enter image description here

A free particle with coordinates as shown has kinetic energy $$T = \frac{1}{2}m\left(\dot r^2 + r^2\dot\theta^2 + r^2\sin^2\theta\dot\phi^2\right)$$ So we see $T$ depends on $\theta$.

Now suppose we rotate our coordinate system such that only one coordinate $\theta$ changes from $\theta$ to $\theta'$ and fix it there as is shown

enter image description here

In this coordinate system the kinetic energy should be the same as before since the kinetic energy should be same in all inertial frames.

However if we substitute values of $r, \theta',\phi$ in the formula for $T$ we will get a different value of $T$ hence a contradiction.

Can anyone please tell me what is wrong?

  • $\begingroup$ So you measure your θs from θ' instead of 0, for every vector on your space. What have you gained? What do you imagine has changed? $\endgroup$ Oct 16, 2021 at 15:33
  • $\begingroup$ Since $\theta$ has changed the kinetic energy might have a different value now according to the formula. $\endgroup$
    – Kashmiri
    Oct 16, 2021 at 15:35
  • 3
    $\begingroup$ Your diagram shows a rotation that will change all the coordinates and not just one. $\endgroup$
    – nasu
    Oct 16, 2021 at 15:58
  • $\begingroup$ I was not able to draw it accurately $\endgroup$
    – Kashmiri
    Oct 16, 2021 at 15:58
  • $\begingroup$ I've deleted the second one as nasu pointed that it was misleading. $\endgroup$
    – Kashmiri
    Oct 16, 2021 at 16:10

4 Answers 4


This is your position vector to the mass

$$\mathbf R= \left[ \begin {array}{c} \cos \left( \varphi \right) \sin \left( \vartheta \right) \\ \sin \left( \varphi \right) \sin \left( \vartheta \right) \\ \cos \left( \vartheta \right) \end {array} \right] $$

from here you got the kinetic energy

$$T=\frac m2\mathbf v^T\,\mathbf v\\ \text{where}\\ \mathbf v=\frac{d}{dt}\mathbf R$$

now if you rotate the position vector with any constant orthonormal rotation matrix $~\mathbf S~$ then

$$\mathbf v\mapsto \mathbf S\,\mathbf v\\ \text{and}\\ \mathbf v^T\,\mathbf v=\mathbf v^T\,\underbrace{\mathbf S^T\,\mathbf S}_{=\mathbf 1}\,\mathbf v=\mathbf v^T\,\mathbf v $$

hence the kinetic energy is unchanged


The rotation matrix that rotate the position vector $~\mathbf R~$ from angle $~\vartheta~$ to angle $~\vartheta+\vartheta_0~$ is ( Rodrigues Rotation matrix)

$$\mathbf S_\vartheta=\mathbf S(\mathbf d,\vartheta_0)$$

where $~\vec d~$ is the rotation axes and $~\vartheta_0~$ is the rotation angle around this axes.

$$\mathbf d=\left[ \begin {array}{c} \sin \left( \varphi _{{0}} \right) \\ -\cos \left( \varphi _{{0}} \right) \\ 0\end {array} \right] $$

with $~\mathbf R\mapsto \mathbf S_\theta\,\mathbf R$ you obtain the "new" kinetic energy which remained unchanged.

now you obtain the "new" kinetic energy with

$~\mathbf R(r~,\varphi~,\vartheta)\mapsto \,\mathbf R(r~,\varphi~,\vartheta+\vartheta_0)~$

this is obviously not the same process .

  • $\begingroup$ Thank you. But could you please tell me the flaw in my argument. Thank you :) $\endgroup$
    – Kashmiri
    Oct 16, 2021 at 16:09
  • $\begingroup$ @Kashmiri The flaw is rotations do not in fact change just under polar angle at a time. Eq. (20) here shows what they do in Cartesian terms. If you apply that beast to this answer's $\mathbf{R}$, you'll see the result is not simply $\mathbf{R}$ again with one angle shifted. $\endgroup$
    – J.G.
    Oct 16, 2021 at 17:31
  • $\begingroup$ That was scary. So you mean there doesn't exist a rotation of coordinate system such that only $\theta$ will change ? $\endgroup$
    – Kashmiri
    Oct 17, 2021 at 6:15
  • $\begingroup$ @Kashmiri see new edit $\endgroup$
    – Eli
    Oct 17, 2021 at 17:02
  • $\begingroup$ I appreciate your help but it doesn't answer why my thinking is flawed. $\endgroup$
    – Kashmiri
    Oct 18, 2021 at 6:42

Basically $\dot\phi$ will change together with $\theta$. Imaging two close points with the same $=r$ and $\theta$, such that $\phi$ differs by really small amount $d\phi$. When you rotate coordinates in $(z,r)$ plane, $d\phi$ changes, and so $\frac{d\phi}{dt}$ does.

Note that you are rotating in the $(z,r)$ plane. All points in this plane will have the same $\phi$ after rotation as needed because it's a plane of constant $\phi$. But points outside of this plane will change their $\phi$ coordinate when you rotate them since planes, parallel to $(z,r)$ are not planes of constant $\phi$ (which are all planes $z$ axis belongs to.)

