In Cartesian Coordinate System, the position of an object is given by a position vector:
$$\vec R=X \hat{i} + Y\hat{j} $$

We move $X$ units in $\hat{i}$ direction and $Y$ units in $\hat {j}$ direction to get the position.

Now while using the spherical coordinate system, why is the position vector just $$\vec R = r \hat{r}$$

and not $$\vec R= r \hat{r} + \theta \hat{\theta}$$

Doesn't it make sense to say that I'm moving $r$ units in $\hat{r}$ direction and $\theta$ units in $\hat{\theta}$ direction $?$


This is the acceleration vector in spherical coordinate system. Upon closer examination, I can't say what this is. There are four components for acceleration in two dimensions:

  1. Linear Acceleration in $\hat{r}$ direction.
  2. Centripetal Acceleration in $-\hat{r}$.
  3. Tangential Acceleration in $\hat{\theta}$ direction.
  4. Coriolis Acceleration in $\hat{\theta}$ direction.

As a whole, it is $$\vec a = (\ddot{r} - r\dot{\theta}^2)\hat{r}+(2\dot{r}\dot{\theta}+r\ddot{\theta})\hat{\theta}$$.

I don't understand is that the frame is non-inertial since rotation occurs and unit vectors are also rotating, then how can we have centripetal force in a non-inertial frame? Shouldn't $r\dot{\theta}^2$ be in $\hat{r}$ direction rather than $-\hat{r}$ direction because the frame rotates, and shouldn't it be centrifugal force?


2 Answers 2


The main thing, I think, to understand about Carteisan coordinates in comparison to other coordinates is that Cartesian coordinates are unique in that they are the same all over space. That is, the unit vector $\hat{x}$ and $\hat{y}$ are independent of position. This is not the case for spherical coordinates, here the unit vectors themselves are functions of the angle. $\hat{r}$ points at different directions depending on where you are! On the one hand, it indeed means that $\vec{R} = r \hat{r}$, which seems very simple, but on the other hand it means that when we take derivatives, we have to derive the unit vector itself! (which we do not need to do in Cartesian coordinates)

So $\dot{\vec{R}} = \dot{r}\hat{r} + r \dot{\hat{r}} = \dot{r}\hat{r} + r \dot{\theta} \hat{\theta} $ where the second term comes out when you look at the derivative of $\hat{r}$ itself.

It is best to convince yourself in that, I believe, by looking at the relation to the fixed Cartesian coordinates $$ \hat{r} = \cos(\theta)\hat{x} + \sin(\theta)\hat{y} \to \dot{\hat{r}} = \dot{\theta}[-\sin(\theta)\hat{x} + \cos(\theta)\hat{y}]=\dot{\theta}\hat{\theta}$$ and you can continue to take the second derivatives as well, to get the expression for the acceleration in spherical coordinates.


Shouldn't $r\dot{\theta}^2$ be in $\hat{r}$ direction rather than $-\hat{r}$ direction because the frame rotates, and shouldn't it be centrifugal force?

Here is an easy way to see why the sign is negative. Consider the radial component $$\vec a = (\ddot{r} - r\dot{\theta}^2)\hat{r}$$

If $r$ is constant, $\ddot{r}=0$. The object is moving in a circle, and the acceleration in circular motion is directed inwards towards the origin, i.e. in the $-\hat{r}$ direction with magnitude $r\dot{\theta}^2$. Therefore, the term is $- r\dot{\theta}^2\hat{r}$.

  • $\begingroup$ ,we are talking about a rotating(non inertial) frame here ,so it shouldn't be centripetal ,it should be centrifugal and in $\hat{θ}$ we have a component with its magnitude as $2\dot{r}\dot{θ}$ ,it's coriolis acceleration ,it should have a negative sign right,if we are considering a non inertial frame ? $\endgroup$
    – Harry Case
    Sep 27, 2021 at 12:59
  • $\begingroup$ @HarryCase Deriving the expression does not imply that we have switched into a rotating frame. We are still in the lab frame. We are just expressing the acceleration in polar coordinates. But we are still in the lab frame. $\endgroup$ Sep 27, 2021 at 13:08
  • $\begingroup$ @HarryCase So we have $\vec a = (\ddot{r} - r\dot{\theta}^2)\hat{r}+(2\dot{r}\dot{\theta}+r\ddot{\theta})\hat{\theta} = \ddot{x}\hat{x}+\ddot{y}\hat{y}+\ddot{z}\hat{z}$. They represent the same vector which is the acceleration in the lab frame. $\endgroup$ Sep 27, 2021 at 13:11
  • $\begingroup$ But how does $Coriolis$ $Acceleration$ arise in a lab frame ?,. $2\dot{r}\dot{\theta}$ -What is the meaning of this term in $\hat{\theta}$ $\endgroup$
    – Harry Case
    Sep 30, 2021 at 8:50
  • $\begingroup$ @HarryCase There is no Coriolis acceleration or any other fictitious acceleration in the lab frame. The extra terms are there because the coordinate lines of constant $r$ and $\theta$ do not correspond to inertial frames (at least one of them are not straight). This means that $\ddot{r}=0$ and $\ddot{\theta}=0$ do not describe straight lines and the must be additional terms. $\endgroup$ Oct 1, 2021 at 12:39

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