In cartesian coordinates we have the coordinates x,y,z, and the position vector is described by r(x,y,z) = $x\hat{x}$+ +$y\hat{y}$ +$z\hat{z}$

However, in polar coordinates, we have the coordinates $r$, $\theta$ but the position vector is r=$r\hat{r}$ and not $r\hat{r}$ + $\theta \hat{\theta}$ as one would expect with the same logic used in cartesian coordinates.

One statement:

"In polar coordinates, the position of a particle A, is determined by the value of the radial distance to the origin, $r$, and the angle that the radial line makes with an arbitrary fixed line, such as the $x$-axis. Thus, the trajectory of a particle will be determined if we know $r$ and $ΞΈ$ as a function of $t$, i.e. $r(t),ΞΈ(t)$."

However in physics textbooks and lectures, to find the velocity of the position vector in polar coordinates we use the derivative of $r\hat{r}$ and not the derivative of $r\hat{r}$+$\theta \hat{\theta}$. (for example https://www.youtube.com/watch?v=3z15i3hjNzo)

Why is this the case?


5 Answers 5


In a Cartesian coordinate system, $$\mathbf{r}=x\mathbf{\hat x}+y\mathbf{\hat y}+z\mathbf{\hat z}$$

where $\mathbf{r}$ is the "position vector." Additionally, $r$ (non bold-faced) is usually taken to equal the distance from the origin to your point of interest, i.e.


which is true in any coordinate system (assuming a Euclidean norm). With these two definitions, it is self-evident that

$$\mathbf{r}=r\mathbf{\hat r}$$

This expression just says that your position vector points in the direction of your normalized position vector, and that its magnitude is $r$. Once again, this expression is true for any coordinate system.

It's really important to note that, if a point of interest $p$ equals $(r,\theta,\phi)$ in spherical coordinates (for example), then $\mathbf{r}$ does not equal $r\mathbf{\hat r}+\theta \mathbf{\hat \theta} +\phi \mathbf{\hat \phi}$. You can see this by simply considering the units of the vectors and the quantities $r$, $\theta$, and $\phi$. Instead, to find $\mathbf{r}$ in spherical coordinates (or any other coordinate system), you will need to use the known expressions for $x$, $y$, $z$, $\mathbf{\hat x}$, $\mathbf{\hat y}$, and $\mathbf{\hat z}$.

It's equally important to realize that the unit vectors $\mathbf{\hat r}$ , $\mathbf{\hat \theta}$, and $\mathbf{\hat \phi}$ change with position (and time). So they're actually functions of $x$, $y$, and $z$ (or $x(t)$, $y(t)$, and $z(t)$).

As an aside, unfortunately, many textbooks also use $r$ to denote the distance from a point to the z-axis in circular / cylindrical coordinates. As somebody else already said, these two definitions are not compatible with each other and can cause quite a bit of confusion. Better notation is $(\rho, \phi)$ or $(s, \phi)$, so that


With each new textbook, paper, etc. it's important to understand how they define their basic coordinates.

Finally, to answer your question, the time derivative of $\mathbf{r}$ is

$$\frac{d\mathbf{r}}{dt}=\frac{d(r\mathbf{\hat r})}{dt}$$

... for the simple reason that $\mathbf{r}$ ALWAYS equals $r\mathbf{\hat r}$. Note that this expression is somewhat difficult to calculate since $\mathbf{\hat r}$ is actually a function of position and time.

In any coordinate system,

$$\frac{d\mathbf{r}}{dt}=\frac{d}{dt}(x\mathbf{\hat x}+y\mathbf{\hat y}+z\mathbf{\hat z})$$

$z = 0$ in polar coordinates, so that

$$\frac{d\mathbf{r}}{dt}=\frac{d}{dt}(x\mathbf{\hat x}+y\mathbf{\hat y})=\frac{dx}{dt}\mathbf{\hat x}+\frac{dy}{dt}\mathbf{\hat y}=\frac{d (\rho \cos{\phi})}{dt}\mathbf{\hat x}+\frac{d (\rho \sin{\phi})}{dt}\mathbf{\hat y}$$

Your quoted expression is correct because it is a simple statement that

$$\mathbf{r}(t)=x(t)\mathbf{\hat x}+y(t)\mathbf{\hat y}$$

which is how we defined the position vector in the first place.

