standard model - Do color-neutral gluons exist?

If I'm correct a quark can change color by emitting a gluon. For example a blue up quark $u_b$ can change into a red up quark by emitting a gluon: $$u_b \longrightarrow u_r + g_{b\overline{r}}$$ (Here, the subscript indicates color and $\overline{r}$ means anti-red). This is needed to keep the color balance (left hand side: $b$, right hand side $r+b+\overline{r}=b$).

My question is then, do color-neutral gluons exist? Eg a gluon that is blue-anti-blue? If it would, it could be created by any quark then: $$u_b \longrightarrow u_b + g_{b\overline{b}}$$

I'm learning about the Standard Model in school, but the text isn't always that clear.

Answer

Color-neutral gluons that have the component blue-antiblue do exist, much like red-antired and green-antigreen. However, the sum of these three possible kinds of gluons is unphysical, so there are only two "diagonal" types of gluons. None of these two types of gluons are "genuinely color-blind" or "completely color-neutral".

This is more manifest if you realize that the color dependence of the gluon field may be written as a traceless $3\times 3$ Hermitian matrix. It is traceless because the gauge group is $SU(3)$ rather than $U(3)$ whose dimension is 8 rather than 9. (There are 3 complex entries strictly above the diagonal, which are copied in the complex conjugate way beneath the diagonal, plus 2 or 3 real entries on the diagonal, depending on whether we require the trace to vanish.)

Completely color-neutral gluons, if they were added, would be proportional to the identity matrix and they would couple to all three colors of the quarks equally. In other words, the interactions mediated by such gluons would only depend on the baryon number of the quarks. Experimentally, this interaction doesn't exist. In beyond-the-Standard-Model physics, one may try to extend $SU(3)$ to $U(3)$ in this way (this is very common in braneworld models) but because no new baryon-charge long-distance interaction is seen, the $U(1)$ in the $U(3)$ has to be spontaneously broken at a pretty high energy scale.