Does Newtonian mechanics work in polar coordinates?

Question

Our teacher suggested that Newtonian Mechanics only applies in cartesian coordinates. Is this true?

He gave this example.

Suppose there a train moving with constant velocity $\vec{v}=v_0\hat{x}$, with initial position vector $\vec{r}=(0, y_0)$, where $v_0,y_0$ are constants. He argued that Newton's second law would not hold in polar coordinates. Any ideas?

(We can assume 2D or 3D cases as well, so spherical or polar, it doesn't really matter)

Answer 1

Your teacher is incorrect. $\vec F = m \vec a$ is valid in any inertial (non-accelerating) coordinate system. You must account for the fact that the unit vectors for position in some coordinate systems (polar for example) do not have constant direction and change with time. See a good physics mechanics text, such as Symom Mechanics, for the correct acceleration $\vec a$ in such coordinate systems, where the time derivatives of the unit position vectors are correctly accounted for.

Answer 2

Actually, Newtonian mechanics can be made to work over any Riemanian manifold of any dimension. It's actually a specialisation of Lagrangian mechanics.

This is called Geometric mechanics as is described in Calin & Chung's book, Geometric Mechanics over Riemannian Manifolds, in particular see section 3.8 where they describe the natural Lagrangian on such a manifold as the difference between kinetic and potential energy. They also describe the work and momentum forms and an expression relating work to momentum.

Then in theorem 3.20 they say a particle on the manifold tracing out a path $c$ is an extremiser of the natural Lagrangian iff it verifies Newton's 2nd law of motion in this context, that is:

$F = \ddot{c} = \nabla_{\dot{c}} \dot{c}$

Here $F$ is the force given by $-\nabla V$ where $V$ is some energy potential on the manifold.

This by the way, means that Newtonian mechanics works in polar coordinates or in any other coordinate system.

It's worth noting here that when the force $F$ vanishes, we get the equation for a geodesic: a curve $c$ is a geodesic when:

$\ddot{c} = 0$

Thus we have the analogue of Newtons first law on Riemannian manifolds: when the force on a particle vanishes then the particle moves on a geodesic. And when this manifold is ordinary affine Euclidean space, these are just straight lines.

All of the above remains true for semi-Riemannian manifolds, in particular Lorentzian manifolds. So one way of describing the geodesic equation in GR is to say this is the path followed by a particle at rest where a particle is at rest when there is no force on it. Usually, it's said there are no absolute rest frames but in this language there is, the frame of light or luxons (massless particles).

Answer 3

Your teacher is definitely incorrect. In fact, the whole point $\vec{F}=m\vec{a}$ is written as a vector equation is to emphasize that the equation does not depend on the coordinate system you choose to represent the vectors.

In fact, let's debunk your teacher's counter example by verifying $\vec{F}=m\vec{a}$ in Cartesian and polar coordinates. Assume the particle is following the trajectory described by your teacher, without loss of generality assume the initial $x$ coordinate is zero and the force acting on the particle is zero at all times.

Cartesian Coordinate System

First write down the trajectory in Cartesian coordinates and verify it satisfies Newton's 2nd law.

The position vector of the particle is (Without loss of generality, assume the initial $x$ coordinate is 0): \begin{align*} \vec{r} = vt \,\hat{x} + y_0 \hat{y} \end{align*} where $y_0\neq 0$

Hence by direct differentiation, one can see that $\vec{a} = \vec{0}$. This satisfies the Newton's 2nd law because by $\vec{F} = m\vec{a}$, when $\vec{F}=0$, $\vec{a}=0$.

Polar Coordinate System

How about polar coordinates? Recall that in polar coordinates, the radial unit vector $\hat{r}$ is always pointing along the displacement vector $\vec{r}$ and the unit vector $\hat{\theta}$ is defined to be $\hat{r}$ rotated $90^o$ clockwise. Now what is $\vec{r}$ represented in terms of the polar coordinates? Well simple, just:

\begin{align*} \vec{r} = r \hat{r} \end{align*} where $r=\sqrt{x^2 + y^2} = \sqrt{(vt)^2 + y_0^2}$

So how about the velocity of the particle? \begin{align*} \vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt} \left(r \hat{r}\right) = \frac{dr}{dt} \hat{r} + r\frac{d\hat{r}}{dt} \end{align*}

Here is the key difference between Cartesian and polar coordinates: In Cartesian coordinates, the basis vector is fixed in space, hence the second term is always zero and we simply need to differentiate the component. However, as one can imagine, in polar coordinates, the unit vector $\hat{r}$ is actually changing direction as the particle move! Therefore we have extra terms in our velocities (actually similarly for acceleration).

