## Definition of the Material Derivative

All very interesting.

To: EMyrone@aol.com

Sent: 31/08/2016 14:26:21 GMT Daylight Time

Subj: Re: Definiiton of the Material DerivativeThis is ok for cartesian coordinates, also eq.(5) of note 356(3). I will see if I can find some other literature references for the corresponding operator in spherical coordinates. It seems to describe the acceleration of fluid particles under translation, rotation and deformation.

Horst

Am 31.08.2016 um 12:25 schrieb EMyrone:

In this definition by Wikipedia, the material derivative has been defined as v dotted in to the gradient of v, more generally A dotted in to the gradient of B. In my opinion the correct definition in Cartesian coordinates is given in Eq. (5) of Note 356(3), because:

v = v sub X i + v sub Y j + v zub Z k

and

del = (partial / partial X) i + (partial /partial Y) j + (partial / partial Z) k

The two definitions are equivalent if and only if partial v sub Y / partial X = 0 and so on. This means that v sub X, v sub Y and v sub Z are linearly independent. With this assumption your method is OK and can be used in the final paper. Wikipedia is often unreliable and / or incomplete.

To: Emyrone

Sent: 30/08/2016 13:21:39 GMT Daylight Time

Subj: Solution: Computation of term (v*gra)v in spherical coordinatesI programmed the formulas for (v*del) v given in

https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinatesAnalysing these formulas, it can be seen that there is a simple term

A_r * partial B_r/partial r

which is the only remaining term if v_r(r), v_theta=0, v_phi=0. So the diff. eq. for the Coulomb electric field is

which is the same as I already found with my simplified terms for (v*del) v. Therefore the solution v_r is proportional to 1/sqrt(r) as described. I agree that this is not a Coulomb potential but a velocity potential so the results may be different. The result for v_r means that there is a flow from or toward the centre. So far all seems to be consistent.Horst

Am 30.08.2016 um 13:24 schrieb Horst Eckardt:

It is not clear to me how the term (v*grad)v is to be transformed to spherical coordinates. You used the comparison to the div operator, I simply multiplied v and del in spherical coordinates and applied to v. In case of the given Coulomb field, the results for the Kambe velocity field v are different:

Method 1 (analogous to div v):

v_r is purely imaginary (I analyzed the result and tried a plot)Method 2 (scalar product v*del):

v_r is proportional to 1/sqrt(r) in the simplest case, depending on integration constant, see Figure.According to VAPS, the expressions grad and div are derived from the tangent vectors and surface flow in the corresponding coordinates. To be sure, we would have to analyze what the geometrical meaning of the operator (v*grad) is and derive this for the corresponding coordinate system.

There is not much info on this operator in the internet. I found the site

https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates

There the operator is denoted “material derivative” and is given. It is much more complicated than both above methods 1 and 2. I will try this operator and see what results.Horst

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