In ##\nabla\cdot\vec{E}## why can ##\nabla## pass through the integral?

[zenterix]

TL;DR Summary

Suppose we wish to calculate $\nabla\cdot\vec{E}$ directly (ie, not by the use of the divergence theorem) when we have a continuous source charge.

We have a dot product between $\nabla$ and the integral of a vector. Why can $\nabla$ pass into the integral?

We have

$$\vec{E}(\vec{r})=\frac{1}{4\pi\epsilon_0}\int_V\frac{\rho(\vec{r}')}{\eta^2}\hat{\eta}d\tau'\tag{1}$$

A few initial observations
1) I am using notation from the book Introduction to Electrodynamics by Griffiths. When considering point charges, this notation uses position vectors $\vec{r}$ and $\vec{r}'$ for field and source charges, respectively, and a separation vector $\vec{\eta}=\vec{r}-\vec{r}'$.

2) In the case I am considering we have a continuous source charge distribution and a fixed field point at which we are computing the electric field due to the source.

3) $d\tau'$ is an infinitesimal volume element. Note that in (1), $\vec{r}$ is the fixed field point at which we are computing the electric field. We are integrating over the source charge and $\vec{r}'$ is the position of $d\tau'$.

4) $\rho(\vec{r}')'d\tau'$ is $dq'$, an infinitesimal charge on the continuous source charge.

My question boils down to: in the dot product $\nabla\cdot\vec{E}$ why can $\nabla$ pass through the integral such that we can write the following?

$$\nabla\cdot\vec{E}=\frac{1}{4\pi\epsilon_0}\int_V \nabla\cdot\left ( \frac{\hat{\eta}}{\eta^2}\right )\rho(\vec{r}')d\tau'\tag{2}$$

 

[Charles Link]

Because the integration is over primed coordinates, and the divergence operator here uses unprimed coordinates in its derivative.

 

[zenterix]

Because the integration is over primed coordinates, and the divergence operator here uses unprimed coordinates in its derivative.

I was just about to amend my question with that interpretation. Though it is simple to state that, I find it quite difficult to grasp at this point.

In $\nabla\cdot\left ( \frac{\hat{\eta}}{\eta^2}\right )$, we are calculating the divergence of a vector function

$$\vec{v}(\vec{r})=\frac{\hat{\eta}}{\eta^2}=\frac{\vec{r}-\vec{r}'}{\lVert \vec{r}-\vec{r}'\rVert^3}$$

This is a function of $\vec{r}$ with $\vec{r}'$ fixed.

So it seems that as far as the $\nabla$ operator is concerned $\frac{\hat{\eta}}{\eta^2}\rho(\vec{r}')$ is a function of $\vec{r}$ with $\vec{r}'$ fixed.

Intuitively, the integral is a summation of infinite terms, each of which is this vector function evaluated at a specific value of $\vec{r}'$ (ie the values at each position in the volume being considered).

Since dot product has the distributive property, then when we apply $\nabla$ to this summation via dot product it distributes over the vector functions in the summation. This is equivalent to applying $\nabla$ to the function and then evaluating it at each of the values of $\vec{r}'$.

After writing it out like this it makes sense to me. Is this correct?

 

[Charles Link]

I think it makes sense. The way I would explain it is that you can first do the sum (integration) over all the primes and take the unprimed derivative after you sum everything, or you can take the unprimed derivative on each term of the sum, and then sum them.

 

[Charles Link]

Perhaps a better explanation than what I just gave you is you can do the integration first for $r= r_o$ and then for $r=r_o+\Delta r$, and then take the difference,(derivative operation), or you can put in the $r_o$ and $r_o+\Delta r$ and take the difference of the terms inside the integral before you do the integration.

 

[vanhees71]

I was just about to amend my question with that interpretation. Though it is simple to state that, I find it quite difficult to grasp at this point.

