nLab
line integral

Line integrals

Idea

A line integral is an integral along a curve. These are also called path integrals (not to be confused with the path integral in quantum field theory, which is integration over a space of curves rather than along a curve in some space) and contour integrals (especially in complex analysis).

By the modern understanding of the integration of differential forms, one integrates 11-forms (cotangent vector fields) along oriented curves, and so this would be the natural way to understand a line integral. However, there are several slightly different line integrals, and not all of them are reducible to integration of 11-forms along oriented curves. Sometimes we have to fall back on more basic notions, ultimately the integration of pseudo-11-forms on a 11-dimensional space.

Definitions

Here we assume familiarity with integration of differential forms and pseudoforms, defining in terms of them the various classical notions of line integral.

In each case, the classical notation (developed before the rigorous treatment of analysis in the 19th century) is not taken literally by the classical definition (developed after the rigorous treatment and still found in calculus texts), and this definition is accompanied by a reparametrisation? theorem. We would like to make sense of the classical notation and eliminate the reparametrisation theorems (or at least make them all special cases of a general theorem to be proved once for all) using differential forms. Often we find that we can relax some of the restrictions in the classical definition as well.

Line integral of a vector field

Classically, we have a Cartesian space XX, a continuous map F:XX\vec{F}\colon X \to X, and a continuously differentiable map C:[a,b]XC\colon [a,b] \to X; the line integral of F\vec{F} along CC is defined as

CFdr a bF(C(t))C(t)dt, \int_C \vec{F} \cdot \mathrm{d}\vec{r} \coloneqq \int_a^b \vec{F}(C(t)) \cdot C'(t) \,\mathrm{d}t ,

where the integral on the right is a Riemann integral and CC' is the derivative of CC componentwise. If ϕ:[e,f][a,b]\phi\colon [e,f] \to [a,b] is a continuously differentiable increasing bijection, then

CϕFdr= CFdr, \int_{C \circ \phi} \vec{F} \cdot \mathrm{d}\vec{r} = \int_C \vec{F} \cdot \mathrm{d}\vec{r} ,

the reparametrisation theorem.

We can start to justify the classical notation by interpreting r\vec{r} as the same map [a,b]X[a,b] \to X as CC; then we have

CFdr a bF(r(t))r(t)dt, \int_C \vec{F} \cdot \mathrm{d}\vec{r} \coloneqq \int_a^b \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) \,\mathrm{d}t ,

in which case the classical notation simply seems to be suppressing some of the notation in the formal definition on the right. In particular, we interpret dr\mathrm{d}\vec{r} as meaning r(t)dt\vec{r}'(t) \,\mathrm{d}t.

But this suggests that we should really be looking at differential forms. Since XX is a Cartesian space, we may identify it with any of its tangent spaces and so identify F\vec{F} with a tangent vector field on XX, or equivalenty a vector-valued 00-form. Since r\vec{r} is only serving to parametrise the curve, it should be interpreted as something trivial, in this case the identity map on XX, also viewed as a vector-valued 00-form. Then dr\mathrm{d}\vec{r} is a vector-valued 11-form (which we can do since a Cartesian space has a trivial connection), and Fdr\vec{F} \cdot \mathrm{d}\vec{r} is an ordinary 11-form. Finally, since the curve CC is given up to an increasing reparametrisation, it is an oriented 11-submanifold of XX. Now we may interpet the notation

CFdr \int_C \vec{F} \cdot \mathrm{d}\vec{r}

literally as the integral of a 11-form. It is now no longer necessary that CC be given by a continuously differentiable parametrisation; the vector-valued 00-form r\vec{r} is continuously differentiable regardless, and so we only need CC to be a rectifiable curve? (although it is a theorem that such a curve must have a parametrisation —by arclength? if nothing else— that is continuously differentiable almost everywhere, so that the classical definition still covers this using a Riemann integral).

However, this operation from F\vec{F} to Fdr\vec{F} \cdot \mathrm{d}\vec{r} is something more fundamental in differential geometry; in fact, Fdr\vec{F} \cdot \mathrm{d}\vec{r} is simply F \vec{F}^\flat, the cotangent vector field that corresponds to the tangent vector field F\vec{F}. Understanding this, we wish to generalise XX to any (pseudo)-Riemannian manifold. In this case, we cannot interpret r\vec{r} literally, but we may still interpret dr\mathrm{d}\vec{r} as a vector-valued 11-form, indeed as the tautological one that (viewing a 11-form as an operation on vector fields) is the identity map on vector fields. With this understanding, the classical notation still makes sense, although it is probably easier to write

CFdr= CF \int_C \vec{F} \cdot \mathrm{d}\vec{r} = \int_C \vec{F}^\flat

as the most general definition of the line integral of a vector field along an oriented curve in a (pseudo)-Riemannian manifold.

