Path/Line/Curvilinear Integral
Path Integral of Scalar Function
Recall the definition of a definite integral (Riemann integral) of real function F(x) defined on interval [a,b]:
I = ∫abF(x) dx = lim Σi∈[1,N] F(xi)·Δxi
where {xi} is partitioning of segment [a,b] into N parts,
and we assume that N→∞, while the maximum width of intervals
This definition pertains to a function F(x) defined on a straight segment [a,b] and traditionally interpreted as an area under a graph of a function on Cartesian XY plane restricted on the left and right by x=a and x=b correspondingly.
Consider a function that is defined not on a straight segment of real numbers [a,b], but on some path (curve) in space we call γ.
Let's assume for definitiveness that our path is in three-dimensional Cartesian space and is defined by a set of three smooth functions
γ={x(s),y(s),z(s)}
of some real parameter s∈[0,S].
Generally speaking, parameterization of a curve in N-dimensional space is just a set of N real functions xi(s) (i∈[1,N]) of real argument s such that
(a) all these functions have the same domain S - a finite or an infinite interval of real numbers;
(b) for each value of parameter s∈S a set of functions xi(s) used as a set of N coordinates correspond to some point on a curve;
(c) the parameterization covers the entire curve: for each point on a curve there is at least one value of parameter s within domain S such that a set of functions xi(s) represent the coordinates of this point.
The parameter s itself does not have any particular meaning. Depending on the problem, it can represent time, distance along the curve, arc length, or simply be an arbitrary convenient parameter.
For each position on this path a function F(x,y,z) is defined.
The path integral of this function F(x,y,z) along path γ is defined as
∫γF(x,y,z)·dγ = ∫[0,S]F(x(s),y(s),z(s))·ds
The expression above defines a new concept of path integral by connecting it to a regular integral based on a parameterization of a curved path.
Thus defined path integral is alternatively named line integral or curvilinear integral.
Obviously, this definition can be expanded to any N-dimensional space by replacing three-dimensional path
Example
Consider a path γ in some mountainous area and a task of measuring the average barometric pressure along this path.
This is a typical problem with the path integral as a solution.
While path is three-dimensional, we can choose a length of a road from the beginning of the path s as a parameter and, for each value of this parameter (that is, for each point on a mountain path
After integration of function H(s) by parameter s we can divide the result by the total length of a path to get the average barometric pressure.
So the most important issue in the path integral of a scalar function is to properly choose the parameter s to define the curved path as a function of a single parameter, thereby reducing the function defined for each point on a three-dimensional path to the function of a single parameter and reducing the path integral to a regular definite integral by a single variable.
Path Integral of Vector Function
Recall a concept of work in physics.
In a simple case of a straight road and a car moving along it from point A to point B, whose engine pulls the car with constant force F directed along the road, the work is a product of a force by a distance
W = F·d
where d is the length of segment AB.
In a more complicated case of a variable but still directed along the straight road force F=F(s) whose magnitude depends on position s on the road from A to B, the work can be approximated by dividing the segment AB into many small intervals, assuming that within each interval the force is practically the same, calculate the work on each interval as above, sum the work on all intervals and start increasing the number of intervals while decreasing their size.
In the limit, when the length of all intervals is infinitesimal, while their number increases to infinity, this calculation will result in an exact amount of work of an engine on a road AB expressed as an integral
W = ∫[0,d]F(s)·ds
Consider now a case of a smooth but curved road from A to B with the force still directed along the road (that is, the direction of the force is always tangential to a road).
This case is no different than the above case where we introduced a path integral.
So, the work of an engine of a car moving along a curved road is just another example of application of path integral to physics.
The work evaluation in cases when the force is not directed along a path of a moving object, however, requires some other technique.
Simple algebraic product of the magnitude of force F(s) by ds and integrating by s would not produce the right result in these cases.
Let's generalize the definition of a path integral to accommodate the calculation of work performed by the force F(s) being a general vector-function not necessarily directed along the path of motion.
To come up with some meaningful formula, let's introduce a system of Cartesian X,Y,Z coordinates on a plane and define the curve by some vector-function r(s)={x(s),y(s),z(s)} parameterized by variable s∈[0,S] with r(0)=A and r(S)=B.
The force, as a vector, would be represented by a vector-function F(r) defined for each point on a curve by its three components
F(r) = {Fx(x,y,z),Fy(x,y,z),Fz(x,y,z)}
In physics the work performed by a constant vector of force F in case it's not colinear with a straight segment of movement s is defined as a product of a projection of the vector of force onto a direction of movement Fproj and the length |s| of the movement.
According to this rule for calculating the work, for the vector of force F and a segment s along which an object moves the work of this force is calculated as
W = |F|·|s|·cos(φ)
where φ is an angle between direction of the force and direction of the segment.
This formula can be written as a scalar (dot) product of two vectors:
W = F·s
Let's return to a task at hand - calculate the work performed by a generally directed force along a curved path.
Dividing, as before, curve AB into n small almost straight intervals, we see that on each interval the force is not directed along it, but makes some angle.
Therefore, for a small interval Δsi and force Fi acting within this small interval the work performed by the force is
ΔWi = Fi · Δsi
where dot between two vectors means scalar product.
The total work for all n intervals is
W = Σi∈[1,n]ΔWi
As the number of intervals increases to infinity and the size of each decreases to zero, this sum tens to some limit. This limit is called the path integral of vector F along path AB:
∫AB[F(r)·dr]
where dot between two vectors means scalar product.
This definition can be rephrased by using the coordinate representation of vectors.
F(r) = F(x,y,z) =
= {Fx(x,y,z),Fy(x,y,z),Fz(x,y,z)}
dr = {dx,dy,dz}
Hence, infinitesimal amount of work our force performs on infinitesimal interval of a curved path equals to
dW = F(r)·dr =
= Fx(x,y,z)·dx+Fy(x,y,z)·dy+Fz(x,y,z)·dz
Considering our curve is defined parametrically as
r = r(s) = {x(s),y(s),z(s)}
we can define the path integral of a vector function above as
∫AB[F(r)·dr] =
= ∫[0,S]Fx(x(s),y(s),z(s))·dx(s) +
+ ∫[0,S]Fy(x(s),y(s),z(s))·dy(s) +
+ ∫[0,S]Fz(x(s),y(s),z(s))·dz(s) =
= ∫[0,S]Fx(x(s),y(s),z(s))·(dx(s)/ds)·ds +
+ ∫[0,S]Fy(x(s),y(s),z(s))·(dy(s)/ds)·ds +
+ ∫[0,S]Fz(x(s),y(s),z(s))·(dz(s)/ds)·ds
thus reducing a path integral to a regular definite integral.
To bring our problem close to physics, we can use time t as a parameter instead of abstract s, having position of a moving object to be a function of time
x = x(t)
y = y(t)
z = z(t)
In this case dx(t)/dt is an X-component of the velocity of a moving object that we can denote as ẋ(t) and similar for other coordinates.
Using time as a parameter that defines the position of a moving object, the path integral that calculates the work would look like
∫AB[F(r)·dr] =
= ∫[0,S]Fx(x(t),y(t),z(t))·ẋ(t)·dt +
+ ∫[0,S]Fy(x(t),y(t),z(t))·ẏ(t)·dt +
+ ∫[0,S]Fz(x(t),y(t),z(t))·ż(t)·dt
Either way, the calculation of a path integral of a vector-function have been reduced to regular integration.
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