About the explaination in "Path independence implies gradient field" part, what if there does not exists a point where f(A) = 0 in the domain of f?

Then lower or rise f until f(A) is 0. Gradient won't change. :)

If there is a way to make sure that a vector field is path independent, I didn't manage to catch it in this article. Is it?, if not, can you please make it? Thanks

If the curl is zero (and all component functions have continuous partial derivatives), then the vector field is conservative and so its integral along a path depends only on the endpoints of that path.

if it is closed loop, it doesn't really mean it is conservative?

It is the vector field itself that is either conservative or not conservative. You can have a closed loop over a field that is conservative, but you could also have a closed loop over a field that is _not_ conservative. You'll talk about this second category more when you get to the later theorems.

Don't forces minimize potential energy? So shouldn't the force be negative of the gradient of potential energy?

Potential energies are defined to be negative, so their gradient would become positive to match the formula for force. For example, gravitational potential energy is given by -GMm/r, the gradient of which is GMm/r^2 which is also the formula for gravitational force.

Main content

Course: Multivariable calculus > Unit 4

Lesson 4: Line integrals in vector fields (articles)

Conservative vector fields

Google Classroom

Especially important for physics, conservative vector fields are ones in which integrating along two paths connecting the same two points are equal.

Background

Fundamental theorem of line integrals, also known as the gradient theorem.

What we're building to

A vector field

F (x, y)

is called a conservative vector field if it satisfies any one of the following three properties (all of which are defined within the article):

Line integrals of $F$ ‍ are path independent.
Line integrals of $F$ ‍ over closed loops are always $0$ ‍.
$F$ ‍ is the gradient of some scalar-valued function, i.e. $F = \nabla g$ ‍ for some function $g$ ‍.

There is also another property equivalent to all these:

F

is irrotational, meaning its curl is zero everywhere (with a slight caveat). However, I'll discuss that in a separate article which defines curl in terms of line integrals.

The key takeaway here is not just the definition of a conservative vector field, but the surprising fact that the seemingly different conditions listed above are equivalent to each other. Madness!

Path independence

Imagine you have any ol' off-the-shelf vector field

F (x, y)

, and you consider the line integrals of

F

of two separate paths,

C_{1}

and

C_{2}

, each starting at a point

A

and ending at a point

B

For almost all vector fields

F

, and almost all choices for the two paths

C_{1}

and

C_{2}

, these integrals will be different.

\begin{array}{r} \int_{C_{1}} F \cdot d s \neq \int_{C_{2}} F \cdot d s \leftarrow Almost always true \end{array}

And this makes sense! Each integral is adding up completely different values at completely different points in space. What's surprising is that there exist some vector fields where distinct paths connecting the same two points will always be equal, no matter the choice of paths (of which there are super-infinitely many).

In the last article, covering the gradient theorem we saw that in the special case of vector fields which are the gradient of some scalar-valued function,

\nabla f

, this magical property is true. The line integrals along distinct paths connecting the same two points

A

and

B

will always evaluate to the same thing:

\begin{array}{r} \int_{C_{1}} \nabla f \cdot d s = \underset{\begin{array}{c} Result of the \\ gradient theorem \end{array}}{\underset{⏟}{f (B) - f (A)}} = \int_{C_{2}} \nabla f \cdot d s \end{array}

Definition: This property is called path independence. Specifically, a line integral through a vector field

F (x, y)

is said to be path independent if the value of the integral only depends on the point where the path starts and the point where it ends, not the specific choice of path in between.

Actually, when you properly understand the gradient theorem, this statement isn't totally magical. This is because line integrals against the gradient of

f

measure the change in the value of

f

. Visualizing this with the graph of

f

, this says that any two paths bringing you from one point to another change your altitude by the same amount.

Khan Academy video wrapper

See video transcript

The takeaway from this result is that gradient fields are very special vector fields. Because this property of path independence is so rare, in a sense, "most" vector fields cannot be gradient fields.

Path independence implies gradient field

Okay, so gradient fields are special due to this path independence property. But can you come up with a vector field

F (x, y)

in which all line integrals are path independent, but which is not the gradient of some scalar-valued function?

I guess I've spoiled the answer with the section title and the introduction: All vector fields in which line integrals are path independent must be the gradient of some function. By why?

Really, why would this be true? Consider an arbitrary vector field

F (x, y)

in which line integrals are path independent, meaning

\begin{array}{r} \int_{C_{1}} F \cdot d s = \int_{C_{2}} F \cdot d s \end{array}

for all paths

C_{1}

and

C_{2}

which connect the same two points

A

and

B

. What is it about this property that ensures the existence of some function

g

such that

\nabla g = F

Challenge question: Can you think of a way to construct such a function

g

in terms of

F

using the fact that

F

is path-independent?

