- #1
meteo student
- 72
- 0
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 ?
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 ?