Path-independent or conservative vector fields

Many physical force fields (vector fields) that you are familiar with are conservative vector fields. The term comes from the fact that some kind of energy is conserved by these force fields. The important consequence for us, though, is that as you move an object from point a to point b, the work performed by a conservative force field does not depend on the path taken from point a to point b. For this reason, we often refer to such vector fields as path-independent vector fields. Path-independent and conservative are just two terms that mean the same thing.

For example, imagine you have to carry a heavy box from your front door to your bedroom upstairs. Because of the gravity (which can be viewed as a force field), you have to do work to carry the box up.

Note: here we mean the scientific definition of work, which as you remember, is force times distance. Although it may feel like work to move the box from one room to another on the same floor, the actual work done against gravity is zero.

Next, imagine that you have two stairways in your house: a gently sloping front staircase, and a steep back staircase. Since the gravitational field is a conservative vector field, the work you must do against gravity is exactly the same if you take the front or the back staircase. As long as the box starts in the same position and ends in the same position, the total work is the same. (In fact, if you decided to first carry the box to your neighbor’s house, then carry it up and down your backyard tree, and then in your back door before taking it upstairs, it wouldn’t make a difference. The net work you performed against gravity would be the same.)

Remember that the line integral of a vector field can be viewed as the total work performed by the force field on an object moving along the path. (OK, one technical point that you can ignore if it doesn’t make sense to you. For the above gravity example, we discussed the work you performed against the gravity field. This work is exactly opposite the work performed by the gravity field. We’d need to multiply the line integral by -1 to get the work you performed against the gravity field, but that’s a technical point we don’t need to worry much about.)

The above discussion implies that if F is a conservative vector field, then the integral

∫ CF ⋅ ds
does not actually depend on the path C. It only depends on the beginning point (say a) and endpoint (say b) of the path C. If we changed the path C to a different path B, then we know that
∫ CF ⋅ ds = ∫ BF ⋅ ds
as long as the path B also starts at the point a and ends at the point b. This property is certainly not true for all vector fields F (as shown below). It is only true for the special class of path-independent vector fields (or conservative vector fields).

The vector field F(x,y) = (x,y) is a path-independent vector field. (You’ll learn how to test for path-independence later. For now, take it on faith.) It is illustrated by the black arrows in the below figure. We want to compute the integral

∫ CF ⋅ ds
where C is a path from the point a = (3,-3) (shown by the cyan square) to the point b = (2, 4) (shown by the magenta square). Since F is path-independent, we don’t need to know anything else about the path C to compute the line integral. Later, you’ll learn to compute that the value of the integral is 1, as shown by the magenta line on the slider below the figure.

To demonstrate the path-independence of F, I’ve shown three paths from a to b below. Path C (shown in blue) is a straight line path from a to b. Paths B (in green) and E (in red) are curvy paths, but they still start at a and end at b. Each path has a colored point on it that you can drag along the path. The similarly colored lines on the slider indicate the line integral along each curve, starting at the point a (the cyan square) and ending at the movable point. If you drag a point up to b (the magenta square), the corresponding colored line on the slider shows the total integral along the path. As you can see, the total integral along each path is equal to 1 (the magenta line on the slider). The vector field appears to be path-independent, as promised. (You’d have to check all the infinite number of possible paths from all points a to all points b to determine that F was really path-independent. Fortunately, you’ll learn some simpler methods.)

Of course, the line integrals to intermediate points can vary widely. By comparing the paths to the vector field, you should be able to explain the behavior of the line integral to intermediate points. In particular, you can determine why the line integral increases or decreases by comparing the direction of motion along the path to the direction of the vector field.

A path-dependent vector field

As mentioned above, not all vector fields are path-independent. If a vector field is not path-independent, we call it path-dependent. The vector field F(x,y) = (y,-x) is an example of a path-dependent vector field. We saw earlier that the vector field represents clockwise circulation around the origin. It turns out, such circulation is the key indicator of path-dependence (however, the circulation may not be as obvious as it is in this example). Again, details will come later.

The vector field and three paths from a = (0,-4) (the cyan square) to b = (0, 4) (the magenta square) as shown in the figure below. Since the vector field is path-dependent, the line integral from a to b isn’t defined until you specify a path from a to b. The value of the line integral will depend on which path you choose.

Path C (shown in blue) is a straight line path from a to b. Paths B (in green) and E (in red) are clockwise and counterclockwise, respectively, circular path from a and to b. As above, you can drag the colored points to b (the cyan square), and the value of the line integrals are shown by the slider below. These paths demonstrate the F is indeed path-dependent, as the three integrals from a to b have different values. Indeed, you only need to show that the integrals along two paths from some point a to some other point b are not the same in order to prove that F was path-dependent (since if F were path-independent, this could never happen).

Path C (in blue) is a straight line up the y-axis. Along the y-axis, F(0,y) = (y, 0) is horizontal. Hence the direction of motion and the vector field are orthogonal, and

∫ CF ⋅ ds = 0.
On the other hand, path B (in green) goes with the clockwise rotation of the vector field. Consequently, the work performed by the vector field is positive, and
∫ BF ⋅ ds > 0.
The same reasoning shows that
∫ EF ⋅ ds < 0,
since the counterclockwise direction of path E (in red) is opposite the direction of the vector field. Since all three paths start at a and end at b, these results confirm that F is path-dependent.

You can read an example of a conservative field.