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Explain $-\int_{b}^{a}\frac{Q}{4 \pi \epsilon \bar{r}^{2}} \cdot d \bar{r}= \left[\frac{Q}{4\pi \epsilon \bar{r}}\right]_{b}^{a}$. What is the term $\frac{1}{\bar{r}^{2}} \cdot d\bar{r}$? I find it hard to think about it.

$r$ is a radius of a ball.

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It is helpful to provide some context for your question. What is $r$? Is it a position vector? Probably. By staring at the integral, maybe you recognize a similarity: $\int \frac{1}{r^2}=-\frac{1}{r}+C$ – Fredrik Meyer Mar 1 '11 at 2:14
I think this is Coulombs Law. – PEV Mar 1 '11 at 2:18
@Fredrik Meyer: yes but there is the confusing dot-product, it is hard for me to think about it. – user7628 Mar 1 '11 at 2:21
This is not a dot product -- the quantity to the left of the dot is a scalar, not a vector, and the quantity on the right-hand side of the equation is also a scalar, so the dot in this case (as far as it should be interpreted as symbolizing multiplication at all, see my answer below) stands for ordinary multiplication between scalars, i.e. the $\bar{r}$ in $\mathrm{d}\bar{r}$ must be intended to mean the scalar radius or the magnitude of a vector, not a vector itself. – joriki Mar 1 '11 at 2:27
up vote 4 down vote accepted

In rigorous mathematics, one does not usually write this as you did with a multiplication dot, as the symbol $\mathrm{d}\bar{r}$ is considered as a formal part of the notation of the integral (indicating the variable of integration) and not as a factor to be multiplied. However, this integral comes from physics, and in physics it's quite usual to treat expressions like $\mathrm{d}\bar{r}$ as if they were "infinitesimal quantities" which can be manipulated algebraically much like ordinary quantities (though not exactly like them). In fact, this was also the viewpoint taken by many in mathematics before analysis was put on a rigorous basis with definitions that don't use such "infinitesimal quantities". However, there's a branch of mathematics called non-standard analysis that does deal with basing analysis on a formal definition of what such "infinitesimal quantities" mean.

Having said all that, in the field where this integral occurs, in physics, $\mathrm{d}\bar{r}$ is often considered as an infinitesimal change in the radius $\bar{r}$, and then $\frac{1}{\bar{r}^2}\mathrm{d}\bar{r}$ means the product of that change with the inverse square of the radius. That is, the integral is considered as "summing" over a large/infinite number of small/infinitesimal radial intervals, and for each interval the product of the change in radius on that interval, $\bar{r}$, with (some proportionality constant times) the inverse square $\frac{1}{\bar{r}^2}$ of the radius is added to the value of the integral. Does that answer your question?

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@joriki: can you deduce it rigorously for non-physicists from left to right? If the dot-product was just a normal product, it would be easy but with dot-product I am stucked. – user7628 Mar 1 '11 at 2:29
@user: See my comment to the question above. It's not a dot product; the dot stands for multiplication of scalars. – joriki Mar 1 '11 at 2:30
@joriki: is it common that physicist mix the symbols like this? Are there more like this? The unclear markings take me a huge amount of time, irritating. – user7628 Mar 1 '11 at 2:33
@user:I think using dots for scalar multiplication is quite common, but it seems rather unusual to use one in this case. The writer may have used the dot in analogy with the path integral $-\int_{\vec{x}_1}^{\vec{x}_2}\vec{x}/r^3\cdot \mathrm{d}\vec{x}=[1/r]_{r_1}^{r_2}$, where there is indeed a dot product between the vector $\vec{x}$ and its infinitesimal increment $\mathrm{d}\vec{x}$. (Here $r = \lvert \vec{x} \rvert$ is the magnitude of $\vec{x}$, i.e. the radius). – joriki Mar 1 '11 at 2:38
@joriki: how do you deduce the path integral? $\int_{\vec{x}_{1}}^{\vec{x}_{2}} \vec{x} / r^{3} \cdot d\vec{x}$ where $\vec{x}=\hat{x} r$ but what is $\hat{x} \cdot d\hat{x}$? – user7628 Mar 1 '11 at 3:35

Maybe, I write up your equation in more conventional (physics) notation. You have the electric field $$\vec{E}(\vec{r}) = \frac{Q}{4 \pi \epsilon_0 r^2} \frac{\vec{r}}{r} \in \mathbb{R}^3$$ at the position $\vec{r}\in \mathbb{R}^3$ of a particle with charge $Q$ at the origin ($r=|\vec{r}|$). Integrating this field over a line gives the potential difference between the endpoints of the line. In your case you want to integrate the electric field from somewhere on the surface with $r=a$ to the surface with $r=b$ (along which path exactly you integrate the field does not matter as the system is radially symmetric and we are in a static situation [mathematically $\nabla \wedge \vec{E}(\vec{r}) =0$]. So we want to integrate (this is a line integral)$$\int_{r=a}^{r=b} \vec{E}(\vec{r}) \cdot d\vec{r} = [-U(r)]_{r=a}^{b} $$ where we introduced the potential $U(r) = \frac{Q}{4 \pi \epsilon_0 r}$ with the property that $E= - \nabla U$ and the gradient theorem.

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what is the $\wedge$ in $\nabla \wedge \vec{E}(\vec{r})=0$? – user7628 Mar 1 '11 at 16:23
$\nabla \wedge$ you can also write as curl. It's the curl of a vector field and it tells you whether a potential exists. – Fabian Mar 1 '11 at 16:49
In American textbooks, curl is usually written $\nabla \times$. – Apprentice Queue Mar 5 '11 at 3:26

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