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It’s for my studies, so only hints please

Many thanks in advance,

- James

- Thread starter ultraviolent
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- #1

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It’s for my studies, so only hints please

Many thanks in advance,

- James

- #2

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And a free gluon can't exist, so that'll form a gluon-jet.

- #3

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I think particles need to have a total charge of zero to exist in isolation, so does this mean it’s an anti quark?

Does anyone know of a good website listing the "rules" for this?

EDIT: Do you have MSN, AIM or IRC? It would help me lots is someone was willing to talk to me about this...

- #4

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Probably.Originally posted by ultraviolent

In the Feynman diagram I have the gluon emits an up quark and an unknown object, which I’ve been asked to work out.

I think particles need to have a total charge of zero to exist in isolation, so does this mean it’s an anti quark?

Remember: conservation of charge. A gluon has charge=0, therefore, so must the [quark, something else] pair that it emits.

Also: conservation of color: if the color-charge of the gluon does not change, then the [quark, something else] pair that it emits must be color-neutral.

As for places to look for the rules: try your textbook or local library...

Good luck!

- #5

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conservation of Q (electric charge)

conservation of Q

conservation of I

conservation of Y {hypercharge)

must all be observed. Find these values for the initial quark that emitted the gluon, and find the values for the object the gluon emitted. Here's a hint; what are the values of these for the standard gluon? Do the math. Here's a bonus hint; the quark flavors are independent of color.

If you come to another road-block, just ask about it.

- #6

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Does the conservation of electric charge imply that the electric charge the gluon + up quark + unknown = 0

As;

Up quark: + 2/3

Down quark: -1/3

Gluon: 0

This means the equation becomes

-2/3 + 2/3 + 0 = 0

The only object with an electric charge of -2/3 is an anti up quark, so this must be the answer?

Please tell me I’m correct

As;

Up quark: + 2/3

Down quark: -1/3

Gluon: 0

This means the equation becomes

-2/3 + 2/3 + 0 = 0

The only object with an electric charge of -2/3 is an anti up quark, so this must be the answer?

Please tell me I’m correct

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I think so, an anti charm and an anti top quark?I can think of two other (anti)quarks that also will have charge -2/3. Can you too?

Because it's not one of the 5 possible answersWhy can/can't it be one of those?

I'm going to assume it's because they have greater mass, and mass is only another form of energy so it would require more energy from the gluon?

That would be the format of the question again (see the diagram)Also... who said that it had to be just one quark?

Is this something to do with conservation of colour charge? How do you work out which colour charge a particular quark has? According to my textbook, an up quark can be red, blue or green!

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Correct!Originally posted by ultraviolent

I think so, an anti charm and an anti top quark?

Sure, that'll work. Just out of curiosity, what are your possible answers?Because it's not one of the 5 possible answers

There is also a physical argument, and this:

is not it.I'm going to assume it's because they have greater mass, and mass is only another form of energy so it would require more energy from the gluon?

Think about the following question: can an electron emit/absorb only a tau-neutrino and nothing else? why/why not?

Yes, a quark can have three colors (obviously this is just a name and not a physical color) and an anti-quark has the three complementary colors. As a result, a gluon can have 8 color-charges (3x3 - 1). The ninth combination would be color-neutral and therefore it does not contribute.How do you work out which colour charge a particular quark has? According to my textbook, an up quark can be red, blue or green!

Already feeling or from looking at the subatomic zoo?

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A. an up quarkJust out of curiosity, what are your possible answers?

B. an anti up quark

C. a down quark

D. an anti down quark

E. an electron

Can I get back to you on that?Think about the following question: can an electron emit/absorb a tau-neutrino? why/why not?

Does this mean they have three colours, they can have one of the three colours at any point, or that it's predefined what colour charge a quark has?Yes, a quark can have three colors

Was this named just to confuse everyone?obviously this is just a name and not a physical color

- #11

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Then I agree with your answer: B.Originally posted by ultraviolent

A. an up quark

B. an anti up quark

C. a down quark

D. an anti down quark

E. an electron

Sure! It has to do with conservation of lepton-number.Can I get back to you on that?

