Differential calculus as plumbing

When talking about quantum or classical physics the word mechanics often comes up.

With fields it's more useful to talk about plumbing.

Differential calculus

in multiple dimensions

  • The gradient, $\myv \grad$, operator: Which way does water flow?
  • Divergence, $\myv \grad \cdot$: Where are the leaks (and sources)?
  • curl, $\myv \grad \times$: How are things swirling?
  • Higher order derivatives.

Slope of a function

For a function $f(x)$ of one variable, the derivative $\frac{df}{dx}$ is the slope.

$f(x)$ vs $x$

Another way of putting this is... $$df = \left(\frac{df}{dx}\right) dx$$

"As you move a small amount $dx$ to the right, the function changes by an amount $df$ equal to that change in $x$ times (the rate of change of $f$ w.r.t. $x$)."

Slope of a function of several variables

Consider the height, $h$, of Earth's surface (say, height=altitude above sea level) as a function of east ($x$) and north ($y$) position. The height $h(x,y)$ is a function of more than one variable, and is often depicted as a contour map: points of equal height are connected by contour lines.

$\del h/ \del x$ is the slope (rate of change of height) as you move east (no change in north/south direction). In the picture it's a positive slope as you move east.

$\del h / \del y$ is the slope as you move north. In the picture, a negative slope in this direction.

Height is literally, one spatial dimension. But we will more often be concerned with a more abstract situation, Some additional quantity, say the temperature, which is a function of all three spatial coordinates.

Rate of change of a function of many variables

A mathematical theorem tells us that a function of many variables, $T(u,v,w,...)$, changes by an amount that depends on the change of each of the variables ...

$$dT = \frac{\del T}{\del u} du + \frac{\del T}{\del v} dv +\frac{\del T}{\del w} dw + ...$$

In particular, the change in height as we move around the landscape is $$dh = \left(\frac{\del h}{\del x}\right) dx + \left(\frac{\del h}{\del y}\right) dy$$

  • The meaning of the partial derivative $\frac{\del T}{\del u}$ is...

    the rate of change of $T$ as you change $u$, while holding all the other variables ($v,w,...$) constant.
  • This is a property of any function of several variables--they don't have to be spatial variables.

The universal gas law $PV=nRT$: Imagine that we have a piston in which we can control $V,n,T$. These three variables determine the pressure of the gas. Solve for $P(V,n,T)$ and find $\frac{\del P}{\del V}$.

Here, $T$ is not some function of position, but rather it is the equilibrium temperature throughout the piston volume, which depends on the pressure, etc.

The gradient

If the function $T$ depends on, say, the three cartesian coordinates, then you can schematically write the change $dT$ as a dot product... $$\begineq dT &=& \frac{\del T}{\del x} dx + \frac{\del T}{\del y} dy +\frac{\del T}{\del z} dz\\ &=&\left(\frac{\del T}{\del x}\uv{x} + \frac{\del T}{\del y}\uv{y} + \frac{\del T}{\del z}\uv{z}\right) \cdot \left(dx\,\uv{x} + dy\,\uv{y} +dz\,\uv{z}\right)\\ dT&=& \myv \grad T \cdot d\myv{l}.\endeq$$

The second vector is our "change in position" vector, $$d\myv l\equiv dx\,\uv x +dy\,\uv y+dz\,\uv z,$$ and the first vector is called the gradient:

$$\myv \grad T \equiv \frac{\del T}{\del x}\uv{x} + \frac{\del T}{\del y}\uv{y} + \frac{\del T}{\del z}\uv{z}.$$

The gradient is a multi-dimension generalization of the "derivative".

Using the coordinate-independent form for the dot product, $$dT = |\myv \grad T| |d\myv{l}| \cos(\theta)$$

So, if we fix the magnitude of $d\myv{l}$, but let it change directions, the change $dT$ will be the greatest when the angle between $d\myv l$ and $\myv \grad T$ is zero. This implies...

  • The maximum possible slope away from the point we're considering is $|\myv \grad T|$,
  • $\myv \grad T$ is pointing "uphill" towards the direction of greatest change in $T$,
  • In the directions $\theta=\pm \pi/2$, the function is not changing ($dT=0$): these are the directions along "contour lines" of constant $T$.

What direction does water flow on a landscape specified by $h(x,y)$? Apparently in the direction $-\myv \grad h$: exactly opposite to the gradient. That is to say...


See also Paul's treatment of the gradient in Analytical Mechanics, using topo maps.

The nabla operator

In the Cartesian expression for the gradient $\myv \grad T$: it looks like there's a vector, "$\myv \grad$" in some sort of relationship with a scalar $T$. $$\myv \grad T \equiv \frac{\del T}{\del x}\uv{x} + \frac{\del T}{\del y}\uv{y} + \frac{\del T}{\del z}\uv{z}.$$

  • We could pretend that there's a thing $\myv \grad$ (called "nabla")...

    $$\myv \grad \equiv \uv{x}\frac{\del}{\del x} +\uv{y} \frac{\del}{\del y} + \uv{z} \frac{\del}{\del z}.$$

  • ...and that "thing" is operating on $T$ in such a way as to produce a vector from the scalar field $T$...
  • We call $\myv \grad$ a vector operator.
  • When it operates on a scalar field $T(x,y,z)$ the result is a vector.