This is a tricky question, but it might help to look back at the gradient theorem for inspiration.

I could just spit out the answer here, but it's more fun and enlightening to walk through how you might discover the answer yourself.

Getting inspiration from the gradient theorem.

The gradient theorem relates the value of a function to the line integral of its gradient:

$\begin{array}{r} \int_{C} \nabla f \cdot d s = f (B) - f (A) \end{array}$ ‍

where

A

and

B

are points in space, and

C

is some curve connecting them.

Suppose it just so happened to be the case that

f (A) = 0

, then this equation would look like

$\begin{array}{r} f (B) = \int_{C} \nabla f \cdot d s \end{array}$ ‍

In other words, for any point

B

in space, you could compute

f

using just its gradient (along with a line integral).

Since our goal is to see how the vector field

F

can be seen as a gradient, this might spark the idea in your mind to replace

\nabla f

with

F

, producing a new scalar value function, which we'll call

g

$\begin{array}{r} g (B) = \int_{C} F \cdot d s \end{array}$ ‍

Usually, we define functions using formulas with respect to the coordinates

(x, y)

of a point, so this might feel like a wacky definition for

g

. To start,

A

is just some arbitrarily chosen point in space. Then the value of

g

at a point

B

is defined by taking an arbitrary curve connecting

A

B

, and evaluating the line integral of

F

along this curve. This is a very indirect definition, but it is a valid definition nevertheless. After all, it gives you a way to compute

g (x, y)

for any point

(x, y)

, doesn't it?

Notice, this definition would not make sense for most vector fields

F

. The choice of which path connects

A

B

would change the value of the integral, leaving the defining value of

g (B)

ambiguous. However, because line integrals are path independent in

F

, this is a real definition.

Defining

g

in this way, it turns out that

$\nabla g = F$ ‍

This should not be too surprising, if you think about what the gradient theorem looks like for

g

. I will not cover the proper proof of this fact here, since I think the details would distract from the main point:

Key takeaway: If line integrals are path independent in a vector field $F$ ‍, you can use those line integrals to define a function $g$ ‍ such that $\nabla g = F$ ‍.

Closed loops

Definition: A path is called closed if it starts and ends at the same point. Such paths are also commonly called closed loops.

For example, the path

C

pictured below starts and ends at

A

If we take a vector field

F

where all line integrals are path independent, the line integral of

F

on any closed loop will be

0

. Why?

Let

C

be a closed path, starting at

A

and ending at

A

, and let

\tilde{C}

be the reverse path. That is,

C

and

\tilde{C}

cover the same points in space, but when you parameterize them you should be walking in opposite directions.

We know that reversing the orientation of a path flips the sign of the line integral:

$\begin{array}{r} \int_{C} F \cdot d s = - \int_{\tilde{C}} F \cdot d s \end{array}$ ‍

This is a general statement about line integrals through a vector field, not specific to conservative vector fields. However, because

F

is path independent, and because both

C

and

\tilde{C}

start at

A

and end at

A

, it must also be true that

$\begin{array}{r} \int_{C} F \cdot d s = \int_{\tilde{C}} F \cdot d s \end{array}$ ‍

The only way both of these can be true is if the integral equals

0

. You can apply this argument to any closed loop, so the line integral over any closed loop must be

0

Take any closed path

C

, which starts and ends at the point

A

. Look at what the gradient theorem says about such a path for some scalar-valued function

f

$\begin{array}{r} \int_{C} \nabla f \cdot d s = f (A) - f (A) = 0 \end{array}$ ‍

Therefore, integrals over closed loops in a gradient field must always be zero. Since we just saw that vector fields in which line integrals are path independent must also be gradient fields, this answers the question.

The converse of this fact is also true: If the line integrals of

F

on all closed loops evaluate to

0

, then all line integrals must be path independent. Why?

Consider a two separate paths

C_{1}

and

C_{2}

, each going from point

A

to point

B

Now define a closed loop

C_{3}

to be a certain combination of these two paths, where you walk from

A

B

along

C_{1}

, then from

B

A

backwards along

C_{2}

The line integral of

F

along this combined path will be

$\begin{array}{r} \int_{C_{3}} F \cdot d s = \int_{C_{1}} F \cdot d s - \int_{C_{2}} F \cdot d s \end{array}$ ‍

The minus sign is because we are going backwards along

C_{2}

. On the other hand, we are assuming that the line integral on all closed loops is

0

, and

C_{3}

is a closed loop, so

$\begin{array}{r} \int_{C_{3}} F \cdot d s = 0 \end{array}$ ‍

Together, these equations imply

$\begin{aligned} \int_{C_{1}} F \cdot d s & - \int_{C_{2}} F \cdot d s = 0 \\ ⇓ \\ \int_{C_{1}} F \cdot d s & = \int_{C_{2}} F \cdot d s \end{aligned}$ ‍

Since

C_{1}

and

C_{2}

were arbitrarily chosen, this shows that all line integrals in

F

must be path independent.

