📂Mathematical Physics

Product Rule Involving the Del Operator

Formulas

Let’s call $f=f(x,y,z)$ a scalar function. Let’s call $\mathbf{A} = A_{x}\hat{\mathbf{x}} + A_{y}\hat{\mathbf{y}} + A_{z}\hat{\mathbf{z}}, \mathbf{B} = B_{x}\hat{\mathbf{x}} + B_{y}\hat{\mathbf{y}} + B_{z}\hat{\mathbf{z}}$ a vector function. Then, the following equations hold.

• Gradient

  (a) $\nabla{(fg)}=f\nabla{g}+g\nabla{f}$

  (b) $\nabla(\mathbf{A} \cdot \mathbf{B}) = \mathbf{A} \times (\nabla \times \mathbf{B}) + \mathbf{B} \times (\nabla \times \mathbf{A})+(\mathbf{A} \cdot \nabla)\mathbf{B}+(\mathbf{B} \cdot \nabla) \mathbf{A}$

• Divergence

  (c) $\nabla \cdot (f\mathbf{A}) = f(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla f)$

  (d) $\nabla \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B})$

• Curl

  (e) $\nabla \times (f\mathbf{A}) = (\nabla f) \times \mathbf{A} + f(\nabla \times \mathbf{A})$

  (f) $\nabla \times (\mathbf{A} \times \mathbf{B}) = (\mathbf{B} \cdot \nabla)\mathbf{A} - (\mathbf{A} \cdot \nabla)\mathbf{B} + \mathbf{A} (\nabla \cdot \mathbf{B}) - \mathbf{B} (\nabla \cdot \mathbf{A})$

Description

Throughout the proof, we use Einstein notation, so be careful not to get confused. That is, if the same index appears twice in an equation, it means the following.

$$x_{i}y_{i}=\sum \limits_{i=1}^{3} x_{i}y_{i}=x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3}$$

Also, you should be familiar with using Kronecker delta and Levi-Civita symbol and know the relationship between them to easily follow the proof.

Proof

(a)

Can be easily shown by the definition of gradient and the properties of differentiation.

$$\begin{align*} \nabla(fg) =&\ \dfrac{\partial (fg)}{\partial x} \hat{\mathbf{x}}+\dfrac{\partial (fg)}{\partial y} \hat{\mathbf{y}} +\dfrac{\partial (fg)}{\partial z} \hat{\mathbf{z}} \\ =&\ \left( g\dfrac{\partial f}{\partial x} + f\dfrac{\partial g}{\partial x} \right) \hat{\mathbf{x}} +\left( g\dfrac{\partial f}{\partial y} + f\dfrac{\partial g}{\partial y} \right) \hat{\mathbf{y}} + \left( g\dfrac{\partial f}{\partial z} + f\dfrac{\partial g}{\partial z} \right) \hat{\mathbf{z}} \\ =&\ g\left( \dfrac{\partial f}{\partial x}\hat{\mathbf{x}} +\dfrac{\partial f}{\partial y}\hat{\mathbf{y}} + \dfrac{\partial f}{\partial z}\hat{\mathbf{z}} \right) + f\left( \dfrac{\partial g}{\partial x}\hat{\mathbf{x}} +\dfrac{\partial g}{\partial y}\hat{\mathbf{y}} + \dfrac{\partial g}{\partial z}\hat{\mathbf{z}} \right) \\ =&\ g\nabla f+ f\nabla g \end{align*}$$

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(b)

Calculating the left side directly, we get the following.

$$\begin{align*} \nabla \left( \mathbf{A}\cdot \mathbf{B} \right) =&\ \frac{ \partial \left( \mathbf{A}\cdot \mathbf{B} \right)}{ \partial x_{1}}\mathbf{e}_{1}+\frac{ \partial \left( \mathbf{A}\cdot \mathbf{B} \right)}{ \partial x_{2}}\mathbf{e}_{2}+\frac{ \partial \left( \mathbf{A}\cdot \mathbf{B} \right)}{ \partial x_{3}}\mathbf{e}_{3} \\ =&\ \sum \limits_{i=1}^{3} \frac{ \partial \left( \mathbf{A}\cdot \mathbf{B} \right)}{ \partial x_{i}}\mathbf{e}_{i} \\ =&\ \sum \limits_{i=1}^{3} \frac{ \partial \left( \sum _{j=1}^{3}A_{j}B_{j} \right)}{ \partial x_{i}}\mathbf{e}_{i} \\ =&\ \sum \limits_{i=1}^{3}\sum \limits_{j=1}^{3} \frac{ \partial \left( A_{j}B_{j} \right)}{ \partial x_{i}}\mathbf{e}_{i} \end{align*}$$

Simplifying with Einstein notation, we get the following.

$$\nabla (\mathbf{A} \cdot \mathbf{B}) = \frac{\partial(A_{j}B_{j})}{\partial x_{i}}\mathbf{e}_{i}=\frac{\partial A_{j}}{\partial x_{i}} B_{j}\mathbf{e}_{i}+A_{j} \frac{\partial B_{j}}{\partial x_{i}} \mathbf{e}_{i}$$

