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I'm studying fluid mechanics, and I got the impression that the material derivative is nothing more than "differentiating along a path" and so I got confused on why do we need it. Basically, let $D\subset \mathbb{R}^3$ be the region containing the fluid and let $f : D\times \mathbb{R}\to \mathbb{R}$ be a time dependent function on $D$.

Suppose then $\gamma : I\subset \mathbb{R}\to D$ is a trajectory on the fluid. The material derivative of $f$ along $\gamma$ is defined by

$$\dfrac{D}{Dt}f(\gamma(t),t) = \dfrac{\partial f}{\partial t}(\gamma(t),t) + (\mathbf{u}\cdot \nabla)f(\gamma(t),t)$$

Where $\mathbf{u}$ is the spatial velocity field of the fluid. But that expression is nothing more nothing less than simply differentiating $f(\gamma(t),t)$ with respect to $t$, or better, differentiating the function $f\circ (\gamma, I)$ where $I$ is the interval in $\mathbb{R}$.

Indeed we have

$$\dfrac{d}{dt}f(\gamma(t),t) = \nabla f(\gamma(t),t)\cdot \gamma'(t) + \dfrac{\partial f}{\partial t}(\gamma(t),t) = \dfrac{D}{Dt} f(\gamma(t),t)$$

So since $\dfrac{d}{dt}= \dfrac{D}{Dt}$ why do we need the material derivative? Why do we define it, since in truth it is just the well know derivative of a composition just? What are the advantages of defining it?

EDIT: I think I got it now. When computing $\dfrac{d}{dt}$ as I said a composition is needed, in other words, we need a path. But since all information is contained in $\mathbf{u}$ we can dispose the path by defining:

$$\dfrac{D}{Dt} f(a,t) = \dfrac{\partial f}{\partial t}(a,t) + (\mathbf{u}\cdot \nabla) f(a,t)$$

And that coincides with $\dfrac{d}{dt}$ if $a = \gamma(t_0)$ for some $\gamma$ and some $t_0$. Is that really the point we are making when defining the material derivative? Is the same we would get if we had a path going through there with velocity $\mathbf{u}$ but we dispose the path since $\mathbf{u}$ already contains all needed info.

I'm studying fluid mechanics, and I got the impression that the material derivative is nothing more than "differentiating along a path" and so I got confused on why do we need it. Basically, let $D\subset \mathbb{R}^3$ be the region containing the fluid and let $f : D\times \mathbb{R}\to \mathbb{R}$ be a time dependent function on $D$.

Suppose then $\gamma : I\subset \mathbb{R}\to D$ is a trajectory on the fluid. The material derivative of $f$ along $\gamma$ is defined by

$$\dfrac{D}{Dt}f(\gamma(t),t) = \dfrac{\partial f}{\partial t}(\gamma(t),t) + (\mathbf{u}\cdot \nabla)f(\gamma(t),t)$$

Where $\mathbf{u}$ is the spatial velocity field of the fluid. But that expression is nothing more nothing less than simply differentiating $f(\gamma(t),t)$ with respect to $t$, or better, differentiating the function $f\circ (\gamma, I)$ where $I$ is the interval in $\mathbb{R}$.

Indeed we have

$$\dfrac{d}{dt}f(\gamma(t),t) = \nabla f(\gamma(t),t)\cdot \gamma'(t) + \dfrac{\partial f}{\partial t}(\gamma(t),t) = \dfrac{D}{Dt} f(\gamma(t),t)$$

So since $\dfrac{d}{dt}= \dfrac{D}{Dt}$ why do we need the material derivative? Why do we define it, since in truth it is just the well know derivative of a composition just? What are the advantages of defining it?

EDIT: I think I got it now. When computing $\dfrac{d}{dt}$ as I said a composition is needed, in other words, we need a path. But since all information is contained in $\mathbf{u}$ we can dispose the path by defining:

$$\dfrac{D}{Dt} f(a,t) = \dfrac{\partial f}{\partial t}(a,t) + (\mathbf{u}\cdot \nabla) f(a,t)$$

And that coincides with $\dfrac{d}{dt}$ if $a = \gamma(t_0)$ for some $\gamma$ and some $t_0$. Is that really the point we are making when defining the material derivative? Is the same we would get if we had a path going through there with velocity $\mathbf{u}$ but we dispose the path since $\mathbf{u}$ already contains all needed info.

