Physics Help Forum Transformations in a rotating frame?

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 Dec 7th 2013, 10:49 AM #1 Junior Member   Join Date: Dec 2013 Posts: 4 Transformations in a rotating frame? A neutral particle moves in a rotating harmonic potential given by V(x,y) = 0.5m { (w1)^2 [xcos(wt)-ysin(wt)]^2 + (w2)^2 [ycos(wt) + xsin(wt)]^2} (the 1 and 2's are supposed to be subscript but I can't get this to display properly) and w is the angular frequency. I need to determine the hamiltonian and the transformed hamiltonian, but I'm unsure how to go about doing this... I know that the Lagrangian is given by the kinetic energy minus the potential, can I use 0.5m(x')^2 and 0.5m(y')^2 and the potential as defined above, or is it different in a rotating frame? Once I've found this, I have an equation that I can use to find the Hamiltonian. If anyone could help, that would be great
Dec 7th 2013, 02:11 PM   #2

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 Originally Posted by Winnie A neutral particle moves in a rotating harmonic potential given by V(x,y) = 0.5m { (w1)^2 [xcos(wt)-ysin(wt)]^2 + (w2)^2 [ycos(wt) + xsin(wt)]^2} (the 1 and 2's are supposed to be subscript but I can't get this to display properly) and w is the angular frequency. I need to determine the hamiltonian and the transformed hamiltonian, but I'm unsure how to go about doing this... I know that the Lagrangian is given by the kinetic energy minus the potential, can I use 0.5m(x')^2 and 0.5m(y')^2 and the potential as defined above, or is it different in a rotating frame? Once I've found this, I have an equation that I can use to find the Hamiltonian. If anyone could help, that would be great
If the neutron is moving down the axis then T is just 1/2 mv^2. If it is off axis then there is motion about the axis so there will be both linear and angular components to T. Looks kinda nasty. Feel free to come back with other questions if you need.

-Dan
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Dec 8th 2013, 05:49 AM   #3
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 Originally Posted by topsquark If the neutron is moving down the axis then T is just 1/2 mv^2. If it is off axis then there is motion about the axis so there will be both linear and angular components to T. Looks kinda nasty. Feel free to come back with other questions if you need.
Ok, thanks . It's moving in 2d so will also have an angular component, will this just be the usual angular kinetic energy 0.5mw^2 r^2 ? The lagrangian would then be 0.5m(v^2 + w^2 r^2) - V, which can be written in terms of x and y and their derivatives.

To find the hamiltonian can I say (p1)=dL/dx and (p2)=dL/dy and then
H= (p1)x' + (p2)y' - L ?

Mar 23rd 2014, 10:54 PM   #4
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 Originally Posted by Winnie Ok, thanks . It's moving in 2d so will also have an angular component, will this just be the usual angular kinetic energy 0.5mw^2 r^2 ? The lagrangian would then be 0.5m(v^2 + w^2 r^2) - V, which can be written in terms of x and y and their derivatives. To find the hamiltonian can I say (p1)=dL/dx and (p2)=dL/dy and then H= (p1)x' + (p2)y' - L ?
No. This is a very easy problem made hard by the worrying about the phrase rotating harmonic potential. The potential is given. If the particle is constrained to move in the xy-plane then the Lagrangian is

L(x, t) = T - V

where T is the kinetic energy given (in two dimensions) as

T = (1/2) mv^2 = (1/2)m[ (dx/dt)^2 + (dy/dt)^2]

You made the mistake of trying to express the speed in polar coordinates. You started in Cartesian coordinates and made no statement about any requirement to change to a change of coordinates. Therefore you mixed coordinate systems. You expressed the kinetic energy in polar coordinates and the potential energy in Cartesian coordinates.

By the way, your notation is wrong. The potential should not be expressed as V(x, y) but as V(x,y ,t ).

Pete

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