# What do Stokes' and Green's theorems represent?

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• NicolaiTheDane
In summary, Stokes' theorem states that the line integral of a vector field around a closed loop is equal to the surface integral of the curl of that vector field.
NicolaiTheDane
I have been spending an embarrassing amount of time, trying to figure out what these two theorems are actually telling me.

As I understand it, it is suppose to tell me, what the "curl" around a boundary is. However there are several examples I can find, where this doesn't make sense. My understanding of curl, is that it is expression of rather there is a tendency for turn or not (vague, hopefully good enough). So a field like $$v = (y^2)*i$$ would have curl, because the further away you get from y = 0, the higher strength the field has.

Now the part that doesn't make sense to me, is when asked to compute the curl around a triangle, with the points (0,0), (0,4), and (0,2), it seems intuitive to me, that the flat line, along the y axis, would provide some, if not most of the curl to the triangle, as the vector field would push more on the outer end of the line, then on the inner end. However when I calculate it, it seems that the one side that to me would provide no curl, provides most of it (the one running along the x axis), while the one along the y axis, provides 0 curl.

This suggest to me, that there is something I simply do not understand. Also I do not understand, why curl around a closed curve, is always zero. To me that seems to suggest curl around a volume is impossible.

I guess what I am asking, if I were to boil it down, is just what is curl. What am I calculating with Greens and Stokes theorem, because it cannot possible be what I think it is.

I think you are confusing the curl with the line integral around a closed loop as in Stokes' theorem. The curl of a vector is a vector field itself. Loosely it can be thought of as a source of circulation. Stokes' theorem relates the line integral of the vector field around a closed loop to a surface integral of the curl of that vector field bound by the loop. In your example the vector field is ##\vec v = y^2~\hat i##. You can calculate its curl to find ##\vec{\nabla}\times \vec v=-2y~\hat k##. Now you need a loop. What you have provided is not a triangle because points (0,0), (0,4) and (0,2) are all on the y-axis. So let's go with (0,0), (0,4) and (2,0) that form a right triangle. It will be instructive for you to verify Stokes' theorem by doing the surface integral ##\int {\vec{\nabla}\times \vec v \cdot \hat n~dA}## over the area of the triangle and compare the result with the line integral ##\oint \vec v \cdot d\vec s## around the perimeter of the triangle. The two should be equal as required by Stokes' theorem.

Note: ##\hat n## is a unit vector normal to the surface. The circulation of the line integral and the direction of ##\hat n## are linked by the right hand rule.

kuruman said:
I think you are confusing the curl with the line integral around a closed loop as in Stokes' theorem. The curl of a vector is a vector field itself. Loosely it can be thought of as a source of circulation. Stokes' theorem relates the line integral of the vector field around a closed loop to a surface integral of the curl of that vector field bound by the loop. In your example the vector field is ##\vec v = y^2~\hat i##. You can calculate its curl to find ##\vec{\nabla}\times \vec v=-2y~\hat k##. Now you need a loop. What you have provided is not a triangle because points (0,0), (0,4) and (0,2) are all on the y-axis. So let's go with (0,0), (0,4) and (2,0) that form a right triangle. It will be instructive for you to verify Stokes' theorem by doing the surface integral ##\int {\vec{\nabla}\times \vec v \cdot \hat n~dA}## over the area of the triangle and compare the result with the line integral ##\oint \vec v \cdot d\vec s## around the perimeter of the triangle. The two should be equal as required by Stokes' theorem.

Note: ##\hat n## is a unit vector normal to the surface. The circulation of the line integral and the direction of ##\hat n## are linked by the right hand rule.

Considering its going to be the same, I'll just do the line integral. Divided into 3 separat intergrals, one for each line I get:

$$\int_0^1 \begin{bmatrix} (0)^2\\ 0 \end{bmatrix}*\begin{bmatrix} 4\\ 0 \end{bmatrix} \, dt + \int_0^1 \begin{bmatrix} (2*t)^2\\ 0 \end{bmatrix}*\begin{bmatrix} -4\\ 2 \end{bmatrix} \, dt + \int_0^1 \begin{bmatrix} (-2*t+2)^2\\ 0 \end{bmatrix}*\begin{bmatrix} 0\\ -2 \end{bmatrix} \, dt = -\frac{16}{3}$$

Now the way I interpret this, is the triangle would be turned clockwise in the field. Or that walking along this line, would require "work", as opposed to provide it.

Is this correct? Also in 3D I assume it would be the same. I think of this, as moving through the field, on the given path, NOT as a surface being turned.

