# Help understanding the Del operator

1. May 21, 2010

### Dawei

I'm an engineering major taking an advanced level physics class. I realize that I really have no clue when it comes to basic mathematics, and it is extremely frustrating. I always just learned *how* to solve equations, never what I was actually doing.

For example the del operator. What exactly does this mean?

And what is the logic for turning it into this?

I feel like I just really need a good lesson starting with the very basics. I have no "feel" for what half of the symbols that I'm using mean.

I need to really understand and use the Navier Stokes equations for my job, and it's just killing me. Every time I try to look up a word or symbol I don't know, the explanation of it ends up having like 3 or 4 more words or symbols in it that I don't know. I took calc 1-3 and diff eq, did well in the classes, so it's extremely frustrating now to feel like I never actually learned anything, just sort of how to find patterns and solve homework problems...

Can anyone recommend a good explanation of the Navier Stokes equations for me?

2. May 21, 2010

### ireland01

del is partial differentiation operator in x/y/z. It's for 3 dimension.

3. May 22, 2010

### HallsofIvy

In Cartesian coordinates $]\nabla$ is the differential operator
$$\frac{\partial}{\partial x}\vec{i}+ \frac{\partial}{\partial y}\vec{j}+ \frac{\partial}{\partial x}\vec{k}$$.

Written that way, not applied to anything, it is not a true "vector", just a symbol, but applied to various kinds of functions, it can be thought of as a vector.

For example, if f(x,y,z) is a scalar valued function we can think of $\nabla f$, also called "grad f" or "gradient of f", as a vector times a scalar- the scalar "multiplies" each part of the vector:
$$\nabla f= \frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}\partial z}\vec{k}$$.

If $\vec{F}(x,y,z)$is a vector function, say, $\vec{F}(x,y,z)= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)\vec{k}$ then we can think in terms of the "dot product" or "cross productZ".

Thinking of "dot product" we have $\nabla\cdot\vec{F}(x,y,z)$, also called "div $\vec{F}$" or "the divergence of $\vec{F}$", equal to
$$\frac{\partial f}{\partial x}+ \frac{\partial g}{\partial y}+ \frac{\partial h}{\partial z}$$.

Thinking of the "cross product" we have $\nabla\times\vec{F}(x,y,z)$, also called "$curl \vec{F}$" or " curl of $\vec{F}$, equal to
$$\left(\frac{\partial h}{\partial y}- \frac{\partial g}{\partial z}\right)\vec{i}- \left(\frac{\partial h}{\partial x}- \frac{\partial f}{\partial z}\right)\vec{j}+ \left(\frac{\partial g}{\partial x}- \frac{\partial f}{\partial y}\right)\vec{k}$$

Now, let's look at
$$\frac{\partial\rho}{\partial t}+ \nabla\cdot(\rho\vec{u})= 0$$

Take $\vec{u}= f\vec{i}+ g\vec{j}+ h\vec{k}$ so that $\rho\vec{u}= \rho f\vec{i}+ \rho\ g\vec{j}+ \rho h\vec{k}$ and
$$\nabla\cdot(\rho\vec{u})= \frac{\partial(\rho f)}{\partial x}+ \frac{\partial(\rho g)}{\partial y}+ \frac{\partial(\rho h)}{\partial z}$$

By the product rule, each of those is
$$\frac{\partial\rho}{\partial x}f+ \rho \frac{\partial f}{\partial x}$$
$$\frac{\partial\rho}{\partial y}g+ \rho \frac{\partial g}{\partial y}$$
$$\frac{\partial\rho}{\partial z}h+ \rho \frac{\partial h}{\partial z}$$

We can write their sum as
$$\frac{\partial\rho}{\partial x}f+ \frac{\partial\rho}{\partial y}g+ \frac{\partial\rho}{\partial z}h$$
$$+ \rho \frac{\partial f}{\partial x}+ \rho\frac{\partial g}{\partial y}+ \rho\frac{\partial h}{\partial z}$$

$$= \nabla\rho\cdot \vec{u}+ \rho\nabla\cdot\vec{u}$$

Now, you will also need to know what
$$\frac{D\rho}{Dt}$$ means. It is NOT just the derivative accidently written with "D" rather than "d". It is the rate of change as we move with the flow. Instead of sticking a, say, thermometer into water and measuring the temperature and rate of change of temperature at that point, we attach it to a float and let float down the stream, measuring the termperature and rate of change of temperature of a little "piece" or water as it flows down the stream.

Specifially, if F(x,y,z, t) is a function of the three space variables and time,
$$\frac{DF}{DT}= \frac{\partial F}{\partial t}+ \nabla F\cdot\vec{v}$$
where "[itex]\vec{v}[itex]" is the velocity vector of the water.

Well, you can see that
$$\frac{\partial\rho}{\partial t}+ \nabla\cdot(\rho\vec{u})$$
becomes
$$\frac{\partial\rho}{\partial t}+ (\nabla\rho\cdot \vec{u}+ \rho(\nabla\cdot\vec{u}))$$
$$=(\frac{\partial\rho}{\partial t}+ \nabla\rho\cdot \vec{u})+ \rho(\nabla\cdot\vec{u})$$
$$= \frac{D\rho}{Dt}+ \rho(\nabla\cdot\vec{u})$$

But how that "u" mysteriously became "v", I cannot say- unless it is a misprint.

Last edited by a moderator: May 23, 2010