A Time differentiation of fluid line integrals

[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
$$\frac{d}{dt}\int_{D(t)}\omega^3_{\rho}=\int_{D(t)}\omega^3_{\frac{\partial\rho}{\partial t}+\mathrm{div}\,(\rho\boldsymbol v)}$$

 

[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

I found the book on line. But pages 115-125 do not talk about this but rather generalities about 1 and 2 dimensional systems. What are the correct pages?

 

[meteo student]

I found the book on line. But pages 115-125 do not talk about this but rather generalities about 1 and 2 dimensional systems. What are the correct pages?

I do not know what version of the book you have but the chapter is M6 - Integral operations.

Search for equations M6.44. In my version of the book it is on page 116. The section is titled Time differentiation of fluid line integrals and the section number is M6.5.1

Dynamics of the Atmosphere - A course in theoretical meteorology by Zdunkowski and Bott

 

[meteo student]

By the way I was looking at another way of going about this and that is the calculus of moving surfaces. Any thoughts on that ?

 

[meteo student]

I found the book on line. But pages 115-125 do not talk about this but rather generalities about 1 and 2 dimensional systems. What are the correct pages?

In the online version of the book those pages are 96-102.

 

[lavinia]

In the online version of the book those pages are 96-102.

I found it. Thanks.

I am still not sure how to derive your vector equation but working on it. Funny the differential form way of doing it is straightforward.
What is the scalar triple product operation that the book is using?

 

[meteo student]

I found it. Thanks.

I am still not sure how to derive your vector equation but working on it. Funny the differential form way of doing it is straightforward.
What is the scalar triple product operation that the book is using?

The definition of the scalar triple product is the same that is used everywhere. It is there on pages 11-13 in your version of the book.

You also need to look at the so called Lamb transformation which is on page 58. I am going to rewrite the Lamb transformation in differential form notation and present it as a verification exercise in this forum.

When you write that "differential form proof is so much easier" you are referring to the one that zwierz has shown or do you have another proof of your own ?

 

[lavinia]

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 ?

I just read through the derivation in your book. The four terms are just the integrals along the boundary of the tiny square. Stokes Theorem relates the sum of the boundary terms - each an integral along one of the four line segments - to an integral over the interior of the square.

The terms $$\Delta t(v.A)_{1,2} $$ are linear approximations. These approximations are legitimate if the vector field $A$ is smooth and if $Δt$ is small enough.

 

[meteo student]

I just read through the derivation in your book. The four terms are just the integrals along the boundary of the tiny square. Stokes Theorem relates the sum of the boundary terms - each an integral along one of the four line segments - to an integral over the interior of the square.

The terms $$\Delta t(v.A)_{1,2} $$ are linear approximations. These approximations are legitimate if the vector field $A$ is smooth and if $Δt$ is small enough.

Thank you ! So if I understand it correctly the linear approximations are themselves derived from definite integrals right ? What is the subscript 1 and 2 mean at the bottom ?

 

[lavinia]

Thank you ! So if I understand it correctly the linear approximations are themselves derived from definite integrals right ? What is the subscript 1 and 2 mean at the bottom ?

If you look at the diagram in the book of the small square you see two boundary line segments in the direction of the fluid velocity $v$. The 1 and 2 denote the end points of these segments at the point 1 and 2.

The boundary integral according to Stokes Theorem is the integral of the inner products $A$ with the tangent vectors to the boundary segments. In the two space directions these are the dot products with $dr$. In the time directions this is $∫_0^{Δt}v(t)⋅Adt$ and for small enough $Δt$ this is $∫_0^{Δt}v(0)⋅A(0)dt$ which is just $Δt(v(0)⋅A(0))$ which your book denotes as $Δt(v⋅A)_{1,2}$

 

[meteo student]

If you look at the diagram in the book of the small square you see two boundary line segments in the direction of the fluid velocity $v$. The 1 and 2 denote the end points of these segments at the point 1 and 2.

The boundary integral according to Stokes Theorem is the integral of the inner products $A$ with the tangent vectors to the boundary segments. In the two space directions these are the dot products with $dr$. In the time directions this is $∫_0^{Δt}v(t)⋅Adt$ and for small enough $Δt$ this is $∫_0^{Δt}v(0)⋅A(0)dt$ which is just $Δt(v(0)⋅A(0))$ which your book denotes as $Δt(v⋅A)_{1,2}$

Beautiful. Now I understand what he meant by phase space. You have two directions in space and time. Thank you very much !

 

[meteo student]

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.

I took your advice seriously and I am reading this book - An Introduction To The Geometry and Topology Of Fluid Flows by RL Ricca. I believes it introduces students of fluid mechanics to the world view of modern mathematics.