Line integral of a scalar field

Classically, we again have a Cartesian space XX, now a continuous map ff from XX to \mathbb{R} or \mathbb{C}, and again a continuously differentiable map C:[a,b]XC\colon [a,b] \to X; the line integral of ff along CC is defined as

Cfds a bf(C(t))C(t)dt, \int_C f \mathrm{d}s \coloneqq \int_a^b f(C(t)) {\|C'(t)\|} \,\mathrm{d}t ,

where again the integral on the right is a Riemann integral and CC' is the derivative of CC componentwise. If ϕ:[e,f][a,b]\phi\colon [e,f] \to [a,b] is a continuously differentiable bijection (whether increasing or decreasing), then

Cϕfds= Cfds, \int_{C \circ \phi} f \mathrm{d}s = \int_C f \mathrm{d}s ,

the reparametrisation theorem.

We cannot interpret ds\mathrm{d}s as the differential of anything; rather, ds\mathrm{d}s is the magnitude of the line integral element dr\mathrm{d}\vec{r} from the previous section. In particular, identifying r\vec{r} with CC again, we have

Cfds a bf(r(t))r(t)dt, \int_C f \mathrm{d}s \coloneqq \int_a^b f(\vec{r}(t)) {\|\vec{r}'(t)\|} \,\mathrm{d}t ,

so ds\mathrm{d}s seems to mean r(t)dt{\|\vec{r}'(t)\|} \,\mathrm{d}t. Using the standard orientation on [a,b][a,b] to change the 11-form dt\mathrm{d}t to the pseudo-11-form dt{|\mathrm{d}t|}, this is consistent with ds=dr\mathrm{d}s = {\|\mathrm{d}\vec{r}\|}.

But for this to really make sense, we have to see what kind of object ds\mathrm{d}s is on XX. It is neither a 11-form (which can be integrated along an oriented curve) nor a pseudo-11-form (which can be integrated along a pseudo-oriented curve, that is a transversely oriented curve); it is instead an absolute 11-form, which can be integrated along an unoriented curve. If we interpret dr\mathrm{d}\vec{r} as the canonical vector-valued 11-form again, then we can take its magnitude to get a positive semidefinite (nonnegative-scalar-valued, and in this case actually definite) absolute 11-form, so ds=dr\mathrm{d}s = {\|\mathrm{d}\vec{r}\|} is literally true.

It would be nice to have notation without fake differentials and with the one piece of structure that actually plays a role: the metric. If we multiply two 11-forms using the symmetric product (instead of the exterior product as usual for differential forms), then we get a symmetric bilinear form, and in this way the (symmetric) square of the arclength element ds\mathrm{d}s is the metric gg. Since ds\mathrm{d}s is positive, we can reasonably call it the principal square root of gg. Thus,

Cfds= Cfg \int_C f \mathrm{d}s = \int_C f \sqrt{g}

defines the line integral of a scalar field along an unoriented curve in a Riemannian manifold.

On a pseudo-Riemannian manifold, gg itself may not be positive and so may not have a square root. In that case, we can take the absolute value of gg first and use

Cfds= Cfg, \int_C f \mathrm{d}s = \int_C f \sqrt{|g|} ,

although this is most intuitive for curves that are consistently timelike or spacelike. It's also possible to keep the previous formula and allow one of these two types of curve to have imaginary? arclength; which one depends on conventions. In any case, a line integral along a lightlike curve is zero.

Contour integral of a complex function

Classically, we have the complex plane \mathbb{C}, a continuous map f:f\colon \mathbb{C} \to \mathbb{C}, and a continuously differentiable map C:[a,b]C\colon [a,b] \to \mathbb{C}; the contour integral (or line integral again) of ff along CC is defined as

Cfdz a bf(C(t))C(t)dt, \int_C f \mathrm{d}z \coloneqq \int_a^b f(C(t)) C'(t) \,\mathrm{d}t ,

where the integral on the right is a Riemann integral and CC' is the derivative of CC. If ϕ:[e,f][a,b]\phi\colon [e,f] \to [a,b] is a continuously differentiable increasing bijection, then

Cϕfdz= Cfdz, \int_{C \circ \phi} f \mathrm{d}z = \int_C f \mathrm{d}z ,

the reparametrisation theorem.

