Line Integral of Scalar Field Along a Curve

[richyw]

Homework Statement

For some scalar field f : U ⊆ Rn → R, the line integral along a piecewise smooth curve C ⊂ U is defined as

\int_C f\, ds = \int_a^b f(\mathbf{r}(t)) |\mathbf{r}'(t)|\, dt

where r: [a, b] → C is an arbitrary bijective parametrization of the curve C such that r(a) and r(b) give the endpoints of C and a<b .

 

[richyw]

sorry, for breaking the format. I will put my attempt in the next post. Just wanted to have the stuff rendered on the screen!

 

[richyw]

so the first part is confusing me quite a bit. What does \mathbf{R}^n\rightarrow \mathbf{R} mean?

when I hear "scalar field" I would think of it as a field of some scalar number in space. Maybe like potential? I know that potential energy is a scalar, so I guess potential (i'm thinking electric potential) would be a scalar as well. So at each point in space there could be a different potential. Is this a "scalar field"?

so anyways I think I get that the piecewise smooth curve C is a subset of U. that just means that the curve is somewhere in the potential field right?

 

[richyw]

basically I'm not asking for someone to explain all of line integrals to me. I have been evaluating them for awhile and have kind of a basic idea of what they are. I'm looking for what this notation is "saying". I am getting lost in all the symbols!

 

[Michael Redei]

$f:U\subseteq\mathbb R^n\to\mathbb R$ means that $f$ maps a subset $U$ of the n-dimensional real space $\mathbb R^n$ to the set $\mathbb R$ of reals. So each vector $\mathbf v\in U$ is mapped onto a scalar $f(\mathbf v)\in\mathbb R$.

Your "potential" example is a good one, and the "field" in this sense is the mapping from $U$ to $\mathbb R$, and not the set $U$ itself, assigning a "potential" to each point in $U$. And the curve $C$ lies somewhere in $U$. It's independent of the field $f$.

 

[Michael Redei]

Second part: if you integrate a function $f:\mathbb R\to\mathbb R$, say from $a$ to $b$, you're forming a kind of sum,
$$
\int_a^bf(x)\,\mathrm dx = \lim_{\delta\to0}\delta\cdot\left(f(a)+f(x+\delta)+f(x+2\delta)+ \ldots +f(b)\right).
$$
There's only one path from $a$ to $b$, and it leads through each intermediate point $a+n\delta$. If you want to integrate from one point in $\mathbb R^n$ to another, you have to specify which path you are going to take. That's what $C$ describes.

 

[richyw]

ok hold on a second. I am still a little bit confused about your first post (and thanks a ton for making it!)

I'm struggling to understand what you mean when you say each vector that is an element of U is mapped onto a scalar field.

 

[richyw]

so like going with that first analogy. The electric field is a vector field right? each point in an electric field also has a potential. so is the electric field somehow mapped onto the potential field?

 

[Michael Redei]

Each vector v in U has some potential, which we call f(v). The "field" here is the function f. In mathematics, a function is also called a "mapping", and each vector v is "mapped" onto the scalar f(v).

Your second example, an electric field ("field" in the physical sense) can be seen as a (mathematical) vector field, i.e. a mapping g that assigns a vector g(v) to each point v in space. This vector g(v) corresponds to the electromagnetic force that would act on a particle at position v.

These two fields, the scalar field f and the vector field g, are related, but not the same. And you can't "map" one to the other, since both are mappings themselves.

 

[Ray Vickson]

so like going with that first analogy. The electric field is a vector field right? each point in an electric field also has a potential. so is the electric field somehow mapped onto the potential field?

Potential is a perfect example: for each point (x,y,z) in R^3 you get a number V(x,y,z). So V:R^3 -> R. This notation just means that for each point p = (p_1, p_2, ..., p_n) in U you get a number f(p) = f(p_1,p_2,...,p_n).