# $\nabla$, $\cdot \nabla$, $\nabla \cdot$, $\nabla^2$ - What do they do? [closed]

I'm trying to teach myself Smoothed Particle Hydrodynamics. Unfortunately, my background is in electronics, so the Navier Stokes equations are somewhat alien to me, as is vector calculus. The video I'm watching is trying to break the equations down into something grokkable by expanding some of the operators into their identities.

Navier Stokes: $$\rho\left[\dfrac{\delta v}{\delta t}+v\cdot\nabla v\right] = \rho g-\nabla p+\mu \nabla^2v$$

Mass Continuity: $$\rho(\nabla\cdot v)=0$$

Convective Acceleration: $$v\cdot\nabla v = \left[v_x\dfrac{\delta v_x}{\delta x},v_y\dfrac{\delta v_y}{\delta y},v_z\dfrac{\delta v_z}{\delta z}\right]$$

Pressure Gradient: $$\nabla p \equiv \left[\dfrac{\delta p}{\delta x},\dfrac{\delta p}{\delta y},\dfrac{\delta p}{\delta z}\right]$$

Pressure: $$p = k( \rho - \rho_0)$$

Diffusion: $$\nabla^2v\equiv\left[\nabla^2v_x,\nabla^2v_y,\nabla^2v_z\right]\nabla^2v_x\equiv\dfrac{\delta^2v_x}{\delta x^2} + \dfrac{\delta^2v_y}{\delta y^2} + \dfrac{\delta^2v_z}{\delta z^2}$$

Mass Continuity: $$\rho(\nabla\cdot v)=0$$

Mass Continuity Solution: $$\nabla\cdot v = \left(\dfrac{\delta v_x}{\delta x} + \dfrac{\delta v_y}{\delta y} + \dfrac{\delta v_z}{\delta z}\right) = 0$$

This to me raises a few questions:

1. $$\nabla$$ seems, from this, to simply be the derivative of each component of the vector.
2. So $$\mathbf{v}\cdot\nabla$$ seems to be the product of those derivatives and $$\mathbf{v}$$
3. But then $$\nabla^2$$ seems to be the sum of the second derivatives of the components, which doesn't seem right.
4. Finally, there's a $$\nabla\cdot$$ operator which seems to be the sum of the components of the first derivatives.

So in the absense of an explanation, I'm somewhat confused as to how the $$\nabla$$, $$\cdot \nabla$$, $$\nabla \cdot$$, and $$\nabla^2$$ operators actually work?

Comments have both suggested I removed the screenshots and advised me of mistakes on the slides, I also don't want to trust my transposition too much. So, I've converted the bulk of the question to MathJax and included the slides for reference.

Reference Slide 1:

Reference Slide 2:

• Please note that you're expected to use Mathjax to display mathematics on this site. Posting images of text and math is very strongly discouraged. Commented Feb 19, 2023 at 0:34
• A few remarks: The notation $\vec v \cdot \nabla \vec v$ is (at least in my experience) quite uncommon. The common way to write this is $(\vec v \cdot \nabla) \vec v$ (which makes the structure of the term quite clear when combined with the convention that differential operators act to the factors right of them). Further, the expansion of that term in components given on the first slide is wrong ... (in the correct version the scalar operator $\vec v \cdot \nabla = v_x \partial_x + v_y \partial_y + v_z \partial_z$ is applied to each component of the vector). Commented Feb 19, 2023 at 1:25
• I'm unsure as to why this has been closed... The question, as is, has produced exactly the answer I needed Commented Feb 19, 2023 at 1:59
• @StephenG-HelpUkraine - I've rewritten the content in MathJax but kept the slides for reference as other comments have indicated there may be mistakes on them Commented Feb 19, 2023 at 3:34
• In re: $\left(\mathbf{v}\cdot\nabla\right)\mathbf{v}$, see physics.stackexchange.com/q/719402/25301, physics.stackexchange.com/q/375826/25301, physics.stackexchange.com/q/160229/25301, and many more. Commented Feb 20, 2023 at 3:35

Treat the $$\nabla$$ symbol as a vector with the components $$(\partial_x,\partial_y,\partial_z)$$ and all of this will follow naturally: $$\nabla\phi=(\partial_x\phi,\partial_y\phi,\partial_z\phi)$$ $$\nabla^2\phi=(\nabla\cdot\nabla)\phi=\partial_x^2\phi+\partial_y^2\phi + \partial_z^2\phi$$ $$\nabla\cdot\mathbf{a}=\partial_xa_x+\partial_ya_y+\partial_za_z$$
• Where you've written $\partial_{x}$, I'm assuming that's shorthand for $\partial/\partial x$? Also is that $\cdot$ in $\nabla\cdot\nabla$ a dot product rather than a straight multiplication? Commented Feb 18, 2023 at 23:28
• Yes – $\partial_x = \frac{\partial}{\partial x}$ is standard notation in the physics literature. And yes, as $\nabla$ is a vector, $\nabla \cdot \nabla$ is a dot-product – there is no "straight multiplication" between vectors (of course there's the cross product – however, $\nabla \times \nabla = 0$ for sufficiently well behaved operands, because partial derviatives commute then). Commented Feb 19, 2023 at 1:10
• This answer does not address the mistake in the slides regarding $\mathbf v\cdot\nabla\mathbf v$. Commented Feb 19, 2023 at 18:44
• $\mathbf{u}\cdot\nabla$ is an abuse of notation? Since when? Commented Feb 21, 2023 at 3:28