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On p.247 of the textbook, the author claimed we have to make some correction to move the gradient operator outside the bracket.

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How does the first equation (6.53) happen?

$$ [\nabla' \rho]_{ret} = \nabla'[\rho]_{ret} - \left[ \frac{\partial \rho}{\partial t'} \right]_{ret} \nabla'(t - R/c)~?\tag{6.53} $$

I saw there was a related Phys.SE answer. But I don't know why this equation provided by the answer is valid.

$$ \left(\sum_i \frac{\partial }{\partial x_i'} \hat{i}\right)[\rho(x_i',x_j',x_k',t')]_{ret} = \left \lbrace \sum_i \left(\frac{\partial x_i'}{\partial x_i'}\frac{\partial }{\partial x_i'} + \frac{\partial x_j'}{\partial x_i'}\frac{\partial }{\partial x_j'} + \frac{\partial x_k'}{\partial x_i'}\frac{\partial }{\partial x_k'} + \frac{\partial t'}{\partial x_i'}\frac{\partial }{\partial t'}\right) \hat{i}\right\rbrace[\rho(x_i',x_j',x_k',t')]_{ret}. $$

Could you give me a step-by-step derivation?

Qmechanic
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IvanaGyro
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1 Answers1

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It's just the chain rule. If $\mathbf{g}$ and $h$ are functions of $\mathbf{r}$ and $t$, and the function $\rho(\mathbf{r}, t)$ means $f(\mathbf{g}(\mathbf{r}, t), h(\mathbf{r}, t)$), then

\begin{equation} \partial_i \rho = (\nabla f)(\mathbf{g}, h) \cdot \partial_i \mathbf{g} + (\partial_t f)(\mathbf{g}, h) \, \partial_i h \end{equation}

You can use this to get any one component of $\nabla' \rho(\mathbf{r}', t - |\mathbf{r} - \mathbf{r}'|)$ and then combine the three components into a single vector expression.

Brian Bi
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