# A Time differentiation of fluid line integrals

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1. Apr 7, 2017

### meteo student

I am looking at a proof from a book in fluid dynamics on time differentiation of fluid line integrals -

Basically I am looking at the second term on the RHS in this equation

$$d/dt \int_L dr.A = \int_L dr. \partial A / \partial t + d/dt \int_L dr.A$$

The author has a field vector A for a line of fluid particles at time t at a position (1,2). After the time increment delta t the particle 1 moves to 1' and the particle 2 moves to 2'. Each particle has moved the distance v delta t. The author applies Stokes integral theorem to the second tem on the RHS of the above equation

$$dS = dr \times v\Delta t$$.

which is surrounded by the close curve $$\Gamma$$ enclosing the points (1,2,2',1',1)

So he substitutes these into Stokes integral theorem

$$\Delta t\int_1^2 (dr \times v) . (\nabla \times A)$$ and it looks like he has a scalar quadruple product. Two cross products multiplied by a dot product.

In the book from which I am learning they show the above integral equal to

$$\Delta t\int_1^2 (dr \times v) . (\nabla \times A)=\int_1^2 (dr . A) + \Delta t(v. A)_2 - \int_{1'}^{2'} (dr' . A) -\Delta t(v.A)_1$$

Can somebody how the four terms on the RHS have been obtained ?

2. Apr 8, 2017

### zwierz

It would be useful to write all this stuff in the modern mathematical terms at last
So consider a vector field $v(x)=(v^1,\ldots,v^m)(x)$ on a smooth manifold $M$ with local coordinates $x=(x^1,\ldots, x^m)$. Let $g^t$ stand for the corresponding phase flow. Assume we have a compact $n-$ dimensional submanifold $N\subset M,\quad n\le m$.
The main fact is as follows. Let $\omega$ be a $n-$ form on $M$ then
$$\frac{d}{dt}\Big|_{t=0}\int_{g^t(N)}\omega=\int_NL_v\omega\qquad (*)$$
here $L_v$ is the Lie derivative. This fact is proved directly by the change of variables (guess which change in the integral and by definition of Lie derivative.
Using the formula $L_v\omega=i_vd\omega+di_v\omega$ and the Stokes theorem (if it is possible) one can rewrite (*) as follows
$$\frac{d}{dt}\Big|_{t=0}\int_{g^t(N)}\omega=\int_Ni_vd\omega+\int_{\partial N}i_v\omega.$$

A case when the vector field $v$ and the coefficients of the form $\omega$ depend on time $t$ is reduced to this one with the help of the following trivial trick. Complete the system $\dot x=v(x,t)$ with differential equation $\dot t=1$. We then obtain an autonomous vector field $\widetilde v=(1,v^1,\ldots,v^m)$ and the form $\omega$ on a manifold $\widetilde M=M\times\mathbb{R}_t$.

The Kelvin-Helmholtz fluid theorems and many other theorems from different branches of physics follow from these two formulas.

Last edited: Apr 8, 2017
3. Apr 11, 2017

### meteo student

Thanks for your solution. It does not help me with my problem does it ?

Can the same result be proved using Levi Cevita's symbol ?

Last edited: Apr 11, 2017
4. Apr 11, 2017

### zwierz

I have answered your questions indeed. But I employed sufficiently advanced mathematical language. Actually it was just a methodological remark. Later when you will study differential forms reread my post.

5. Apr 11, 2017

### meteo student

Thank you for your answer. Unfortunately I have had any exposure to Lie Derivatives. Differential forms yes but not Lie Derivatives. Oh well one day hopefully.

6. Apr 20, 2017

### lavinia

So in the OP's case what is the differential form $ω$ and what is the submanifold $N$?

7. Apr 20, 2017

### zwierz

In OP's case $\widetilde M=\mathbb{R}^3\times\mathbb{R}_t$ and the submanifold $N$ is the curve $L$. Introduce in $\mathbb{R}^3$ standard Cartesian frame $xyz$ and let $(A_x,A_y,A_z)$ be the coordinates of OP's vector field $A$ all the components are functions of $(x,y,z,t)$. Then $\omega= A_xdx+A_ydy+A_zdz$. If $u=(u_x,u_y,u_z)$ is the velocity field of the fluid then put $v=(u_x,u_y,u_z,1)$.

8. Apr 20, 2017

### meteo student

What is $$R_t$$ in that cross product ? ?

9. Apr 20, 2017

### zwierz

$\mathbb{R}_t$ stands for the line $\mathbb{R}$ marked with coordinate $t$
$\mathbb{R}^3\times \mathbb{R}_t=\{(x,y,z,t)\in\mathbb{R}^4\}$

10. Apr 21, 2017

### lavinia

This from the OP's original post seems wrong

$d/dt \int_L dr.A = \int_L dr. \partial A / \partial t + d/dt \int_L dr.A$

11. Apr 21, 2017

### meteo student

I copied the equation as is from the book. So I can check it again.

