Why is this closed line integral zero?

• I
• kmot
In summary, the Kelvin circulation theorem states that if density "rho" is a function of only pressure "p", then closed line integral of grad(p) / rho(p) equals zero. This theorem is implied by the gradient theorem, and can be proven using the following vector identity. Additionally, it can be shown that g(F(x)) * grad(F(x)) is also zero, by using the gradient identity and the Stokes theorem.

kmot

TL;DR Summary
By gradient theorem, if f(x)=grad(F(x)), where F(x) is conservative field, then closed line integral of f is zero. Books tell me that also g(F(x)) * grad(F(x)) is zero. I can't derive why.
This problem comes from fluid dynamics where Kelvin circulation theorem states, that if density "rho" is a function of only pressure "p", then closed line integral of grad(p) / rho(p) equals zero. It seems so trivial, so that no one ever gives reason for this claim.

When trying to solve it, I've generalized it to the following:
If f(x)=grad(F(x)), then closed line integral of f(x) is zero, by gradient theorem. From this, it somehow follows that g(F(x)) * grad(F(x)) is also zero.

I tried to rewrite:
And somehow show that the expanded form can be rewritten as a gradient of some function. I did not succeed.

Can somebody help me?

Thank you

What is g here, and what exactly does the multiplication by grad(F(x)) mean?

kmot said:

So $f$ is the vector field which is the gradient of the scalar field $F$? We would then say that $f$ is the conservative field.

then closed line integral of f is zero. Books tell me that also g(F(x)) * grad(F(x)) is zero. I can't derive why.

It's the gradient of $G(\mathbf{x}) = \int_0^{F(\mathbf{x})} g(t)\,dt$.

This problem comes from fluid dynamics where Kelvin circulation theorem states, that if density "rho" is a function of only pressure "p", then closed line integral of grad(p) / rho(p) equals zero. It seems so trivial, so that no one ever gives reason for this claim.

I don't think that's what the theorem actually says.

But in any event, $\nabla p / \rho$ can be written as $\nabla G = \frac{dG}{dp}\nabla p$ by taking $$\frac{dG}{dp} = \frac 1{\rho(p)}.$$

WWGD and Delta2
pasmith said:
So $f$ is the vector field which is the gradient of the scalar field $F$? We would then say that $f$ is the conservative field.

It's the gradient of $G(\mathbf{x}) = \int_0^{F(\mathbf{x})} g(t)\,dt$.

I don't think that's what the theorem actually says.

But in any event, $\nabla p / \rho$ can be written as $\nabla G = \frac{dG}{dp}\nabla p$ by taking $$\frac{dG}{dp} = \frac 1{\rho(p)}.$$
That's wonderful. Thank you.

OK, Kelvin theorem doesn't say that per se, but it is implied. It says:
For barotropic fluid, i.e. fluid where ##\rho## is a function of only ##p##
##\frac{D\Gamma}{Dt} = \frac{D}{Dt}(\oint \vec{v} \,dl ) = 0##
After some manipulations, mainly inserting Newton's second law of motion it implies:
##\oint \nabla{p} \cdot \frac{1}{\rho} \,dl = 0##

I've found yet another way to prove this.
Convert line integral to surface integral using Stokes theorem:
##\oint{\frac{1}{\rho} \cdot \nabla p \, dl } = \iint_{A}{\nabla \times \left( \frac{1}{\rho} \cdot \nabla p \right) \cdot \vec{n}\, dA } = \iint_{A}{\left(\nabla\frac{1}{\rho}\right) \times \left( \nabla p \right) \cdot \vec{n}\, dA } ##
Apply the following observation:
If ##\rho## depends only on ##p##, then their gradients point in the same direction. To see this it helps me to imagine their isosurfaces, which must be parallel.

Since cross product of parallel vectors is zero, then ##\left(\nabla\frac{1}{\rho}\right) \times \left( \nabla p \right) = 0 ##

Last edited:
Office_Shredder said:
What is g here, and what exactly does the multiplication by grad(F(x)) mean?
g is a generic function of which we know just that it depends only on F(X)

g is a function from ##\mathbb{R}\to \mathbb{R}##?

Office_Shredder said:
g is a function from ##\mathbb{R}\to \mathbb{R}##?
Yes. It's a density depending on pressure.

It looks like you solved it, but I believe I have another way that is very simple. There is a vector identity
## \nabla \times (u \vec{a})=u \nabla \times \vec{a}-\vec{a} \times \nabla u ##.
Let ## \vec{a}=\nabla F(\vec{x}) ##, and ## u=g(F(\vec{x})) ##.
The first term on the right side is zero because it is the curl of the gradient. With the second term, I think it gets ## \vec{a} \times \vec{a}=0 ## when you take ## \nabla u =g'(F(\vec{x})) \nabla F(\vec{x})=g'(F(\vec{x})) \vec{a} ##.
By Stokes the line integral ## \oint (u \vec{a}) \cdot dl=\iint \nabla \times (u \vec{a}) \cdot dS=0 ##.

Last edited:
Delta2