Now what is the time derivative of $\hat{r}$ (and similarly $\hat{\theta}$ since we will have that term in acceleration anyway). The common way to find it is to first convert $\hat{r}$ and $\hat{\theta}$ into Cartesian and take time derivative. However, since the reasoning might sounds circular by doing so, let's consider the problem geometrically

First, when will $\hat{r}$ and $\hat{\theta}$ change and how would it change? Well, some imagination will tell us that since $\hat{r},\hat{\theta}$ are both unit vectors, the only way the can change is when they rotate. Furthermore, since $\hat{r}$ points in the same direction as $\vec{r}$ and $\hat{\theta}$ is "locked" relative to $\hat{r}$, we know the two will only change when the displacement vector rotate about the origin by some angle $d\theta$.

Next, let's consider the following figure:

As you can see, when $d\theta$ is small, $d\hat{r}$ is along the direction of $\hat{\theta}$ and $d\hat{\theta}$ is along the direction of $-d\hat{r}$. Furthermore, if $d\theta$ is in radian, the length of these vectors are all $d\theta \times 1$ (i.e. the arc length of the arc sweep out by $\hat{r}$ and $\hat{\theta}$). Hence, we obtain the following equation:

\begin{align*} d\hat{r} &= d\theta \hat{\theta}\\ d\hat{\theta} &= -d\theta \hat{r} \end{align*}

Therefore, dividing both side by $dt$, we have: \begin{align*} \dot{\hat{r}} &= \dot{\theta}\hat{\theta}\\ \dot{\hat{\theta}} &= -\dot{\theta}\hat{r} \end{align*}

Using these equations and simply by differentiating $\vec{r}$, we will have:

\begin{align*} \vec{v} &= \dot{r} \hat{r} + r\dot{\theta} \hat{\theta}\\ \vec{a} &= (\ddot{r} - r\dot{\theta}^2)\hat{r} + (r\ddot{\theta} + 2\dot{r} \dot{\theta}) \hat{\theta} \end{align*}

Now what is $\theta(t)$ and $r(t)$ in your teacher's example? Well, by definition: \begin{align*} r(t) &= \sqrt{x^2 + y^2} = \sqrt{(vt)^2 + y_0^2}\\ \theta(t) &= \arctan(y/x) = \arctan(y_0/vt) \end{align*}

So we just have:

\begin{align*} \dot{r} &= \frac{v^2 t}{r}\\ \ddot{r} &= \frac{v^2}{r} -\frac{(v^2 t)^2}{r^3}\\ \dot{\theta} &= -\frac{vy_0}{r^2}\\ \ddot{\theta} &= \frac{2 t v^3 y_0}{r^4} \end{align*}

Substituting everything back to the acceleration formula we derived: \begin{align*} \vec{a} = \left[\frac{v^2}{r} -\frac{(v^2 t)^2}{r^3} - r \left(-\frac{vy_0}{r^2}\right)^2\right] \hat{r} + \left[r\left(\frac{2 t v^3 y_0}{r^4}\right) +2\left(\frac{v^2 t}{r}\right)\left(-\frac{vy_0}{r^2}\right)\right] \hat{\theta} \end{align*}

After some algebra (you can let WolframAlpha do the heavy lifting) both components are $0$. So the acceleration measured in polar coordinate is actually 0, agreeing with Newton's law of motion.

Answer 4

This is an example of how operators do not in general commute. That is: if $x$ and $y$ are variables, $xy=yx$, but if $f$ and $g$ are operators, $fg$ does not generally equal $gf$. An operator is a set of instructions for what to do to the expression that follows it. Consider as a simple example $f =$"add 5" and $g =$ "multiply by 10". Then $fgx = 10x+5$ and $gfx = 10x + 50$. If we want to reverse the operators, we need a third operator, which has the effect of undoing the consequence of the order reversal. Suppose we started with $gx$ and wanted to operate on $x$ with $f$. In this case, we could introduce $h =$ "subtract 45". Then $fgx = hgfx$.

Or we could introduce an operator that undid $g$, using $g^{-1} $="divide by 10". Then we can use the identity $gfg^{-1}gx=gfx$.

Here, "convert Cartesian to polar" and "take the time derivative" are operators. Newton's mechanics are formulated in Cartesian, so if we want to operate with the time derivative and get Newtonian results, we need either a Cartesian coordinate expression, or a third operator. That is: either "convert polar to Cartesian" as $ g^{-1}$ or "undo the consequence of operating on 'convert Cartesian to Polar' with 'take the time derivative'" as $h$.