  • 1
    $\begingroup$ Perfect, thank you $\endgroup$
    – qubitz
    Mar 10, 2022 at 0:01
  • 3
    $\begingroup$ Your first statement isn't strictly true. In any coordinate system, $\vec{r}=r\hat{r}$. The vectors $\hat{x}$, $\hat{y}$ and $\hat{z}$ only explicitly exist in Cartesian coordinates, and don't have any particular significance in spherical coordinates. It might be true that $\vec{r}=x\hat{x}+y\hat{y}+z\hat{z}$, if you define the three unit vectors appropriately, but doesn't have to be. I think you're trying to say something like "in Euclidean space, Cartesian coordinates can always be used". $\endgroup$
    – MichaelS
    Mar 10, 2022 at 7:27
  • 1
    $\begingroup$ @AntoniosSarikas The answer is actually a bit subtle. In general, the number of unit vectors in a space is equal to the dimension of the space. So in 3D space, you simply have your three basis vectors $\mathbf{\hat x, \hat y}$, and $\mathbf{\hat z}$. In curvilinear coordinates, its convenient to refer to $\mathbf{\hat r}$ as a basis vector that depends on a particular particle's position, but it's important to realize that $\mathbf{\hat r}$ is linearly dependent on $\mathbf{\hat x, \hat y}$, and $\mathbf{\hat z}$. So, in a way, it's not a new basis vector. Continued below... $\endgroup$
    – Dr. Momo
    Jul 24, 2022 at 12:05
  • 1
    $\begingroup$ @AntoniosSarikas Now... you need 3N numbers to describe the positions of N particles in 3D space whether you are working in Cartesian coordinates or curvilinear coordinates. Of course, you can define different basis vectors with respect to different particles: $\mathbf{\hat r_1}$, $\mathbf{\hat r_2}$, ..., $\mathbf{\hat r_n}$. The important point, however, is that all of these basis vectors (and their phi and theta counterparts) all depend on $\mathbf{\hat x, \hat y}$, and $\mathbf{\hat z}$. They do share the space space, after all. Continued below... $\endgroup$
    – Dr. Momo
    Jul 24, 2022 at 12:13
  • 1
    $\begingroup$ @AntoniosSarikas With all of that, I can finally answer your question directly. The minimum number of basis vectors in a space is that space's dimension. But there is no maximum number, because you can always define linear combinations of existing basis vectors to form new basis vectors. You can define basis vectors with respect to certain particle's positions (as we often do in E&M out of convenience), but you can also define basis vectors in an infinite number of additional ways. $\endgroup$
    – Dr. Momo
    Jul 24, 2022 at 12:17

As you imply, the position vector, $\mathbf r$, can be expressed as the sum of three cartesian components: $$\mathbf r=x\mathbf {\hat x}+y\mathbf {\hat y}+z\mathbf {\hat z}$$ This can't be done in polars. The problem is that there don't exist unit vectors $\mathbf{\hat r}, \mathbf {\hat \theta}, \mathbf {\hat \phi}$ that are constant vectors, in the same way that $\mathbf {\hat x}, \mathbf {\hat y}$ and $\mathbf {\hat z}$ are constant vectors. [We can express $d\mathbf r$ in terms of local unit vectors $\mathbf{\hat r}, \mathbf {\hat \theta}, \mathbf {\hat \phi}$, but this can't be extended to finite displacement vectors.]

So if we want to use polars, we use the co-ordinates, $r, \theta, \phi$, but we don't see them as scalar coefficients of unit vectors.