In $\nabla\cdot\left ( \frac{\hat{\eta}}{\eta^2}\right )$, we are calculating the divergence of a vector function

$$\vec{v}(\vec{r})=\frac{\hat{\eta}}{\eta^2}=\frac{\vec{r}-\vec{r}'}{\lVert \vec{r}-\vec{r}'\rVert^3}$$

This is a function of $\vec{r}$ with $\vec{r}'$ fixed.

So it seems that as far as the $\nabla$ operator is concerned $\frac{\hat{\eta}}{\eta^2}\rho(\vec{r}')$ is a function of $\vec{r}$ with $\vec{r}'$ fixed.

Intuitively, the integral is a summation of infinite terms, each of which is this vector function evaluated at a specific value of $\vec{r}'$ (ie the values at each position in the volume being considered).

Since dot product has the distributive property, then when we apply $\nabla$ to this summation via dot product it distributes over the vector functions in the summation. This is equivalent to applying $\nabla$ to the function and then evaluating it at each of the values of $\vec{r}'$.

After writing it out like this it makes sense to me. Is this correct?

Indeed the expression $\vec{\nabla} \cdot \vec{v}(\vec{r},\vec{r}')$ means to take the partial derivatives with respect to the components $(x_1,x_2,x_3)$ of $\vec{r}$ considering the components $(x_1',x_2',x_3')$ of $\vec{r}'$ as constant.

If you work in Cartesian coordinates for $\vec{r} \neq \vec{r}'$ you get
$$\vec{\nabla} \cdot \frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^3}=\frac{1}{|\vec{r}-\vec{r}'|^3} \vec{\nabla} \cdot (\vec{r}-\vec{r}') + (\vec{r}-\vec{r}') \cdot \vec{\nabla} \frac{1}{|\vec{r}-\vec{r}'|^3}.$$
Now
$$\vec{\nabla} \cdot (\vec{r}-\vec{r}')=3$$
and
$$\vec{\nabla} \frac{1}{|\vec{r}-\vec{r}'|^3} = -\frac{3}{|\vec{r}-\vec{r}'|^4} \vec{\nabla} |\vec{r}-\vec{r}'|= -\frac{3 (\vec{r}-\vec{r}')}{|\vec{r}-\vec{r}'|^5}.$$
Putting everything together you get
$$\vec{\nabla} \frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^3}=0.$$
That's to be expected since
$$\vec{E}(\vec{r})=\frac{Q}{4 \pi \epsilon_0} \frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|^3}$$
is the electric field of a point charge at $\vec{r}'$. The charge density is thus 0 everywhere, except at $\vec{r}=\vec{r}'$, where it is undefined or rather infinite.

In the sense of distributions a point charge at $\vec{r}'$ is represented by the charge density
$$\rho(\vec{r})=Q \delta^{(3)}(\vec{r}-\vec{r}'),$$
and in this same sense you thus find
$$\vec{\nabla} \cdot \vec{E} = \frac{Q}{4 \pi \epsilon_0} \vec{\nabla} \cdot \frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|}=\frac{Q}{\epsilon_0} \rho(\vec{r}) \delta^{(3)}(\vec{r}-\vec{r}'),$$
i.e.,
$$\vec{\nabla} \cdot \frac{\vec{r}-\vec{r}'}{|\vec{r}-\vec{r}'|}=4 \pi \delta^{(3)}(\vec{r}-\vec{r}').$$
Now you can also express everything in terms of the electrostatic potential,
$$\Phi(\vec{r})=\frac{Q}{4 \pi \epsilon_0 |\vec{r}-\vec{r}'|}.$$
Since $\vec{E}=-\vec{\nabla} \Phi$ you get
$$-\Delta \frac{1}{|\vec{r}-\vec{r}'|} =4 \pi \delta^{(3)}(\vec{r}-\vec{r}'),$$
i.e.,
$$G(\vec{r},\vec{r}')=\frac{1}{4 \pi |\vec{r}-\vec{r}'|}$$
is the Green's function of $-\Delta$, i.e., you can write the solution for
$$-\Delta \Phi(\vec{r}) = \vec{\nabla} \cdot \vec{E}=\frac{1}{\epsilon_0} \rho$$
as
$$\Phi(\vec{r})=\int_{\mathbb{R}^3} \mathrm{d}^3 r' G(\vec{r},\vec{r}') = \int_{\mathbb{R}^3} \mathrm{d}^3 r' \frac{\rho(\vec{r}')}{4 \pi \epsilon_0 |\vec{r}-\vec{r}'|}$$
and then you get again
$$\vec{E}(\vec{r}) = -\nabla \Phi(\vec{r}) = -\int_{\mathbb{R}^3} \mathrm{d}^3 r' \vec{\nabla}\frac{\rho(\vec{r}')}{4 \pi \epsilon_0 |\vec{r}-\vec{r}'|} = \int_{\mathbb{R}^3} \mathrm{d}^3 r' \frac{\rho(\vec{r}') (\vec{r}-\vec{r}')}{|\vec{r}-\vec{r}'|^3}.$$

