a color and an anti-color!
These gluons are responsible for holding both individual particles -- like protons, neutrons, and pions -- together, as well as for binding larger atomic nuclei together.
Image credit: CERN / European Organization for Nuclear Research, http://www.physik.uzh.ch/.
How does this work? With three colors (red, green, and blue) and three anti-colors (anti-red = cyan, anti-green = magenta, and anti-blue = yellow), you might think there are nine types of gluons that you can get from matching each color with each anti-color. That's a good first thought, and it's almost right.
Imagine you're a red quark, and you emit a red/anti-green gluon. You're going to turn the red quark into a green quark, because color is conserved like that, and then that gluon is going to find a green quark, and turn it red. In this fashion, colors get exchanged.
That turns out to be a good explanation for six of the gluons: red/anti-green, red/anti-blue, green/anti-red, green/anti-blue, blue/anti-red, and blue/anti-green.
But what about the others: there should also be red/anti-red, green/anti-green, and blue/anti-blue, right?
Almost, it turns out. Because each of those has no inherent color, those quantum states are allowed to mix together. In quantum physics, whenever mixing isn't forbidden, it happens. So you get mixtures of red/anti-red, green/anti-green, and blue/anti-blue states.
But one of them -- the one that's an equal mixture of all three color/anti-color pairs -- is truly colorless, and doesn't physically exist. So there are only eight physical gluons.
And it's the exchange of these gluons between the quarks and antiquarks that keep protons, neutrons, mesons, baryons, and all other atomic nuclei together. There is much more to the strong interactions than what I've described here, and if you want to go deeper, I recommend this jaunt by Nobel Laureate Frank Wilczek. Whether you do or not, the strong force is what holds every atomic nucleus together; without it, we'd simply be a lifeless sea of fundamental particles, too repulsive to hold together in any meaningful fashion.
And yet, here we are, so much more than a cosmic soup, with galaxies, stars, planets, heavy elements, molecules, life, and you and me. It's the strongest force in the Universe, and without it, none of this would be possible. Enjoy!
Just a quick correction. Your picture of the neutron shows 2 red (-1/3) and one green (+2/3) but the math underneath the picture is incorrect. Should be 2x -1/3 plus 2/3 = 0.
Thanks Mike (and Rich, via phone); corrected, and that sure doesn't look good for University of Alberta!
If we can never see a color charged particle outside of confinement how do we know they exist? Couldn't this be instead a useful abstraction?
I've never understood conceptually why there are 8 gluons, not only 6. (mathematically, it's clear - the group SU(3) has 8 generators...) None of the explanations I've read so far was satisfactory - and that includes yours. The crucial sentence is this:
But one of them — the one that’s an equal mixture of all three color/anti-color pairs — is truly colorless,...
In what sense are the other two *not* "truly colorless"?
Can you please explain how exchanging particles creates force? In electromagnetism, the photon exchanged is the same regardless of the charges, so why is the force sometimes attractive and sometimes repulsive? And, once more, thanks for all the great explanations in your blog.
Bjoern sort of nails what I was going to say; this post is pretty much as good an explanation as you can get at the mathematical level of this weblog, but to really understand chromodynamics you absolutely must sit down and learn some Lie group theory. It all looks so much simpler once you have that language available.
Bjoern and John, consider a proton trying to interact with a neutron. If you exchange one of the six "net color" gluons, you'll be left with a proton and a neutron each with a net color: a dangerous, unstable and forbidden game. But you can exchange a gluon with no net color. Although I apologize for not having adequate words for it, only the combination of color-anticolor pair that are completely symmetric across all three colors is what I meant by "truly colorless", the other colorless gluons all have attractive couplings to quarks and antiquarks.
I think your summary of the number of charges that source the forces is incorrect. You say that gravity is sourced by 1 charge, electromagnetism by 2, and strong force by 3. Firstly, in GR gravity is sourced by both energy and momentum, so there are 4 sources of gravity. In electromagnetism, its better to just say that there is 1 charge which can be either a particle or anti-particle, rather than saying there are 2. If you insist to call it 2, then to be consistent you should say there are 6 types of color, counting for red, green, blue, and particle plus anti-particle. By your counting, the weak force has how many charges? 2.5? You see, it doesn't really work the way you've done the counting.
