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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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    ohh, yeah correct. Thanks for clearing that out2011-05-07

1 Answers 1

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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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    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.2011-05-08