We start to justify the classical notation by interpreting zz as the same map [a,b][a,b] \to \mathbb{C} as CC; then we have

Cfdz a bf(z(t))z(t)dt, \int_C f \mathrm{d}z \coloneqq \int_a^b f(z(t)) z'(t) \,\mathrm{d}t ,

in which case the classical notation seems again to be suppressing some of the notation in the formal definition. In particular, dz\mathrm{d}z is interpreted as z(t)dtz'(t) \,\mathrm{d}t.

But again, we should really be looking at differential forms. We may identify the space \mathbb{C} with the scalar field and so identify ff with a scalar field on \mathbb{C}, or equivalenty a 00-form. Since zz is only serving to parametrise the curve, it should be interpreted as the identity map on \mathbb{C}, also viewed as a 00-form. Then dz\mathrm{d}z is a 11-form, and fdzf \mathrm{d}z is a 11-form. Finally, since the curve CC is given up to an increasing reparametrisation, it is an oriented 11-submanifold of XX. Now we may interpet the notation

Cfdz \int_C f \mathrm{d}z

literally as the integral of a 11-form. It is now no longer necessary that CC be given by a continuously differentiable parametrisation; the vector-valued 00-form zz is continuously differentiable regardless, and so we only need CC to be a rectifiable curve? (although it is a theorem that such a curve must have a continuously differentiable parametrisation after all).

Absolute contour integral of a complex function

Classically, we have the complex plane \mathbb{C}, a continuous map f:f\colon \mathbb{C} \to \mathbb{C}, and a continuously differentiable map C:[a,b]C\colon [a,b] \to \mathbb{C}; the absolute contour integral (or whatever one calls it) of ff along CC is defined as

Cfdz a bf(C(t))C(t)dt, \int_C f {|\mathrm{d}z|} \coloneqq \int_a^b f(C(t)) {|C'(t)|} \,\mathrm{d}t ,

where the integral on the right is a Riemann integral and CC' is the derivative of CC. If ϕ:[e,f][a,b]\phi\colon [e,f] \to [a,b] is a continuously differentiable bijection (whether increasing or decreasing), then

Cϕfdz= Cfdz, \int_{C \circ \phi} f {|\mathrm{d}z|} = \int_C f {|\mathrm{d}z|} ,

the reparametrisation theorem.

As before, if zz is interpreted as the same map [a,b][a,b] \to \mathbb{C} as CC; then we have

Cfdz a bf(z(t))z(t)dt, \int_C f {|\mathrm{d}z|} \coloneqq \int_a^b f(z(t)) {|z'(t)|} \,\mathrm{d}t ,

so dz{|\mathrm{d}z|} seems to mean z(t)dt{|z'(t)|} \,\mathrm{d}t. As before, if we use the standard orientation on [a,b][a,b] to change the 11-form dt\mathrm{d}t to the pseudo-11-form dt{|\mathrm{d}t|}, then dz{|\mathrm{d}z|} is the absolute value of dz\mathrm{d}z from before.

Of course, we should really be looking at differential forms. Again, ff is a scalar field on \mathbb{C}, or equivalenty a 00-form. Again zz is the identity map on \mathbb{C}, viewed as a 00-form. Then dz\mathrm{d}z is a 11-form, its absolute value dz{|\mathrm{d}z|} is a (positive definite) absolute 11-form, and fdzf {|\mathrm{d}z|} is an (arbitrary) 11-form. Since CC is given up to an arbitrary reparametrisation, it is an unoriented 11-submanifold of XX, which is just what we need for an absolute 11-form. Now we may interpet the notation

Cfdz \int_C f {|\mathrm{d}z|}

literally as the integral of an absolute 11-form.

This is actually a special case of the line integral of a scalar field, since

dz=d(x+iy)=dx+idy=(dx) 2+(dy) 2=g=ds, {|\mathrm{d}z|} = {|\mathrm{d}(x + \mathrm{i}y)|} = {|\mathrm{d}x + \mathrm{i} \,\mathrm{d}y|} = \sqrt{(\mathrm{d}x)^2 + (\mathrm{d}y)^2} = \sqrt{g} = \mathrm{d}s ,

since dx 2+dy 2\mathrm{d}x^2 + \mathrm{d}y^2 is the standard metric on \mathbb{C}.

References

Here are a couple of old Usenet posts that explain how line integrals of scalar fields should be viewed in terms of forms and pseudoforms.

These are obsolete with the concept of absolute forms, but they contain more explicit calculations.

Revised on March 1, 2013 23:57:27 by Toby Bartels (64.89.53.120)