As to why there are 8 instead of 9 gluons is due to the model. You can't just reason it by saying the other one is "truly colorless". Instead it is a property of the model. To be technical, one has to decide whether nature is described by a U(3) symmetry or an SU(3) symmetry. If it is the former, then there are 9 gluons, if it is the latter, then there are 8 gluons. Nature has chosen the latter, we don't really know why.
Bob, there's no momentum in relativity.
It's all frames of reference, and you can't get a different result just from changing from one inertial frame to another.
And you're wrong, the electron is entire, but the proton isn't, therefore two distinct particles.
Your basis for disagreement has a flawed premise.
David, exchanging particles exchanges energy/momentum, therefore imparts an impulse: force.
The photon is its own antiparticle but also travels in no time in its own frame of reference, making it difficult to discuss in words how it does it. You really need advanced university level maths to work the detail out. Easiest to just take it as it comes. Short answer: the photons aren't the same.
Gravity doesn't give a fig about charge. If ever we have anti gravity, it would appear similarly to photons.
Wow, if you've opened up any book on relativity you will see that momentum plays a central role in relativity. It is true that it is frame dependent, but so is energy. This does not stop it from acting as a source of gravity. There is a related quantity called the stress energy tensor, which carries these 4 conserved quantities, which is really the source of gravity. It has 10 components, from which we can construct 4 conserved quantities, energy and momentum, in the flat space limit.
The proton is different from the electron. So what? There are 6 different quarks as well, called up, down, charm, strange, bottom, top. Does that mean there are 18 different colors? No. The number of charges is not given by the number of particles that exist.
Although I apologize for not having adequate words for it, only the combination of color-anticolor pair that are completely symmetric across all three colors is what I meant by “truly colorless”, the other colorless gluons all have attractive couplings to quarks and antiquarks.
As I said: from the mathematical standpoint, that's clear - but can you give an intuitive argument *why* this is so? Why do two combinations of the "colorless" gluons have couplings to the quarks, but the third, the completely symmetric one, hasn't?
It doesn't play a role in gravity, nor in relativity, bob.
Energy is not gravity. And an electron doesn't gain more energy because *I* sped up. I can get more energy from it to my frame of reference (by bumping into it for example).
You've read the words but are applying them incorrectly.
"On the largest scales in the Universe, gravitation is not only the most important force, but arguably the only important force in play."
-- uhh Ethan... isn't it you who is always telling us how dark energy is (almost certainly) real? :-)
[a little off-topic, admittedly]
"There’s no “anti-mass” or “anti-energy” that causes some objects to be gravitationally repelled while others are gravitationally attracted."
I read that you can get a repulsive effect between a mass of matter and a mass of negative matter. Newton's Universal Gravitation formula would agree. Does Relativity or some other more modern physics predict differently?
Don't worry, Ethan, I know what "truly colorless" means. I'm not even blaming you for not having better words to explain at the level that you're using to write this weblog. Some things are really just *hard*, and you really do need some heavy math to explain them right. I think you've done about as well as anyone could without breaking out Lie theory.
wow @11:47 - thanks for trying. but you didn't really answer my question, and your reply has an error. In electromagnetism, the exchanged particle (photon) is the same regardless of the charges carried by the fermions. Unlike what you said, there is only one kind of photon, and all photons are the same. The identical properties of identical particles is a basic principle. Additionally, if the effect on the fermions comes from an "impulse" imparted by the photon, then you could not possibly explain attraction (unless you propose a photon carrying momentum in a direction opposite to its direction of travel). Your "impulse" explanation is BS. So how does exchange of a photon produce attractive force?
David, the photon is its own antiparticle. I was using a metaphor badly.
Tty this way of looking at it: positive charges react in the opposite way to photons than negative ones. The negative charge emits the antiphoton, the positive charge tho photon.