Then, using the Kronecker delta, the equation can be expressed as follows.

$$\begin{align*} &&\frac{\partial A_{j}}{\partial x_{i}} B_{j}\mathbf{e}_{i}+A_{j} \frac{\partial B_{j}}{\partial x_{i}} \mathbf{e}_{i} =&\ {\color{blue}\delta_{jm}}\frac{ \partial {\color{blue}A_{j}}}{ \partial x_{i}} {\color{blue}B_{m}} \mathbf{e}_{i} + {\color{blue}\delta_{jm} A_{m}}\frac{ \partial {\color{blue}B_{j}} }{ \partial x_{i} }\mathbf{e}_{i} \\ && =&{\color{red}\delta_{il}}{\color{blue}\delta_{jm}}\frac{ {\color{red}\partial} {\color{blue}A_{j}}}{ {\color{red}\partial x_{i}} } {\color{blue}B_{m}} {\color{red}\mathbf{e}_{l}} + {\color{red}\delta_{il}}{\color{blue}\delta_{jm} A_{m}} {\color{red}\frac{ \partial {\color{blue}B_{j}} }{ \partial x_{i} }} {\color{red}\mathbf{e}_{l}} \\ && =&{\color{red}\delta_{jl}}{\color{blue}\delta_{jm}} \left( {\color{red}\frac{ \partial {\color{blue}A_{j}}}{ \partial x_{i} }} {\color{blue}B_{m}}\color{red} {\mathbf{e}_{l}} + {\color{blue}A_{m}}{\color{red}\frac{ \partial {\color{blue}B_{j}} }{ \partial x_{i} } \mathbf{e}_{l} }\right) \\ \implies && \nabla \left( \mathbf{A} \cdot \mathbf{B} \right) =&\ \delta_{jl}\delta_{jm} \left( \frac{ \partial A_{j} }{ \partial x_{i} } B_{m} \mathbf{e}_{l} + A_{m}\frac{ \partial B_{j} }{ \partial x_{i} } \mathbf{e}_{l} \right) \end{align*}$$

Furthermore, since $\epsilon_{ijk} \epsilon_{klm} = \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}$, we can expand the equation as follows.

$$\begin{align*} \nabla(\mathbf{A} \cdot \mathbf{B}) =&\ (\epsilon_{ijk} \epsilon_{klm} + \delta_{im} \delta_{jl}) \left(\frac {\partial A_{j}}{\partial x_{i}}B_{m} \mathbf{e}_{l} + A_{m} \frac{\partial B_{j}}{\partial x_{i}}\mathbf{e}_{l}\right) \\ =&\ \epsilon_{ijk} \epsilon_{klm } \frac{\partial A_{j}}{\partial x_{i}}B_{m} \mathbf{e}_{l} + \epsilon_{ijk} \epsilon_{klm} A_{m} \frac{\partial B_{j}}{\partial x_{i}} \mathbf{e}_{l} + \delta_{im} \delta_{jl} \frac{\partial A_{j}}{\partial x_{i}} B_{m} \mathbf{e}_{l} + \delta_{im} \delta_{jl}A_{m} \frac{\partial B_{j}}{\partial x_{i}} \mathbf{e}_{l} \end{align*}$$

Here, by the definition of the Levi-Civita symbol, since $\epsilon_{ijk} \dfrac{\partial A_{j}}{\partial x_{i}}=(\nabla \times \mathbf{A})_{k}$ and $\epsilon_{ijk} \dfrac {\partial B_{j}}{\partial x_{i}}=(\nabla \times \mathbf{B})_{k}$, we obtain the following result.

$$\begin{align*} \nabla (\mathbf{A} \cdot \mathbf{B}) =&\ \epsilon _{ klm }(\nabla \times \mathbf{A})_{ k }B_{ m }\hat { \mathbf{e}_{ l } }+\epsilon _{ klm }A_{ m }(\nabla \times \mathbf{B})_{ k }\hat { \mathbf{e}_{ l } }+\frac { \partial A_{ j } }{ \partial x_{ i } }B_{ i }\hat { e_{ j} }+A_{ i }\frac { \partial B_{ j } }{ \partial x_{ i } }\hat { \mathbf{e}_{ j } } \\ =&\ \mathbf{B}\times (\nabla \times \mathbf{A})+\mathbf{A} \times (\nabla \times \mathbf{B})+(\mathbf{B} \cdot \nabla )\mathbf{A}+(\mathbf{A} \cdot \nabla )\mathbf{B} \end{align*}$$

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(c)​

$$\begin{align*} \nabla \cdot (f \mathbf{A}) =&\ \delta _{ ij }\nabla _{ i }(fA_{ j }) \\ =&\ \delta _{ ij }(\nabla _{ i }f)A_{ j }+\delta _{ ij }f(\nabla _{ i }A_{ j }) \\ =&\ (\nabla _{ i }f)A_{ i }+f(\nabla _{ i }A_{ i }) \\ =&\ (\nabla f)\cdot \mathbf{A}+f(\nabla \cdot \mathbf{A}) \end{align*}$$

