# Why does Taylor take the derivative or only $r \bf{\hat r}$ for the velocity/derivative of the position vector?

Why does Taylor take the derivative or only $$r \bf{\hat r}$$ for the velocity/derivative of the position vector? If the position of an object in Polar coordinates is given by (r, theta), why is the 'position vector' here just $$r \bf{\hat r}$$ ? What of the angle the vector makes with the horizontal? Isn't that like saying the y-axis position is given because you somehow know the x position, in Cartesian coordiantes?

I am trying to derive the full equation for acceleration in Polar coordinates. If I took $$\bf r = r \bf\hat r + \theta \bf{\hat{\theta}}$$, I get a wrong answer; I get two extra terms in the r-direction for acceleration, at least.

I think I understand the difference between r, the distance radially away from the origin, and r, the position of the particle in the coordinate system (a vector), so what gives?

Page below (pg 27):

I looked in a class's lecture notes and it is the same, I do not understand:

• What are the definitions of $r$ and $\hat{r}$? Sep 14, 2022 at 21:49
• $\hat{r}$ is a unit vector that points from to origin in the direction of $(r, \theta)$. I.e. it points from the origin towards the particle and has unit length. If you multiply this by the scalar $r$, you get a vector that point from the origin in the direction of the particle, and the length is equal to $r$, which is the distance from the particle to the origin. Hence the result is a vector that points from the origin to the particle's location. This is $\vec{r}$. Sep 14, 2022 at 21:51
• Refer to Kleppner and Kolenkow Sep 15, 2022 at 6:43

The information of angle $$\theta$$ is entirely "encoded" in the position unit vector $$\mathbf{\hat{r}}$$, even though it isn't encoded in the radial variable $$r$$. You can see that if you convert to cartesians vectors: $$\mathbf{\hat{r}} = \cos(\theta) \mathbf{\hat{i}} + \sin(\theta) \mathbf{\hat{j}} \Longrightarrow \mathbf{r} = r(\cos(\theta) \mathbf{\hat{i}} + \sin(\theta) \mathbf{\hat{j}}) = x\mathbf{\hat{i}} + y \mathbf{\hat{j}}$$ You can check using trig that these definitions agree to what you call polar coordinates, that is, $$x = r \cos(\theta)$$ and $$y = r \sin (\theta)$$.

• You could similarly describe a point in space using $\mathbf{r} = r \mathbf{\hat{r}}$ where (using zenith angle as $\theta$ starting from the top and azimuthal angle as $\phi$) $\mathbf{\hat{r}} = \sin \theta \cos \phi \mathbf{\hat{i}} + \sin \theta \sin \phi \mathbf{\hat{j}} + \cos \theta \mathbf{\hat{k}}$ where similarly $r$ encodes just "which sphere" it lies on and $\theta, \phi$ precisely the point of that sphere. The situation is pretty similar. Sep 14, 2022 at 22:05
• I think I understood what you mean. Take the 2D polar coordinates example: you could argue there's some vector $\mathbf{v} = r \mathbf{\hat{r}} + \theta \mathbf{\hat{\theta}} = (r, \theta)$ that "encodes" your information (and if you program it into some code our something, you probably should make such vector). However that is not the position vector! For the position vector was implicitly assumed to be $\mathbf{r} = x\mathbf{i} + y\mathbf{j}$ and the particular choice of taking $r$ as the distance from the origin (that is, $r = \sqrt{x^2 + y^2}$) limits what you can now make Sep 14, 2022 at 22:12
• If you make, as above, $r = \sqrt{x^2 + y^2}$, then $r = |\mathbf{r}|$, that is, $r$ is the same thing as the "length" of position vector (that's why we chose the label $r$ to begin with, but really could be anything else). The notation is also nice because the definition of unit vector is $\mathbf{\hat{r}} = \mathbf{r}/ r$. But now we must make $x = \mathbf{r \cdot \mathbf{i}} = r \mathbf{\hat{r} \cdot \mathbf{i}}$ and $y = \mathbf{r \cdot \mathbf{j}} = r \mathbf{\hat{r} \cdot \mathbf{j}}$ Sep 14, 2022 at 22:19
• $\mathbf{\hat{r}}$ is unit vector, so we impose (taking square of norm) $(\mathbf{\hat{r} \cdot \mathbf{i}})^2 + (\mathbf{\hat{r} \cdot \mathbf{j}})^2 = 1$ and really, you could make any $\mathbf{\hat{r}} = f(r, \theta) \mathbf{\hat{i}} + g(r, \theta) \mathbf{\hat{j}}$ work, as long as $f^2(r, \theta) + g^2(r,\theta) = 1$. But "to make sense of it", we usually choose something such as $f(r, \theta) = \cos \theta$ and $g(r, \theta) = \sin \theta$ because "it makes sense" as polar coordinates... Sep 14, 2022 at 22:23
• TLDR: once you (implicitly) chose $r$ to mean "distance from the origin", there are limits on how you can describe your position using $r$ and $\theta$... when $\theta$ means "polar angle starting from $x$ axis", it has the form described above... Sep 14, 2022 at 22:28

Vector is a thing that has direction and length. $$\hat{\boldsymbol{r}}$$ is unit vector in the direction of $$\boldsymbol{r}$$ and encodes all the information about its direction, while r is its length. Thus all the information is encoded in $$\boldsymbol{r}=r\hat{\boldsymbol{r}}.$$

With forces, you could do the same. Take unit vector in the direction of the force and multiply it by its length $$\boldsymbol{F}=F\hat{\boldsymbol{F}},$$ but this is not always usefull. Usually you work in some coordinate basis with unit vectors $$\hat{\boldsymbol{x}}, \hat{\boldsymbol{y}}$$ in the case of cartesian coordinates, or $$\hat{\boldsymbol{r}}, \hat{\boldsymbol{\theta}}$$ in the case of polar coordinates. In that case, the direction of the force might be different than the direction of your basis vectors and thus the force will be linear combination of both vectors $$\boldsymbol{F}=F_r\hat{\boldsymbol{r}}+F_{\theta}\hat{\boldsymbol{\theta}}$$.

