Writing Hamiltonian: Classical Mechanics

[WWCY]

Homework Statement

I'm having some issues understanding a number of concepts in this section here. I attached the corresponding figure at the end of the post for reference.

Issue 1)

1st of all, I understand that a Hamiltonian can be written as such
$$H = T_2 - T_0 + U$$
whereby $T_2$ and $T_0$ are Kinetic Energy terms quadratically dependent and linearly independent on generalised velocities respectively.

However, the text above writes that $T_2 = \frac{ma^2}{2 \dot{\theta} ^2}$ and $T_0 = \frac{1}{2}m\omega ^2 a^2 \sin ^2 \theta$.

Since we defined $r,\theta , \phi$ as our spherical coordinates, and $\omega = \dot{\phi}$, why was the term identified as $T_0$ not considered to be quadratic in velocity dependence?

What constitutes a generalised coordinate/velocity, and what constitutes "something else"?

Issue 2)

So far, I have learned that the azimuthal angle ($\theta$ in this case) is to be defined from the "North" end of the $z$ axis. In this case, $\theta$ is defined from the "South" end. Is there any appreciable difference in both definitions?

Assistance is greatly appreciated!

Homework Equations

The Attempt at a Solution

 

[Orodruin]

Since we defined r,θ,ϕr,θ,ϕr,\theta , \phi as our spherical coordinates, and ω=˙ϕω=ϕ˙\omega = \dot{\phi}, why was the term identified as T0T0T_0 not considered to be quadratic in velocity dependence?

The only generalised coordinate you have is $\theta$. Both $r$ and $\phi$ are given by constraints. Your configuration space is one-dimensional.

So far, I have learned that the azimuthal angle (θθ\theta in this case) is to be defined from the "North" end of the zzz axis. In this case, θθ\theta is defined from the "South" end. Is there any appreciable difference in both definitions?

You can use whatever coordinates you want. If you are uncomfortable with theta going from the ”south”, let it go from the ”north” and introduce a new coordinate $\vartheta = \pi -\theta$.

 

[WWCY]

Thanks for the response.

The only generalised coordinate you have is $\theta$. Both $r$ and $\phi$ are given by constraints. Your configuration space is one-dimensional.

Would I be right to say that generalised coordinates / velocities are those that are not given by a problem, and thus are allowed to vary?

 

[Orodruin]

Generalised coordinates are the coordinates that you need to specify in order to know the configuration of a system. For a particle moving on a sphere (or any two-dimensional surface) the configuration space is two-dimensional and requires two coordinates. Whether or not something has been stated in a problem is not really relevant. You should be able to identify the configuration space regardless. For example, I could ask you what the configuration space of a double planar pendulum is.

 

[WWCY]

Generalised coordinates are the coordinates that you need to specify in order to know the configuration of a system. For a particle moving on a sphere (or any two-dimensional surface) the configuration space is two-dimensional and requires two coordinates. Whether or not something has been stated in a problem is not really relevant. You should be able to identify the configuration space regardless. For example, I could ask you what the configuration space of a double planar pendulum is.

I'm afraid I still don't quite follow. In the example I quoted, wouldn't i need all of $r, \phi$ and $\theta$ to specify a mass' position in space?

Thanks for assisting.

 

[Orodruin]

I'm afraid I still don't quite follow. In the example I quoted, wouldn't i need all of $r, \phi$ and $\theta$ to specify a mass' position in space?

Thanks for assisting.

No. The mass is restricted to move on the ring. If you specify $\theta$, that is enough to know where it is. In order to know its 3D position you would also need to know the configuration of the ring, but its motion is already known. The system you are looking at is only the mass’ position on the ring.

 

[WWCY]

No. The mass is restricted to move on the ring. If you specify $\theta$, that is enough to know where it is. In order to know its 3D position you would also need to know the configuration of the ring, but its motion is already known. The system you are looking at is only the mass’ position on the ring.

Is this to say that because the motion and radius has already been defined, all that's left is to define $\theta$, hence we use it as a generalised coordinate?

 

[Orodruin]

Is this to say that because the motion and radius has already been defined, all that's left is to define $\theta$, hence we use it as a generalised coordinate?

I do not like to phrase it this way, it seems to somehow suggest that it is the coordinates we put on something that we use to define its motion, it is not. Coordinates is something we use to describe the motion. Given the physical setup of motion on the rong, it is sufficient with one coordinate to fully describe the motion, the position on the ring.

If you want the actual spatial coordinates of the mass, then you will also need to know the orientation and position of the ring, but the configuration of the motion itself is one-dimensional.

 

[WWCY]

If you want the actual spatial coordinates of the mass, then you will also need to know the orientation and position of the ring, but the configuration of the motion itself is one-dimensional.

Could you elaborate a little on the idea of configuration space?

Thanks for assisting!