I'm recently working on material derivatives, defined as follows.

$$\frac{D }{Dt}=\frac{\partial }{\partial t}+v\cdot \nabla $$

It is easy to show that a material(or convective) derivative of an infinitesimal vector can be shown as

$$\frac{D \delta x}{Dt}=\delta x \cdot \nabla v $$

Now, if there is a surface element defined as $\delta S = \delta x \times \delta y $, I have to show that this has the following relationship

$$\frac{D \delta S}{Dt}=(\nabla \cdot v)\delta S-(\nabla v) \cdot \delta S $$

I am trying to solve this by directly using the levi-civita symbol which gives

$$(\frac{D \delta S}{Dt})_{i} = (\frac{D \delta x}{Dt} \times \delta y + \delta x \times \frac{D \delta y}{Dt})_{i} = \varepsilon_{ijk}[(\delta x\cdot \nabla v)_{j} \delta y_{k}+\delta x_{j}(\nabla v \cdot \delta y)_{k}] = \varepsilon_{ijk}[\nabla v_{jl}\delta y_{k}\delta x_{l}+\nabla v_{kl}\delta y_{l}\delta x_{j}]$$

where $\nabla v_{ij} = \frac{\partial v_{j}}{\partial q_{i}}$

I could not go further, so I wrote down the result in the similar form,

$$(\frac{D \delta S}{Dt})_{i}=(\nabla \cdot v)\delta S_{i}-((\nabla v) \cdot \delta S)_{i} = \nabla v_{jj}\delta S_{i}-\nabla v_{ji}\delta S_{j} $$

and this is all I have for now. Could anyone give me a clue to proving the equation above?

  • $\begingroup$ you are done, nothing else to do $\endgroup$
    – lurscher
    Sep 4 '21 at 3:04
  • $\begingroup$ @lurscher Thanks for your comment! Could you elucidate please? I don't find how the two equations could be the same at this point. $\endgroup$
    – HANEUL
    Sep 4 '21 at 3:07
  • $\begingroup$ the former is in geometric vector notation, the latter is in component vector notation. Both represent the same mathematical object $\endgroup$
    – lurscher
    Sep 4 '21 at 17:41
  • $\begingroup$ To clarify the common type of vector notations typically used in textbooks: en.wikipedia.org/wiki/Vector_notation#Rectangular_vectors $\endgroup$
    – lurscher
    Sep 4 '21 at 17:43
  • $\begingroup$ @lurscher Well I am sorry, I suppose I have misexplained what I wanted to say. The last graybox is what I had written down directly from "result"(or the RHS of what I want to prove), which ought to be connected with the equation in component-vector-notation right above. I believe some terms need to be contracted or modulated so that the two equations meet, and that connection is what I am looking for. $\endgroup$
    – HANEUL
    Sep 5 '21 at 4:32

It can be solved by using the fact that $\varepsilon_{ijk} a_j b_k = - \varepsilon_{ijk} a_k b_j$.

\begin{align*} \frac{D \delta S_i}{D t} &= \varepsilon_{ijk}(\nabla v_{jl} \delta y_k \delta x_l + \nabla v_{kl} \delta y_l \delta_j) \\ &= \varepsilon_{ijk}(\nabla v_{jl} \delta y_k \delta x_l - \nabla v_{jl} \delta y_l \delta_k) \\ &= \varepsilon_{ijk}(\delta_{ln} \delta_{km} - \delta_{kn} \delta_{lm}) \nabla v_{jl} \delta x_n \delta y_m \\ &= \varepsilon_{ijk}\varepsilon_{plk}\varepsilon_{pnm} \nabla v_{jl} \delta x_n \delta y_m \\ &= (\delta_{ip} \delta_{jl} - \delta_{il} \delta_{jp}) \nabla v_{jl} \delta S_p \\ &= \nabla v_{jj} \delta S_i - \nabla v_{ji} \delta S_j \end{align*}


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