Edit: Here is other way to think about it. If you make transformation $\theta'=\theta + \delta\theta$ keeping $r$ and $\phi$ constant, then Cartesian coordinates will change to the following:

$$x = r \cos\phi \sin(\theta'-\delta\theta)$$ $$y = r \sin\phi \sin(\theta'-\delta\theta)$$ $$z = r \cos(\theta'-\delta\theta)$$

But that's not a spherical coordinate system anymore, so the formula for kinetic energy doesn't apply anymore. Basically it's not a rotation!

  • $\begingroup$ Then do you mean there is no way to rotate and keep r, phi the same? $\endgroup$
    – Kashmiri
    Oct 16, 2021 at 16:53
  • $\begingroup$ Yep. Imaging that your point is in $(z,y)$ plane initially, so you want to rotate around $x$ axe, so that it's $r$ and $\phi$ won't change. But consider any point in $(z,x)$ plane, that in located not on axes, e.g. $x=10,y=0,z=10$. After rotation it has to go out of the $(z,x)$ plane, so it's $\phi$ will change. $\endgroup$
    – Yuras
    Oct 16, 2021 at 17:00
  • $\begingroup$ To make is completely clear: you can rotate keeping $r$ and $\phi$ the same for all points in some plane only. If all physics happens in this plane, then $\dot\phi=0$, and the formula for kinetic energy works. But if something is moving out of that plane, then $\dot\phi$ will change. $\endgroup$
    – Yuras
    Oct 16, 2021 at 17:08

A free particle with coordinates as shown has kinetic energy

$$T = \frac{1}{2}m\left(\dot r^2 + r^2\dot\theta^2 + r^2\sin^2\theta\dot\phi^2\right)$$

So we see $T$ depends on $\theta$.

Maybe you think that the new calculation for T is: $$T = \frac{1}{2}m\left(\dot r^2 + r^2\dot\theta'^2 + r^2\sin^2\theta'\dot\phi^2\right)$$ because $r$ and $\phi$ are unchanged. As $\theta' \neq \theta$, the values would be numerically different.

The issue is that a rotation would not let $\phi$ unchanged. It is easy to visualize imaging a new North Pole shifted some degrees to Asia and following the Greenwich meridian. At any point in this meridian it is true that $\theta' = \theta + \delta$, and $r$ and $\phi$ are the same. But it is not difficult to see that there are several points at other meridians that would keep the same latitude in both systems for example. So the coordinate transformation is:

$r' = r$
$\theta' = f(\theta , \phi)$
$\phi' = g(\theta, \phi)$

Now it is clear that $T$ is not necessarily numerically different. It is necessary to do a lot of calculations to express it in the new coordinates.

  • $\begingroup$ "The issue is that a rotation would not let ϕ unchanged", So you mean that we can't rotate in such a way so that the $\theta$ changes but $r, \phi$ remain the same? $\endgroup$
    – Kashmiri
    Oct 17, 2021 at 6:01
  • $\begingroup$ Yes, we can't. The meridians from the new North pole don't match the old ones. And more: A rotation of the coordinates in $\theta$ doesn't mean that for all points the new $\theta$ coordinate changes in the same way. Some of them are even unchanged. It is a new grid. $\endgroup$ Oct 17, 2021 at 13:44

Kinetic energy should be the same as before since the kinetic energy should be same in all inertial frames.

This thinking/thought is actually incorrect. Kinetic energy is a relative quantity. The kinetic energy changes as per chosen coordinate frame.

The energy remains conserved only, when you are measuring all energies from a paticular coordinate frame not like kinetic energy from one frame, or other energy from other frame and then using in conservation equation.

Think it this way: Lets say we are sitting on a chair, according to earth's frame you are stationary that is $KE=0$. But if you look from some other planet's frame, you will see earth rotating and revolving around sun, so you are actually moving and indeed have $KE\ne0$.

If you look from other star system (or other body outside our solar system) frame, you would see we are rotating, revolving around sun and as well as around the black hole in center of our galaxy, so they will get some different $KE$ value for you.

So, changing $\theta \rightarrow \theta'$ will might change the kinetic energy or not. Rest remains is calculation

Hope it helps you out!

  • $\begingroup$ What do you mean? Is there any flaw in my answer? $\endgroup$ Oct 16, 2021 at 15:44
  • $\begingroup$ It might or might not change. Main doubt of questionairre was that whether different values of kinetic energy was contradiction which I clarified. Yes I used, rotating frames but explanation is in Layman's terms. Examples are just for clarification. Should I add new or remove this example? $\endgroup$ Oct 16, 2021 at 15:54
  • 1
    $\begingroup$ Well, I think the kinetic energy won't change. The kinetic energy doesn't depend on whether you are laying flat or sitting up (different $\theta$), since the speed (distance per unit time) isn't changing - it would depend if the frame was translating or rotating, which it is doing neither. $\endgroup$
    – Anu3082
    Oct 16, 2021 at 15:59

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