When talking about quantum or classical physics the word mechanics often comes up.
With fields it's more useful to talk about plumbing.
in multiple dimensions
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. $x$)." Say the coordinates of the red dot are $\left({\color{red}x_0,f(x_0)}\right)$. The red line is tangent to the blue graph: Its slope, $m$, is the same as the first derivative of the function, $\left.\frac{df}{dx}\right|^{x=x_0}$ evaluated at $x=x_0$. Near $x_0$, the red and blue lines are very close, and the function can be approximately with the equation for the straight, red line: $${\color{blue}f(x)}\approx \color{red}f(x_0)+\left.\frac{df}{dx}\right|^{x=x_0} * (x-x_0).$$ These are the first two terms of the function's Taylor series. 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 $h$ depends on, say, two cartesian coordinates, then you can schematically write the change $dh$ as a dot product... $$\begineq dh =& \frac{\del h}{\del x} dx + \frac{\del h}{\del y} dy \\ =&\left(\frac{\del h}{\del x}\uv{x} + \frac{\del h}{\del y}\uv{y} \right) \cdot \left(dx\,\uv{x} + dy\,\uv{y} \right)\\ dh=& \myv \grad h \cdot d\myv{l}.\endeq$$ The second vector is our "change in position" vector, $$d\myv l\equiv dx\,\uv x +dy\,\uv y,$$ and the first vector is called the gradient: In three dimensions, the gradient of a scalar function $T(x,y,z)$ of 3 Cartesian coordinates is: $$\myv \grad T \equiv \frac{\del T}{\del x}\uv{x} + \frac{\del T}{\del y}\uv{y} + \frac{\del T}{\del z}\uv{z}.$$ The gradient is a multi-dimension generalization of the "derivative". It's a vector with components consisting of the partial derivatives in the $\uv x$, $\uv y$, etc directions. Using the coordinate-independent form for the dot product, $$dh = \myv \grad h \cdot d\myv l=|\myv \grad h| |d\myv{l}| \cos(\theta)$$ 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... Downhill!! If $h(x,y)$ is the height of landscape features on a topographical map, what are the partial derivatives $\del h/\del x$ and $\del h/\del y$ at position corresponding to the top of a (rounded) hill? What is the gradient at the top of the hill? 1.11 b - Find the gradient of $$f(x,y,z)=x^2y^3z^4$$ 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? Estimating the divergence of a vector field - Narrated: 2 min. 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 $\myv V(\myv r) = -y \uv{x} +x \uv{y} \equiv -y \uv{x} +x \uv{y}+ 0 \uv z$, 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? Graphically evaluating the curl of a 2-d vector field - Narrated: 2 min. 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 The curl of the gradient of a scalar field $$\myv \grad \times \myv \grad U(x,y,z) = 0$$ is always zero. It turns out that, going the other way, if we find that the curl of a vector field $\myv F(x,y,z)$, then this implies that the vector field can be written as the gradient of a scalar field, that is: $$\myv \grad \times \myv F=0 \Leftrightarrow \myv F=\myv \grad \phi(x,y,y) \Leftrightarrow \oint \myv F\cdot d\myv l=0.$$ We used this fact in Analytical Mechanics to test whether a force field is "conservative" (that is, can be expressed as the gradient of a potential energy) or not. The last property means that the path integral of $\myv F \cdot d\myv l$ around *any* closed path is zero. Now, you'll prove it (in Cartesian coordinates) in problem 1.27. Image credits Limajo, ign, John Kay, Felix if (! $homepage){ $stylesheet="/~paulmr/class/comments.css"; if (file_exists("/home/httpd/html/cment/comments.h")){ include "/home/httpd/html/cment/comments.h"; } } ?>
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. $x$)."
Say the coordinates of the red dot are $\left({\color{red}x_0,f(x_0)}\right)$. The red line is tangent to the blue graph: Its slope, $m$, is the same as the first derivative of the function, $\left.\frac{df}{dx}\right|^{x=x_0}$ evaluated at $x=x_0$.
