Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Question about vector calculus

  1. Aug 11, 2017 #1
    particles in plane polar coordinates

    r = rcosθ i + rsinθ k

    F = Fer + F

    r/∂r =|∂r/∂r|er = (cos2θ + sin2θ)½er = er

    r/∂θ =|∂r/∂θ| = (r2cos2θ + r2sin2θ)½ = r

    I understand that ∂r/∂θ = -rsinθ + rcosθ but why r/∂θ = (r2cos2θ + r2sin2θ)½
  2. jcsd
  3. Aug 11, 2017 #2


    User Avatar
    Science Advisor
    Gold Member

    Please use LaTeX for typing formulae. Then everything gets much better readable!

    Let's start from scratch. As you correctly write in terms of polar coordinates you have
    $$\vec{r}=r \cos \theta \vec{i}+r \sin \theta \vec{j}.$$
    If you have a parametrization of the position vector with some generalized parameters like here ##r## and ##\theta##, you can define a new basis at every point in the vector space by using the tangent vectors to the coordinate lines as any point. They are given by the partial derivatives with respect to the parameters:
    $$\vec{b}_r=\partial_r \vec{r}=\cos \theta \vec{i} + \sin \theta \vec{j}, \quad \vec{b}_{\theta}=\partial_{\theta} \vec{r}=-r \sin \theta \vec{i} + r \cos \theta \vec{j}.$$
    Now you realize that you have a particularly nice case of parameters here, namely socalled "curvilinear orthogonal coordinates", which means that the basis vectors are perpendicular to each other. Indeed, as you can easily check by direct calculation, using ##\vec{i} \cdot \vec{i}=\vec{j} \cdot \vec{j}=1## and ##\vec{i} \cdot \vec{j}=0##,
    $$\vec{b}_r \cdot \vec{b}_{\theta}=0.$$
    In such cases, it is convenient to use normalized basis vectors. So let's calculate the lengths of the basis vectors
    $$g_r^2=\vec{b}_r \cdot \vec{b}_r=1, \quad g_{\theta}^2=\vec{b}_{\theta} \cdot \vec{b}_{\theta}=r^2.$$
    Thus you can introduce orthonormal vectors by just normlizing these (socalled holonomous) basis vectors,
    $$\vec{e}_r=\frac{1}{g_r} \vec{b}_r=\vec{b}_r, \quad \vec{e}_{\theta} = \frac{1}{g_{\theta}} \vec{b}_{\theta}=\frac{1}{r} \vec{b}_{\theta}.$$
    Thus you indeed find
    $$\frac{\partial \vec{r}}{\partial r}=\vec{e}_r, \quad \frac{\partial \vec{r}}{\partial \theta}=r \vec{e}_{\theta}.$$
    The advantage to use the normalized orthogonal basis vectors instead of the holonomous ones is also easy to understand. You can very easily calculate the coordinates of any vector ##\vec{V}## with respect to the orthonormal basis,
    $$V_r=\vec{e}_r \cdot \vec{V}, \quad V_{\theta}=\vec{e}_{\theta} \cdot \vec{V}.$$
    It's easy to show that indeed
    $$\vec{V}=V_r \vec{e}_r + V_{\theta} \vec{e}_{\theta}$$
    by expressing everything again in Cartesian coordinates and the corresponding basis vectors ##\vec{i}## and ##\vec{j}##.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?