# Defining a Lagrangian in an rotating reference frame frame

1. Oct 21, 2009

### mtak0114

Hi
I'm trying to define a Newtonian lagrangian in an
rotating reference frame (with no potential)

Something to note is that the time derivative of in a rotating reference frame must be corrected for by:

$$\frac{d {\bf B}}{dt} \rightarrow \frac{d {\bf B}}{dt} + {\bf \omega} \times {\bf B}$$

where B is some vector, this can be found in wikipedia

Therefore I get something like

$$L = \frac{1}{2} m(\dot{{\bf x}}+{\bf \omega} \times {\bf x})^2$$

where the dot is the time derivative
and Im expecting to get three ficticious forces: Centrifugal, centripetal and euler forces.
but this does not appear what am I doing wrong?

I believe the answer should be

$$m(\ddot{{\bf x}}+2{\bf \omega} \times \dot{{\bf x}}+\dot{{\bf \omega}} \times {\bf x}+{\bf\omega \times (\omega} \times {\bf x}))=0$$

I get this by taking the newtonian lagrangian in a non rotating frame and calculating the euler lagrange equations of motion, and then transforming into the rotating frame. but the two results do not agree?

any help would be greatly appreciated

2. Oct 21, 2009

### hamster143

What is the equation of motion that you're getting?

3. Oct 21, 2009

### mtak0114

This is the equation of motion I'm getting is:

$$m(\ddot{{\bf x}}+2{\bf \omega} \times \dot{{\bf x}}+\dot{{\bf \omega}} \times {\bf x}+{\bf\omega \times (\omega} \times {\bf x}))+m {\bf \omega} \times(\dot{{\bf x}}+{\bf \omega} \times {\bf x})=0$$

thanks

M

4. Oct 22, 2009

### mtak0114

Ive work it out

the temporal derivative is the ordinary temporal derivative not the corrected one because this is the velocity which is already defined in the rotating frame of reference

thanks again

M

5. Oct 22, 2009

### mtak0114

I think ive work it out:

the temporal derivative is the ordinary temporal derivative not the corrected one because this is the velocity which is already defined in the rotating frame of reference.
now this gives the correct equations of motion

Now if I choose an arbitrary frame of reference i.e. one that is rotating and translating
this does not work, i.e. if I choose a lagrangian:

$$L = \frac{1}{2} m(\dot{{\bf x}}+{\bf \omega} \times {\bf x}+\tau)^2$$

but this does not work.

I think this is because the translation term $$\tau$$ is an object that does not transform like the coordinates x when in a rotating frame... The correct result should be the original plus

$$\ddot{\tau}$$

should there be a lagrangian for such a frame?

cheers

M

Last edited: Oct 22, 2009