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(d)

$$\begin{align*} \nabla \cdot (\mathbf{A} \times \mathbf{B}) &= \delta _{ ij }\nabla _{ i }(\mathbf{A} \times \mathbf{B})_{ j } \\ &= \delta _{ ij }\nabla _{ i }(\epsilon _{ jkl }A_{ k }B_{ l }) \\ &= \delta _{ ij }\epsilon _{ jkl }\nabla _{ i }(A_{ k }B_{ l }) \\ &= \delta _{ ij }\epsilon _{ jkl }(\nabla _{ i }A_{ k })B_{ l }+\delta _{ ij }\epsilon _{ jkl }A_{ k }(\nabla _{ i }B_{ l }) \\ &= (\epsilon _{ jkl }\nabla _{ j }A_{ k })B_{ l }+A_{ k }(\epsilon _{ jkl }\nabla _{ j }B_{ l }) \\ &= (\nabla \times A)_{ l }B_{ l }-A_{ k }(\nabla \times B)_{ k } \\ &= (\nabla \times \mathbf{A})\cdot \mathbf{B}-\mathbf{A}\cdot (\nabla \times \mathbf{B}) \end{align*}$$

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(e)​

$$\begin{align*} \nabla \times (f\mathbf{A}) &= \epsilon _{ ijk }\nabla _{ i }(fA_{ j })\mathbf{e}_{k} \\ &= \epsilon _{ ijk }(\nabla _{ i }f)A_{ j }\mathbf{e}_{k}+\epsilon _{ ijk }f(\nabla _{ i }A_{ j })\mathbf{e}_{k} \\ &= (\nabla f)\times \mathbf{A}+f\epsilon _{ ijk }(\nabla _{ i }A_{ j })\mathbf{e}_{k} \\ &= (\nabla f)\times \mathbf{A}+f(\nabla \times \mathbf{A}) \\ &= f(\nabla \times \mathbf{A})-\mathbf{A} \times (\nabla f) \end{align*}$$

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(f)

If you’re not familiar with Einstein notation, it may be difficult to follow the proof.

$$\begin{align*} & \nabla \times (\mathbf{A}\times \mathbf{B}) \\ =&\ \epsilon_{ijk} \nabla_{i} \left(\mathbf{A}\times \mathbf{B}\right)_{j} \mathbf{e}_{k} \\ =&\ \epsilon_{ijk} \nabla_{i} (\epsilon_{jlm} A_{l} B_{m})\mathbf{e}_{k} \\ =&\ \epsilon_{ijk} \epsilon_{jlm} \nabla_{i} (A_{l} B_{m}) \mathbf{e}_{k} \\ =&\ \epsilon_{jki} \epsilon_{jlm} \left[ B_{m}(\nabla_{i} A_{l}) \mathbf{e}_{k} + A_{l} (\nabla_{i} B_{m}) \mathbf{e}_{k} \right] \\ =&\ (\delta_{kl} \delta_{im} - \delta_{km} \delta_{il} ) [ B_{m} (\nabla_{i} A_{l} ) \mathbf{e}_{k} + A_{l} ( \nabla_{i} B_{m} ) \mathbf{e}_{k} ] \\ =&\ \delta_{kl} \delta_{im} B_{m} ( \nabla_{i} A_{l}) \mathbf{e}_{k} - \delta_{km} \delta_{il} B_{m} (\nabla_{ i} A_{l} ) \mathbf{e}_{k} + \delta_{kl} \delta_{im} A_{l} ( \nabla_{i} B_{m} ) \mathbf{e}_{k} - \delta_{km} \delta_{il} A_{l} ( \nabla_{i} B_{m} ) \mathbf{e}_{k} \\ =&\ B_{i} ( \nabla_{i} A_{k}) \mathbf{e}_{k} - B_{k} (\nabla_{i} A_{i} ) \mathbf{e}_{k} + A_{k} ( \nabla_{i} B_{i} ) \mathbf{e}_{k} - A_{i} ( \nabla_{i} B_{k} ) \mathbf{e}_{k} \\ =&\ (\mathbf{B}\cdot \nabla )\mathbf{A}-(\nabla \cdot \mathbf{A})\mathbf{B}+\mathbf{A}(\nabla \cdot \mathbf{B})-(\mathbf{A}\cdot \nabla )\mathbf{B} \\ =&\ (\mathbf{B}\cdot \nabla )\mathbf{A}-(\mathbf{A}\cdot \nabla )\mathbf{B}+\mathbf{A}(\nabla \cdot \mathbf{B})-\mathbf{B}(\nabla \cdot \mathbf{A}) \end{align*}$$

The fourth line holds due to $\epsilon_{jki} \epsilon_{jlm} = \delta_{kl} \delta_{im} - \delta_{km} \delta_{il}$. The seventh line holds due to Einstein Notation.

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