# Writing Kinetic Energy in Cylindrical Coordinates

My Textbook wrote the Kinetic Energy while teaching Hamiltonian like this:

(in Cylindrical coordinates) $$T=\frac{m}{2}[(\dot{\rho})^2 +(\rho \dot{\phi} )^2+ (\dot{z})^2]$$

I know to find velocity in Cartesian coordinates $$position = x+y+z$$ $$velocity = \dot{x}+\dot{y}+\dot{z}$$ But in this case, even though the components of position in Cylindrical coordinates are $$position = \rho +\phi + z$$ the velocity is $$Velocity =\dot{\rho} +\rho \dot{\phi} + \dot{z}$$

where did the extra $\rho$ come from?

• This is about cylindrical coordinates (polar is two-dimensional). The 'extra' radius makes the units commensurate. Can't add m/sec quantities to radians/sec quantities! It comes from distance measurement. – Whit3rd Dec 1 '16 at 9:25
• $position \neq x+y+z$. It is basically a vector: $position = (x,y,z)$ and then $velocity = v = (\dot{x},\dot{y},\dot{z})$ and $v^2 = \dot{x}^x+\dot{y}^2 + \dot{z}^2$ is the inner product of these vectors. – user1583209 Dec 1 '16 at 11:01
• Similarly for polar coordinates, $position\neq \rho + \phi + z$ (How would you even add an angle to a length?) – user1583209 Dec 1 '16 at 11:08
• Probably related: physics.stackexchange.com/q/183882 – Kyle Kanos Dec 1 '16 at 14:26
• @user1583209 It's a good mnemonic but that's not the "reason" that there is an extra term. You could add any constant with the right units to fix the units... They result from the fact that the unit vectors are not fixed... – Michael Angelo Dec 6 '16 at 8:26

Imagine an object which it a fixed distance $\rho$ from the origin in the xy-plane (z=0).

The angle between the x-axis and the line joining the origin to the object is $\phi$.

Now change the angle to $\phi + \Delta \phi$ in a time $\Delta t$

Using your notation the angular speed $\phi^* = \frac {\Delta \phi}{\Delta t}$.

This distance moved by the object in time $\Delta t$ is $\rho\; \Delta \phi$ and so the linear speed of the object is $\rho \frac {\Delta \phi}{\Delta t} = \rho \phi^*$ which has to be added to the two other linear speeds $\rho^*$ and $z^*$.

• So I need to express the angular velocity in terms of linear velocity. But is there ever a case where you can use angular velocity in the lagrangian? For example, using $I \omega$ instead of mv? – Jess L Dec 1 '16 at 9:13

Kinetic energy is always $\frac{1}{2} m \dot{\vec{r}}^2$. What is important here is the vector sign for the position vector $\vec{r}$. In cartesian coordinates, $\vec{r}=x\vec{e_x}+y\vec{e_y}+z\vec{e_z}$ and therefore $\dot{\vec{r}}=\dot{x}\vec{e_x}+\dot{y}\vec{e_y}+\dot{z}\vec{e_z}$. Here,$\vec{e_x},\vec{e_y},\vec{e_z}$ are the unit vectors in $x,y,z$ direction. This is not as simple in any coordinate system - for cylindrical coordinates, $\vec{r}=\rho\vec{e_{\rho}}+z\vec{e_z}$ and $\dot{\vec{r}}=\dot{\rho}\vec{e_{\rho}}+\rho\dot{\vec{e_{\rho}}}+\dot{z}\vec{e_z}$. As $\dot{\vec{e_{\rho}}}=\dot{\varphi} \vec{e_{\rho}}$, we arrive at your formula given above.
The important thing is to always start from $\dot{\vec{r}}^2$ in non cartesian coordinates and you will often see extra terms appear.

The key is to always start in Cartesian coordinates, and remember that $T=\frac{m}2\left(\dot{x}^2+\dot{y}^2+\dot{z}^2\right)$. You can always start in Cartesian because the kinetic energy is a scalar and thus independent of the coordinate system in which you choose to evaluate it, although scalar products are most easily computed in Cartesian coordinates.

From the expression of $T$ and the definition of your coordinates: \begin{align} x&=\rho\cos(\phi)\, ,\\ \dot{x}&=\dot{\rho}\cos(\phi)-\rho\sin(\phi)\dot{\phi}\, ,\\ y&=\rho\sin(\phi)\, ,\\ \dot{y}&=\dot{\rho}\sin(\phi)+\rho\cos(\phi)\dot{\phi}\, ,\\ \dot{z}&=\dot{z}\, . \end{align} Squaring the dotted terms and summing gives $$T=\frac{m}{2}\left(\dot{\rho}^2+\rho^2\dot{\phi}^2+\dot{z}^2\right)$$ as requested. This procedure of course will work for any coordinate system.