# What does it mean if something Transforms Covariantly ? (Quantum Field Theory)

1. Oct 31, 2011

### Elwin.Martin

What does it mean if something "Transforms Covariantly"? (Quantum Field Theory)

Referring to an isospinor, $\psi$ which transforms as $\psi$(x)→ $\psi$'(x)=S(x) $\psi$(x) (S(x) being an n by n matrix)

I'm told that it is clear that $∂_{μ}$$\psi$ does not transform covariantly.

Now, correct me if I'm wrong, but it would appear that $∂_{μ}$$\psi$ ' can be found by the product rule to be S($∂_{μ}$$\psi$) + ($∂_{μ}$S)$\psi$.

What is meant by that it doesn't transform covariantly?

I know what covariant vs. contravariant indices are, but I don't know what it means for something to transform covariantly. I understand that we *want* a covariant derivative, but I don't understand why =|

Any and all help would be great!

**from Ryder, in case anyone was wondering

2. Oct 31, 2011

### dextercioby

Re: What does it mean if something "Transforms Covariantly"? (Quantum Field Theory)

I don't a copy of Ryder handy, however I can tell you that $\partial_{\mu}\psi$ does not transform covariantly with respect S(x), because you have another term - $(\partial_{\mu} S)\psi$ - which spoils covariance. If the gradient of psi had transformed covariantly, this extra term would not have been there. Covariant comes from co-variant, in other words, the derivative of the spinor should have transformed like the spinor, i.e. with an identical matrix.