Dot product in the line integral of a vector field

Question

Why does the integrand of a line integral consist of a dot product between the vector filed $\vec F$ and the unit tangent vector $\hat T$( or normal $\hat n$ )? Does this follow from physical examples( Work done, for an example ) or is there a particular reason behind considering the dot product?

Answer 1

Higher dimensions always present the difficulty of how to adapt tools to the new situations. The integral is one of those tools, and the generalization is not unique: actually, there are two kind of extensions. Multiple integration (usually carried on with measure theory) and vector calculus (generalized by the calculus on differential forms). Lets focus on the second one. How to generalize the integral from a scalar function on an interval to a vector valued function on a curve? The definition comes up naturally from some considerations:

Lets start with the easy possibility: straight lines. To fix the ideas, we will consider an axis (parametrized resp, the x one: it is natural to consider the line integral as $\int_{\mathbb{R}}F_x$

The passage from one axis to a general line is easy: the geometrical meaning stays the same (projection onto the line) and thus the calculation.

Now, in order to generalize to curves, we can split up in two steps:

• Considering the arc parametrization, the definition is quite natural: you are only restricting yourself to the component of the function "near" to the curve (in mathematical terms, tangent).

• We have now reached an important result: the integral is now extended to vector-valued functions on curves arc-parametrized. The geometrical ideas, though, stays the same even if the parametrization is changed: what is relevant is the substain. Thus, we would like the integral to be parametrization-invariant: from easy computation, we see that the definition of line-integral is what we are looking for

Just to have a very vague idea, the fundamental theorem of calculus is generalized in two ways, depending on the integral:

With the lebesgue one,we have the so called "Lebesgue differentiation theorem"

$\begin{equation} \lim_{\mu(B_x)\to 0}\frac{1}{\mu(B_x)}\int_{B}f(y)\ d\mu(y)\to_{a.e.}f(x) \end{equation}$

While with the one we defined, we have the "Fundamental theorem for line integrals" (generalized by Stokes-Cartan theorem)

$\begin{equation} \int_\gamma \nabla f=f(\gamma(b))-f(\gamma(a))\ \ \ \ \ \Bigg(\int_Md\omega=\int_{\partial M}\omega\Bigg) \end{equation}$

Obviously the line integral, as a lot of mathematical tools, is very useful in physics

Answer 2

Assume you have ahomogeneous, i.e. constant, vector field ${\bf v}$ and a segment path $$\sigma:\quad t\mapsto(1-t){\bf a}+t{\bf b}\qquad(0\leq t\leq 1)\ .$$ In such a case you might be interested in the scalar quantity $q:={\bf v}\cdot({\bf b}-{\bf a})$, or in the vector quantity ${\bf F}:={\bf v}\times({\bf b}-{\bf a})$, or in the quantity ${\bf M}:={\bf v}\,|{\bf b}-{\bf a}|$. Other such primitive paradigmatic cases can be cooked up.

When the vector field ${\bf v}$ is not constant and the curve is not a simple segment the above more or less bilinear concepts morphe to integrals of a characteristic appearence. The first becomes the line integral used to calculate work done. It is not so that we "always" have a dot product in integrals created in this way.