Swapping particle and antiparticle gives attractive force.
Swapping the same type gives repulsive.
Its going to have to be a metaphor because you really need the maths. I neither went far enough into EPP nor have the capability on a blog to give you the maths.
You really are going to have to get a university grade course book on it.
Although it can be proved 1 + 1 = 2, we don't teach kids logic and set theory so they can see and understand the proof themselves. Unless you're a postgraduate physics major, you either have to do the learning or just take it on faith.
PS yes, you can explain attractive forces with impulses. You're forgetting in your eagerness to snark that we're dealing with virtual particles not real ones.
Look at evanescent waves or what happens to a particle able to quantum tunnel out of a potential well.
Imaginary products have physical manifestation when presented a venue of expression.
See also casmir effect.
Wow, Energy is frame dependent. Have you even studied high school physics? Do you know the formula for kinetic energy, 1/2 m v^2 ? What do you think happens to it when different observers measure different v? Ok, you don't know any physics...
Ethan, this is just to say thank you for the blog. For all entries, I read every one of them. For a non-specialist, what you write about is like the best adventure book ever written, an SF adventure in a real Universe. I learn a lot, and wonder like a child.
Many, many thanks for what you are doing -- never stop.
>And an electron doesn’t gain more energy because *I* sped up.
Yes it does. You're assuming that the electron has a 'true' energy which implies that the universe has a special inertial frame in which energies should be measured. This is contrary to special relativity.
That's what I said. It's dependent on your frame of reference.
What do you think v is measured from?
A FRAME OF REFERENCE.
Like I said, you seem to have read the words but lacked the comprehension of what they mean.
(bloody shortcut keys...)
A FRAME OF REFERENCE.
The gravitational attraction of the earth doesn't get higher because something moves past it quicker.
Yes, Bob. Different observers measure different v from each other. They also measure a different m. The measured mass depends on the rest mass, AND the measured velocity.
Bjoern, John Armstron, Ethan and others who understand QCD somewhat
Let me join you discussion about color.
Let me paraphrase my understanding, and ask a few question. NOTE: I'm not trying to invent a new QCD theory; I'm just trying to (metaphorically) understand QCD (I don't know Lie group theory yet).
The primary color analogy from optics is not an except analogy. Hence Bjoern's first question.
Now Ethan's chart of the three mixed states and his explanation me thinking.
If (R/aR + B/aB + G/aG)/ sqroot 3 = 0 i.e. truely colorless
.... and if (R/aR - B/aB)/ sqroot 2 is not= 0 i.e. is not truelly collorless
THEN R/aR "must not=" B/aB and neither are "truelly colorless" nor is G/aG. So it seems to me.
Is this thinking correct?
Let me stop here before proceeding. I know QCD is not grade school algebra like I've shown above. But "truelly colorless" means (I assume) no QCD color (we're not talking about optical color) in the quantum wave function. And it seems that
only (R/aR + B/aB + G/aG)/ sqroot 3 is "truelly colorless" Is this correct?
(by the way, is there any simple introduction to Lie theory for QCD that you can point to?)
So R/aR and B/aB and G/aG must have tree distinct yet almost colorless wave functions. Is this correct? But the three of them together add up to truelly colorless, i.e. (R/aR + B/aB + G/aG)/ sqroot 3 = 0 i.e. truely colorless.
Now, let me proceed with my thought. So there are eight gluons
...... R/aB (this is quantum mixture) and R/aG
...... B/aR and B/aG
...... G/aR and G/aB and
....... (R/aR - B/aB)/ sqroot 2 (this is a quantum mixture) and (R/aR + G/aG - 2B/aB)/ sqroot 6 (this is a quantum mixture)
So each of the eight qluons is a quantum mixture of states involving 3 colors and 3 anticolors.
Nice I think I'm unstanding. I think this paragraph is correct.
