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I was reading about line integrals of a scalar field and line integrals of a vector field on Wikipedia. You can find the links here: line integral for a scalar field and line integral for a vector field.

In the second link the sentence: "A line integral of a scalar field is thus a line integral of a vector field where the vectors are always tangential to the line."

However I don't really understand why this is true. Also, because the scalar field outputs scalars whereas the vector field outputs vectors.

How do we explain the sentence above and more in general, what is the relationship between a line integral of a vector field and the line integral of a scalar field?

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2 Answers 2

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A line integral of a scalar field $g $ is $\displaystyle\int g\left(\overrightarrow{r}(t)\right)\left|\overrightarrow{r}'(t)\right|\,\mathrm dt$ along the curve. A line integral of a vector field $\overrightarrow{F}$ is $\displaystyle\int \overrightarrow{F}\left(\overrightarrow{r}(t)\right)\bullet\overrightarrow{r}'(t)\,\mathrm dt$. By a property of dot products, this equals $\displaystyle\int \left|\overrightarrow{F}\left(\overrightarrow{r}(t)\right)\right|\left|\overrightarrow{r}'(t)\right|\cos\left(\theta (t)\right)\,\mathrm dt$, where $\theta (t) $ is the angle between $ \overrightarrow{F}\left(\overrightarrow{r}(t)\right)$ (the direction of the vector field) and $\overrightarrow{r}'(t)$ (the direction of the curve).

If we choose $$\overrightarrow{F}\left(\overrightarrow{r}(t)\right)=g\left(\overrightarrow{r}(t)\right)*\frac{\overrightarrow{r}'(t)}{\left|\overrightarrow{r}'(t)\right|}\text{,}$$ then $\overrightarrow{F}$ has magnitude $|g|$ and direction: either the same as $\overrightarrow{r}'(t)$ or directly opposite (precisely when $g <0$). These directions make $\cos\left(\theta (t)\right) = \pm 1$ according to the sign of $g $. This way, $\left|\overrightarrow{F}\left(\overrightarrow{r}(t)\right)\right|\left|\overrightarrow{r}'(t)\right|\cos\left(\theta (t)\right)=g\left(\overrightarrow{r}(t)\right)\left|\overrightarrow{r}'(t)\right|$, so the vector field integral becomes the scalar field integral.

In summary, by choosing $\overrightarrow{F}$ so that it goes in the same direction of the curve (up to a minus sign), we can make a vector field integral that equals the scalar field integral.

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  • $\begingroup$ This was really illuminating. But I am having hard time figuring out why this holds: reversing path changes sign for line integral over vector field, but reversing path does not change integral over scalar field, despite integral over scalar field is special case of line integral over vector field. $\endgroup$
    – Silent
    Feb 6, 2019 at 7:53
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    $\begingroup$ @Silent Reversing path turns $r(t)$ into $r(-t)$ (and changes the bounds on $t$ accordingly). By the chain rule, the derivative gives you $-r'(-t)$. The integral over a scalar field has length/norm bars around $r'$ so the negative sign here doesn't change that. But the integral over a vector field has no bars and the minus sign can be pulled out of the dot product and put of the integral (Intuitively, a scalar field integral adds up the numbers along the curve, so direction doesn't matter, but a vector field integral calculates things like physical work so there it does (towards/away gravity)) $\endgroup$
    – Mark S.
    Feb 6, 2019 at 13:01
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If $\vec{F}= \pmatrix{P\\Q}$ is your vector field, then: $$ \underbrace{\int_C \vec{F} \cdot d\vec{r}}_{\text{vector field line integral}} = \int_C \pmatrix{P\\Q}\cdot \pmatrix{dx\\dy} = \int_C P\; dx + Q\; dy = \underbrace{\int_C P\; dx}_{\text{scalar field line integral}} +\underbrace{\int_C Q\; dy}_{\text{scalar field line integral}} $$

So in other words, your initial vector field line integral is the sum of two scalar field line integrals, in the direction of the $x$ and $y$ axes, respectively.

Note that if $\vec{r}(t)$ is a parametrization of $C$, you can define the line integral of a vector field as follows: $$ \int_C \vec{F} \cdot d\vec{r} = \int_C \vec{F} \cdot\vec{T}\; ds = \int_t \vec{F}(\vec{r}(t)) \cdot\frac{\vec{r}'(t)}{\mid \mid \vec{r}'(t) \mid \mid}\;\mid \mid \vec{r}'(t) \mid \mid dt = \int_t \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t)\; dt $$ The last integral is also a scalar field integral.

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    $\begingroup$ Thank you for your answer. So yes it's the sum of two scalar field line integrals (does this even have anything to do with Greens'?) but I don't understand how we deduce that "the vectors are always tangential to the lien". I mean if the vectors are tangential to the line, then surely the dot product would have $\cos(\theta)=1$ and hence would just be the product of their absolute values. But it's not what we get. Or is it? $\endgroup$ Mar 30, 2017 at 16:26
  • $\begingroup$ Green says that if $C$ is closed and orientated counter clockwise, then $$ \int_C P dx + Q dy = \iint_D \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \; dA $$ $\endgroup$
    – Kuifje
    Mar 30, 2017 at 17:15
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    $\begingroup$ @Euler_Salter We’re not trying to prove that the vectors are always tangential to the curve. The point of the statement in the article is that if the vector field is tangent to the curve, then the integrand is (up to sign) $\|\vec F\|d\vec r$, which is what you’d get for the scalar field $\|\vec F\|$. $\endgroup$
    – amd
    Mar 30, 2017 at 19:45
  • $\begingroup$ @amd oh alright it makes sense. Still somehow it seems weird that two different objects, if integrated along the same path, give the same answer, by setting the direction of the vectors tangential to the curve. I wonder if you can do the "same" for a surface integral and a tensor field $\endgroup$ Mar 30, 2017 at 23:00
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    $\begingroup$ @Euler a surface integral is usually involving a normal vector, so you could do the same with a surface integral and a vector field, by making the vector field point perpendicular to the surface. However, you might be able to define a weird "surface integral of a matrix" that doesn't involve the vector field, I haven't thought about it. $\endgroup$
    – Mark S.
    Mar 30, 2017 at 23:12

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