# Expression of kinetic energy in polar coordinates

Expression for kinetic energy in Cartesian coordinate:

Expression for kinetic energy in polar coordinate (applying the transformation of coordinates):

Why can't we express it in the following terms by taking the time derivative of each degree of freedom:

• You have an obvious problem of dimensionality for the second term of your last expression, so it is false. Sep 25, 2013 at 17:26

If you express velocity in polar (planar) coordinates you get:$$\mathbf v = \dot r \hat r + +r\dot \theta \hat \theta,$$ so a correct expression for the kinetic energy would be:$$T=\frac{m}{2}(\dot r ^2 + r^2\dot \theta ^2).$$

To find the expression of the velocity in polar coordinates you can work in different ways. I'll suggest you one, very straightforward in my point of view.

First of all, as you noted, we have $$\mathbf r = r(\cos \theta, \sin \theta)$$ (in the first part of the post I'm simply defining $\hat r:=(\cos \theta, \sin \theta)$). One differentiation yelds:$$\dot {\mathbf r} = \dot r (\cos \theta, \sin \theta) + r \dot \theta (- \sin \theta,\cos \theta),$$ and here we call $$\hat \theta=(-\sin \theta,\cos \theta).$$ You can easily check that $\hat\theta$ is perpendicular to $\hat r$. Also note that the norm of both $\hat r$ and $\hat \theta$ is $1$, hence the norm of $\dot {\mathbf r}$ is:$$|\dot {\mathbf r}|=(\dot r ^2 + r^2 \dot \theta ^2)^\frac{1}{2}.$$

I wish I was able to add also a geometric derivation of the result, it's very easy and nice to compare with the one above. Surely you'll be able to find one on some good mechanic's book.

• Perhaps you could show the derivation of $\mathbf{v}$ in polar from Cartesian? That might help the OP in realizing his error. Sep 25, 2013 at 19:31
• You implicitly assumed that kinetic energy is simply expressed in cartesian coordinate. And your implicit understanding of physics don't let you write the 3rd equation in my question. You use transformation equation to change coordinate. Why only adding the time derivative of degrees of freedom gives you a wrong answer as in equation 3? Sep 26, 2013 at 14:26
• Why should it give the correct answer? I don't know much of Lagrangian mechanics, but I think that the kinetic energy is, in it's most general form, something like: $$T=\sum _{i,j} \frac{A_{i,j} (\mathbf q)}{2}\dot q_i \dot q_j.$$ There's no reason why simply setting $A_i = \delta _{i,j}$ should give the correct answer. Also, dimensional analysis should be the first alarm bell. Sep 27, 2013 at 8:20

If the particle is near the origin, say r=1, then a change in $\alpha$ corresponds to a small physical distance. If the particle is far away, r=1000, then the same change in $\alpha$ corresponds to a much larger change in physical location, about 1000 times as large. A term like ${1 \over 2}m \dot\alpha^2$ doesn't account for that. But one like ${1 \over 2}m (r\dot\alpha)^2$ does.

• You are using $s=r \theta$ Oct 8, 2013 at 0:59