NicolaiTheDane said:
Is this correct?
I can't tell. You are using a method for doing line integrals that is unfamiliar to me. Can you post a picture of the triangle that you used? I will then do the line integral my way and compare my results with yours.
NicolaiTheDane said:
Considering its going to be the same ...
If you mean that the surface integral of the curl is the same as the line integral, it will be best for you not to assume that if you are uncertain about Stokes' theorem. The idea is to see how the two sides are put together by calculating each separately and then verifying that Stokes' theorem holds.

kuruman said:
I can't tell. You are using a method for doing line integrals that is unfamiliar to me. Can you post a picture of the triangle that you used? I will then do the line integral my way and compare my results with yours.

If you mean that the surface integral of the curl is the same as the line integral, it will be best for you not to assume that if you are uncertain about Stokes' theorem. The idea is to see how the two sides are put together by calculating each separately and then verifying that Stokes' theorem holds.

This is the triangle. Now what I'm doing, is doing the line integral from A to B, then B to C, and lastly C to A, parameterizing my path to the variable t. So for example with the first integral, my path would be

$$r = \begin{bmatrix} 4*t \\ 0 \end{bmatrix}$$

The first I say 4*t instead of t, is simply to make the integral easier. Now the way I think of it, is that

$$r = \begin{bmatrix} x \\ y \end{bmatrix} \Rightarrow dr = \begin{bmatrix} dx \\ dy \end{bmatrix}$$
Using ##a=\frac{a}{b}*b##
$$|dr| = \sqrt{(x'(t)*dt)^2+(y'(t)*dt)^2} = \sqrt{x'(t)^2*(dt)^2+y'(t)^2*(dt)^2} = \sqrt{(x'(t)^2+y'(t)^2)*(dt)^2} = \sqrt{x'(t)^2+y'(t)^2}*dt \Rightarrow dr = \begin{bmatrix} x'(t) \\ y'(t) \end{bmatrix}*dt$$

Now the line integral becomes

\begin{align} \int_0^t v(r(t))*dr &= \int_0^t v\left(\begin{bmatrix} x \\ y \end{bmatrix}\right)*\begin{bmatrix} x'(t) \\ y'(t) \end{bmatrix} dt \\ &= \int_0^1 v\left(\begin{bmatrix} 4*t \\ 0 \end{bmatrix}\right)*\begin{bmatrix} 4 \\ 0 \end{bmatrix} dt \\ &= \int_0^1 \begin{bmatrix} (0)^2 \\ 0 \end{bmatrix}*\begin{bmatrix} 4 \\ 0 \end{bmatrix} dt \\ &= \int_0^1 (0^2*4+0*0) dt \\ &= \int_0^1 0 dt \\ &= 0 \end{align}

I'm not saying its pretty, but that was the idea.

Now again my problem isn't so much with understand why Stokes is equal to a given line integral, its understanding what exactly these results mean. I keep getting things mixed up in my head. Like in a scalar field, a line integral tells you the area beneath the curve. Simple. But for some reason in a vector field, I keep thinking I'm getting an area, or a line length, or some other thing. I know its ridiculous, but that is what happens. I'm guessing I figured it out, and that a line integral in a vector field, basically just tells you, rather the vector field is working against the movement, or not.

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Thank you for posting the triangle picture. I sort of understand your method of using parameter ##t##, but I don't understand why introduce the added complication. Be that as it may, your answer for the line integral in #5 is ##0## whereas in #3 it is ##-\frac{16}{3}.## This discrepancy shows that (a) using your method may lead to mistakes and (b) you might have picked up the error if you did the surface integral of the curl to check your answer. Here is the way I do line integrals.
1. You start with ##\oint \vec v \cdot d\vec s## and find an expression for the integrand.
##\vec v \cdot d\vec s=(y^2~\hat i)\cdot (dx~\hat i +dy ~\hat j)=y^2dx##
2. You break the total integral into 3 legs (as you did) and calculate each separately.
##I_{AB}=0## because ##y=0## on the x-axis.
To do ##I_{BC}## note that the straight line giving the hypotenuse is ##y = -\frac{1}{2}x+2##. Then
##I_{BC}=\int_4^0 (-\frac{1}{2}x+2)^2dx=-\frac{16}{3}.##
##I_{CA}=0## because ##dx=0## along the y-axis.
3. Add the results from each leg.
The answer is ##\oint \vec v \cdot d\vec s=0+(-\frac{16}{3})+0=-\frac{16}{3}.##

The curl is ##\vec{\nabla}\times \vec v=-2y~\hat k##. According to the right hand rule, the normal to the surface is ##\hat n = \hat k## because the line integration was done counterclockwise. Note the connection between circulation and direction. Then ##(\vec{\nabla}\times \vec v) \cdot \hat n~dA=(-2y~\hat k)\cdot \hat k~dA=-2y~dx~dy.## The surface integral then is $$\int_A (\vec{\nabla}\times \vec v) \cdot \hat n~dA=\int_0^4{dx}\int_0^{-\frac{1}{2}x+2}(-2y)dy=-\frac{16}{3}.$$
NicolaiTheDane said:
... basically just tells you, rather the vector field is working against the movement, or not.
I don't understand what you mean by movement. What is moving? In this example you can imagine the curl of ##\vec v## as a source of circulation of ##\vec v## field lines. Over the area of the triangle the strength of the circulation source at each point varies with position. If you add all the sources over the area of the triangle, you get a measure of the overall circulation of the field around the boundary of the area. Doing the line integral is another way to get the same measure. The negative sign here indicates that the circulation in this case is clockwise, opposite to ABCA.