Hmm, now since the 7th and 8th gluon are not special (different than the first 6) then in Lie theory there must be some kind of transformation that transforms a
..... R/aR gluon of the Red antiRed quantum mixture) into for example the (R/aR - B/aB)/ sqroot 2 (quantum mixture)
Meaning that the Red frame of reference so to speak is arbitrary. We could define Red' (prime) to be equal to (R/aR - B/aB)/ sqroot 2 (this is a quantum mixture)
And remembering of course each gluon is it's own antiparticle and there are eight gluons.
This further suggests that we could somehow in Lie theory have picked a different reference frame, for example with these
..... 3 colors : R', B, G
..... and 3 anti colors aR', aB, aG
where we have the following 8 gluons:
..... B/aR and B/aG
...... G/aR and G/aB
..... R'/B and R'/G where R' = (R/aR - B/aB)/ sqroot 2)
..... (R'/aR' - B/aB)/ sqroot 2 and (R'/aR' + G/aG - 2B/aB)/ sqroot 6
So the mathematics of Lie theory has some kind of almost symmetry. It's almost like 3 optical colors but the symmetry of 3 optical colors is somehow different than the symmetry of 3 QCD colors (i.e. Lie theory). Thus the analogy with 3-optical colors is only an approximate analogy.
Now let me keep pushing my thought.
Protons and neutrons each have 3 quarks and need 3 colors because experiment tells us that we need 3 colors and eight gluons (and I don't argue with experiment).
But experiment also tells us, that we can't distinquishing a red from a blue quark. e.g. a Meson is made a quark and an antiquark has no specific color. Which means that the meson shuffles through the three colors. i.e. is a superposition of all colors.
Thus the dynamic neutron (uud) (above Image credit: Wikipedia / Wikimedia Commons user Qashqaiilove) in which the quarks are continuously emitting a gluon then changing color; and absorbing a gluon then changing.
But!!!! this wonderful image from Qashqaiilove leaves something out. Its leaves out.
First it only uses three gluons: e.g. if these three R/aB, B/aG, G/aR
then it leaves out these three R/aG, B/aR, G/aB
OK, maybe I can accept this omission as a simplification.
But what about the 7th and the 8th gluons,
i.e. (R/aR - B/aB)/ sqroot 2 / anti:(R/aR + G/aG - 2B/aB)/ sqroot 6 and (R/aR + G/aG - 2B/aB)/ sqroot 6 / anti: (R/aR - B/aB)/ sqroot 2
Now the 3 quarks of a neutron need to quantum transition between 3 different QCD colors. And that seems to require only 3 gluons. Is this correct?
So my question; in the full QCD
... how many gluons are needed for a neutron or proton (3, 6, or 8)
... if 3, are the 3 necessary gluons for a neutron different than the 3 necessary gluons for an anti-neutron
... if in Qashqaiilove image a 7th or 8th gluon is emitted or absorbed e.g. a Red quark emits a (R/aR - B/aB)/ sqroot 2 QCD gluon what QCD color quark do I get?
If I misunderstand, please educate me. thanks.
If my questions make sense, answers will be appreciated.
Any clarification in words will be much appreciated.
Any pointing to a good simple intorduction to Lie theory algebra for QCD will be appreciated.
One last point of confusion. I follow Ethan's link title "eight physical gluons" which brings me to a wiki paragraph titled "Eight gluon colors".
So does this mean that in the full QCD that
.... gluons have 8 gluon colors (these color octet things)?? and that
.... quarks have 3 QCD colors and 3 QCD anti-colors??
Are these words (metaphorically) correct?
Which means again, can you point me to the best beginners explanation of Lie theory for QCD?
Aside from the usual typos and errors; how did I get that happy face in my entry I meant to type number eight??
So let me if this eight 8 uploads as a happy face or an eight.
It's because in the new internet there is never ever the need to end a parenthetical statement with an eight.
Art majors pretending to be programmers, basically.
When did the Canadians learn the New Math? Anyway, I don't think it looks bad for the U. of Alberta - these are the sort of mistakes that make it through all the proof readers and which may be pointed out by someone months or years later. I remember one of my Physical Chemistry texts years ago had so many mistakes in places that I was cursing while reading through it and yet it was one of the better texts.