I checked it again. It is as it is there in the book. I only omitted one thing. The L limit of the integral is actually L(t). I did not how to do that using MathJax.

$$d/dt ( \int_L dr.A )= \int_L dr. \partial A / \partial t + d/dt \int_L dr.A$$

Reference - Dynamics of the atmosphere - A Bott. Pages 115-125

12. Apr 21, 2017

### lavinia

Why aren't the left term and the second term on the right the same?

13. Apr 21, 2017

### meteo student

I added the parenthesis on the left hand side and the book states that is RHS is an instance of the product rule of calculus.

The first term is change with time of a vector field A for a line fixed in space at time t while the second term on the RHS refers to vector field A being fixed but there is a deformation and displacement of the line during the time Delta t.

14. Apr 21, 2017

### lavinia

So $∫_{L(t)}ω = ∫_{L(t)}dr.A$

15. Apr 21, 2017

### zwierz

Yes, but to prove something it is better to write it formally
In the space $(x,y,z,t)$ we have a system $$\frac{d x}{d\tau}=u_x(x,y,z,t),\quad \frac{d y}{d\tau}=u_y(x,y,z,t),\quad \frac{d z}{d\tau}=u_z(x,y,z,t),\quad \frac{d t}{d\tau}=1$$ and $g^\tau$ is a phase flow of this system. So we consider the integral
$\int_{g^\tau(L)}\omega$

Last edited: Apr 21, 2017
16. Apr 21, 2017

### zwierz

by the way, second integral formula of #2 implies also the fundamental result from Hamiltonian mechanics : $\oint_{\mathcal L(t)}p_idq^i-Hdt=const$

17. Apr 22, 2017

### lavinia

I think t would be helpful for the OP if you broke down your differential form formulation into the vector calculus terms that he is working with.

Last edited: Apr 22, 2017
18. Apr 22, 2017

### meteo student

That would be great. Even if he could relate the Lie derivative to my vector calculus terms that would be great as well.

19. Apr 23, 2017

### zwierz

I will try to write something but it hardly be clear, it is very incomplete and it just to encourage OP to start reading a modern textbook.

So let $\boldsymbol A=(A_x,A_y,A_z),\quad \boldsymbol B=(B_x,B_y,B_z)$ be vector fields in $\mathbb{R}^3$ with standard Cartesian frame $x,y,z$.And let $f(x,y,z)$ be a function.

Construct the following correspondences
$$\boldsymbol A\mapsto \omega^1_{\boldsymbol A}=A_xdx+A_ydy+A_zdz,\quad \boldsymbol A\mapsto\omega^2_{\boldsymbol A}=A_xdy\wedge dz+A_ydz\wedge dx+A_zdx\wedge dy,\quad f\mapsto\omega^3_f=fdx\wedge dy\wedge dz.$$
Then it follows that $d\omega^1_{\boldsymbol A}=\omega^2_{\mathrm{rot}\,\boldsymbol A},\quad d\omega^2_{\boldsymbol A}=\omega^3_{\mathrm{div}\,\boldsymbol A}$
and
$$i_{\boldsymbol B}\omega^1_{\boldsymbol A}=(\boldsymbol A,\boldsymbol B),\quad i_{\boldsymbol B}\omega^2_{\boldsymbol A}=\omega^1_{\boldsymbol A\times\boldsymbol B},\quad i_{\boldsymbol B}\omega^3_f=f\omega^2_{\boldsymbol B}.$$

Now turn to OP's case. Let $l(t)$ be a curve with ends $1,2$. This curve is an image of some curve $l_0$ under the flow with vector field $\boldsymbol v(x,y,z,t)$ From above formulas we get
$$\frac{d}{dt}\int_{l(t)}\omega^1_{\boldsymbol A}=\int_{l(t)}\omega^1_{\frac{\partial \boldsymbol A}{\partial t}+(\mathrm{rot}\,\boldsymbol A)\times \boldsymbol v}+(\boldsymbol A,\boldsymbol v)\Big|_2-(\boldsymbol A,\boldsymbol v)\Big|_1.$$

20. Apr 23, 2017

### meteo student

Good post to clarify my doubts on differential forms.

I understand if you differentiate a 1 form you get a 2-form and if you differentiate a 2-form you get a three form and finally differentiating a 3-form gives you back a 1-form.

I presume rot refers to curl and div for divergence. So that part seems straightforward.

may I know what is 'i' is ?

21. Apr 23, 2017

### zwierz

22. Apr 23, 2017

### zwierz

yes
no. differential of3-form in $\mathbb{R}^3$ is equal to zero

23. Apr 23, 2017

### meteo student

I am going to get this notation clarified in my own words.

From the definition of the inner product from Wikipedia

In the first one you are having an interior product of a differential 1-form of the vector field A with the vector field B and the result is a inner product of A and B.

Second case you have an interior product of 2-form of the vector field A with the vector field B and the result is a 1-form but I am not sure what the AXB subscript notation refers to.

Third one you have an interior product of a 3-form with a B vector field and the result is a 2-form of the vector field B.

24. Apr 23, 2017

### zwierz

the cross product of two vectors https://en.wikipedia.org/wiki/Cross_product

25. Apr 23, 2017

### meteo student

How can a 1-form be a cross product ? I am not getting the notation.