# Talk:Material derivative

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## Thank you

Probably not the correct place for this, but the swimmer description of this derivative wrt lake temperature was wonderful and *really* clarified this for me. Thank you! 133.87.57.32 (talk) 13:06, 23 June 2012 (UTC)

## 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)

## 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)

## 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.

## 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))

## Same as "total" derivative

Assume that

${\displaystyle \phi =\phi (x,y,z,t)}$

By the chain rule

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

dividing both side by ${\displaystyle dt}$, we get

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

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

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

Hence, we see that ${\displaystyle {\frac {d\phi }{dt}}}$ and ${\displaystyle {\frac {D\phi }{Dt}}}$ 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)

In transport phenomena the partial derivative and the material derivative (the latter also called the substantial derivative) are both special cases of the total derivative. There are three cases to consider with respect to the terms ${\displaystyle dx/dt}$, ${\displaystyle dy/dt}$, and ${\displaystyle dz/dt}$ which describe the motion of the observer and which appear in the definition of the total derivative:
• In general, the motion of the observer may be an arbitrary function of time, defined by ${\displaystyle d{\boldsymbol {r}}(t)/dt=[dx(t)/dt,dy(t)/dt,dz(t)/dt]}$. Note that this motion of the observer ${\displaystyle d{\boldsymbol {r}}(t)/dt}$ is independent of the motion of the fluid and does not necessarily establish an inertial frame of reference. In this general case, the rate of change in ${\displaystyle \phi }$ as observed by the arbitrarily moving observer is given by the total derivative ${\displaystyle d\phi /dt=\partial \phi /\partial t+(\partial \phi /\partial x)(dx/dt)+(\partial \phi /\partial y)(dy/dt)+(\partial \phi /\partial z)(dz/dt)}$, which may be interpreted as the stationary rate of change (given by the partial derivative; see the next case below) plus corrections due to the motion of the observer.
• If the observer is stationary, ${\displaystyle dx/dt=0}$, ${\displaystyle dy/dt=0}$, and ${\displaystyle dz/dt=0}$. The rate of change in ${\displaystyle \phi }$ as observed by the stationary observer is then given simply by the partial derivative ${\displaystyle \partial \phi /\partial t}$.
• If the motion of the observer follows the motion of the fluid, then ${\displaystyle dx/dt=v_{x}}$, ${\displaystyle dy/dt=v_{y}}$, and ${\displaystyle dz/dt=v_{z}}$, where the vector ${\displaystyle {\boldsymbol {v}}(t,x,y,z)=[v_{x}(t,x,y,z),v_{y}(t,x,y,z),v_{z}(t,x,y,z)]}$ describes the motion of the fluid. The rate of change in ${\displaystyle \phi }$ as observed by the observer drifting along with the moving fluid is given by the substantial derivative ${\displaystyle D\phi /Dt=\partial \phi /\partial t+v_{x}(\partial \phi /\partial x)+v_{y}(\partial \phi /\partial y)+v_{z}(\partial \phi /\partial z)}$.
It appears that the commenter above mistook ${\displaystyle dx/dt}$, ${\displaystyle dy/dt}$, and ${\displaystyle dz/dt}$ to mean always the velocity of the fluid, and arrived at the incorrect conclusion that the total derivative and the material (substantial) derivative are the same.
Ydw (talk) 06:56, 24 November 2010 (UTC)

## 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)

## A subtlety

There is a subtlety that the proof has missed. Consider the point ${\displaystyle \mathbf {r} =\mathbf {r} (t)}$ and ${\displaystyle \phi (\mathbf {x} ,t)}$ for ${\displaystyle \mathbf {x} }$ a position coordinate independent of ${\displaystyle t}$, then the total derivative ${\displaystyle d\phi (\mathbf {r} (t),t)/dt}$ may be found
${\displaystyle {\frac {d\phi (\mathbf {r} (t),t)}{dt}}=\left.{\frac {\partial \phi (\mathbf {x} ,t)}{\partial t}}\right|_{\mathbf {x} =\mathbf {r} }+{\frac {d\mathbf {r} }{dt}}\cdot \left[\nabla \phi (\mathbf {x} ,t)\right]_{\mathbf {x} =\mathbf {r} }}$

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

139.80.48.19 (talk) 22:29, 19 March 2008 (UTC)

## Rename to Material derivative

1. Convective derivative is ambiguous, since it is also used to denote only the spatial part, v•∇.
2. Material derivative is more commonly used. A Google Books search for the exact phrase gives:
phrase hits
material derivative 837
substantial derivative 693
convective derivative 650 (incl. uses for only the spatial part)
Lagrangian derivative 490
substantive derivative 407
derivative following the motion 321
particle derivative 177
hydrodynamic derivative 162
Stokes derivative 106