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

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.

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

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

• The thing in square brackets is $\text{div}(V\,\vec{E})$. So, apply the divergence theorem. WHat you wrote as $\vec{E}\nabla V$ in the last line is wrong, it should just be $\vec{E}\cdot \nabla V$ (the standard inner product of $\vec{E}$ with $\nabla V$). Commented May 19, 2021 at 22:48
• Far as I understand the left side of equal is a scalar while the right side is a vector. Are you sure the formula is right? Commented May 19, 2021 at 22:51
• Oh right, but how did you translate $E^i \nabla_i (V)$ into $\vec{E} \cdot \nabla V$, I don't see the contraction for divergence Commented May 19, 2021 at 22:53
• can you write out the definition of $E^i$, $\nabla_iV$ and of the standard inner product on the LHS? btw the proof of this vector identity is given in Griffiths (see the product rule section in the beginning) Commented May 19, 2021 at 22:57
• Hm so $\nabla_i V=- Z^{ik} E^i e_k$, I am not sure what you mean by the definition of $E^i$ .. I guess I would define it as the component of $E$ w.r.t some basis. I don't see a contraction with the covariant metric tensor here for a dot product Commented May 19, 2021 at 23:00

## 3 Answers

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.

• Okie, this helps but I'm still stuck on how to do write it in a general coordinate system Commented May 20, 2021 at 8:16

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.

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.