Just testing Eight parentheses 8) to see if it makes a smile-face.
As usual Wow must likely is correct in the detail; so if this comment has no smiley faces then I'm doing something wrong.
Well one out of three ain't so bad 8) but three 8) out of four 8) is better
You see to the art major there's no need for 8 then ) because that's *maths). And EVERYONE wants a smiley cool face in their messages, because
A- kids use emoticons and that's cool
B- its patented, so it must be serious
C- this is the internet where it's hip and cool and geeks are outnumbered and sidelined again, as nature intended.
In any case, where's the face's nose? Surely it should be 8 - ) for the eyes an 8, for the nose an -, and for the smiley mouth a ).
But kids don't type any more than they hav 2. So all these smiling faces have no nose (how do they smell?)
Smiley faces are silly.
But!!!!! I really want to learn more about color gluons and quarks.... Can someone take a stab at some of my questions. I'm not looking for exact perfect answers; just a little bit better way to think about QCD and yes readable primer on QCD with Lie Algebra would be nice. I'd like to have a good guidebook leading me into the tough stuff. Much thanks.
A here's how I think of it answer (but you know it isn't prefectly correct) would be just fine.
The Feynman papers are going to be some of your best sources.
Heck just go to your library and ask for everything hes's written. They often say of someone "he wrote the book", well when it comes to QCD and Feynman, it's not hyperbole...
John McC what's your point?
OKthen, there is no reason why nature chose 8 instead of 9 gluons, it just did. It could have chosen the group U(3), but it didn't, it chose SU(3). Ethan's claims that he knows why are silly. Even the inventors of qcd don't know why. It judt matches the data, that's all anyone can say.
Just to reiterate:
Electromagnetism has 1 charge
Weak force has 2 charges
Strong force has 3 charges
Gravity has 4 charges.
Ethan's claims are incorrect, as can be checked in any book, where he claims:
Electromagnetism has 2 charges - wrong, its an abelian theory
Weak force has 2.5 charges - a silly value between 2 & 3
Strong force has 3 charges and 8 gluons, the 9th is not present because it is an SU(3), not U(3) gauge theory
Gravity has 1 charge - wrong, gravity is sourced by the energy-momentum tensor
Bob's wrong as can be inferred from his statement "it can be found in any book" without even a reference to it.
Oh, and his insistence that momentum is important in general relativity when working out gravitational strength...
Oh, and his insistence that momentum is important in general relativity when working out gravitational strength…
So you dispute that in general relativity, the energy-momentum-tensor is the source of gravity...?
Gravity has 4 charges ... gravity is sourced by the energy-momentum tensor
Well, since the energy-momentum-tensor has 10 independent component, shouldn't one say then that gravity has 10 "charges"?
It could have chosen the group U(3), but it didn’t, it chose SU(3). Ethan’s claims that he knows why are silly. Even the inventors of qcd don’t know why.
Well, I've always thought that the reason is this: U(3) can be decomposed into U(1)xSU(3). But an U(1)-gauge transformation leads to electromagnetism, so only the SU(3)-group is left for QCD.
@OKthen: Without maths, I can't answer your questions, sorry - if I could, I would not have asked Ethan for an intuitive explanation! (which he apparently can't provide - or did he miss my post on June 28, 1:10 pm?)
For learning the math behind that, you could perhaps try "An introduction to quarks and partons" by F. E. Close; if I remember correctly, I found that quite helpful when working on my PhD thesis many years ago...
Bjorn - yes?
In the way bob wants it to be the imprimateur of gravity, yes, I dispute it.
It's rather like saying that because you get blue shifted light moving toward the source that momentum creates photons.
Pop back earlier and you can see where I say that Bob's read the words but didn't understand them.
Creationist scientists do the same when trying to use science to prove god.
Bjoern thanks. I will get “An introduction to quarks and partons” by F. E. Close
This wiki page is a discussion page, a working page that is not part of the wikipedia yet.
It gives the best answer that I've found to why there are 8 gluons instead of 9 and that distinquishes between the concepts of color neutral and colorless as two distinct concepts. I did not know.
read the section titled Resolution to the 8 gluon problem.
the section Color neutral gluons is also interesting
If I find something else interesting I'll post it.