I'm studying fluid mechanics, and I got the impression that the material derivative is nothing more than "differentiating along a path" and so I got confused on why do we need it. Basically, let $D\subset \mathbb{R}^3$ be the region containing the fluid and let $f : D\times \mathbb{R}\to \mathbb{R}$ be a time dependent function on $D$.

Suppose then $\gamma : I\subset \mathbb{R}\to D$ is a trajectory on the fluid. The material derivative of $f$ along $\gamma$ is defined by

$$\dfrac{D}{Dt}f(\gamma(t),t) = \dfrac{\partial f}{\partial t}(\gamma(t),t) + (\mathbf{u}\cdot \nabla)f(\gamma(t),t)$$

Where $\mathbf{u}$ is the spatial velocity field of the fluid. But that expression is nothing more nothing less than simply differentiating $f(\gamma(t),t)$ with respect to $t$, or better, differentiating the function $f\circ (\gamma, I)$ where $I$ is the identity in $\mathbb{R}$.

Indeed we have

$$\dfrac{d}{dt}f(\gamma(t),t) = \nabla f(\gamma(t),t)\cdot \gamma'(t) + \dfrac{\partial f}{\partial t}(\gamma(t),t) = \dfrac{D}{Dt} f(\gamma(t),t)$$

So since $\dfrac{d}{dt}= \dfrac{D}{Dt}$ why do we need the material derivative? Why do we define it, since in truth it is just the well know derivative of a composition just? What are the advantages of defining it?

EDIT: I think I got it now. When computing $\dfrac{d}{dt}$ as I said a composition is need, in other words, we need a path. But since all information is contained in $\mathbf{u}$ we can dispose the path by defining:

$$\dfrac{D}{Dt} f(a,t) = \dfrac{\partial f}{\partial t}(a,t) + (\mathbf{u}\cdot \nabla) f(a,t)$$

And that coincides with $\dfrac{d}{dt}$ if $a = \gamma(t_0)$ for some $\gamma$ and some $t_0$. Is that really the point we are making when defining the material derivative? Is the same we would get if we had a path going through there with velocity $\mathbf{u}$ but we dispose the path since $\mathbf{u}$ already contains all needed info.

I'm studying fluid mechanics, and I got the impression that the material derivative is nothing more than "differentiating along a path" and so I got confused on why do we need it. Basically, let $D\subset \mathbb{R}^3$ be the region containing the fluid and let $f : D\times \mathbb{R}\to \mathbb{R}$ be a time dependent function on $D$.

Suppose then $\gamma : I\subset \mathbb{R}\to D$ is a trajectory on the fluid. The material derivative of $f$ along $\gamma$ is defined by

$$\dfrac{D}{Dt}f(\gamma(t),t) = \dfrac{\partial f}{\partial t}(\gamma(t),t) + (\mathbf{u}\cdot \nabla)f(\gamma(t),t)$$

Where $\mathbf{u}$ is the spatial velocity field of the fluid. But that expression is nothing more nothing less than simply differentiating $f(\gamma(t),t)$ with respect to $t$, or better, differentiating the function $f\circ (\gamma, I)$ where $I$ is the interval in $\mathbb{R}$.

Indeed we have

$$\dfrac{d}{dt}f(\gamma(t),t) = \nabla f(\gamma(t),t)\cdot \gamma'(t) + \dfrac{\partial f}{\partial t}(\gamma(t),t) = \dfrac{D}{Dt} f(\gamma(t),t)$$

So since $\dfrac{d}{dt}= \dfrac{D}{Dt}$ why do we need the material derivative? Why do we define it, since in truth it is just the well know derivative of a composition just? What are the advantages of defining it?

EDIT: I think I got it now. When computing $\dfrac{d}{dt}$ as I said a composition is needed, in other words, we need a path. But since all information is contained in $\mathbf{u}$ we can dispose the path by defining:

$$\dfrac{D}{Dt} f(a,t) = \dfrac{\partial f}{\partial t}(a,t) + (\mathbf{u}\cdot \nabla) f(a,t)$$

And that coincides with $\dfrac{d}{dt}$ if $a = \gamma(t_0)$ for some $\gamma$ and some $t_0$. Is that really the point we are making when defining the material derivative? Is the same we would get if we had a path going through there with velocity $\mathbf{u}$ but we dispose the path since $\mathbf{u}$ already contains all needed info.

