Kinetic energy depends on ##\theta## but this argument says otherwise

[PeroK]

So we can't rotate in such a way that changes theta but keeps r and phi the same?

No. You can keep $\phi$ fixed on a single line of longitude, but $\phi$ will changes for points not on that line.

 

[wrobel]

I don't have a formula. I'm just imagining that we rotate our coordinate system by a finite rotation with the origin fixed where it was initially in such a manner that all other variables except remain the same.

I hope I'm clear?

Yes you are
but this is not a rotation of the space
$\varphi\mapsto \varphi+c$ is a rotation
$\theta\mapsto\theta+c$ is not

 

[Kashmir]

Yes you are
but this is not a rotation of the space
$\varphi\mapsto \varphi+c$ is a rotation
$\theta\mapsto\theta+c$ is not

That means a change of coordinate system such that only theta changes is not a rotation?

 

[wrobel]

That means a change of coordinate system such that only theta changes is not a rotation?

exactly, just take two different points and trace how they move as $\theta$ changes

 

[PeroK]

That means a change of coordinate system such that only theta changes is not a rotation?

It's a change to the z-axis only. It would leave the x and y axes unchanged. Note that $\tan \phi = \frac y x$.

 

[vanhees71]

Okay, but then $\theta' = \theta$. That's just regular spherical coordinates.

No! If you direct the polar axis in another direction, it's another angle.

 

[vanhees71]

It's a change to the z-axis only. It would leave the x and y axes unchanged. Note that $\tan \phi = \frac y x$.

Be careful with this "sloppy formula". In the programming language you have the function atan2 for what you want (in Fortran it's atan2(y,x) in C atan2(x,y) ;-))).

A formula with the cut of the polar angle at $\pm \pi$ is
$$\phi=\text{sign} \, y \arccos \left (\frac{x}{\sqrt{x^2+y^2}} \right), \quad \phi \in (-\pi,\pi).$$

 

[PeroK]

No! If you direct the polar axis in another direction, it's another angle.

The OP's diagram indicated a rotation of all three axes, but with the false assumption that $\phi$ was unchanged for all points.

 

[Lnewqban]

I am not keeping theta the same. I'm keeping the other two same and vary theta

That is not what I wrote, sorry.
If you increase theta by 90 degrees, for example, the value of the other angle respect to vector r must change.

As you tilt the z axis, you are simultaneously rotating it around the x-axis and around the y axis.
Because of that, the x-y plane tilts about an axis that crosses the origin and is perpendicular to vector r.

When all the above happens, the length of the projection of vector r on x-y plane changes, making change the projections of vector r on the x-axis and on the y axis, even when the spatial position of vector r and its absolute magnitude remain the same.

An example would be to assign x, y and z axes to three edges of a carboard box.
Looking directly from above, you will see and angle of 90 degrees between axes x and y.
Now, as you tilt edge z back diagonally to edges x and y, you could see how the angle between edges x and y increases until reaching 180 degrees when edge z reaches the ground: your phi angle has naturally changed.

 

[Lnewqban]

So we can't rotate in such a way that changes theta but keeps r and phi the same?

The value of r remains unchanged, but its projections on the three planes change (please, see attached animation).
The value of phi is affected by the rotation of the three axes, as well as it happened to theta.

Perhaps I am wrong, but I believe that:
1) The particle with mass m is located at the origin of coordinates x-y-z.
2) Vector r is the spatial velocity of mass m.
3) Theta and phi are angles that can define the spatial location of r because those angles are contained in planes that are perpendicular to each other.

Please, look for "6.4 Choosing Generalized Coordinates" in the following link:
https://courses.physics.ucsd.edu/2010/Fall/physics200a/LECTURES/CH06.pdf