Colour neutral: equal red, green and blue.
Or opaque (actual white rather than simulated)
In the way bob wants it to be the imprimateur of gravity, yes, I dispute it.
Bob never did say anything about "imprimateur". He talked about "charges"; a better choice of word would perhaps been "sources".
So, let's try this again, without obfuscation about "imprimateur" this time, please: do you really dispute that in general relativity, the energy-momentum-tensor is the source of gravity?
It’s rather like saying that because you get blue shifted light moving toward the source that momentum creates photons.
Err, no, it's nothing like that, and I have no idea why you think that would be an appropriate analogy.
Pop back earlier and you can see where I say that Bob’s read the words but didn’t understand them.
Well, although I think that Bob does express his thoughts not in the best way, nevertheless I'd say that he understands better what he read than you... (or you also express yourself rather poorly, and you simply talk past each other)
@OKThen: The proposed solution at that page looks nice - but unfortunately it is wrong.
First, the distinction between colour neutral and colourless which is made does in no way correspond to the math in QCD - there, these two things are the same.
Second, the 7th and 8th gluons which he proposes are *not* the ones which are usually used in QCD - see
Ethan's text above for how the usual ones look like!
Third, the argument which he gives for why th 9th gluon does not exist ("This is because colourless particles are the only stable particles and if a colourless gluon existed it should have been detected long ago.") makes no sense on several levels.
"Err, no, it’s nothing like that, and I have no idea why you think that would be an appropriate analogy."
And I have no idea why you say "it's nothing like that". Bob says that momentum causes gravity because of the energy-momentum tensor. And that's wrong.
Bob says that momentum causes gravity because of the energy-momentum tensor. And that’s wrong.
That's not wrong, that's simply badly expressed. The energy-momentum tensor is the source of gravity, hence gravity depends in a sense on momentum. Hey, look at the standard models in cosmology - there you have pressure, which is nothing else than the flux of momentum, acting as a source of gravity!
Bjoern 6:09 am above
Thanks for that clarification that there is no difference between color neutral and colorless. I defer to you .
But let me add a point of clarification and a question.
Above I said,
"But what about the 7th and the 8th gluons,
i.e. (R/aR – B/aB)/ sqroot 2 / anti:(R/aR + G/aG – 2B/aB)/ sqroot 6 and (R/aR + G/aG – 2B/aB)/ sqroot 6 / anti: (R/aR – B/aB)/ sqroot 2"
And I think this is wrong. (confused myself, it happens)
I now think that the 7th and 8th gluon are simply
..... (R/aR – B/aB)/ sqroot 2 is the 7th gluon
..... (R/aR + G/aG – 2B/aB)/ sqroot 6 is the 8th gluon.
Is this correct?
What happens if, for example, a Red quark absorbs a 7th or 8th gluon?
.....Does a Red quark becomes an antiBlue or antiGreen quark? or
.....Does a Red quark become a Blue or Green quark?
.....or something else?
I assume that the 7th and 8th gluons can be absored or emitted by some of the color or anticolor quarks. So can anyone give an example. or a hypothesis or two
I also assume that nature doesn't think of the 7th and 8th gluon as different than the other 6 gluons; otherwise we'd have a different name for them like "neutral gluon" or maybe universal gluon, maybe???? Let me not guess. does anyone know what happens when the 7th or the 8th gluon is absorbed or emitted by a quark? If you'd like to explain how the 7th or 8th gluon participates in glueballs; that's fine too.
OH, I think I understand it!! Maybe?
First let me summarize the first six gluons.
A Red quark and emit a R/aB to become Blue or R/aG gluon to become Green
A Red quark can absorb a B/aR to become Blue or a G/aR gluon to become green
But a Red quark can NOT absorb or emit a B/aG or a G/aB gluon
Yes, Maybe! speculation on 7th and 8th gluon follows.
The 7th gluon (R/aR – B/aB)/ sqroot 2 is the supersition of two gluons.
