Talk:Convective derivative

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1 Notation consistancy
2 Two of a kind
3 Identity
4 Parentheses
5 Same as "total" derivative
6 What has been proven?
7 A subtlety

[edit] Notation consistancy

Am I confused, or does this article change notation between the introduction and the proof? Is D/Dt the same operator as d/dt? Tom Duff 19:19, 30 January 2006 (UTC)

Yes I have fixed it. d/dt is used in many texts but I think D/Dt makes it clear that we are discussing a property in a vector field. Rex the first ^{talk | contribs} 01:00, 27 February 2007 (UTC)

Also, what does the hat on the B in the "proof" section refer to? A hat can have many meanings, there should be a line saying "where B-hat indicates that [whatever it indicates]" 140.184.21.115 13:35, 19 September 2007 (UTC)

[edit] Two of a kind

Looks to me like this article ought to be rolled in with substantive derivative. Linuxlad 14:19, 22 June 2006 (UTC)

[edit] Identity

The identity given for taking the material derivative of an integral is the Reynolds Transport Theorem, though written in a form that is dissimilar to the one listed in the article concerning that theorem. This is also a varient of the Liebnitz Rule.

[edit] Parentheses

Is there any special reason for the parentheses used on the RHS in the definitions? —DIV (128.250.204.118 09:04, 6 April 2007 (UTC))

[edit] Same as "total" derivative

Assume that

φ = φ(x, y, z, t)

By the chain rule

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

dividing both side by $d t$ , we get

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

since $\frac{dx}{dt}=u$ , $\frac{dy}{dt}=v$ and $\frac{dz}{dt}=w$ , the above equation becomes

$\frac{d\phi}{dt} = \frac{\partial \phi}{\partial t} + u \frac{\partial \phi}{\partial x} + v \frac{\partial \phi}{\partial y} + w \frac{\partial \phi}{\partial z} = \frac{\partial \phi}{\partial t} + (\mathbf{u}\cdot\nabla)\phi$

Hence, we see that $\frac{d\phi}{dt}$ and $\frac{D\phi}{Dt}$ are one and the same. Therefore, the substantial derivative is nothing more than a total derivative with respect to time. The only advantage of the substantial derivative notation is that it higlights more of the physical significance (time rate of change following a moving fluid element).

I think that the terminology "substantial derivative" and "total derivative" are unnecessarilly confusing (As far as I know, this terminology is mainly prevalent in fluid dynamics) The wikipedia article should explain that they are different way to express the same thing.

199.212.17.130 13:47, 31 August 2007 (UTC)

[edit] What has been proven?

The first section of this article claims to define the convective derivative. The next section offers a proof. How can a definition be proven? I am confused. Is the proof intended to show that the convective derivative is the partial derivative with respect to time in a frame that moves with material particles? That requires some reasoning, I think, not just direct application of the chain rule.

155.37.79.216 14:27, 7 September 2007 (UTC)

[edit] A subtlety

There is a subtlety that the proof has missed. Consider the point $\mathbf{r}=\mathbf{r}(t)$ and $\phi(\mathbf{x},t)$ for $\mathbf{x}$ a position coordinate independent of

t

, then the total derivative $d\phi(\mathbf{r}(t), t) / dt$ may be found

$\frac{d\phi(\mathbf{r}(t), t)}{dt} = \left. \frac{\partial \phi(\mathbf{x},t)}{\partial t} \right|_{\mathbf{x}=\mathbf{r}} + \frac{d \mathbf{r}}{dt} \cdot \left[\nabla \phi(\mathbf{x},t) \right]_{\mathbf{x}=\mathbf{r}}$

If and only if $\mathbf{r}$ is a Lagrangian point, so $d \mathbf{r}(t) / dt = \mathbf{u}$ , does the total time derivative equal the convective derivative.