I'm studying fluid mechanics, and I got the impression that the material derivative is nothing more than "differentiating along a path" and so I got confused on why do we need it. Basically, let $D\subset \mathbb{R}^3$ be the region containing the fluid and let $f : D\times \mathbb{R}\to \mathbb{R}$ be a time dependent function on $D$.

Suppose then $\gamma : I\subset \mathbb{R}\to D$ is a trajectory on the fluid. The material derivative of $f$ along $\gamma$ is defined by

$$\dfrac{D}{Dt}f(\gamma(t),t) = \dfrac{\partial f}{\partial t}(\gamma(t),t) + (\mathbf{u}\cdot \nabla)f(\gamma(t),t)$$

Where $\mathbf{u}$ is the spatial velocity field of the fluid. But that expression is nothing more nothing less than simply differentiating $f(\gamma(t),t)$ with respect to $t$, or better, differentiating the function $f\circ (\gamma, I)$ where $I$ is the identity in $\mathbb{R}$.

Indeed we have

$$\dfrac{d}{dt}f(\gamma(t),t) = \nabla f(\gamma(t),t)\cdot \gamma'(t) + \dfrac{\partial f}{\partial t}(\gamma(t),t) = \dfrac{D}{Dt} f(\gamma(t),t)$$

So since $\dfrac{d}{dt}= \dfrac{D}{Dt}$ why do we need the material derivative? Why do we define it, since in truth it is just the well know derivative of a composition just? What are the advantages of defining it?

EDIT: I think I got it now. When computing $\dfrac{d}{dt}$ as I said a composition is need, in other words, we need a path. But since all information is contained in $\mathbf{u}$ we can dispose the path by defining:

$$\dfrac{D}{Dt} f(a,t) = \dfrac{\partial f}{\partial t}(a,t) + (\mathbf{u}\cdot \nabla) f(a,t)$$

And that coincides with $\dfrac{d}{dt}$ if $a = \gamma(t_0)$ for some $\gamma$ and some $t_0$. Is that really the point we are making when defining the material derivative? Is the same we would get if we had a path going through there with velocity $\mathbf{u}$ but we dispose the path since $\mathbf{u}$ already contains all needed info.

I'm studying fluid mechanics, and I got the impression that the material derivative is nothing more than "differentiating along a path" and so I got confused on why do we need it. Basically, let $D\subset \mathbb{R}^3$ be the region containing the fluid and let $f : D\times \mathbb{R}\to \mathbb{R}$ be a time dependent function on $D$.

Suppose then $\gamma : I\subset \mathbb{R}\to D$ is a trajectory on the fluid. The material derivative of $f$ along $\gamma$ is defined by

$$\dfrac{D}{Dt}f(\gamma(t),t) = \dfrac{\partial f}{\partial t}(\gamma(t),t) + (\mathbf{u}\cdot \nabla)f(\gamma(t),t)$$

Where $\mathbf{u}$ is the spatial velocity field of the fluid. But that expression is nothing more nothing less than simply differentiating $f(\gamma(t),t)$ with respect to $t$, or better, differentiating the function $f\circ (\gamma, I)$ where $I$ is the identity in $\mathbb{R}$.

Indeed we have

$$\dfrac{d}{dt}f(\gamma(t),t) = \nabla f(\gamma(t),t)\cdot \gamma'(t) + \dfrac{\partial f}{\partial t}(\gamma(t),t) = \dfrac{D}{Dt} f(\gamma(t),t)$$

So since $\dfrac{d}{dt}= \dfrac{D}{Dt}$ why do we need the material derivative? Why do we define it, since in truth it is just the well know derivative of a composition just? What are the advantages of defining it?

EDIT: I think I got it now. When computing $\dfrac{d}{dt}$ as I said a composition is need, in other words, we need a path. But since all information is contained in $\mathbf{u}$ we can dispose the path by defining:

$$\dfrac{D}{Dt} f(a,t) = \dfrac{\partial f}{\partial t}(a,t) + (\mathbf{u}\cdot \nabla) f(a,t)$$

And that coincides with $\dfrac{d}{dt}$ if $a = \gamma(t_0)$ for some $\gamma$ and some $t_0$. Is that really the point we are making when defining the material derivative? Is the same we would get if we had a path going through there with velocity $\mathbf{u}$ but we dispose the path since $\mathbf{u}$ already contains all needed info.