---So if a Red quark absorbs a 7th gluon (R/aR – B/aB)/ sqroot 2 ; the Red quark becomes an antiBlue quark.
---and if a Red quark emits a 7th gluon it becomes an antiBlue quark.
this 7th gluon can only interact with Red, Blue, antiRed, and antiBlue gluons; but not Green or antiGreen gluons
Now the 8th gluon
So if a Red quark absorbs an 8th gluon (R/aR + G/aG – 2B/aB)/ sqroot 6; it can become a Green quark (1/3 of tijme) or an antiBlue quark (2/3 of time)
If a Red quark emits an 8th gluon; it becomes a Green quark (1/3 of time) and an antiBlue quark (2/3 of time)
The 8th gluon interacts with all 3 color and all 3 anticolor quarks.
Now there seems to be an assymetry regarding the 7th and 8th gluons. And I assume that that is because the 7th and 8th gluon nomenclature is really more complicated than shown. I suspect the for example all three permutations of the letters R, B, G equally describe the 7th gluon and similarly the 8th gluon.
OK, tell me if I've really seen a pattern here or I've just gone bonkers. Silence doesn't mean that I am correct; it just means nobody out here knows for sure either.
Or the binding energy is such that the configuration is unstable and cannot be said to exist, OKThen.
C26 would be a non-QCD version. Good luck keeping it together long enough to be called an atom...
I now think that the 7th and 8th gluon are simply .......... Is this correct?
What happens if, for example, a Red quark absorbs a 7th or 8th gluon?
It stays red. You could say that it absorbs the R/aR-"part" of the gluon, and that obviously does make it red again.
You know about quantum superposition, right? Well, the 7th gluon is simply a superposition of the gluon R/aR and the gluon B/aB; an interaction with a quark corresponds in a sense to a "measurement" of the gluon, and the gluon wave function "collapses" to either R/aR or B/aB. Either way, this leads to no color change of the quark. (and the same holds for the 8th gluon) That's another sense in which these gluons can be called "colourless" - an interaction with them does not change the colour of a quark.
BTW: Do you know how one multiplies a matrix with a vector? That's essentially all you need to know if you want to get a rough impression of how one writes the interaction of a gluon with a quark mathematically...
I assume you correct as usual and will digest. Much appreciated.
Yes I know how to multiply a matrix by a vector.
I'm looking forward to the book you recommended. I ordered it.
Daaaaaaamn! Physics be crazy, yo!
Yes I know how to multiply a matrix by a vector.
Well, then it's not too hard. (keep in mind that this is still not 100% correct, but it should suffice to get an idea)
You could picture the three colours of the quarks as three (column) unit vectors: R = (1, 0, 0), G = (0, 1, 0), B = (0, 0, 1). The gluons then correspond to matrices, where the row corresponds to the colour and the column to the anti-colour. For example, the gluon R/aG would correspond to the matrix (I give each row separately):
(0 1 0) (0 0 0) (0 0 0)
If you want to have the action of a gluon on a quark, you then simply multiply this matrix with the corresponding quark colour vector. For example, the action of the gluon R/aG on a quark with colour G would be given by the multiplication of the matrix given above with the (column) vector (0, 1, 0), and the result is obviously the (column) vector (1, 0, 0), hence a quark with colour R - as expected. (if the result is the zero vector, you could interpret that as "the gluon does not interact with that colour")
The matrices which are usually used in QCD correspond almost (but not exactly) with the gluons Ethan listed above; you can find more about them here:
I’m looking forward to the book you recommended. I ordered it.
Good luck, I hope it will be helpful to you...
@OKthen: Oh, I should add that the Gell-Mann matrix number three on the Wiki page corresponds to the 7th gluon in Ethan's text, and number 8 to the 8th gluon. I think you can puzzle out the rest for yourself then. ;-)
Bjoern says "Well, since the energy-momentum-tensor has 10 independent component, shouldn’t one say then that gravity has 10 “charges”?"
