Deriving product rule for divergence of a product of scalar and vector function in tensor notation

Question

On page-94 of the 4th edition in the international version of Griffith's Electrodynamic, the following identity is used:

$$ \int \left[ V(\nabla \cdot \vec{E} ) + \vec{E} \cdot \nabla V \right]dV= \oint V \vec{E} \cdot dA$$

Where, $ \vec{E}$ is a vector function and $V$ is a scalar function.

My goal is to prove the above identity using tensor calculus notation.

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The equation reminded me a bit of how the covariant derivative acted in Tensor calculus, so I tried that:

$$ \nabla_i (VE^i) = E^i \nabla_i (V) + V \nabla_i (E^i) \tag{1}$$

This equation in vector in vector notation is:

$$ \nabla \cdot( VE) = \text{?}+ V \nabla \cdot \vec{E}$$

Now, I can't figure out how to get the dot product in the $E^i \nabla_i (V)$ term

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Related post: Has the worked out proof of the above identity, but I can't seem to figure how to simplify equation (2) from it

Answer 1

Seems you're getting tangled up in notation. Your integrand, in a Euclidean frame at least and with and a slight change of notation, is \begin{equation*} \sum_i\ \biggl( v \frac{\partial E^i}{\partial x^i} + E^i\frac{\partial v}{\partial x^i} \biggr). \end{equation*} The term in parentheses is $\partial(v E^i)/\partial x^i$, the divergence of $vE$. Apply the divergence theorem for the integral.

Answer 2

OK, I finally figured it out. The thing is that the gradient is naturally identified with a convector, hence the product is the dot product, here it is explicitly:

$$ \vec{E} \cdot \nabla V=(E^i e_i) \cdot (\nabla_j V e^j) = E^i \nabla_j V \delta_i^j= E^i \nabla_i V$$

Hence, the result.

Answer 3

Also in differential forms, where $\tilde{E}$ is the electric field two-form:

$$ \int_{\partial \omega} V \tilde{E}= \int_{\omega}d (V \tilde{E}) = \int_{ \omega} dV \wedge \tilde{E} + V d \tilde{E} $$

Now, we use the identites:

$$ V d \tilde{E} = V (\nabla \cdot E) dx_1 \wedge dx_2 \wedge dx_3$$

$$ dV \wedge \tilde{E} = -|E|^2 dx_1 \wedge dx_2 \wedge dx_3$$

Because $dV = - \star E$. We recover the required.