Acceleration of a fluid particle

[coverband]

1. The velocity vector which describes the motion of a particle (point) in a fluid is $$\vec {u} = \vec {u}(\vec{x},t),$$ so that the particle follows the path on which $$\frac {dx}{dt}=\vec{U}(t)=\vec{u}[\vec{x}(t),t].$$
Write $$\vec{x}\equiv (x,y,z)$$ and $$\vec{u} \equiv (u,v,w)$$ (in rectangular Cartesian coordinates), and hence show that the acceleration of the particle is $$\frac {d\vec{U}}{dt} = \frac {\partial \vec{u}}{\partial t} + (\vec {u} . \nabla) \vec {u} \equiv \frac {D\vec {u}}{Dt}$$, the material derivative

Homework Equations

Acceleration of particle = $$\frac{d^2 x}{dt^2} = \frac {dU}{dt}=\frac{du}{dt}[\frac {\partial x}{\partial t}, \frac {\partial y}{\partial t}, \frac {\partial z}{\partial t}, 1]$$

$$(\vec{u}.\nabla)\vec{u} = uu_x\vec{i} + uv_x\vec{j} + uw_z\vec{k} + vu_y\vec{i} + vv_y \vec{j} + vw_y \vec {k} + wu_z \vec{i} + wv_z \vec{j} + ww_z \vec {k}$$

 

[jdwood983]

What do you know about the total derivative?

 

[coverband]

Got it! Thanks..

 

[coverband]

Wait! The total derivative. How does that help ?

$$\frac {dU}{dt} = \frac {\partial U}{\partial t} + \frac {\partial U}{\partial x} \frac {dx}{dt} + \frac {\partial U}{\partial t} \frac {dt}{dt}$$

 

[jdwood983]

Wait! The total derivative. How does that help ?

$$\frac {dU}{dt} = \frac {\partial U}{\partial t} + \frac {\partial U}{\partial x} \frac {dx}{dt} + \frac {\partial U}{\partial t} \frac {dt}{dt}$$

This isn't the total derivative, this is:

$$\frac{dU}{dt}=\frac{\partial u}{\partial t}+\frac{dx}{dt}\frac{\partial u}{\partial x}+\frac{dy}{dt}\frac{\partial u}{\partial y}+\frac{dz}{dt}\frac{\partial u}{\partial z}$$

Note that the last three terms look like:

$$\frac{dx}{dt}\frac{\partial }{\partial x}+\frac{dy}{dt}\frac{\partial }{\partial y}+\frac{dz}{dt}\frac{\partial }{\partial z}=u_x\frac{\partial}{\partial x}+u_y\frac{\partial}{\partial y}+u_z\frac{\partial}{\partial z}=\mathbf{u}\cdot\nabla$$

Which is what you're looking for:

$$\frac{dU}{dt}=\frac{\partial u}{\partial t}+\left(\mathbf{u}\cdot\nabla\right)u$$

This last formula is also called the convective derivative or the material derivative.

 

[coverband]

Very helpful thank you. Just one quick question. Why does U become u in your total derivative expressions. Thanks again.

$$\frac{dU}{dt}=\frac{\partial u}{\partial t}+\frac{dx}{dt}\frac{\partial u}{\partial x}+\frac{dy}{dt}\frac{\partial u}{\partial y}+\frac{dz}{dt}\frac{\partial u}{\partial z}$$

 

[jdwood983]

Very helpful thank you. Just one quick question. Why does U become u in your total derivative expressions. Thanks again.

Because you have $U(t)$ defined as $U(t)=u\left[x(t),t\right]$ in the problem. If $U(t)$ is only a function of $t$, then $\left(\mathbf{u}\cdot\nabla\right)u=0$ and $dU/dt=\partial U/\partial t$, which is not what you want for this problem.


Glad I can help.