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Our physics prof wrote the following equation:

$\int\frac{\vec{r}}{r^3}d\vec{r} = \int\frac{1}{r^2}dr$

This is logical as long as I argue that $\vec{r}$ and $d\vec{r}$ are parallel, which is why the dot product evaluates as $|\vec{r}||d\vec{r}| = r dr$ However then i tried to do it by hand:

$\vec{r}d\vec{r} = \left(\begin{array}{c}x\\y\\z\\ \end{array}\right)\left(\begin{array}{c}dx\\dy\\dz\\ \end{array}\right) = xdx + ydy + zdz$

but this is nowhere near

$rdr = \sqrt{x^2+y^2+z^2}\sqrt{dx^2 + dy^2 + dz^2}$

which is why I would like to ask you what i am doing wrong.

Thanks in advance

ftiaronsem

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Your $d\vec{r}$ is not necessarily radial (and thereby parallel to $\vec{r}$). –  Fabian May 7 '11 at 12:09
    
hmm, why is that? I thought that this would always be true in clyndrical/spherical coordinates??? –  ftiaronsem May 7 '11 at 13:13
    
but your $\displaystyle{d\vec{r} =\left(\begin{array}{c}dx\\dy\\dz\\ \end{array}\right) }$ is neither in cylindrical nor in spherical coordinates. –  Fabian May 7 '11 at 13:25
    
ahh, yeah, ok convinced ^^. But even in that case I thought it would be true. Can you give a counter example? –  ftiaronsem May 7 '11 at 14:58
    
The line integral in general depends on the whole line along which you integrate it (and not only on the endpoints). In general, $d\vec{r}$ points along the line and is not radially. –  Fabian May 7 '11 at 15:18

1 Answer 1

up vote 2 down vote accepted

Maybe the following helps (it seems to be sloppy notation of your prof):

The integral $$\int_{\mathbf{r}_a}^{\mathbf{r}_b}\frac{\vec{r} \cdot d\vec{r}}{r^3}$$ is a line-integral. The vector field $$\vec{E}(\vec{r})=\frac{\vec{r}}{r^3}$$ is the gradient of a scalar field $\phi(\vec{r}) = -r^{-1}$, i.e., $$\vec{E}(\vec{r}) = \vec{\nabla} \phi(\vec{r}).$$ Therefore, the line-integral is path independent and the result is given by $$\int_{\mathbf{r}_a}^{\mathbf{r}_b} \vec{E}(\vec{r}) \cdot d\vec{r} = \int_{\mathbf{r}_a}^{\mathbf{r}_b} \vec{\nabla} \phi(\vec{r}) \cdot d\vec{r} = \phi(\vec{r}_b) - \phi(\vec{r}_a) = r_a^{-1} - r_b^{-1},$$ which coincides with the result of your prof.

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thanks, for your answer, this is a nice way of solving the integral. I know nearly nothing about vector calculus so could you please explain why $\int_{\mathbf{r}_a}^{\mathbf{r}_b} \vec{\nabla} \phi(\vec{r}) \cdot d\vec{r} = \phi(\vec{r}_b) - \phi(\vec{r}_a)$? I tried to do it, which resulted at integrating $\int_{\mathbf{r}_a}^{\mathbf{r}_b}3d\phi$ after evaluating the dot product. –  ftiaronsem May 7 '11 at 13:16
    
@ftiaronsem: this is called Gradient theorem (en.wikipedia.org/wiki/Gradient_theorem). It is the fundamental theorem of line-integral (and replaces thereby the fundamental theorem of calculus). –  Fabian May 7 '11 at 13:28
    
ahh, I see. Thanks a lot. Originally I planned on rewriting the integral and solving it afterwards. I understand your way of solving the problem now, but can you help spot the mistake in my consideration? Would gladly accept your answer then ;-) –  ftiaronsem May 7 '11 at 15:50
    
@ftiaronsem: (some formulas in your post don't make a lot of sense) To perform a line integral you should define a contour $\gamma:\vec{r}(t)$ with $t\in[a,b]$ along which you want to integrate $\vec{r}/r^3$. Then the definition of $\int_\gamma \frac{\vec{r}}{r^3}d\vec{r}$ reads $\displaystyle \int_\gamma \frac{\vec{r}}{r^3}\cdot d\vec{r} = \int\frac{\vec{r}}{r^3} \cdot \vec{r}'(t) dt$. See en.wikipedia.org/wiki/… for more information. In general, the integral depends on $\gamma$. However, in your case it does not. –  Fabian May 7 '11 at 15:56
    
Ok, thanks for this answer. I see now how it is supposed to be done. You said that there are some formulas that don't make a lot of sense in my post. Could you point out where? I would really appreciate to know where i am making a mistake in the above considerations. –  ftiaronsem May 8 '11 at 9:01

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