Funky notation for closed-loop integrals.

You will sometimes see a line integral over a closed loop

C

written as

\begin{array}{r} \oint_{C} F \cdot d r \end{array}

Don't worry, this is not a new operation that needs to be learned. It is just a line integral, computed in just the same way as we have done before, but it is meant to emphasize to the reader that

C

is a closed loop.

Potential energy

In the article introducing line integrals through a vector field, I mentioned briefly how in physics, the work done by a force on an object in motion is computed by taking a line integral of the force's vector field along the path of motion.

\begin{array}{r} W = \int_{C} F \cdot d s \end{array}

A force is called conservative if the work it does on an object moving from any point

A

to another point

B

is always the same, no matter what path is taken. In other words, if this integral is always path-independent. Fundamental forces like gravity and the electric force are conservative, and the quintessential example of a non-conservative force is friction.

This has an interesting consequence based on our discussion above: If a force is conservative, it must be the gradient of some function.

F = \nabla U

Moreover, according to the gradient theorem, the work done on an object by this force as it moves from point

A

to point

B

can be computed just by evaluating this function

U

at each point:

\begin{aligned} W & = \int_{C} F \cdot d s \\ = \int_{C} \nabla U \cdot d s \\ = U (B) - U (A) \end{aligned}

As the physics students among you have likely guessed, this function

U

is potential energy. For example, if you take the gradient of gravitational potential or electric potential, you will get the gravitational force or electric force respectively. This is why computing the work done by a conservative force can be simplified to comparing potential energies.

It also means you could never have a "potential friction energy" since friction force is non-conservative.

Escher

Moving from physics to art, this classic drawing "Ascending and Descending" by M.C. Escher shows what the world would look like if gravity were a non-conservative force.

Closed loop perspective:

Imagine walking clockwise on this staircase. With each step gravity would be doing negative work on you. So integrating the work along your full circular loop, the total work gravity does on you would be quite negative. However, that's an integral in a closed loop, so the fact that it's nonzero must mean the force acting on you cannot be conservative.

Path independence perspective

Imagine walking from the tower on the right corner to the left corner. If you get there along the clockwise path, gravity does negative work on you. If you get there along the counterclockwise path, gravity does positive work on you. Since both paths start and end at the same point, path independence fails, so the gravity force field cannot be conservative.

Gradient perspective:

In the real world, gravitational potential corresponds with altitude, because the work done by gravity is proportional to a change in height. What makes the Escher drawing striking is that the idea of altitude doesn't make sense. Many steps "up" with no steps down can lead you back to the same point. This corresponds with the fact that there is no potential function $U$ ‍ such that $\nabla U$ ‍ give the gravity field.

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wcyi56
Posted 8 years ago. Direct link to wcyi56's post “About the explaination in...”
About the explaination in "Path independence implies gradient field" part, what if there does not exists a point where f(A) = 0 in the domain of f?
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(8 votes)
Answer
- alek aleksander
  Posted 6 years ago. Direct link to alek aleksander's post “Then lower or rise f unti...”
  Then lower or rise f until f(A) is 0. Gradient won't change. :)
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  (5 votes)
Hemen Taleb
Posted 8 years ago. Direct link to Hemen Taleb's post “If there is a way to make...”
If there is a way to make sure that a vector field is path independent, I didn't manage to catch it in this article. Is it?, if not, can you please make it? Thanks
Button navigates to signup pageComment on Hemen Taleb's post “If there is a way to make...”
(4 votes)
Answer
- T H
  Posted 8 years ago. Direct link to T H's post “If the curl is zero (and ...”
  If the curl is zero (and all component functions have continuous partial derivatives), then the vector field is conservative and so its integral along a path depends only on the endpoints of that path.
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  (7 votes)
Andrea Menozzi
Posted 7 years ago. Direct link to Andrea Menozzi's post “any exercises or example ...”
any exercises or example on how to find the function g?
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(3 votes)
Answer
- Ardra
  Posted 13 days ago. Direct link to Ardra's post “Yes, Sal has covered in t...”
  Yes, Sal has covered in this course earlier. Go to multivariable calculus course and unit 4.
  
