# Converting Velocity-Delta Products to Divergences of Dyads ## The Problem When we compute $\partial_t \boldsymbol{g}$ (the time derivative of momentum density), we get: $\partial_t \boldsymbol{g} = \sum_i m_i \, \dot{\boldsymbol{v}}_i \, \delta_i + \sum_i m_i \, \boldsymbol{v}_i \, \partial_t \delta_i$ The first term is easy (acceleration). The second term involves $\boldsymbol{v}_i \, \partial_t \delta_i$, which looks messy. This note shows how to turn it into a clean divergence of a dyadic tensor. --- ## The Goal Show that: $\boxed{\sum_i m_i \, \boldsymbol{v}_i \, \partial_t \delta_i = -\nabla \cdot \sum_i m_i \, \boldsymbol{v}_i \boldsymbol{v}_i \, \delta_i}$ where $\boldsymbol{v}_i \boldsymbol{v}_i$ is the outer product (dyad): in components, $(\boldsymbol{v}_i \boldsymbol{v}_i)_{\alpha\beta} = v_{i\alpha} v_{i\beta}$. --- ## The Derivation — Three Steps ### Step 1: Use the moving delta identity From [[Moving-Delta-identity|The moving delta identity]], we have: $\partial_t \delta_i = -\boldsymbol{v}_i \cdot \nabla_{\boldsymbol{r}} \delta_i$ Substitute: $\sum_i m_i \, \boldsymbol{v}_i \, \partial_t \delta_i = -\sum_i m_i \, \boldsymbol{v}_i \, (\boldsymbol{v}_i \cdot \nabla_{\boldsymbol{r}} \delta_i)$ --- ### Step 2: The Key Insight — $\nabla_{\boldsymbol{r}}$ Does Not Act on $\boldsymbol{v}_i$ This is crucial: When we write $\nabla_{\boldsymbol{r}}$, we mean the gradient with respect to the *field coordinate* $\boldsymbol{r}$ (the spatial point where we observe the field). - $\boldsymbol{v}_i(t) = \dot{\boldsymbol{r}}_i(t)$ is the Lagrangian velocity of particle $i$. It depends on the particle index $i$ and time $t$, but *not* on the field coordinate $\boldsymbol{r}$. - When we take $\nabla_{\boldsymbol{r}}$ of something, we differentiate only w.r.t. $\boldsymbol{r}$. - Therefore: $\nabla_{\boldsymbol{r}} \boldsymbol{v}_i = 0$. > [!tldr] TL;DR > Treat $\boldsymbol{v}_i(t)$ as a *constant vector* when we're doing spatial derivatives. The only spatial dependence is in the delta spikes. --- ### Step 3: Convert to Divergence of Dyad Use the vector identity: For any constant vector $\boldsymbol{c}$ and any scalar field $f(\boldsymbol{r})$: $\boldsymbol{c} (\boldsymbol{c} \cdot \nabla f) = \nabla \cdot (\boldsymbol{c} \boldsymbol{c} \, f)$ Why? Use the product rule on the RHS: $\nabla \cdot (\boldsymbol{c} \boldsymbol{c} \, f) = (\nabla \cdot \boldsymbol{c} \boldsymbol{c}) f + \boldsymbol{c} \boldsymbol{c} : \nabla f$ The first term vanishes (the divergence of a constant dyad is zero), so: $\nabla \cdot (\boldsymbol{c} \boldsymbol{c} \, f) = \boldsymbol{c} \boldsymbol{c} : \nabla f = \boldsymbol{c}(\boldsymbol{c} \cdot \nabla f)$ (Here $:$ denotes the double-dot product: $\boldsymbol{A} : \nabla f = A_{\alpha\beta} \partial_\beta f$ summed over $\alpha, \beta$.) Apply with $\boldsymbol{c} = \boldsymbol{v}_i$ and $f = \delta_i$: $\boldsymbol{v}_i (\boldsymbol{v}_i \cdot \nabla \delta_i) = \nabla \cdot (\boldsymbol{v}_i \boldsymbol{v}_i \, \delta_i)$ Substitute back: $\sum_i m_i \, \boldsymbol{v}_i \, \partial_t \delta_i = -\sum_i m_i \, \nabla \cdot (\boldsymbol{v}_i \boldsymbol{v}_i \, \delta_i) = -\nabla \cdot \sum_i m_i \, \boldsymbol{v}_i \boldsymbol{v}_i \, \delta_i$ --- ## Component View (for those who prefer indices) In index notation: Let $\delta_i = \delta(\boldsymbol{r} - \boldsymbol{r}_i)$. Then: $\partial_t g_\alpha = \sum_i m_i \, \dot{v}_{i\alpha} \delta_i + \sum_i m_i \, v_{i\alpha} \, \partial_t \delta_i$ Use $\partial_t \delta_i = -v_{i\beta} \partial_\beta \delta_i$: $\sum_i m_i \, v_{i\alpha} \, \partial_t \delta_i = -\sum_i m_i \, v_{i\alpha} v_{i\beta} \, \partial_\beta \delta_i = -\partial_\beta \left( \sum_i m_i \, v_{i\alpha} v_{i\beta} \, \delta_i \right)$ Define the kinetic momentum-flux tensor (as a matrix): $K_{\alpha\beta}(\boldsymbol{r}, t) = \sum_i m_i \, v_{i\alpha} v_{i\beta} \, \delta_i$ Then: $\partial_t g_\alpha + \partial_\beta K_{\alpha\beta} = \sum_i m_i \, \dot{v}_{i\alpha} \, \delta_i$ Or in vector form: $\partial_t \boldsymbol{g} + \nabla \cdot \boldsymbol{K} = \sum_i m_i \, \dot{\boldsymbol{v}}_i \, \delta_i$ --- ## Interpretation: What is the Kinetic Momentum Flux? $\boldsymbol{K}_{\alpha\beta} = \sum_i m_i \, v_{i\alpha} v_{i\beta} \, \delta_i$ Physical meaning: This tensor describes the *transport of momentum* due to particle motion. - Diagonal component $K_{xx}$ = flux of $x$-momentum in the $x$-direction (normal stress) - Off-diagonal component $K_{xy}$ = flux of $x$-momentum in the $y$-direction (shear) When particles move with velocity $\boldsymbol{v}$, they carry momentum $m\boldsymbol{v}$ with them. If the velocity varies from point to point (e.g., shear flow), then momentum fluxes arise—and those are captured by $\boldsymbol{K}$. Example (1D, shear flow): If $v_x(y) = \dot{\gamma} y$ (linear shear), then: - Particles at height $y$ move faster than particles at height $y - \lambda$. - Fast particles carry momentum downward and slow particles carry momentum upward. - The net *transport* of $x$-momentum in the $y$-direction is $\boldsymbol{K}_{xy}$. - This is exactly where viscosity comes from! (See [[2-What-is-Viscosity]].) --- ## Common Confusion: Why Is This a Dyad, Not a Vector? Question: $\boldsymbol{v}_i$ is a vector. Why does $\boldsymbol{v}_i \boldsymbol{v}_i$ (the outer product) give a tensor? Answer: When we write $\boldsymbol{v}_i \boldsymbol{v}_i$ in bold, it's shorthand for the outer product. The component form makes it clear: $(\boldsymbol{v}_i \boldsymbol{v}_i)_{\alpha\beta} = v_{i\alpha} \, v_{i\beta}$ This is a 3×3 matrix (or 2×2 in 2D). When we take the divergence $\nabla \cdot \boldsymbol{K}$, we get: $(\nabla \cdot \boldsymbol{K})_\alpha = \partial_\beta K_{\alpha\beta} = \partial_\beta \left( \sum_i m_i v_{i\alpha} v_{i\beta} \delta_i \right)$ which is a vector (summing over the second index). --- ## Summary | Term | Type | Meaning | |---|---|---| | $\boldsymbol{v}_i(t)$ | Vector | Particle velocity (doesn't depend on field point $\boldsymbol{r}$) | | $\nabla_{\boldsymbol{r}}$ | Operator | Gradient w.r.t. field coordinate $\boldsymbol{r}$ | | $\boldsymbol{v}_i \boldsymbol{v}_i$ | Dyad (tensor) | Outer product; components $(v_{i\alpha} v_{i\beta})$ | | $\boldsymbol{K} = \sum_i m_i \boldsymbol{v}_i \boldsymbol{v}_i \delta_i$ | Tensor field | Kinetic momentum-flux density | | $\nabla \cdot \boldsymbol{K}$ | Vector field | Momentum transport due to bulk and thermal motion | --- ## The Takeaway > [!tldr] Summary > The key algebraic move is recognizing that $\boldsymbol{v}_i (\boldsymbol{v}_i \cdot \nabla \delta_i)$ can be rewritten as $\nabla \cdot (\boldsymbol{v}_i \boldsymbol{v}_i \delta_i)$ because $\boldsymbol{v}_i$ is constant w.r.t. the field gradient $\nabla_{\boldsymbol{r}}$. > > This gives the microscopic momentum-flux tensor $\boldsymbol{K}$, which—when averaged—contains both thermal and bulk contributions to stress. ---