I Force fields in curvilinear coordinate systems

[calculo2718]

I am trying to solve problems where I calculate work do to force along paths in cylindrical and spherical coordinates.

I can do almost by rote the problems in Cartesian: given a force $\vec{F} = f(x,y,z)\hat{x} + g(x,y,z)\hat{y}+ h(x,y,z)\hat{z}$ I can parametricize my some curve $\gamma: \vec{r}(t) = x(t)\hat{x} + y(t)\hat{y} + z(t)\hat{z}$ then take $\int_\gamma \vec{F}\cdot\vec{r}'(t)dt$

I know it should work similarly in cylindrical, if I have $\vec{F}(R,\phi) = f(R,\phi)\hat{R} + g(R,\phi)\hat{\phi}$, The problem is I am having trouble thinking through the curve parametrization in cylindrical and similarly in spherical. For example, If I want to integrate along the path where $\phi = 0$, from some radius $r_1$ to $r_2$ I would say
$\gamma_1: \vec{r}(t) = t\hat{R} + 0\hat{\phi}$

If I then want to go from along the quarter circle path from $\phi = 0$ to $\phi = \frac{\pi}{2}$ then

$\gamma_2: \vec{r}(t) = r_2\hat{R} + r_2\cdot t\hat{\phi}$

If I want to go back down to $r_1$
$\gamma_3: \vec{r}(t) = t\hat{R} + \frac{\pi}{2}\hat{\phi}$

This is where my confusion gets in the way, the $\gamma_2$ curve makes it look like I can in general parametricize a curve in cylindrical as $\vec{r}(t) = R(t)\hat{R} + \phi(t)R(t)\hat{\phi}$ where on this curve $R(t) = r_2$ and $\phi(t) = t$, however the $\gamma_3$ curve is not of this form and if I try to parametricize it that way($\vec{r}(t) = t\hat{R} + \frac{\pi}{2}t\hat{\phi}$) it is incorrect(I tested it with a Force only in the $\hat{\phi}$ direction which should give me 0, but I get a non-0 integral).

How can I parametricize a general curve in cylindrical and spherical the way I have done above for cartesian? What am I missing when trying to do these problems?

In addition how can I derive from my $\vec{r}(t)$ in cylindrical the line element in cylindrical $d\vec{r} = dR\hat{R} + rd\phi\hat{\phi}$. I feel like understand how this is done will help me see more clearly why I am confused

 

[jambaugh]

I am going to give you a quick general overview which I think will answer most of your questions:

To start there are two things going on when we speak in terms of coordinate systems. There are the linear coordinate systems, including Cartesian coordinates where the coordinates are the components of a position vector in a specific coordinate system, \vec{r}=x\hat{\imath} + y\hat{\jmath}+z\hat{k}.

You then have non-linear (curvilinear) coordinates (polar, cylindrical, spherical, etc). When using general curvilinear coordinates you should express positions either as coordinate positions (coordinate tuples which are not considered a vector) or using the standard basis which is effectively giving the cartesian coordinate as function of these non-linear coordinates. Example:
\vec{r} = R\cos(\phi)\hat{\imath} + R\sin(\phi)\hat{\jmath}+z\hat{k}
in cylindrical coordinates (R,\phi,z)

Now if you are considering other vectors besides the position vector (such as velocity or force or differential position for a parameterized curve) then you can express it either in the standard basis but with components still a function of the curvilinear coordinates, or in a coordinate basis, or in a normalized coordinate basis.

Example: In a (2-dimensional) central force problem where the force is toward the origin and with magnitude k/r^2 where r is distance from the origin. (I am going to use (r,\theta) polar coordinates in the plain since that's the familiar convention to me.)

I can express my force vector in purely rectangular coordinates and in the standard basis as:
\vec{F} = \frac{-k x}{(x^2+y^2)^{3/2}}\hat{\imath}+\frac{-k y}{(x^2+y^2)^{3/2}}\hat{\imath}
of course this is rather awkward and we can instead express the force in terms of the polar coordinates but still in the standard basis:
\vec{F} = \frac{-k \cos(\theta)}{r^2}\hat{\imath}+\frac{-k \sin(\theta)}{r^2}\hat{\imath}
even better we might decide to use a normalized coordinate basis. By a coordinate basis we mean that we use the gradient of each coordinate as our basis. These are not necessarily unit vectors so if we want a normalized basis we would then make each into a unit vector to get a normalized coordinate basis (which is then not the coordinate basis as such!).
In polar coordinates:
r = \sqrt{x^2 + y^2} , \nabla r = \frac{x}{r} \hat{\imath}+ {y}{r}\hat{\jmath} = \hat{r}
it happens to already be a unit vector. In fully polar coordinates it is \nabla r = \hat{r} = \cos(\theta)\hat{\imath} + \sin(\theta)\hat{\jmath}.
\theta = \tan^{-1}(y/x) + \text{const}, \nabla \theta = \frac{-y}{r^2}\hat{\imath} + \frac{x}{r^2}\hat{\jmath}
it is not a unit vector but r\nabla \theta is so we can use:
\hat{\theta} = r\nabla\theta = \frac{-y}{r}\hat{\imath} + \frac{x}{r}\hat{\jmath} = -\sin(\theta)\hat{\imath} + \cos(\theta)\hat{\jmath}

We can now express the force above in both polar coordinates and a normalized coordinate basis:
\vec{F} = \frac{-k}{r}\hat{r}

Note that the curvilinear coordinate basis and normalize coordinate basis can in general depend on the coordinates which can complicate mightily the differential calculus. That's a whole chapter of a textbook by itself. But for the parametric curve case it is not so bad.

If you are dealing with a parametric curve you are giving your position coordinates (in whatever coordinate system) as functions of the parameter. But do remember that for curvilinear case where you are using a coordinate or normalized coordinate basis, you must remember that the basis vectors are also changing with the parameter. You simply apply the product rule. E.g.

\dot{\vec{F}} = \frac{d}{dt}\left(\frac{-k}{r^2}\right)\hat{r} + \frac{-k}{r^2}\frac{d}{dt}\hat{r} = ...

I hope this has clarified the subject for you a bit.

 

[calculo2718]

thank you! this was helpful.