Ryder's SU(2) Example in Quantum Field Theory

[ShayanJ]

In section 3.5 of his textbook Quantum Field Theory, Ryder discusses an example of a non-Abelian gauge theory. He considers a 3D internal space and rotations in this space.

At first he shows that the fields in this internal space transform like $\delta \vec \phi=-\vec \Lambda \times \vec \phi$ under a rotation $\vec \Lambda$ in the internal space. Then he shows that $\delta(\partial_\mu \vec \phi)=-\vec \Lambda \times (\partial_\mu \vec \phi)-(\partial_\mu \vec \Lambda)\times \vec \phi$.

He then introduces the covariant derivative $D_\mu \vec \phi=\partial_\mu \vec \phi+g\vec W_\mu \times \vec \phi$ and demands that $\delta(D_\mu \vec \phi)=-\vec \Lambda \times (D_\mu \vec \phi)$ which gives him the transformation rule for $\vec W_\mu$ which is $\delta \vec W_\mu=-\vec \Lambda \times \vec W_\mu+\frac 1 g \partial_\mu \vec \Lambda$.

Now here is where my confusion begins. He writes $\delta(D_\mu \vec \phi)=\delta(\partial_\mu \vec \phi)+g(\delta \vec W_\mu)\times \vec \phi+g\vec W_\mu \times (\delta \vec \phi)$ and then using the above expressions, he verifies that $\delta(D_\mu \vec \phi)=-\vec \Lambda \times (D_\mu \vec \phi)$.
But it seems to me that without using $\delta(D_\mu \vec \phi)=\delta(\partial_\mu \vec \phi)+g(\delta \vec W_\mu)\times \vec \phi+g\vec W_\mu \times (\delta \vec \phi)$, I still should be able to get the same conclusion with just replacing the transformed quantities in the expression for the covariant derivative, so I write:
$(D_\mu \vec \phi)_{transformed}=\partial_\mu \vec \phi-\vec \Lambda \times (\partial_\mu \vec \phi)-(\partial_\mu \vec \Lambda)\times \vec \phi+g\left( \vec W_\mu-\vec\Lambda\times \vec W_\mu+\frac 1 g \partial_\mu \vec \Lambda \right) \times \left( \vec\phi-\vec\Lambda\times \vec \phi \right)$.

But when I do the calculations, the result is far from what Ryder gets(which should already be obvious from my expression).

What's wrong with this method?
Thanks

 

[nrqed]

In section 3.5 of his textbook Quantum Field Theory, Ryder discusses an example of a non-Abelian gauge theory. He considers a 3D internal space and rotations in this space.

At first he shows that the fields in this internal space transform like $\delta \vec \phi=-\vec \Lambda \times \vec \phi$ under a rotation $\vec \Lambda$ in the internal space. Then he shows that $\delta(\partial_\mu \vec \phi)=-\vec \Lambda \times (\partial_\mu \vec \phi)-(\partial_\mu \vec \Lambda)\times \vec \phi$.

He then introduces the covariant derivative $D_\mu \vec \phi=\partial_\mu \vec \phi+g\vec W_\mu \times \vec \phi$ and demands that $\delta(D_\mu \vec \phi)=-\vec \Lambda \times (D_\mu \vec \phi)$ which gives him the transformation rule for $\vec W_\mu$ which is $\delta \vec W_\mu=-\vec \Lambda \times \vec W_\mu+\frac 1 g \partial_\mu \vec \Lambda$.

Now here is where my confusion begins. He writes $\delta(D_\mu \vec \phi)=\delta(\partial_\mu \vec \phi)+g(\delta \vec W_\mu)\times \vec \phi+g\vec W_\mu \times (\delta \vec \phi)$ and then using the above expressions, he verifies that $\delta(D_\mu \vec \phi)=-\vec \Lambda \times (D_\mu \vec \phi)$.
But it seems to me that without using $\delta(D_\mu \vec \phi)=\delta(\partial_\mu \vec \phi)+g(\delta \vec W_\mu)\times \vec \phi+g\vec W_\mu \times (\delta \vec \phi)$, I still should be able to get the same conclusion with just replacing the transformed quantities in the expression for the covariant derivative, so I write:
$(D_\mu \vec \phi)_{transformed}=\partial_\mu \vec \phi-\vec \Lambda \times (\partial_\mu \vec \phi)-(\partial_\mu \vec \Lambda)\times \vec \phi+g\left( \vec W_\mu-\vec\Lambda\times \vec W_\mu+\frac 1 g \partial_\mu \vec \Lambda \right) \times \left( \vec\phi-\vec\Lambda\times \vec \phi \right)$.

But when I do the calculations, the result is far from what Ryder gets(which should already be obvious from my expression).

What's wrong with this method?
Thanks

Nothing is wrong with your method if you keep in mind that we are supposed to work in first order of the transformation parameter $\Lambda$. If you drop all your terms with two powers of $\Lambda$ and then look at the difference between your expression and $D_\mu \phi$(because you want the variation $\delta D_\mu \phi$) then you should get his expression.