Answer 5

those are the NEWTON equations in cartesian coordinates (your case)

$$m\,\ddot x=0\\ m\,\ddot y=0$$

and the initial conditions $\dot x(0)=v_0~,\dot y(0)=0~$ and $~x(0)=0~,y(0)=y_0$

with :

$$x=r\,\cos(\phi)\tag 1$$ $$y=r\,\sin(\phi)\tag 2$$

you transfer the EOM's to polar space and obtain two differential equations $~\ddot r=\ldots~,\ddot\phi=\ldots$

you have to transfer also the initial conditions

from

$$x_0=0=r_0\,\cos(\phi_0)\\ y_0=r_0\,\sin(\phi_0)\quad\Rightarrow\\ r_0=y_0~,\phi_0=\pi/2$$

and from $$\dot x=v_0=\sin(\phi_0)\,\dot r_0+r\,\cos(\phi_0)\,\dot\phi_0\\ \dot y=0=\cos(\phi_0)\,\dot r_0-r_0\,\sin(\phi_0)\,\dot\phi_0\quad\Rightarrow\\ \dot\phi_0=\frac{v_0}{y_0}~,\dot r_0=0$$

you can solve now the polar EOM's and obtain $~r(t)~,\phi(t)~$ and with equations (1) ,(2) $~x(t)~,y(t)$

---

General solution \begin{align*} &\text{position vector in polar space}\\ &\mathbf{R}=\begin{bmatrix} x \\ y \\ \end{bmatrix}= \left[ \begin {array}{c} r\cos\left( \phi \right) \\ r\sin\left( \phi \right) \end {array} \right]\\ &\text{the EOM's in polar space }\\\\ &m\,\mathbf{J}^T\,\mathbf J\,\mathbf{\ddot{q}}=\mathbf{J}^T\,(\mathbf{F}-m\,\mathbf{Z})\\ &\text{where}\\ &\mathbf{{q}}=\begin{bmatrix} r \\ \phi \\ \end{bmatrix}\quad ,\mathbf{J}=\frac{\partial \mathbf R}{\partial \mathbf q} \quad, \mathbf{Z}= \frac{\partial \left(\mathbf{J}\,\mathbf{\dot{q}}\right)}{\partial \mathbf q}\,\mathbf{\dot{q}}\quad, \mathbf{F}=\begin{bmatrix} F_x(\mathbf q~,\mathbf{\dot{q}}) \\ F_x(\mathbf q~,\mathbf{\dot{q}}) \\ \end{bmatrix}\\\\ &\text{Transfer the Initial conditions} \\ & \begin{bmatrix} x_0 \\ y_0 \\ \end{bmatrix}= \left[ \begin {array}{c} r_0\cos \left( \phi_0 \right) \\ r_0\sin \left( \phi_0 \right) \end {array} \right]\quad\Rightarrow r_0~,\phi_0\\\\ & \begin{bmatrix} \dot x(0) \\ \dot y(0) \\ \end{bmatrix}= \mathbf{J}\,\mathbf{\dot{q}}\bigg|_{r_0~,\phi_0}\quad \Rightarrow\ \dot{r}(0)~,\dot{\phi}(0) \end{align*}

\begin{align*} &\text{your example}\\\\ &\mathbf{J}=\left[ \begin {array}{cc} \cos \left( \phi \right) &-r\sin \left( \phi \right) \\ \sin \left( \phi \right) &r\cos \left( \phi \right) \end {array} \right] \\ &\mathbf{Z}= \left[ \begin {array}{c} -\dot\phi \, \left( 2\,\sin \left( \phi \right) {\dot{r}}+r\cos \left( \phi \right) \dot\phi \right) \\ \dot\phi \, \left( 2\,\cos \left( \phi \right) {\dot{r}}-r\sin \left( \phi \right) \dot\phi \right) \end {array} \right] \\ &\mathbf{F}=\mathbf{0}\\\\ &\text{EOM's}\\ &\ddot{r}-r\dot{\phi}^2=0\\ &\ddot{\phi}\,r+2\dot{\phi}\dot{r}=0\\ &\text{with the initial conditions}\quad r(0)=y_0~,\dot{r}(0)=0~,\phi(0)=\frac{\pi}{2}~, \dot{\phi}(0)=-\frac{v_0}{y_0}\\ &\text{you obtain the solution}\\ &r(t)=\sqrt{y_0^2+v_0^2\,t^2}\\ &\phi(t)=\arctan\left(\frac{y_0}{r(t)},\frac{v_0\,t}{r(t)}\right)\\ &\Rightarrow\\ &\begin{bmatrix} x(t) \\ y(t) \\ \end{bmatrix}=r(t)\begin{bmatrix} \cos(\phi(t)) \\ \sin(\phi(t)) \\ \end{bmatrix}=\begin{bmatrix} v_0\,t \\ y_0 \\ \end{bmatrix} \end{align*}

you obtain the same solution with cartesian coordinates

$$m\ddot x=0~,x(0)=0~,\dot x(0)=v_0\quad \Rightarrow x(t)=v_0\,t\\ m\ddot y=0~,y(0)=y_0~,\dot y(0)=0\quad \Rightarrow y(t)=y_0$$

conclusion

you obtain the same solution in cartesian coordinates and polar coordinates only if you can map the initial conditions. this is the case if

$$\det(\mathbf J)\bigg|_{r(0),\phi(0)}\ne 0$$

with $$\det(\mathbf J)=r\bigg|_{r(0)}\\ r(0)=\sqrt{x_0^2+y_0^2}$$ thus if $~x_0=0~$ and $~y_0=0~$ you can't map the initial conditions to polar coordinates in this case you can't map the solutions