A quark has a fixed color, but this color can change when it has an interaction with something. Typically, the gluon that couples to the quark 'takes away' the color charge on the quark and 'adds' a new one (determined by the color charge of the gluon itself). That's why a gluon does not have a pure color, but it's always a mixed state of two different color charges.Does this mean they have three colours, they can have one of the three colours at any point, or that it's predefined what colour charge a quark has?

Of course! If physics were to seem easy to everyone, then physicists wouldn't earn such huge salaries... O wait, we don't!Was this named just to confuse everyone?

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I just wanted to elaborate on this for future reference. The hypercharge of the quarks and gluon is also just as important in determining the unknown as the electrical charges are. From the five choices given in the problem, it is clear which one is correct, and so far you are all doing great at that. But the question by suyver does raise some issues. In the real world, where your choices are not limited to the given five, anti-charm and anti-top would satisfy the charge-conservation requirement, yet they would not be correct. Here's why; hypercharge-conservation and isospin-conservation.Originally posted by suyver

In the electroweak-regime, the properties that determine the flavours of particles are related to the electric charge by;

[tex]

Q = I_z + Y_{weak}

[/tex]

where [itex]Q[/itex] is the electrical charge, [itex]I_z[/itex] is the isospin eigenvalue, and [itex]Y_{weak}[/itex] is the weak hypercharge associated with the flavour of the particle. In conserving [itex]I_z[/itex] and [itex]Y_{weak}[/itex], you automatically conserve electric charge. So I will run through the math of it here since you have already found the correct answer.

The hypercharge is, itself, a composite of several inmportant quantum numbers;

[tex]

Y_{weak} = \frac {b + S + C + B + T}{2}

[/tex]

obtains for all quarks and their composites, with [itex]b[/itex] the baryon number, [tex]S[/itex] the strangeness, [itex]C[/itex] the charmness, [itex]B[/itex] the bottomness, and [itex]T[/itex] the topness for a given particle. For the known quarks, and in this case the ones we are looking at, the properties in each of these is as follows;

for the anti-up quark;

[tex]

I_z = -\frac {1}{2}, b = -\frac {1}{3}, S = C = B = T = 0

[/tex]

for the anti-charm quark;

[tex]

I_z = 0, b = -\frac {1}{3}, S = B = T = 0, C = -1

[/tex]

for the anti-top quark;

[tex]

I_z = 0, b = -\frac {1}{3}, S = C = B = 0, T = -1

[/tex]

for the up quark;

[tex]

I_z = +\frac {1}{2}, b = +\frac {1}{3}, S = C = B = T = 0

[/tex]

for the gluon;

[tex]

I_z = 0, b = 0, S = C = B = T = 0

[/tex]

Putting these figures of the known objects into a conservation equation gives us the appropriate numbers for the unknown;

[tex]

I^{unknown}_z = I^{g}_z - I^{u}_z = 0 - \frac {1}{2} = -\frac {1}{2}

[/tex]

[tex]

Y^{unknown}_{weak} = Y^{g}_{weak} - Y^{u}_{weak} = 0 - \frac {1}{6} = -\frac {1}{6}

[/tex]

Which quark or antiquark has [itex]Y_{weak} = -\frac {1}{6}[/itex] and [itex]I_z = -\frac {1}{2}[/itex] ? It cannot be either the anti-charm or anti-top quarks because of the isospin number. This leaves only the anti-up quark, which has both [itex]I_z = -\frac {1}{2}[/itex] and [itex]Y_{weak} = -\frac {1}{6}[/itex]. This is also in accordance with a general rule that photons, Z particles, and gluons can only produce particle-antiparticle pairs. So the anti-up quark is clearly the identity of the unknown, all others being excluded in one way or another.

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- #14

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Yes, thanks for the reply, I now have some idea what hypercharge and isospin are, but I didn’t understand much else :)

I’m sure it will be useful to me in the future

Going back to suyver's question "can an electron emit/absorb a tauon neutrino? why/why not?"