Hmmm, doesn't it kinda make you wonder if there is something like a dot product or a cross product involving $\myv \grad$.??

  • $\myv \grad T$: A vector known as the gradient of a scalar field $T$,
  • $\myv \grad \cdot \myv V$: A scalar known as the divergence of a vector field $\myv V$, and
  • $\myv \grad \times \myv V$: A vector known as the curl of a vector field $\myv V$.

The Divergence

Where are the leaks?

The dot product of the $\myv \grad$ operator with a vector function $\myv v$ should result in a scalar: a single number, at a particular location in space. $$\begineq \myv \grad \cdot \myv v &=& \uv{x} \frac{\del}{\del x} +\uv{y} \frac{\del}{\del y} +\uv{z} \frac{\del}{\del z} \cdot \left(v_x\uv{x} + v_y\uv{y} + v_z\uv{z}\right)\\ &=&\frac{\del v_x}{\del x} + \frac{\del v_y}{\del y} + \frac{\del v_z}{\del z}.\endeq$$

For the following sketches of 2-dimensional vector fields, $\myv V(x,y)$, what [pos, neg or 0] is the 2-d divergence, $\frac{\del V_x}{\del x}+\frac{\del V_y}{\del y}$ at the positions marked in purple?

One useful intuitive way to think about the divergence for a 2-d vector field, is to imagine that the arrows represent the 2-d flow of an incompressible fluid (water!). Imagine injecting a "cloud" of purple dye at a location you'd like to investigate for "leaks" in your system.

A place where the cloud gets smaller corresponds to a location of a negative divergence: a "leak" (or sink, or drain) where water is leaving your 2-d system

...leaving your 2d system **into another dimension**!

Places where the cloud gets bigger have positive divergence. There must be a "source" (or faucet) of water **from another dimension**(!).

The Curl

How are things swirling?

The cross-product-like object involving $\myv \grad$ is the curl:

$$\begineq\myv \grad \times \myv V &=& \begincv \uv{x}& \uv{y}& \uv{z}\\ \frac{\del}{\del x} & \frac{\del}{\del y} & \frac{\del}{\del z}\\ V_x& V_y& V_z\endcv \\ &=& \uv{x}\left(\frac{\del V_z}{\del y}-\frac{\del V_y}{\del z}\right) + \uv{y}\left(\frac{\del V_x}{\del z}-\frac{\del V_z}{\del x}\right) +\uv{z}\left(\frac{\del V_y}{\del x}-\frac{\del V_x}{\del y}\right)\endeq.$$


For $\myv V(\myv r) = -y \uv{x} +x \uv{y}$, sketch $\myv V$ in the $x-$, $y-$ plane ($z=0$)...

$$\myv \grad \times \myv V = \begincv \uv{x}& \uv{y}& \uv{z}\\ \frac{\del}{\del x}&\frac{\del}{\del y}&\frac{\del}{\del z}\\ -y& +x& 0\endcv = 2\uv{z}.$$

Notice that this has the same value everywhere, not just at the origin.

Extending our water flow metaphor: You can think of placing a paddle-wheel with axis oriented in the $\uv{z}$ direction in the water flow represented by the vector function $\myv V$ and seeing whether the paddle-wheel turns counter-clockwise (positive curl) or clockwise (negative curl).

In a 3-d water flow field, we could also try orienting the paddle wheel towards $\uv{y}$ and $\uv{x}$ to check for components of the curl in all possible directions.

This function has equation: $\myv V = x\uv{y}$:

What about these two? Assume that they are not changing as you change $z$. Does the curl of each field have a $z$- component or not?


Note the product rules--discussed in section 1.2.6 and noted on the inside front cover of your text.

Second derivatives involving $\myv \grad$

Of all the possible combinations of second derivatives, the two most useful will be...

  • $\myv \grad \cdot \myv \grad $, and
  • $\myv \grad \times \myv \grad $.

The Laplacian

$$\begineq \grad^2 T\equiv \myv \grad \cdot \myv \grad T = \frac{\del^2T}{\del x^2} + \frac{\del^2T}{\del y^2} + \frac{\del^2T}{\del z^2}.\endeq$$

The curl of the gradient

$$\myv \grad \times \myv \grad U(x,y,z) = 0.$$

It's always zero... A fact we used in Analytical Mechanics to test whether a force field, $\myv T (x,y,z)$, is "conservative" (that is, can be expressed as the gradient of a potential energy) or not.

Now, prove it! (Problem 1.28)

Image credits

Limajo, ign, John Kay, Felix