Identity, vector product and gradient

[Telemachus]

Hi there. I was following a deduction on continuum mechanics for the invariant nature of the first two laws of thermodynamics. The thing is that this deduction works with an identity, and there is something I'm missing to get it.

I have the vector product: ##\vec \omega \times grad \theta##, where θ is a scalar field, ω is a vector field, and ##grad \vec \omega=0##

Now, the book uses this identity: ##\vec \omega \times grad \theta=-curl (\theta \vec \omega)##

The thing is that I've tried to demonstrate the identity, but I couldn't, I get an extra term, and I don't see why it should be zero.

I have that ##\displaystyle curl (\theta \vec \omega)_i=\varepsilon_{ijk}\frac{\partial (\theta \omega_k)}{\partial x_j}=\varepsilon_{ijk} \omega_k \frac{\partial \theta}{\partial x_j}+\varepsilon_{ijk} \theta \frac{\partial \omega_k}{\partial x_j}\rightarrow \displaystyle curl (\theta \vec \omega)=-\vec \omega \times grad \theta+ \theta curl \vec \omega##

So, to get the identity that the book uses I should have that ##\theta curl \vec \omega=0##, and I don't see why that has to be accomplished.

 

[Fredrik]

Doesn't the assumption that ##\operatorname{grad} \vec\omega=0## mean that all partial derivatives of all components of ##\vec\omega## are zero? That would make your second term zero.

 

[Telemachus]

Yes, I think you are right. I wasn't sure about that the null gradient would imply a null curl, but now I think it does.

Actually, if ##\displaystyle grad \vec \omega=0 \Rightarrow \frac{\partial \omega_k}{\partial x_j}=0 \Rightarrow \varepsilon_{ijk}\frac{\partial \omega_k}{\partial x_j}=0=curl \vec \omega##

It was silly :p