Generalized momentum and Hamiltonian over a non inertial reference frame

[Telemachus]

Hi there. I need help to work this out.

A particle with mass m is studied over a rotating reference frame, which rotates along the OZ axis with angular velocity [tex]\dot\phi=\omega[/tex], directed along OZ. It is possible to prove that the potential (due to inertial forces) can be written as:
[tex]V=\omega \cdot L-\frac{1}{2}m(\omega\times r)^2[/tex]
L denotes the angular momentum round the origin O. Determine:
a) The generalized moment taking as generalized coordinates the cartesian coordinates (X,Y,Z) taken over the rotating system.
b) The generalized moment taking as generalized coordinates the cylindrical coordinates [tex](\rho,\phi,Z)[/tex] taken over the rotating system.
c) Use the corresponding Legendre transformation, assuming there are no additional forces to find the Hamiltonian. Demonstrate that the Hamiltonian is:
[tex]H=H_0-\omega \cdot L[/tex]
Where H0 is the hamiltonian for a free particle.

Excuse my english :P

I don't know how to start. I've tried making a transform from x', y',z' inertial coordinates, using a rotation. Let's say:
[tex]x'=X \cos\phi-Ysin\phi[/tex]
[tex]y'=Y\cos\phi+X\sin\phi[/tex]
[tex]z'=Z[/tex]

Should I just use this transformation to get the kinetic energy and then just set L=T-V?

Thanks for your help :)

 

[Telemachus]

Delete.

 

[Telemachus]

Alright, it was easier than what I thought. The fictitious potential contained all the terms that appear considering the velocity over the non inertial reference frame. So all what I had to do was [tex]T=\frac{m}{2}(\dot x^2+\dot y^2)[/tex]

I had to use the transformation to realize about it, but I think that what I did is quiet correct. Anyway, I couldn't completely verify the equality in c), I neither did b). On c), I get something that looks pretty much like what it gives, but I've probably made some algebra mistake somewhere, and I get an extra term [tex]H=H_0-\omega \cdot L+\frac{m \omega^2}{2}(x^2+y^2)[/tex]

Anyone?

From the transformation I got:
[tex]\dot x^2+\dot y^2=\dot x'^2+\dot y'^2-2\omega \dot x'^2y'+2\omega \dot y'x'+\omega^2(x'^2+y'^2)[/tex]

Thats the square of the velocity for a particle moving on the rotating frame with a speed [tex]\dot x+\dot y[/tex] with respect to the rotating frame, measured from the inertial reference frame.

 

[Telemachus]

I finally got what I was looking for, but I'm not sure why. I had to use in the first place the lagrangian obtained using the fictitious potential. From this lagrangian I've obtained the generalized momentums with respect to the rotating frame. That cofuesed me a little bit, because I had moments with respect to both reference frames, I wasn't sure to which corresponded the ones that appeared in the fictitious potential, but now I'm pretty sure those correspond to the moments taken in the inertial reference frame. Once I got the generalized moments, using the lagrangian with the fictitious potential, I had to construct the Hamiltonian, using those moments, but when considering the lagrangian I just had to consider the lagrangian without the fictitious potential. I think that information is already given in the generalized moments. When I was constructing the hamiltonian that I have previously posted, I was considering the Lagrangian as the one with the fictitious potential included. But the result is obtained by considering the lagrangian without that potential, with the kinetic energy just as the sum of the components:
[tex]\dot x+\dot y[/tex]

That is: [tex]H=p_x \dot x+p_y \dot y-\frac{m}{2}(\dot x^2+\dot y^2)[/tex]
Where the momentums are obtained from the Lagrangian with the fictitious potential.

Assuming there is no other potential energy. In that way I obtain the result given by the exercise. I realized about it because just in that way I don't get the extra terms, but I didn't get the more profound reasoning on why it must be done this way, on a deeper physical sense. In the first place I thought of using the fictitious potential in the lagrangian for this hamiltonian, but it's like some combination of things, which confuses me a bit.

 

[paranormal]

I am happy to assist you with your question. Let's break down the problem into smaller parts to better understand it.

First, let's define some terms. Generalized momentum is a concept in classical mechanics that describes the momentum of a particle in terms of generalized coordinates, rather than just the traditional Cartesian coordinates. In a non-inertial reference frame, the concept of momentum becomes more complicated as there are additional forces acting on the particle due to the rotation of the reference frame.

The Hamiltonian is a function that describes the total energy of a system in terms of its generalized coordinates and momenta. It is a key concept in Hamiltonian mechanics, a branch of classical mechanics that deals with systems with many degrees of freedom.

Now, let's look at the problem at hand. We are studying a particle with mass m over a rotating reference frame. The potential due to inertial forces can be written as V=\omega \cdot L-\frac{1}{2}m(\omega\times r)^2, where \omega is the angular velocity of the rotating reference frame and L is the angular momentum of the particle around the origin O.

a) To determine the generalized momentum in terms of Cartesian coordinates, we can use the equations of motion in a rotating reference frame, which take into account the additional forces due to the rotation. These equations can be derived from the Lagrangian formalism, which is a mathematical framework for describing the dynamics of a system. The resulting generalized momentum will be in terms of the Cartesian coordinates (X, Y, Z).

b) To determine the generalized momentum in terms of cylindrical coordinates, we can use the same approach as in part a), but with the cylindrical coordinates (\rho, \phi, Z) instead. This will result in a different expression for the generalized momentum.

c) To find the Hamiltonian, we can use the Legendre transformation, which relates the Lagrangian to the Hamiltonian. This transformation takes into account the generalized momenta and coordinates. By assuming there are no additional forces acting on the particle, we can use the Hamiltonian to describe the total energy of the system. In this case, the Hamiltonian will be given by H=H_0-\omega \cdot L, where H_0 is the Hamiltonian for a free particle.

To solve this problem, you will need to use the equations of motion and the Lagrangian formalism to derive the generalized momenta in terms of the