In general any vector can be written as a linear combination of all basis vectors. Thus you can also write $$\boldsymbol{r}=r_r\hat{\boldsymbol{r}}+r_\theta\hat{\boldsymbol{\theta}},$$ but the component $$r_\theta$$ is zero for the position vector, because the unit vector $$\hat{\boldsymbol{\theta}}$$ is always perpendicular to it, as can be seen from the picture you provided.

The same can happen to a force. For example, gravitational force by which some source at the origin of coordinate system acts on some other object is always in the direction of position vector $$\boldsymbol{F}_G=F_r\hat{\boldsymbol{r}}=-G\frac{mM}{r^2}\hat{\boldsymbol{r}}$$

• What is the point of theta-hat, in polar coordinates, since r-hat can encode direction, and theta-hat is by definition orthogonal to it? Is r-vec just a special vector, that happened to be defined in the most convenient way possible, that is, pointing along one of the coordinate axes? Sep 14, 2022 at 23:37
• @RukiyaMeria first of all check this out math.meta.stackexchange.com/questions/5020/… Sep 15, 2022 at 0:04
• @RukiyaMeria $\hat{\boldsymbol{r}}$ encodes one direction, $\hat{\boldsymbol{\theta}}$ encodes another direction. Both of them can generate any conceivable vector in 2D space by taking their linear combination. You need two basis vectors for describing 2D linear space, so you cannot do with just one, except in special cases in which quantity under consideration is parallel with one of the basis vectors. Sep 15, 2022 at 0:25
• @RukiyaMeria And there is one-one correspondance between coordinate axes and (holonomic) basis. Specific coordinates generate unique basis vectors (more correctly - vector fields) and (holonomic) basis generate coordinates. This is usually silently glossed over as it is more advanced math that you can ignore in the case of cartesian coordinates and intuitively manage in the case of simple curvilinear coordinates like the polar ones. Anyway there is nothing that special about $\boldsymbol{\hat{r}}$ pointing along coordinate axis, thats pretty much how basis/coordinate axes duo always work. Sep 15, 2022 at 0:34
• the mathematical discipline dealing with this is called differential geometry. Its pretty advanced I think, but its very elegant and powerful piece of math. If you are interested, I can recommend Differential geometry textbook by Marian Fecko. The first two chapters should be enough for you. Sep 16, 2022 at 6:32

$$x=r\cos(\theta)$$

$$y=r\sin(\theta)$$

Use the Chain Rule and Product Rule.

$$\dot{x}= \dot{r} \cos{\theta} -r \sin(\theta) \dot{\theta}$$

$$\dot{y}= \dot{r} \sin{\theta} +r \cos(\theta) \dot{\theta}$$

$$\ddot{x}= \ddot{r} \cos{\theta} -\dot{r} \sin(\theta) \dot{\theta}-\dot{r} \sin(\theta) \dot{\theta}-r\cos(\theta)\dot{\theta}^2-r\sin(\theta)\ddot{\theta}$$

$$\ddot{y}= \ddot{r} \sin{\theta} +\dot{r} \cos(\theta) \dot{\theta}+\dot{r} \cos(\theta) \dot{\theta}-r\sin(\theta)\dot{\theta}^2+r\cos(\theta)\ddot{\theta}$$

$$\text{acceleration}=(\ddot{x},\ddot{y})=\ddot{x} \hat{i}+\ddot{y} \hat{j}=f(r,\theta) \hat{r} + g(r,\theta) \hat{\theta}$$

Use the fact that basis vectors transform like covariant tensors of rank 1.

$$\hat{r}= \frac{dx}{dr} \hat{i} + \frac{dy}{dr} \hat{j} \\$$

$$\hat{\theta}= \frac{dx}{d\theta} \hat{i}+\frac{dy}{d\theta} \hat{j}$$

And

$$\hat{r} \cdot \text{acceleration} = f(r,\theta)$$

$$\hat{\theta} \cdot \text{acceleration} = g(r,\theta)$$.

Therefore,

$$f(r,\theta)=\cos(\theta)[\ddot{r} \cos{\theta} -\dot{r} \sin(\theta) \dot{\theta}-\dot{r} \sin(\theta) \dot{\theta}-r\cos(\theta)\dot{\theta}^2 -r\sin(\theta)\ddot{\theta}]+\\ \sin(\theta)[\ddot{r} \sin{\theta} +\dot{r} \cos(\theta) \dot{\theta}+\dot{r} \cos(\theta) \dot{\theta}-r\sin(\theta)\dot{\theta}^2 +r\cos(\theta)\ddot{\theta}] =\ddot{r}-r\dot{\theta}^2$$

(using $$\cos^2 (\theta) + \sin^2 (\theta)=1)$$