Kinetic energy depends on $\theta$ but symmetry argument says otherwise

Question

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

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?

Answer 1

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

edit

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 .

Answer 2

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!

Answer 3

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.

Answer 4

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!