Near $x_0$, the red and blue lines are very close, and the function can be approximately with the equation for the straight, red line: $${\color{blue}f(x)}\approx \color{red}f(x_0)+\left.\frac{df}{dx}\right|^{x=x_0} * (x-x_0).$$ These are the first two terms of the function's Taylor series.
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.
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 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.
If the function $h$ depends on, say, two cartesian coordinates, then you can schematically write the change $dh$ as a dot product... $$\begineq dh =& \frac{\del h}{\del x} dx + \frac{\del h}{\del y} dy \\ =&\left(\frac{\del h}{\del x}\uv{x} + \frac{\del h}{\del y}\uv{y} \right) \cdot \left(dx\,\uv{x} + dy\,\uv{y} \right)\\ dh=& \myv \grad h \cdot d\myv{l}.\endeq$$
The second vector is our "change in position" vector, $$d\myv l\equiv dx\,\uv x +dy\,\uv y,$$ and the first vector is called the gradient:
In three dimensions, the gradient of a scalar function $T(x,y,z)$ of 3 Cartesian coordinates is: $$\myv \grad T \equiv \frac{\del T}{\del x}\uv{x} + \frac{\del T}{\del y}\uv{y} + \frac{\del T}{\del z}\uv{z}.$$
The gradient is a multi-dimension generalization of the "derivative".
It's a vector with components consisting of the partial derivatives in the $\uv x$, $\uv y$, etc directions.
Using the coordinate-independent form for the dot product, $$dh = \myv \grad h \cdot d\myv l=|\myv \grad h| |d\myv{l}| \cos(\theta)$$
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...
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...
Downhill!!
If $h(x,y)$ is the height of landscape features on a topographical map, what are the partial derivatives $\del h/\del x$ and $\del h/\del y$ at position corresponding to the top of a (rounded) hill? What is the gradient at the top of the hill?
1.11 b - Find the gradient of $$f(x,y,z)=x^2y^3z^4$$
See also Paul's treatment of the gradient in Analytical Mechanics, using topo maps.
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 $T$. $$\myv \grad T \equiv \frac{\del T}{\del x}\uv{x} + \frac{\del T}{\del y}\uv{y} + \frac{\del T}{\del z}\uv{z}.$$
$$\myv \grad \equiv \uv{x}\frac{\del}{\del x} +\uv{y} \frac{\del}{\del y} + \uv{z} \frac{\del}{\del z}.$$
Hmmm, doesn't it kinda make you wonder if there is something like a dot product or a cross product involving $\myv \grad$.??
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?
Estimating the divergence of a vector field - Narrated: 2 min.
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**(!).
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 $\myv V(\myv r) = -y \uv{x} +x \uv{y} \equiv -y \uv{x} +x \uv{y}+ 0 \uv z$, 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?
Graphically evaluating the curl of a 2-d vector field - Narrated: 2 min.
Note the product rules--discussed in section 1.2.6 and noted on the inside front cover of your text.
Of all the possible combinations of second derivatives, the two most useful will be...
$$\begineq \grad^2 T\equiv \myv \grad \cdot \myv \grad T = \frac{\del^2T}{\del x^2} + \frac{\del^2T}{\del y^2} + \frac{\del^2T}{\del z^2}.\endeq$$
The curl of the gradient of a scalar field $$\myv \grad \times \myv \grad U(x,y,z) = 0$$ is always zero.
It turns out that, going the other way, if we find that the curl of a vector field $\myv F(x,y,z)$, then this implies that the vector field can be written as the gradient of a scalar field, that is: $$\myv \grad \times \myv F=0 \Leftrightarrow \myv F=\myv \grad \phi(x,y,y) \Leftrightarrow \oint \myv F\cdot d\myv l=0.$$ We used this fact in Analytical Mechanics to test whether a force field is "conservative" (that is, can be expressed as the gradient of a potential energy) or not. The last property means that the path integral of $\myv F \cdot d\myv l$ around *any* closed path is zero.
Now, you'll prove it (in Cartesian coordinates) in problem 1.27.
Limajo, ign, John Kay, Felix if (! $homepage){ $stylesheet="/~paulmr/class/comments.css"; if (file_exists("/home/httpd/html/cment/comments.h")){ include "/home/httpd/html/cment/comments.h"; } } ?>