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IamZack
IamZack
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Let´s go:

[1] From the DIS (Deep Inelastic Scattering) of eletron-proton, we can imagine that the photon exchanged in te process "sees" a parton (possible constituent of the proton) distribuition.

We can imagine a cross-section of photons and that constituents of the proton. And we can analyze two situations:

From a cross-section of longitudinal (scalar) helicity photons, $\sigma_{s}$ and a cross-section of transverse helicity photons, $\sigma_{t}$ we can stabilish a relation:

$$ \frac{\sigma_{s}}{\sigma_{t}} $$

And experimentally, we know its values. From the theory, this rate goes to infinityinfinite if we are talking about a spinless constituent, and goes to zero (at high frequencies of the scattered photon) for a spin-half constituent. This is how we know that quarks are spin-half. (this answers 4)

[2] There is a excited state of the proton, called $\Lambda^{++}$, this. This particle is composed by 3 up quarks. As we know the principle of exclusion, we couldn't have 3 fermions in the same state unless there wereis a additional degree of freedom we are not taking into account.

This additional degree of freedom is the Colour charge. But the fact that we need 3 colours (3 kinds of charge) cames from the choice of the gauge group that describes the strong interactions, the SU(3).

And by this very choice, we have a non-abelian gauge symmetry wich means that our gauge bosons (in this case gluons) interact with each other, because in the non-abelian case:

$$ F_{\mu \nu} = {\partial}_{\mu} A_{\nu} - {\partial}_{\nu} A_{\mu} - iq[A_{\mu},A_{\nu}] $$

The last term (the commutator) of $A_{\mu}$ and $A_{\nu}$ does not vanish and so the Lorentz invariant term in the Lagrangian $ F_{\mu \nu}F^{\mu \nu} $ gives 3-field and 4-field interactions of the gauge bosons. (this answers 3 and 2)

Let´s go:

[1] From the DIS (Deep Inelastic Scattering) of eletron-proton, we can imagine that the photon exchanged in te process "sees" a parton (possible constituent of the proton) distribuition.

We can imagine a cross-section of photons and that constituents of the proton. And we can analyze two situations:

From a cross-section of longitudinal (scalar) helicity photons, $\sigma_{s}$ and a cross-section of transverse helicity photons, $\sigma_{t}$ we can stabilish a relation:

$$ \frac{\sigma_{s}}{\sigma_{t}} $$

And experimentally, know its values. From the theory, this rate goes to infinity if we are talking about a spinless constituent, and goes to zero (at high frequencies of the scattered photon) for a spin-half constituent. This is how we know that quarks are spin-half. (this answers 4)

[2] There is a excited state of the proton, called $\Lambda^{++}$, this particle is composed by 3 up quarks. As we know the principle of exclusion, we couldn't have 3 fermions in the same state unless there were a additional degree of freedom we are not taking into account.

This additional degree of freedom is the Colour charge. But the fact that we need 3 colours (3 kinds of charge) cames from the choice of the gauge group that describes the strong interactions, the SU(3).

And by this very choice, we have a non-abelian gauge symmetry wich means that our gauge bosons (in this case gluons) interact with each other, because in the non-abelian case:

$$ F_{\mu \nu} = {\partial}_{\mu} A_{\nu} - {\partial}_{\nu} A_{\mu} - iq[A_{\mu},A_{\nu}] $$

The last term (the commutator) of $A_{\mu}$ and $A_{\nu}$ does not vanish and so the Lorentz invariant term in the Lagrangian $ F_{\mu \nu}F^{\mu \nu} $ gives 3-field and 4-field interactions of the gauge bosons. (this answers 3 and 2)

Let´s go:

[1] From the DIS (Deep Inelastic Scattering) of eletron-proton, we can imagine that the photon exchanged in te process "sees" a parton (possible constituent of the proton) distribuition.

We can imagine a cross-section of photons and that constituents of the proton. And we can analyze two situations:

From a cross-section of longitudinal (scalar) helicity photons, $\sigma_{s}$ and a cross-section of transverse helicity photons, $\sigma_{t}$ we can stabilish a relation:

$$ \frac{\sigma_{s}}{\sigma_{t}} $$

And experimentally, we know its values. From the theory, this rate goes to infinite if we are talking about a spinless constituent, and goes to zero (at high frequencies of the scattered photon) for a spin-half constituent. This is how we know that quarks are spin-half. (this answers 4)

[2] There is a excited state of the proton, called $\Lambda^{++}$. This particle is composed by 3 up quarks. As we know the principle of exclusion, we couldn't have 3 fermions in the same state unless there is a additional degree of freedom we are not taking into account.

This additional degree of freedom is the Colour charge. But the fact that we need 3 colours (3 kinds of charge) cames from the choice of the gauge group that describes the strong interactions, the SU(3).

And by this very choice, we have a non-abelian gauge symmetry wich means that our gauge bosons (in this case gluons) interact with each other, because in the non-abelian case:

$$ F_{\mu \nu} = {\partial}_{\mu} A_{\nu} - {\partial}_{\nu} A_{\mu} - iq[A_{\mu},A_{\nu}] $$

The last term (the commutator) of $A_{\mu}$ and $A_{\nu}$ does not vanish and so the Lorentz invariant term in the Lagrangian $ F_{\mu \nu}F^{\mu \nu} $ gives 3-field and 4-field interactions of the gauge bosons. (this answers 3 and 2)

Source Link
IamZack
IamZack
  • 568
  • 2
  • 13

Let´s go:

[1] From the DIS (Deep Inelastic Scattering) of eletron-proton, we can imagine that the photon exchanged in te process "sees" a parton (possible constituent of the proton) distribuition.

We can imagine a cross-section of photons and that constituents of the proton. And we can analyze two situations:

From a cross-section of longitudinal (scalar) helicity photons, $\sigma_{s}$ and a cross-section of transverse helicity photons, $\sigma_{t}$ we can stabilish a relation:

$$ \frac{\sigma_{s}}{\sigma_{t}} $$

And experimentally, know its values. From the theory, this rate goes to infinity if we are talking about a spinless constituent, and goes to zero (at high frequencies of the scattered photon) for a spin-half constituent. This is how we know that quarks are spin-half. (this answers 4)

[2] There is a excited state of the proton, called $\Lambda^{++}$, this particle is composed by 3 up quarks. As we know the principle of exclusion, we couldn't have 3 fermions in the same state unless there were a additional degree of freedom we are not taking into account.

This additional degree of freedom is the Colour charge. But the fact that we need 3 colours (3 kinds of charge) cames from the choice of the gauge group that describes the strong interactions, the SU(3).

And by this very choice, we have a non-abelian gauge symmetry wich means that our gauge bosons (in this case gluons) interact with each other, because in the non-abelian case:

$$ F_{\mu \nu} = {\partial}_{\mu} A_{\nu} - {\partial}_{\nu} A_{\mu} - iq[A_{\mu},A_{\nu}] $$

The last term (the commutator) of $A_{\mu}$ and $A_{\nu}$ does not vanish and so the Lorentz invariant term in the Lagrangian $ F_{\mu \nu}F^{\mu \nu} $ gives 3-field and 4-field interactions of the gauge bosons. (this answers 3 and 2)