Path integrals (like Work)

Work done by a force field, $\myv F(x,y,z)$ on a particle is an example of a path integral: $$W=\int_{\cal{P}}\myv F(x,y,z)\cdot d\myv r$$ depends on the path $\cal{P}$:

• Depends on the starting and ending points, $\myv r_i$ and $\myv r_f$,
• But there are an infinite number of possible paths connecting any two points, so we also need a sketch or a parametric equation or something else to specify which path.

How to solve this integral?

Strategies from Calc III...

$$W=\int_{\cal P}\myv F(x,y,z)\cdot d\myv r$$

The main strategy is to write $d\myv r$ ($d\myv r\equiv d\myv l$ in Griffiths' notation for Electrodynamics) in terms of its Cartesian components, $d\myv r = dx\,\uv x+dy\,\uv y+ dz\,\uv z$, and then write out the dot product as a sum of Cartesian products:

$$W=\int_{\cal P}\myv F(x,y,z)\cdot d\myv r= \int_{\cal P} F_x\,dx +\int_{\cal P} F_y\,dy+\int_{\cal P} F_z\,dz $$ where $F_x=F_x(x,y,z)$, etc.

If you are integrating along a path, ${\cal P}$, that is parallel to one of the Cartesian directions, this is all you need. E.g. if you are integrating in the positive $x$ direction, then $d\myv r=dx\,\uv x+0\,\uv y+0\,\uv z$ that is, $dy=0$ and $dz=0$, and you know that 2 of the 3 integrals have to vanish.

• See the green path in this example , and red paths II and III.

If ${\cal P}$ is a more complicated path, try to parameterize the path: If you can describe the path in terms of a parameter $t$ as $\myv r(t)=x(t)\uv x+y(t)\uv y+z(t)\uv z$ when $t_i\lt t \lt t_f$, then you can write out one of the three path integrals as: $$\int_{\cal P} F_x\,dx=\int_{t_i}^{t_f} F_x(x(t),y(t),z(t))\frac{dx}{dt}\,dt,$$ and similarly for the integrals of the $y-$ and $z-$components.

• See the blue path, and red path I in this example
• Another example - See particularly the 3rd path along the diagonal.

Conservative fields

For conservative fields, only the endpoints matter, since the integral is independent of path. Remember that you can test for a conservative field by checking that the curl vanishes: $$\myv \grad \times \myv F = 0.$$ Then, you can *guess* a potential function, if you can find one such that (in physics terms): $$-\myv \grad U=\myv F,$$ then the integral is given in terms of the initial and final coordinates of ${\cal P}$ as $$U(\myv r_i)-U(\myv r_f)$$

• Saved by a conservative field

But it's not always easy / possible to guess the potential. There is a recipe to find it... Integrate to find the potential:

Pick a reference point (often the origin, but it can be any point) and integrate from your reference point to a final position $(x,y,z)$ using any path you like (since the integral depends only on the end points): $$U(x,y,z)-U(0,0,0)=-\int_{(0,0,0)}^{(x,y,z)}\myv F(x',y', z')\cdot d\myv r'.$$

• An example of the recipe to find the potential. (But here we're integrating $\myv F\cdot d\myv r$ rather than -$\myv F\cdot d\myv r$, so we get a "potential function" (calculus) rather than the "potential energy" (physics).