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NicolaiTheDane and FactChecker
kuruman said:
Thank you for posting the triangle picture. I sort of understand your method of using parameter ##t##, but I don't understand why introduce the added complication. Be that as it may, your answer for the line integral in #5 is ##0## whereas in #3 it is ##-\frac{16}{3}.## This discrepancy shows that (a) using your method may lead to mistakes and (b) you might have picked up the error if you did the surface integral of the curl to check your answer.

No I got the same in both posts. I just only did 1 line of the triangle in the second post, in order to be very specific with what I did. If I did it for last to lines of the triangle, the result would be the same, for the same reasons you find. :)

kuruman said:
Here is the way I do line integrals.

From what I can see, we are essentially doing the same thing, except you don't automatically parameterize to a common variable. This I usually do, simply because it makes things easier for me, but I prefer your way.

Doing things with Stokes, is definitely something I'd need some practice in. Luckily that is what old exam sets are for. But fundamentally my problem was understanding how to think of it, not so much the calculating part. I'm simply unsure, as to what the result represents.

Perhaps you might visualize Stokes' theorem better with a concrete example from physics if you are familiar with Faraday's law. In differential form it is
$$\vec{\nabla}\times \vec{E}=-\frac{\partial \vec B}{\partial t}$$ where ##\vec E## and ##\vec B## are respectively the electric and magnetic field at a point in space. This is a law in the sense that it describes how Nature works, but in the form shown it's not very informative. Now suppose we define an area ##A## and a closed loop ##C## bounding the area. I can perform the following valid algebraic steps to Faraday's law:
1. Take the dot product on both sides with directed area element ##\hat n~dA## (##\hat n## is locally normal to ##dA##.)$$(\vec{\nabla}\times \vec{E})\cdot \hat n~dA=-\frac{\partial}{\partial t}(\vec B\cdot \hat n~dA)$$
2. Integrate over area ##A##$$\int_A (\vec{\nabla}\times \vec{E})\cdot \hat n~dA=-\frac{\partial}{\partial t}\int_A \vec B\cdot \hat n~dA$$
3. Apply Stokes' theorem to the left side to get$$\oint_C \vec{E} \cdot d\vec l=-\frac{\partial}{\partial t}\int_A \vec B\cdot \hat n~dA$$
Now this form of Faraday's law is more informative. It says that a time-varying magnetic flux ##(\vec B\cdot \hat n~dA)## through the area is the same as the line integral of the electric field bounding the area. It is a law of Nature, it is established experimentally, and indicates that the electric and magnetic fields are related; a changing magnetic flux generates a circulating magnetic field. This is one of Maxwell's equations. There are three more equations, but I won't go into them. You might wish to read about them here https://en.wikipedia.org/wiki/Maxwell's_equations.

fresh_42

## 1. What is Stokes' theorem and what does it represent?

Stokes' theorem is a fundamental theorem in vector calculus that relates a surface integral of a vector field to a line integral of the same vector field along the boundary of the surface. It represents the fundamental connection between the line and surface integrals and is used to evaluate the circulation of a vector field around a closed path.

## 2. How is Stokes' theorem related to Green's theorem?

Green's theorem is a special case of Stokes' theorem, which specifically applies to two-dimensional vector fields. It relates a double integral over a region in the xy-plane to a line integral around the boundary of that region. In this way, Green's theorem is a simplified version of Stokes' theorem for two-dimensional problems.

## 3. What is the significance of Green's theorem in physics?

Green's theorem is commonly used in physics to solve problems involving conservative vector fields, which are those that have a potential function. It allows for the calculation of work done by a conservative force, such as gravity, and can be applied in many areas of physics, including electromagnetism and fluid mechanics.

## 4. Can Stokes' theorem be applied to any surface?

Stokes' theorem can be applied to any surface that is bounded by a simple closed curve, meaning that the boundary of the surface does not intersect itself. It can also be extended to more complex surfaces by breaking them down into smaller, simpler surfaces and applying the theorem to each one separately.

## 5. How is Stokes' theorem used in real-world applications?

Stokes' theorem has numerous applications in physics and engineering, such as calculating the flow of a fluid through a closed loop or determining the force on a moving object in a magnetic field. It is also used in more abstract mathematical concepts, such as differential forms and de Rham cohomology.

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