As has already been ably pointed out, leaving the displacement vector as $\mathbf r$ is not committing to any co-ordinate system.


Fix a vector space $V$ of dimension $n$. Lets also fix a Cartesian and polar coordinate system.

Now, the distinction that you are looking at is basically due to the fact that radial direction in polar coordinates is dependent on all the Cartesian directions and so we expect it to have $n$ directions of freedom (dof). But the normalised radial vector is restricted to lie on the unit sphere and so it has a single dof less.

Thus the radial vector multiplied by some scale allows it to cover the space. This is what you see Any other coordinate system with the same property would be able to do the same.

Now the vectors ($\hat x_i$) as well as ($\hat r, \hat \theta_i$), although vectors, do not live in the given vector space $V$. They lie in $TV$, the tangent space of $V$. This is also a vector space and of twice the dimension of $V$ and so it has dimension $2n$. Properly speaking, they are tangent vectors. And in fact, more properly speaking, they are tangent fields. It's only when you specify a position $p$ (which here is a vector, since our underlying space is a vector space) that you get a tangent vector, say $\hat r(p)$ and this lives in the tangent space $T_pV := TV[p]$.

Now, the expression:

$x^i(p).\hat x_i(p)$

is then just a general tangent vector at $p$.

as is the expression:

$r(p).\hat r(p) + \theta^i(p).\hat \theta_i(p)$

The two families of tangent vectors ${\hat x_i(p)}$ and ${\hat r(p),\hat \theta_i(p)}$ are both bases of $T_pV$ and the transformation matrix between them changes tangent vectors (and not position vectors!) expressed in one coordinate system into another.

In particular, the expression:

$x^i(p).\hat x_i(p)$

doesn't in general - that is over a curved space - make sense. Why? Because $x^i(p)$ is the components of a vector and not a tangent vector and so we shouldn't be summing this over the tangent basis. The reason why we can do this is that the tangent spaces of vector space have a canonical isomorphism with the underlying vector space. So in this expression we can also think of $\hat x_i(p)$ as also lying in the vector space and so the sum makes sense.


Position is given as the vector $\vec r$. Using polar coordinates $\vec r$ is specified using $r$ and $\theta$. Let $\hat n$ be a unit vector that specifies the direction of $\vec r$; $\vec r = r \hat n$. (You use $\hat r$ for the unit vector, I use $\hat n$ to make the difference between $\hat r$ and $\hat n$ clearer.) $\hat n$ has magnitude one but is not fixed in direction; $\hat n$ depends on $\theta$, so $\vec r = r \hat n(\theta)$.

The velocity $\vec v$ is given by: $\vec v = {d \over dt} \vec r = {dr \over dt} \hat n + r {d \hat n \over dt} = {dr \over dt} \hat n + r{d\hat n \over d\theta}{d\theta \over dt}$.

Let $\hat l$ be a unit vector in the $\theta$ direction. (You use $\hat \theta$ for the unit vector, I use $\hat l$.) ${d \over d\theta} \hat n = \hat l$. So, $\vec v = {dr \over dt} \hat n + r {d \theta \over dt} \hat l $.

Information added based on OP comment. $\vec r \ne r\hat n + \theta\hat l$; position does not depend on $\hat l$. Velocity and acceleration both depend on $\hat l$ as well as $\hat n$. See a good physics mechanics book.