  In short, you integrate the P(x,y) function, which is the x-component of your vector field with respect to x (assume y= constant) and add a pure function of y like f(y) because it could go to zero when differentiated to find the gradient (remember that we are trying to find the scalar function whose gradient field is the vector function that we possess), do the same for Q(x,y) and see if you can write a U(x,y) so that when you differentiate partially with respect to x you get P(x,y) and Q(x,y) when with y.
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  (1 vote)
012010256
Posted 7 years ago. Direct link to 012010256's post “Just curious, this curse ...”
Just curious, this curse includes the topic of The Helmholtz Decomposition of Vector Fields?
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(2 votes)
Answer
Aravinth Balaji R
Posted 8 years ago. Direct link to Aravinth Balaji R's post “Can I have even better ex...”
Can I have even better explanation Sal? I would love to understand it fully, but I am getting only halfway. Could you please help me by giving even simpler step by step explanation?
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(3 votes)
Answer
- Ad van Straeten
  Posted 8 years ago. Direct link to Ad van Straeten's post “Have a look at Sal's vide...”
  Have a look at Sal's video's with regard to the same subject!
  Have fun!
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  (0 votes)
Jonathan Sum AKA GoogleSearch@arma2oa
Posted 7 years ago. Direct link to Jonathan Sum AKA GoogleSearch@arma2oa's post “if it is closed loop, it ...”
if it is closed loop, it doesn't really mean it is conservative?
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(1 vote)
Answer
- Will Springer
  Posted 7 years ago. Direct link to Will Springer's post “It is the vector field it...”
  It is the vector field itself that is either conservative or not conservative.
  
  You can have a closed loop over a field that is conservative, but you could also have a closed loop over a field that is not conservative. You'll talk about this second category more when you get to the later theorems.
  Button navigates to signup page
  (3 votes)
Naman
Posted a year ago. Direct link to Naman's post “Don't forces minimize pot...”
Don't forces minimize potential energy? So shouldn't the force be negative of the gradient of potential energy?
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(1 vote)
Answer
- Ryan Lee
  Posted a year ago. Direct link to Ryan Lee's post “Potential energies are de...”
  Potential energies are defined to be negative, so their gradient would become positive to match the formula for force. For example, gravitational potential energy is given by -GMm/r, the gradient of which is GMm/r^2 which is also the formula for gravitational force.
  Button navigates to signup page
  (2 votes)
jp2338
Posted 2 years ago. Direct link to jp2338's post “quote > this might spark ...”
quote > this might spark the idea in your mind to replace \nabla f∇fdel, f with \textbf{F}Fstart bold text, F, end bold text, producing a new scalar value function, which we'll call g

I don't understand why this idea comes out at all. why ? what is the purpose if this replacement ? very ambiguous
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(1 vote)
Answer
White
Posted 7 years ago. Direct link to White's post “All of these make sense b...”
All of these make sense but there's something that's been bothering me since Sals' videos.

If the integral over a gradient field of a closed loop is zero. Why did Sal's example evaluate to -2*pi and Grant's previous example evaluated to 2*pi?

A circle is a closed loop. What gives?
-----
Edit:
Oh wait, silly me.
-yi^ + xj^ isn't a gradient field now is it?
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(1 vote)
Answer
- Rubén Jiménez
  Posted 7 years ago. Direct link to Rubén Jiménez's post “no, it can't be a gradien...”
  no, it can't be a gradient field, it would be the gradient of the paradox picture above. If the arrows point to the direction of steepest ascent (or descent), then they cannot make a circle, if you go in one path along the arrows, to return you should go through the same quantity of arrows relative to your position, but in the opposite direction, the same work but negative, the same integral but negative, so that the entire circle is 0
  Comment on Rubén Jiménez's post “no, it can't be a gradien...”
  (1 vote)
John Smith
Posted 3 years ago. Direct link to John Smith's post “Correct me if I am wrong,...”
Correct me if I am wrong, but why does he use F.ds instead of F.dr ? ds is a tiny change in arclength is it not? if it is a scalar, how can it be dotted?

Secondly, in the Path independence implies gradient part, doesn't defining g(B) the way it is, already use the fundamental theorem, implying that F is already a gradient field?
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(1 vote)
Answer
- adam.ghatta
  Posted 2 years ago. Direct link to adam.ghatta's post “dS is not a scalar, but r...”
  dS is not a scalar, but rather a small vector in the direction of the curve C, along the path of motion.
  
  For the gradient of a potential function U, the vector field f created from grad(U) is path independent by definition. The fundamental theorem simply relies on the fact, that gradient fields are path-independent. The fundamental gradient theorem that allows us to use f(B) - f(A) only suffices if the gradient of the potential function f exists.
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  (1 vote)

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Multivariable calculus