 

[PeroK]

I was just about to amend my question with that interpretation. Though it is simple to state that, I find it quite difficult to grasp at this point.

In $\nabla\cdot\left ( \frac{\hat{\eta}}{\eta^2}\right )$, we are calculating the divergence of a vector function

$$\vec{v}(\vec{r})=\frac{\hat{\eta}}{\eta^2}=\frac{\vec{r}-\vec{r}'}{\lVert \vec{r}-\vec{r}'\rVert^3}$$

This is a function of $\vec{r}$ with $\vec{r}'$ fixed.

So it seems that as far as the $\nabla$ operator is concerned $\frac{\hat{\eta}}{\eta^2}\rho(\vec{r}')$ is a function of $\vec{r}$ with $\vec{r}'$ fixed.

Intuitively, the integral is a summation of infinite terms, each of which is this vector function evaluated at a specific value of $\vec{r}'$ (ie the values at each position in the volume being considered).

Since dot product has the distributive property, then when we apply $\nabla$ to this summation via dot product it distributes over the vector functions in the summation. This is equivalent to applying $\nabla$ to the function and then evaluating it at each of the values of $\vec{r}'$.

After writing it out like this it makes sense to me. Is this correct?

If you replace the integral with a finite sum, then you have the additive property of the derivative for a sum of functions of $r$. In this case $r'$ would be replaced by an integer index over a finite range.

The question is whether this simple process can be extended to the case where the finite sum is replaced by an integral, which is the limit of a sequence of finite sums - with $r'$ replacing the integer dummy index. For appropriately well-behaved functions the answer is yes. It's also possible to find more exotic functions where the order of the derivative and the integral cannot be interchanged.

 

[fresh_42]

For appropriately well-behaved functions the answer is yes. It's also possible to finite more exotic functions where the order of the derivative and the integral cannot be interchanged.

I think it needs an extra argument at $r'=r.$ The "Feynman trick" (exchanging differentiation and integration) works fine if the integrand is continuously differentiable along the differentiating variable. If not, then there are counterexamples:
https://www.physicsforums.com/insights/the-art-of-integration/#The-Feynman-Trick-–-Parameter-IntegralsBut that's only the mathematical point of view. I was always wondering how physicists deal with the singularity at the center of such potentials, be it electromagnetic or gravitational.

 

[vanhees71]

As usual, physicists don't bother much and just integrate. In this case they are lucky enough that it works ;-). It's amazing how far you get with such a sloppy attitude, but sometimes you get into trouble. The examples that come to my mind are rather in quantum mechanics, where the domain of self-adjoint operators is usually not considered. Also here you get amazingly far. One of my math professors said that's, because the (separable) Hilbert space is almost as goodmannered as finite-dimensional vector spaces ;-)).