How can I calculate the integral for 1/r grad 1/r?

In summary, the integral can be calculated using two methods - brute force and clever guessing. A "physics proof" derivation involves using the identity ##\nabla \frac{1}{r} = -4 \pi \delta^3(\mathbf y - \mathbf c)## and the spherical form of the divergence in c-space. The result is ##2\pi \mathbf{\hat{c}}##, which is independent of the distance between the two points.
  • #1
Jano L.
Gold Member
1,333
75
I would like to calculate the integral

\begin{equation}
\int \frac{1}{|\mathbf y|} \nabla \frac{1}{|\mathbf y -\mathbf c|}~d^3\mathbf y,
\end{equation}

but I do not know how. I was wondering whether somebody has encountered such integral already. Do you know how to calculate it? Do you know some resource where similar integrals are calculated or tabulated ?
 
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  • #2
There are basically two techniques for any integrals.
1. brute force and 2. clever guessing.
The integral you' posted can be solved with either method.

Since I'm too lazy to type a lot, I'll give you a slick
"physics proof" derivation. (which in retrospect still requires
too much typing!) This solution requires that you believe
the identity
$$
\nabla \frac{1}{r} = - 4 \pi \delta^3({\vec r}),
$$
familiar from electrostatics.

Take your integral which is a function of the vector c, and rewrite it.
$$
\vec{f}({\vec c}) = \int d^3y \frac{1}{y} \nabla \frac{1}{|{\vec y} - {\vec c}|}
= -\int d^3y \frac{1}{y} \nabla_c \frac{1}{|{\vec y} - {\vec c}|},
$$
where [itex]\nabla_c[/itex] has the meaning of differentiating as
if c were variable.

Now calculate the c-divergence of f. We find
$$ \nabla_c \cdot {\vec f}
= -\int d^3y \frac{1}{y} \nabla_c^2 \frac{1}{|{\vec y} - {\vec c}|}
= -\int d^3y \frac{1}{y} (-4 \pi \delta^3({\vec y} - {\vec c}))
= 4 \pi \frac{1}{c}.
$$

now from the symmetry of the problem we can argue that
$${\vec f}({\vec c}) = f_c({\vec c}) {\widehat c}$$

And thus we use the spherical form of the divergence (in c-space)
to find fc.
$$\frac{1}{c^2}\frac{\partial }{\partial c} (c^2 f_c) = 4 \pi \frac{1}{c}$$
and the solution with [itex]f_c[/itex] well defined as [itex]c\rightarrow 0[/itex] is
$$f_c = 2 \pi, $$
and so the original integral is
$$ \int d^3 y \frac{1}{y} \nabla \frac{1}{|{\vec y}- {\vec c}|}
= 2 \pi {\widehat c}
= \frac{2 \pi}{c} {\vec c}.$$
 
  • #3
That is very nice way to calculate it. Thank you. I might add that for ##\mathbf c = \mathbf 0##, the integral is zero.
The result is quite interesting, if we interpret it as scalar product of potential and electric field due to two point charges. It seems that this product is independent of the distance between the two.
 

1. What is "Integral 1/r grad 1/r"?

Integral 1/r grad 1/r is a mathematical formula used in physics and engineering to calculate the electric field created by a charged particle. It is also known as the Coulomb potential and is used to describe the behavior of electric fields and forces.

2. How is "Integral 1/r grad 1/r" calculated?

The Integral 1/r grad 1/r formula is calculated by taking the inverse square of the distance between two charged particles and multiplying it by the gradient of the charge. This equation is used to determine the strength and direction of the electric field at a specific point.

3. What is the significance of "Integral 1/r grad 1/r" in physics?

Integral 1/r grad 1/r is significant in physics because it helps us understand the behavior of electric fields and the forces that they create. It is an important tool for studying the behavior of charged particles and their interactions with each other.

4. Can "Integral 1/r grad 1/r" be used for other types of fields besides electric fields?

Yes, Integral 1/r grad 1/r can be used for other types of fields, such as gravitational fields. In this case, the formula is known as the Newtonian potential and is used to calculate the gravitational pull between two masses. However, the concept of the formula remains the same.

5. Are there any limitations to using "Integral 1/r grad 1/r" in calculations?

Yes, there are limitations to using Integral 1/r grad 1/r in calculations. It assumes that the charged particles are point charges and do not have any physical dimensions. It also does not take into account the effects of relativity. Therefore, it is only suitable for calculating electric fields and forces in simple and idealized situations.

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