How to take time derivative of gradient in deriving the retarded scalar potential?

Question

In Griffiths E&M 4th edition, equation 10.28 reads $$\nabla \rho = \dot{\rho}\nabla t_r$$ This is in the context of computing the gradient of the retarded potential $$V(\vec{r'},tr)=\frac{1}{4\pi \epsilon_0}\int \frac{\rho(\vec{r'},t_r)}{\mathit{r}_1}d\tau'$$ where $$t_r \equiv t-\frac{r}{c}\\ \mathit{r}_1 \equiv |\vec{r}-\vec{r'}|.$$

The step is clear, using the product rule for the gradient $$\nabla V= \frac{1}{4\pi \epsilon_0}\int\left[\left(\nabla \rho \right )\frac{1}{\mathit{r}_1} + \rho \nabla \left(\frac{1}{\mathit{r}_1} \right ) \right ]d\tau'.$$

Then the problem I am having is understanding the next step. Without explanation or reference, states that $$\nabla \rho = \dot{\rho}\nabla t_r.$$

I can follow the rest of it, but I don't see how he goes from $\nabla \rho(r,t_r)$ to $\dot{\rho}\nabla t_r$.

It seems like a small matter in such a complex subject, but I could use some help explaining this. BTW I get that $\dot{\rho}$ is the time derivative of $\rho$.

Answer 1

I had a brief look at Griffiths. Not sure what $\int\dots d\tau'$ is, but it looks like integration with respect to $\mathbf{r}'$, i.e. the position of the source ($\int\dots d\tau'=\int\dots d^3r'$). Which would make sense.

So then you have:

$$ \boldsymbol{\nabla}\int d^3 r' \dots\rho\left(\mathbf{r}',\,t_r \right)=\int d^3 r' \dots \boldsymbol{\nabla}\rho\left(\mathbf{r}',\,t_r \right) $$

Now you have to understand $\boldsymbol{\nabla}$ is - it is derivative with respect to $\mathbf{r}$ - the position of the observer. That position is independent from what of the source, i.e. $\boldsymbol{\nabla}r'_i=0$ for all $i$. The only way $\rho\left(\mathbf{r}',\,t_r \right)$ depends on the position of the observer is through the retarded time. Hence, by chain rule:

$$ \boldsymbol{\nabla}\rho\left(\mathbf{r}',\,t_r \right)=\frac{\partial \rho\left(\mathbf{r}',\,t\right)}{\partial t}\Bigg|_{t=t_r}\cdot\boldsymbol{\nabla}t_r $$

Finally, Griffiths uses notation:

$$ \frac{\partial \rho\left(\mathbf{r}',\,t\right)}{\partial t}\Bigg|_{t=t_r}=\dot{\rho}\left(\mathbf{r}',\,t_r\right) $$