Intuitive understanding of path integral formula

Question

I have learned a general formula for a path/line integral

$$ \int_a^b f\left(\mathbf{r}(t)\right) \|\mathbf{r}'(t)\|\ dt \tag{1} $$

and I'm trying to better understand it. Specifically, I'm wondering what's not right about this:

$$ \int_a^b f\left(\mathbf{r}(t)\right)\ dt \tag{2} $$

If I try to work out what's happening in $(2)$, I reason that $f(\mathbf{r}(t))$ is the value of the field $f$ at the point $\mathbf{r}(t)$ for some $t$. As you vary $t$ from $a$ to $b$, the sum of those values $f(\mathbf{r}(t))$ seems to me like the value of the whole path, i.e. the path integral. Clearly, there's a gap in my understanding.

I do feel that I understand the similar formula for the arc length $s$ for the same path

$$ s = \int_a^b \|\mathbf{r}'(t)\|\ dt \tag{3} $$

from geometry, so I understand extending $(3)$ to $(1)$ by simply including the scalar field $f$. But, I think knowing what $(2)$ does mean might help me really take things to the next level.

Answer 1

This got me thinking to come up with a good example to demonstrate the problem with (2).

Lets say we want to find $\int_{-1}^1x^2\>dx$

But we want to do it the hard way: by a path integral. So we take: $\mathbf{r}(t) = 2x^{n} - 1$ (this just goes from -1 to 1 as t goes from 0 to 1) and solve the following integral instead:

$\int_{0}^1(2x^{n} - 1)^2\>dt$

This is your (2). Now note what happens for big k (this is the graph for $\mathbf{r}(t) = 2x^{101} - 1$):

Now you see that this will spend 'most of it's time' at practically -1. So if we evaluate (2) with this $\mathbf{r}(t)$, we will basically just compute:

$\int_{0}^1(2x^{101} - 1)^2\>dt = \int_{0}^1(-1)^2\>dt = 1$

So now you see why it's important to take into consideration the derivative of $\mathbf{r}(t)$, because when it's close to zero, we really aren't moving the region that we are integrating over, so we don't want to count it.

Answer 2

If we want the integral of $f$ over a curve to depend on the curve as geometrical object (subset of the space) instead of a mapping from a real interval, we want the integral to be invariant under reparametrizations of the curve. The most natural parametrization is the arc length parametrization, and we want the integral to be (2) in this case. From these two principles the form (1) for the integral follows.

Answer 3

When $$\gamma:\quad t\mapsto{\bf x}(t)\qquad(a\leq t\leq b)\tag{1}$$ is just an arbitrary parametric representation of the curve $\gamma$ then your integral $(2)$ has no intrinsic geometric meaning. If, however, $t$ is time, and $(1)$ describes the exact movement of a particle in time then the integral $(1)$ may very well have a meaning of physical significance.

Assume that $(1)$ describes the path of an American student in Europe $E$ and that $f({\bf x})$ measures the inspirational level of the location ${\bf x}\in E$. Then the integral $$\int_\gamma f({\bf x})\>dt=\int_a^b f\bigl({\bf x}(t)\bigr)\>dt$$ encodes the intellectual intake of this student along his journey.