I've found some information in my textbook (always a good place to look).

I’m not sure how point 3 is relevant though, perhaps you could shed some light on this?

- James

I’m sure it will be useful to me in the future

Going back to suyver's question "can an electron emit/absorb a tauon neutrino? why/why not?"

I've found some information in my textbook (always a good place to look).

So, in answer to your question I would say no - because electric charge would not be conserved.In all weak interactions:

1. Electric charge is conserved

2. The number of quarks minus the number of antiquarks is conserved;

3. The number of leptons minus the number of anti leptops is conserved;

4. The flavour changing of either quarks or leptons is allowed, as long as these three rules are obeyed.

I’m not sure how point 3 is relevant though, perhaps you could shed some light on this?

- James

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Why do you say this? A neutrino does not have a charge!Originally posted by ultraviolent

So, in answer to your question I would say no - because electric charge would not be conserved.

Actually #3 directly gives the answer to my question. Hint: there are 3 lepton-numbers, one for each of the families (electron, muon & tau)...

- #16

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suyver is correct. In fact, this can be characterized by the leptonic hypercharge which falls into the same form of equation as the one for quark composites;

[tex]

Q = I_z + Y_L

[/tex]

where [itex]I_z[/itex] is, once again, the familiar isospin eigenvalue for the particular lepton, and [itex]Y_l[/itex] is the leptonic hypercharge. As you can guess by now, this hypercharge is also a composite of the lepton number [itex]L[/itex], [itex]\mu[/itex] number, and [itex]\tau[/itex] number in the same fashion that the flavour numbers worked in the other one. Namely;

[tex]

Y_L = \frac {L + \mu + \tau}{2}

[/tex]

The effect is that [itex]I_z[/itex] and [itex]Y_L[/itex] are equal in both magnitude and sign for the electron and positron each, adding to [itex]Q = -1[/itex] and [itex]Q = +1[/itex] respectively. For the electron-neutrino and its conjugate, the signs of these are inverse, so that [itex]Q = 0[/itex] obtains for both. For the heavier leptons, [itex]I_z = 0[/itex] and the lepton number and family number are equal in sign and magnitude, so that once again [itex]Q = \pm 1[/itex] for each lepton family set. The neutrinos for each have their family number opposite in sign to their lepton number, so that [itex]Q = 0[/itex] once again for each. Hence, the leptons of each family will only interact with neutrinos of the same family in the weak field (flavour) regime.

Basically, this is just the long way for justifying suyver's answer. Just thought you might enjoy seeing it for yourselves.

[tex]

Q = I_z + Y_L

[/tex]

where [itex]I_z[/itex] is, once again, the familiar isospin eigenvalue for the particular lepton, and [itex]Y_l[/itex] is the leptonic hypercharge. As you can guess by now, this hypercharge is also a composite of the lepton number [itex]L[/itex], [itex]\mu[/itex] number, and [itex]\tau[/itex] number in the same fashion that the flavour numbers worked in the other one. Namely;

[tex]

Y_L = \frac {L + \mu + \tau}{2}

[/tex]

The effect is that [itex]I_z[/itex] and [itex]Y_L[/itex] are equal in both magnitude and sign for the electron and positron each, adding to [itex]Q = -1[/itex] and [itex]Q = +1[/itex] respectively. For the electron-neutrino and its conjugate, the signs of these are inverse, so that [itex]Q = 0[/itex] obtains for both. For the heavier leptons, [itex]I_z = 0[/itex] and the lepton number and family number are equal in sign and magnitude, so that once again [itex]Q = \pm 1[/itex] for each lepton family set. The neutrinos for each have their family number opposite in sign to their lepton number, so that [itex]Q = 0[/itex] once again for each. Hence, the leptons of each family will only interact with neutrinos of the same family in the weak field (flavour) regime.

Basically, this is just the long way for justifying suyver's answer. Just thought you might enjoy seeing it for yourselves.

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