Ideal Fluids in Motion — Question 6

Question 6

Steadiness depends on the observation frame

A flow in laboratory coordinates is $v_x=U+a(x-Ut)$ and $v_y=-ay$, with constants $U\ne0$ and $a>0$. Introduce a frame translating at velocity $U\hat i$, so $x^{\prime}=x-Ut$, $y^{\prime}=y$ and $v_x^{\prime}=v_x-U$.

1. Write the velocity field in the moving frame. Is the field steady in each frame?

2. Find the laboratory material acceleration and express it using $x^{\prime},y^{\prime}$. Explain why the two inertial frames agree on acceleration.

Question 6 - Solution

Steadiness depends on the observation frame

(a) Derivation and calculation

$$x^{\prime}=x-Ut,\quad y^{\prime}=y,\quad v_x^{\prime}=v_x-U.$$$$\boxed{\mathbf v^{\prime}=(ax^{\prime},-ay^{\prime})},\qquad\partial\mathbf v^{\prime}/\partial t=0.$$$$\left.\frac{\partial v_x}{\partial t}\right|_x=-aU\ne0.$$The moving-frame flow is steady; the laboratory field is unsteady.

(b) Result and interpretation

$$\frac{Dv_x}{Dt}=\frac{\partial v_x}{\partial t}+v_x\frac{\partial v_x}{\partial x}=-aU+a[U+a(x-Ut)]=a^2x^{\prime}.$$$$\frac{Dv_y}{Dt}=v_y(-a)=a^2y^{\prime}.$$$$\boxed{\mathbf a=(a^2x^{\prime},a^2y^{\prime})=\mathbf a^{\prime}}.$$A constant-velocity coordinate transformation leaves acceleration unchanged.

Physical check

The local time derivative changes between frames, but the full material derivative does not.