What is a velocity field and its relationship to a fluid element's motion?

[mathmari]

Hey!

Let the fluid occupies the space $D \subset \mathbb{R}^n, n=2 \text{ or } 3$.
$\overrightarrow{x}$ is a point of $D$.
We consider the element of the fluid that is at the position $\overrightarrow{x}$ at the time $t$ , and moves along the trajectory $\Gamma$.

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Let $\overrightarrow{u}(\overrightarrow{x}, t)$ the velocity of this element. For a given time $t$, $\overrightarrow{u}(\cdot , t)$ is a vector field over $D$, and is called velocity field.

Could you explain to me the last part:

"For a given time $t$, $\overrightarrow{u}(\cdot , t)$ is a vector field over $D$, and is called velocity field."

?? (Wondering)

 

[I like Serena]

Hey!

Let the fluid occupies the space $D \subset \mathbb{R}^n, n=2 \text{ or } 3$.
$\overrightarrow{x}$ is a point of $D$.
We consider the element of the fluid that is at the position $\overrightarrow{x}$ at the time $t$ , and moves along the trajectory $\Gamma$.



Let $\overrightarrow{u}(\overrightarrow{x}, t)$ the velocity of this element. For a given time $t$, $\overrightarrow{u}(\cdot , t)$ is a vector field over $D$, and is called velocity field.

Could you explain to me the last part:

"For a given time $t$, $\overrightarrow{u}(\cdot , t)$ is a vector field over $D$, and is called velocity field."

?? (Wondering)

Hi! (Blush)

Formally, we would say that the velocity field $\overrightarrow v$ is a function of position to velocity, that is, $\overrightarrow v: D \to \mathbb{R}^n$, given by $\overrightarrow v(\overrightarrow{x}) = \overrightarrow{u}(\overrightarrow{x}, t)$.
The latter can also written as $\overrightarrow v(\cdot) = \overrightarrow{u}(\cdot, t)$, without changing the meaning, where $\cdot$ is an arbitrary symbol. (Nerd)To abbreviate it, we would like to say that $\overrightarrow{u}$ is the velocity field, but that would not be correct, since $\overrightarrow{u}$ takes 2 arguments and we need one argument with a fixed $t$.
So we abbreviate it as $\overrightarrow{u}(\cdot, t)$ with the understanding that $\cdot$ is the placeholder for the implicit argument. (Wasntme)

 

[mathmari]

I see... Thanks for the explanation! (flower)