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I know it's surely a beginner's question but I don't see why you can write \begin{align} \frac{\vec{r}}{r^3} = \frac{1}{r^2}\cdot \frac{|\vec{r}|}{r} \end{align} Could someone explain it please? It would help understand quite a few things ...

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  • $\begingroup$ You are missing the position unit vector on the right side of your equation. $\endgroup$ – Aaron Stevens May 14 '18 at 17:02
  • $\begingroup$ What do you mean? Or better: how (and why) can I rewrite $\frac{\vec{r}}{r^3}$. $\endgroup$ – offline May 14 '18 at 17:05
  • $\begingroup$ You cannot have a vector equal to a scalar. You either need to replace the vector with its magnitude as @HiddenBabel does, or you need to add the unit r vector to the right side. $\endgroup$ – Aaron Stevens May 14 '18 at 17:10
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What you wrote is not true. The $r$ without an arrow is only a scalar - the length of a vector. So the right-hand side is a scalar. The left-hand side is a vector $\vec{r}$ divided by its length cubed. So the result is still a vector. However, the length of the resulting vector in the left is $1/r^2$. This is because the $\vec{r}$ carries a length of $r$ with it. So if you take the norm, you get

$$\frac{|\vec{r}|}{r^3} = \frac{r}{r^3} = \frac{1}{r^2}$$

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  • $\begingroup$ Ah ok, well our teacher literally always writes it the way I wrote it in the question! How can I rewrite $\frac{\vec{|r|}}{r}$? $\endgroup$ – offline May 14 '18 at 17:08
  • $\begingroup$ That is the unit vector, which is $\vec{r}$ scaled to length 1. It is written as $\vec{\hat{r}}$. If you put that on the right hand side of your first equation, then the equation is correct. $\endgroup$ – HiddenBabel May 14 '18 at 17:13
  • $\begingroup$ I'm confused. I thought the unit vector is $\hat{r} = \frac{\vec{r}}{|\vec{r}|}$. $\endgroup$ – offline May 14 '18 at 17:19
  • $\begingroup$ A vector can be represented as, $\vec{r}= |\vec{r}|\hat{r}$. $|\vec{r}|$ shows the magnitude of the vector and $\hat{r}$ shows the direction. $\endgroup$ – Mitchell May 14 '18 at 17:20
  • $\begingroup$ Oh I misread what you​ wrote. $|\vec{r}|/r$ is just 1, then. $\endgroup$ – HiddenBabel May 14 '18 at 17:21
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$$\frac{\vec{r}}{r^3} = \frac{1}{r^2}\frac{\vec{r}}{r}=\frac{1}{r^2} \vec{u_r}$$ where $\vec{u_r}$ is an unitary vector with the direction of $\vec{r}$

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Another possibility is: being $\overline r= r\mathbf{\widehat r}$, then

$$\frac{\overline r}{r^3}=\frac{r\mathbf{\widehat r}}{r^3}=\frac{\not r\mathbf{\widehat r}}{r^{\not 3\,2}}=\frac{\mathbf{\widehat r}}{r^{2}}.$$

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