Answer 6

Newtonian mechanics is independent of any coordinate system.

It is, however, far easier to write the equations down in Cartesian coordinates than in polar coordinates.

Answer 7

It depends on what one means by "work in these coordinates". In two-dimensional Cartesian coordinates, an object moving at constant velocity has coordinates $(x_0+v_xt,y_0+v_yt)$. If one wishes to translate this to polar coordinates as simply $(r_o+v_rt,\theta_0+v_{\theta}t)$, this will indeed "not work". The form of the equations of Newtonian mechanics will not be the same in polar coordinates, and thus the Cartesian coordinate forms of those equations will "not work".

Keep in mind that a coordinate system is simply a system for assigning numbers to points in space (or, in relativity, to events in spacetime). A coordinate system, by definition, has some way of assigning numbers to everything that happens, it's just that the math is simpler in some coordinate systems than others. The space exists independently of the coordinate system, and so do the physics. The only thing that differs between coordinate system is the numbers that are assigned to points, so the only thing that could "not work" is stuff with those numbers, not the mechanics themselves.

For instance, in Cartesian coordinates, the numbers that represent the sum of two vectors can be obtained by taking the sum of the numbers that represent the two vectors being added, but this does not work in polar coordinates. In Cartesian coordinates, if you have the position vector (which isn't really a vector, it's actually in affine space, but things are complicated enough already without getting into that) is given in terms of a function giving the x coordinate and another function giving the y coordinate, the velocity vector can be obtained by just taking the derivatives of those functions, but that doesn't work in polar coordinates. However, the numbers representing a point is not the same as the point itself, and facts about the former shouldn't be confused for facts about the latter.

Answer 8

The advantage of cartesian coordinates is that we can take vectors as simply indexed functions. But there is an additional feature that can not be forgotten when changing variables: coordinate basis.

The velocity vector of the OP example can be expressed in its complete form as:

$v_1(X^1, X^2) = V^1B_{11} + V^2B_{21}$
$v_2(X^1, X^2) = V^1B_{12} + V^2B_{22}$

In cartesian coordinates, $X^1 = x$, $X^2 = y$, $V^1 = v_x$, $V^2 = v_y$, $B_{11} = 1$, $B_{12} = 0$, $B_{21} = 0$, $B_{22} = 1$

That way:
$v_1(x,y) = V^1$ and $v_2(x,y) = V^2$ are the familiar cartesian components of the indexed function (and vector) $v(x,y)$.

When transforming to polar coordinates, it is possible to derive $B_{ab}$ and $V^a$ such that $v_b$ don't change. Here:
$B_{11} = cos(\theta)$, $B_{12} = sin(\theta)$, $B_{21} = -Rsin(\theta)$, $B_{22} = Rcos(\theta)$, $X^1 = R$, $X^2 = \theta$, $V^1 = v_R$, $V^2 = v_{\theta}$

So:
$v_1(R, \theta) = v_Rcos(\theta) - v_{\theta}Rsin(\theta)$
$v_2(R, \theta) = v_Rsin(\theta) + v_{\theta}Rcos(\theta)$

In the example, $v_1 = v_0$ and $v_2 = 0$. It is easy to realize that to keep the same values:

$$v_R = v_0cos(\theta)$$ and $$v_\theta = -v_0\frac{sin(\theta)}{R}$$

In polar coordinates, the components change with time for this vector be constant with time as desired. The vector equations of the Newton laws are valid, but the notion of what is a vector must be closely understood.

Answer 9

Newton's laws are vector relations, which are independent on the coordinate systems. It is likely that the OP misinterprets the statement made by the professor. E.g., one of the following could be the case:

• That addition of components of vectors in curvilinear coordinates (such as polar coordinates) is not as simple as in in the rectangular coordinates
• That the Newton laws do not work in a rotating frame of reference.

It is also possible that the professor said what they actually said, simply to guard off the predictable errors that most students make (alas, after a year or two of teaching the same course the errors and questions are very predictable), but this simplification made the statement indeed incorrect, when examined more rigorously.