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enter image description hereI came across these expressions in my book. And the book says that all these are different from each other. The expressions are : $\int$$d$$|\vec r|$, $\int$$|$$d$$\vec r$|, and $|$$\int$$d$$\vec r|$

Are they different from each other?

I know that

  • $|$$\int$$d$$\vec r|$ means magnitude of displacement,

  • $\int$$|$$d$$\vec r$| means total distance,

  • But what about $\int$$d$$|\vec r|$?

I think it should mean total distance too, but I’m not sure if $\int$$d$$|\vec r|$ and $\int$$|$$d$$\vec r$| have the same meaning, do they?

Edit : Some of the answers say that the question is not very clear, and that a little more explanation would help. I’m not sure what else to add, so I’m attaching a picture of that page

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  • $\begingroup$ The name of this book? $\endgroup$ Feb 16, 2022 at 18:59

4 Answers 4

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enter image description here


In the above Figure-01 we see the quantities $\;\mathrm d\mathbf{r}, \Vert\mathrm d\mathbf{r}\Vert,\mathrm d\Vert\mathbf{r}\Vert\;$ for an infinitesimal displacement from point $\;\boldsymbol{1}\;$ to point $\;\boldsymbol{2}$.

Note that

  1. $\mathrm d\mathbf{r}\;$ is an infinitesimal vector
  2. $\Vert\mathrm d\mathbf{r}\Vert\;$ is an always non-negative infinitesimal scalar representing the magnitude of $\;\mathrm d\mathbf{r}$, and
  3. $\mathrm d\Vert\mathbf{r}\Vert\;$ is a real infinitesimal scalar representing the change of the magnitude of the position vector of point $\;\boldsymbol{1}\;$ to point $\;\boldsymbol{2}$. So, it's positive if increasing, negative if decreasing and zero if constant.

Let see now the integrals of these quantities along a curve $\;C\;$ as in Figure-02.

enter image description here

\begin{equation} \Delta \mathbf{r} \boldsymbol{=}\int\limits_{1}^{2}\!\!\!\!\!\!C\mathrm d \mathbf{r}\boldsymbol{=} \mathbf{r}_2\boldsymbol{- } \mathbf{r}_1 \tag{01}\label{01} \end{equation} \begin{equation} \Delta\Vert \mathbf{r}\Vert \boldsymbol{=}\int\limits_{1}^{2}\!\!\!\!\!\! C\mathrm d \Vert \mathbf{r}\Vert \boldsymbol{=} \Vert \mathbf{r}_2\Vert\boldsymbol{-} \Vert \mathbf{r}_1\Vert \tag{02}\label{02} \end{equation}

So, these quantities depend on the start and end points $\;\boldsymbol{1},\boldsymbol{2}\;$ and are independent of the curve joining them. I don't think that these quantities are useful in any case.

Now, \begin{equation} s_{1}^{2} \boldsymbol{=}\int\limits_{1}^{2}\!\!\!\!\!\! C\Vert \mathrm d \mathbf{r}\Vert \boldsymbol{=}\text{lenght of curve between points }\boldsymbol{1},\boldsymbol{2} \tag{03}\label{03} \end{equation}


Note :

This textbook is confusing. On one hand it's written that we have a vector \begin{equation} \vec s \boldsymbol{=}\Delta \vec r \boldsymbol{=}\lim_{\Delta \vec r_{i}\boldsymbol{\rightarrow} 0} \sum\limits_{\boldsymbol{i=1}}^{\boldsymbol{i=n}}\Delta \vec r_{i}\boldsymbol{=} \int d\vec r \tag{A}\label{A} \end{equation} and then this same integral of the rhs is used for a scalar, the length of the line joining points 1,2 \begin{equation} \int d\vec r\boldsymbol{=}\text{length of the line joining point 1 to 2}\boldsymbol{=}\vert \Delta \vec r\vert \tag{B}\label{B} \end{equation}

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  • $\begingroup$ The two diagrams that you have used for the explanation make the difference between these three notations very clear. Thanks for taking the time to answer the question. So $\int d|\vec r|$ is basically a scalar, and is equal to the difference between the magnitudes of $\vec r_1$ and $\vec r_2$ But does it have a physical meaning? $\endgroup$
    – 4d_
    Feb 6, 2019 at 11:01
  • $\begingroup$ @π times e : I don't think that $\int d|\vec r|$ has any physical meaning. $\endgroup$
    – Frobenius
    Feb 6, 2019 at 12:26
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Looks like an integral over radial distance.

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  • $\begingroup$ This probably could use more exposition to be a useful answer. $\endgroup$
    – Kyle Kanos
    Feb 3, 2019 at 12:23
  • $\begingroup$ I'd edit my question and attach a picture to make it clearer $\endgroup$
    – 4d_
    Feb 3, 2019 at 13:21
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It is the radial part of the total distance along some path. For example, for a circle there is an angular part of the total distance ($2\pi r$), but there is no any radial part contribution: $dr=0$.

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  • $\begingroup$ Is this radial path (or radial distance) part of 'circular motion' ? I haven't studied circular motion yet, currently I'm staying motion in one dimension & motion in a plane. I'm not sure why it would show up in the book in the chapter I am studying. Do you think this expression can mean something else? Something related to what I am studying right now? $\endgroup$
    – 4d_
    Feb 3, 2019 at 13:20
  • $\begingroup$ A motion in a plane may be circular: a point makes a two dimensional path around some point staying always at a certain distance $r=R$ from it. $\endgroup$ Feb 3, 2019 at 18:51
  • $\begingroup$ My bad! Yeah circular motion is an example of 2D motion too. I meant to say that, I'm not studying circular motion right now. It is the next chapter. But this thing showed up in the current chapter that I am studying $\endgroup$
    – 4d_
    Feb 3, 2019 at 19:00
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Notation matters. You have probably seen $\int d|\vec{r}|$ written as $\int dr$, without vectors.

  • In 1D, this is the same as $\int dx$. The letter for the disttance is not relevant.
  • In more than 1 dimension, it's the integral along the radius. For example, in a circle, you'd integrate for all angles and also for all radii from 0 to $R$, for example. There you'd find that integral. $S=\int_0^{2\pi}d\varphi \int_0^R dr$
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  • $\begingroup$ I just attached the pic, it’s exactly the same notation as I have used in my question. Now that I’ve attached the pic, can you please have another look at it? Thanks $\endgroup$
    – 4d_
    Feb 3, 2019 at 14:26
  • $\begingroup$ Your picture is helpful to understand the two former cases, but you ask for a third case, $d|\vec{r}|$, which is not in the picture. It doesn't change anything, does it? $\endgroup$
    – FGSUZ
    Feb 3, 2019 at 20:06
  • $\begingroup$ Actually...yeah, the third notation hasn't been talked about in the picture. It's just mentioned. Hence I asked it on here. Thanks for the explanation, I guess it is radial distance and will show up again in other chapters $\endgroup$
    – 4d_
    Feb 4, 2019 at 3:05

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