Time differentiation of fluid line integrals

[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 ?

 

[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.

 

[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 ?

 

[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.

 

[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.

 

[lavinia]

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.

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

 

[zwierz]

So in the OP's case what is the differential form ωω and what is the submanifold NN?

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 $v=(u_x,u_y,u_z,1)$.

 

[meteo student]

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

 

[zwierz]

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

 

[lavinia]

This from the OP's original post seems wrong

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

 

[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

 

[lavinia]

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

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

 

[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.

 

[lavinia]

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 $v=(u_x,u_y,u_z,1)$.

So $∫_{L(t)}ω = ∫_{L(t)}dr.A$

 

[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
$\int_{g^\tau(L)}\omega$

 

[zwierz]

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

 

[lavinia]

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
$\int_{g^\tau(L)}\omega$

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.

 

[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.

 

[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
$$\frac{d}{dt}\int_{l(t)}\omega^1_{\boldsymbol A}=\int_{l(t)}\omega^1_{\frac{\partial \boldsymbol A}{\partial t}+(\mathrm{rot}\,\boldsymbol A)\times \boldsymbol v}+(\boldsymbol A,\boldsymbol v)\Big|_2-(\boldsymbol A,\boldsymbol v)\Big|_1.$$

 

[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 ?

 

[zwierz]

https://en.wikipedia.org/wiki/Interior_product

 

[zwierz]

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

yes

and finally differentiating a 3-form gives you back a 1-form.

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

 

[meteo student]

$$i_{\boldsymbol B}\omega^1_{\boldsymbol A}=(\boldsymbol A,\boldsymbol B),\quad i_{\boldsymbol B}\omega^2_{\boldsymbol A}=\omega^1_{\boldsymbol A\times\boldsymbol B},\quad i_{\boldsymbol B}\omega^3_f=f\omega^2_{\boldsymbol B}.$$

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.

Please correct as you wish.

 

[zwierz]

nd 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 r

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

 

[meteo student]

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

 

[zwierz]

Cross product is a vector see correspondences above

 

[meteo student]

The way I understand the notation which you introduced is that a 1-form can be associated with a cross product of vectors. Is that correct ?
Inner product of vector B with the 2-form of vector A gives you a 1-form of a cross product. Did I get the explanation correct ?
My understanding of differential forms is that only a 2-form can be associated with a cross product. Is that correct ? But in your notation you are associating a differential 1-form with a cross product. That is the part I cannot understand.

 

[zwierz]

1-form is associated with a vector and 2-form is associated with a vector by another manner. Cross product is a vector. So you can associate 1-form with cross product and you can associate 2-form with cross product

 

[meteo student]

OK. Thanks for patiently answering my questions.

You are right a 1-form can be associated with a cross product. But that would mean that one of the terms of the cross product is actually a basis 0-form and the second term is a basis 1-form. Correct ?

 

[zwierz]

I do not understand that. Look $\omega^1_{\boldsymbol A\times \boldsymbol B}=(A_yB_z-B_yA_z)dx+(-A_xB_z+B_xA_z)dy+...$

 

[meteo student]

You are the Master. My final questions are the third part of your detailed explanation. The one -form you have there on the first term of the RHS - you are now assuming the vector field is a function of time as well right or at least A(x(t), y(t), z(t),t).

So the A subscript of the 1 form on the first term of the RHS transforms as the material derivative(total). Is that correct?

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
$$\frac{d}{dt}\int_{l(t)}\omega^1_{\boldsymbol A}=\int_{l(t)}\omega^1_{\frac{\partial \boldsymbol A}{\partial t}+(\mathrm{rot}\,\boldsymbol A)\times \boldsymbol v}+(\boldsymbol A,\boldsymbol v)\Big|_2-(\boldsymbol A,\boldsymbol v)\Big|_1.$$

 

[zwierz]

Yes. The term "material derivative" is usually used for functions. The material derivative of an arbitrary tensor (of differential forms for our case) is called the Lie derivative.

 

[meteo student]

Thank you.

So a material derivative transforms as a Lie derivative for an autonomous vector field(tensor).

What about the final two terms on the RHS ?
Do they evolve from the interior product of v and A ?

 

[zwierz]

The last difference is the term $\int_{\partial N}$ from general formula. In our case it is actually a definition of the integral over 0-boundary of 0-form.

To this end let's bring other three dimensional corollaries from the integral formulas of #2

1) Let $\Sigma$ be compact two dimensional manifold in $\mathbb{R}^3$ then
$$\frac{d}{dt}\int_{\Sigma(t)}\omega^2_{\boldsymbol A}=\int_{\Sigma(t)}\omega^2_{\boldsymbol B},\quad \boldsymbol B=\frac{\partial\boldsymbol A}{\partial t}+\boldsymbol v\mathrm{div}\,\boldsymbol A+\mathrm{rot}\,(\boldsymbol A\times \boldsymbol v);$$

2) Let $D$ be a compact domain in $\mathbb{R}^3$ and $\rho(t,x,y,z)$ be a scalar function then
$$\frac{d}{dt}\int_{D(t)}\omega^3_{\rho}=\int_{D(t)}\omega^3_{\frac{\partial\rho}{\partial t}+\mathrm{div}\,(\rho\boldsymbol v)}$$

 

[meteo student]

Yea beautiful technique. I can't appreciate all the nuances you employed to derive this but as I get more comfortable with differential forms, manifolds and Lie derivatives, Cartan's identity, exterior and interior derivative I maybe able to appreciate your methodology a lot more.

For right now let me think over your proofs and cutting edge math rather than ask more questions. Maybe when I think about the Lie derivative and the definition of the integral I may ask a separate question. For right now this thread can be closed as ANSWERED.

The last difference is the term $\int_{\partial N}$ from general formula. In our case it is actually a definition of the integral over 0-boundary of 0-form.

To this end let's bring other three dimensional corollaries from the integral formulas of #2

1) Let $\Sigma$ be compact two dimensional manifold in $\mathbb{R}^3$ then
$$\frac{d}{dt}\int_{\Sigma(t)}\omega^2_{\boldsymbol A}=\int_{\Sigma(t)}\omega^2_{\boldsymbol B},\quad \boldsymbol B=\frac{\partial\boldsymbol A}{\partial t}+\boldsymbol v\mathrm{div}\,\boldsymbol A+\mathrm{rot}\,(\boldsymbol A\times \boldsymbol v);$$

2) Let $D$ be a compact domain in $\mathbb{R}^3$ and $\rho(t,x,y,z)$ be a scalar function then
$$\frac{d}{dt}\int_{D(t)}\omega^3_{\rho}=\int_{D(t)}\omega^3_{\frac{\partial\rho}{\partial t}+\mathrm{div}\,(\rho\boldsymbol v)}$$