No. That would be like saying that simce the electromagnetic current has 4 componenets, then electromagnetism has 4 charges, or that qcd has 4*3=12 charges. No, the number of charges is the number of conserved quantities. For EM it is 1. For weak interaction it is 2. For qcd it is 3. For gravity it is 4, corresponding to energy and momentum. Think anout it, it is called the "energy-momentum" tensor for a reason. There are 4 consrved quantities that arise from a 10 component matrix of various types of densities as required by lorentz summetry.
Bjoern says "Well, I’ve always thought that the reason is this: U(3) can be decomposed into U(1)xSU(3). But an U(1)-gauge transformation leads to electromagnetism, so only the SU(3)-group is left for QCD."
The point I am making is that you don't have to do this. Although U(3) can be wriiten as SU(3)*U(1), one can decide to treat it as a single unified group with a single gauge coupling, and that the quarks transform in the fundamental rep of U(3). This is a different theory to having the quarks transform seperately under SU(3) and U(1) with different gauge couplings and alllowing arbitrary U(1) charge assignments.
Also, to all the people here who are arguing that the 9th gluon is "truly colorless" or "color neutral" or whatever, and therefore it does not couple to quarks... This is all nonsense. For example, the photon is completely neutral, however you look at it, but it surely couples to charges. So this argument is just plain wrong.
Let me reiterate, the reason there is no 9th gluon is because nature chose the group SU(3), not U(3). They are both sensible theories with 8 or 9 interacting gluons, respectively. You can never use group theory, etc, to "explain" it, you will fail; thats why all your explanations havn't worked. To explain this, would be like explaining why there are 3 colors instead of 4... No-one knows. Its just because nature selected it, and we know that because the 8 gluon theory matches the data, not the 9 gluon theory.
Because the photon has an electric field, bob.
An electric field affects charged particles but aren't themselves charged particles.
And we do have people who can see four colours. Most common in women. As to why, it's because of three types of photo receptors that have a narrow spectral response that we see three colours.
You may as well ask why a colour DSLR has three colours:RGB. Answer: because that's what we put in them.
@bob: Thanks. I've realized that myself already in the meantime, sorry for the glitch. ;-)
I'm late to the party, so I don't know if you'll read this, but thank you, thank you, thank you!
I'm the beginner who has always been fascinated by how the universe works, and by what else is out there when I look up at the stars, or inward in the abstract, but I could never really tell you what quarks and photons and mesons actually are, and where they are precisely, and the forces they represent. Anti-matter was just the Great Satan of bad science fiction movies, so you can probably imagine my surprise (though I confess that dark matter and dark energy remain more poetry than science, so far).
As an artist, where I'm not a beginner, it was thrilling to discover that the two things I do understand -- the RGB of my computer screen versus the printer's CKMY -- are so close to the heart of everything!
So it's onward to the Higgs Bosun, now with some real hope of understanding what confirming its existence means to physics, even if I'll never get the math. I appreciate this plain English lesson more than I can say.
NWhy no 9th gluon? Think of it this way: red charge (or blue, or green) are not specifically intrinsic to a given particle. Unlike electric charge, which any given particle is stuck with, any quarkcan be red at one time, then green another, then blue, or some linear combination. Like space can be described with x, y, and z axes, but there is no defined coordinate system in the universe. The thing that makes that ninth gluon special is that if you rotate the red, green, and blue, the linear combination does not change. In a sense, gluons must not be rotationally invariant in RGB space.
so I'm counting six gluons that are not made up of a particle antiparticle pair, then there is the "red anti red plus blue anti blue", the "red anti red plus green anti green" and that would leave 'green anti green plus blue anti blue" 9 types of gluon according to the rules given so what gives? why is "green anti green plus blue anti blue" ruled out?
Ethan: I like the diagram of the gluon holding the proton and neutron together. I would like to use that diagram. Is it public, or does one need permission? I looked at the web page you cite (CERN, University of Zurich), but I don't see that diagram there. Any idea whose permission is needed to use the diagram?
You're going to want to send your question to startswithabang at gmail dot com. I'm not sure Ethan will check this old thread.