The figure below explains how the same position vector $\vec r$ can be expressed using the polar coordinate unit vectors $\hat n$ and $\hat l$, or using the Cartesian coordinates unit vectors $\hat i$ and $\hat j$, unit vectors along the Cartesian x and y axes, respectively. $\hat n$ and $\hat l$ are not fixed in directions, they move as $\theta$ changes. $\hat i$ and $\hat j$ are fixed in directions along their respective fixed axes.

enter image description here

  • $\begingroup$ So because $\hat{n}$ depends on theta there is no need to use the theta component of the vector r in polar coordinates ? In polar coordinates, a vector is described by r(r,$\theta$) = r$\hat{r}$+$\theta$$\hat{\theta}$. Why to describe the position vector we use $\hat{n}$($\theta$) instead of using the theta component itself directly ? When we use the normal r$\hat{r}$ is $\hat{r}$ also a function of theta just like n is? $\endgroup$
    – qubitz
    Mar 9, 2022 at 23:07
  • $\begingroup$ You seem to be asking the same question over and over again in the comments. A vector $\vec v$ can be written as $\vec v = v_r \hat r + v_\theta \hat \theta$. When that vector $\vec v$ happens to be the position vector of interest, we have $v_r = r$ and $v_\theta = 0$. The position vector points from the origin to the position of interest and clearly has no non-radial component. $\endgroup$
    – hft
    Mar 9, 2022 at 23:12
  • $\begingroup$ See my updated response. $\endgroup$
    – John Darby
    Mar 9, 2022 at 23:24
  • $\begingroup$ Thank you. π‘Ÿβƒ— β‰ π‘Ÿπ‘›Μ‚ +πœƒπ‘™ but a point of interest 𝑝 can be identified by (π‘Ÿ,πœƒ). In conclusion, if a point p is located at (π‘Ÿ,πœƒ) in polar coordinates it is not true that the position vector π‘Ÿβƒ— =π‘Ÿπ‘›Μ‚ +πœƒπ‘™ but it is true in cartesian coordinates. Could you confirm? Thank you so much! $\endgroup$
    – qubitz
    Mar 10, 2022 at 2:16
  • $\begingroup$ @qubitz I added a picture that should help. $\endgroup$
    – John Darby
    Mar 10, 2022 at 3:41

Your $r$'s are playing multiple roles.

Instead, let $\vec R$ be the position vector ($\vec R=R\hat R$), and let $r$ be the radial coordinate in (say) $(r,\theta,\phi)$.

So, now, the position vector $\vec R$ and its time-derivative is the velocity $\vec V=\frac{d}{dt}\vec R$,
which could be expressed in any set of coordinates,
e.g., rectangular $(x,y,z)$ or spherical $(r,\theta,\phi)$ or cylindrical $(s,\theta,z)$.

  • 1
    $\begingroup$ R expressed in rectangular coordinates is R(x,y,z) but in polar coordinates it is R(r) and not R(r,theta) why is that the case. I would expect that a vector expressed in a particular coordinate system would utilize all of the coordinates of that system $\endgroup$
    – qubitz
    Mar 9, 2022 at 22:31
  • $\begingroup$ @qubitz In 3-d space, we need three numbers to locate the tip of the vector. So, $\vec R(r)$ is insufficient... all that suggests is that the position vector $\vec R$ has magnitude $r$ so the tip lies on a sphere of radius $r$. Which point on that sphere requires the specification of two more coordinates: the angular coordinates $\theta$ and $\phi$ or a latitude-and-longitude. $\endgroup$
    – robphy
    Mar 9, 2022 at 22:37
  • $\begingroup$ youtube.com/watch?v=3z15i3hjNzo , in this video and in classical mechanics by John R taylor they use the position vector R = r$\hat{r}$ in polar coordinates. Why is there no theta component ? is it not necessary that the position vector have a theta component too ? $\endgroup$
    – qubitz
    Mar 9, 2022 at 22:50
  • $\begingroup$ @qubitz $\hat r$ (a unit vector) is different from $r$. The unit-vector $\hat r$ has three coordinates... with its radial component equal to 1. $\endgroup$
    – robphy
    Mar 9, 2022 at 22:52
  • $\begingroup$ @qubitz My GeoGebra visualization might help: geogebra.org/m/sjzxecxm (not so much for clarifying notation but for visualizing the spherical coordinates). $\endgroup$
    – robphy
    Mar 9, 2022 at 22:55

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