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Next: 6 Restricted circular three-body Up: Lagrangian and Hamiltonian mechanics Previous: 4 Hamiltonian mechanics


5 Motion of a body in a rotating coordinate system


5.1 Time-dependent, linear change of variables

It will be convenient to first consider arbitrary linear changes of variables. Let L(q,$\dot{q}$,t) be a Lagrangian and let us perform the change of variables

q = B(t)Q (28)

where B(t) : $\reals^{n}_{}$$\to$$\reals^{n}_{}$ any linear transformation (matrix).

5.2 Newtonian case

Let us first consider what happens to Newton's equation m$\ddot{q}$ = f (q) under the change of variables 28. We have
$\displaystyle\dot{q}$ = $\displaystyle\dot{B}$Q + B$\displaystyle\dot{Q}$.      (29)
It will be convenient to introduce the matrix $\Omega$ = B - 1$\dot{B}$ , so that the following first-order matrix differential equation is satisfied:

$\displaystyle\dot{B}$ = B$\displaystyle\Omega$. (30)

Using this equation, we may write

$\displaystyle\dot{q}$ = B$\displaystyle\Omega$Q + B$\displaystyle\dot{Q}$ = B($\displaystyle\dot{Q}$ + $\displaystyle\Omega$Q). (31)

Differentiating the second time we obtain:
$\displaystyle\ddot{q}$ = B($\displaystyle\ddot{Q}$ + $\displaystyle\Omega$$\displaystyle\dot{Q}$ + $\displaystyle\dot{\Omega}$Q) + $\displaystyle\dot{B}$($\displaystyle\dot{Q}$ + $\displaystyle\Omega$Q)   
  = B($\displaystyle\ddot{Q}$ + $\displaystyle\Omega$$\displaystyle\dot{Q}$ + $\displaystyle\dot{\Omega}$Q) + B$\displaystyle\Omega$($\displaystyle\dot{Q}$ + $\displaystyle\Omega$Q)   
  = B($\displaystyle\ddot{Q}$ + 2$\displaystyle\Omega$$\displaystyle\dot{Q}$ + $\displaystyle\dot{\Omega}$Q + $\displaystyle\Omega^{2}_{}$Q). (32)
We multiply this equation by m and obtain

f (q) = B(m$\displaystyle\ddot{Q}$ + 2m$\displaystyle\Omega$$\displaystyle\dot{Q}$ + m$\displaystyle\dot{\Omega}$Q + m$\displaystyle\Omega^{2}_{}$Q). (33)

We introduce new ``force''

F(Q) = B - 1f (BQ). (34)

(This formula means to a physicist that force is realy a vector, i.e. it transforms under coordinate changes as above.) We multiply equation 33 by B - 1 and obtain:

F(Q) = m$\displaystyle\ddot{Q}$ + 2m$\displaystyle\Omega$$\displaystyle\dot{Q}$ + m$\displaystyle\dot{\Omega}$Q + m$\displaystyle\Omega^{2}_{}$Q. (35)

5.2.1 Newton's equations of motion in new coordinates

Summarizing the calculations of the previous section, Newton equations of motion m$\ddot{q}$ = f (q) after a linear time-dependent change of coordinates q = B(t)Q become (see 35):

m$\displaystyle\ddot{Q}$ = F(Q) - 2m$\displaystyle\Omega$$\displaystyle\dot{Q}$ - m$\displaystyle\dot{\Omega}$Q - m$\displaystyle\Omega^{2}_{}$Q (36)

where $\Omega$ = B - 1$\dot{B}$ .

5.2.2 Rotating coordinate system

This is a special case when n = 3 and the matrix B is orthogonal of determinant 1. In this case the matrix $\Omega$ is skew-symmetric. There is a vector $\omega$ which represents the angular velocity of the rotating coordinate system, such that for all Q $\in$ $\reals^{n}_{}$ we have

$\displaystyle\Omega$Q = $\displaystyle\omega$ x Q (37)

where x denotes the usual cross product of vectors. Indeed, if $\omega$ = ($\omega_{1}^{}$,$\omega_{2}^{}$,$\omega_{3}^{}$) then the matrix of the linear transformation Q $\mapsto$ $\omega$ x Q is $\Omega$ . On the other hand
\omega\times Q=
0 & -\omega_3 &...]
Q_1\ Q_2\ Q_3\end{array}\right].\end{equation}(38)
Thus, we have

$\displaystyle\Omega$ = $\displaystyle\left[
0 & -\omega_3 & \omega_2 \ \omega_3 & 0 & -\omega_1 \ -\omega_2 & \omega_1 & 0\end{array}\right].$ (39)

5.2.3 Newton's equations in rotating coordinates

According to the previous section for n = 3 the equations of motion are:

m$\displaystyle\ddot{Q}$ = F(Q) - 2m$\displaystyle\omega$ x $\displaystyle\dot{Q}$ - m$\displaystyle\dot{\omega}$ x Q - m$\displaystyle\omega$ x ($\displaystyle\omega$ x Q). (40)

The interpretation of these equations is that the fact that in a rotating coordinate system there are additional forces acting upon the body, represented by the three additional terms in the right-hand side of this equation. They have their names:
- 2m$\omega$ x $\dot{Q}$ is called the Coriolis force;
- m$\omega$ x ($\omega$ x Q) is simply the centripetal force;
- m$\dot{\omega}$ x Q is the force of inertia; this force is 0 if the rotation is uniform, i.e. $\Omega$ is constant.

5.3 Lagrangian case

Let L(q,$\dot{q}$,t) be a Lagrangian. According to 20 the change of variables q = B(t)Q leads to motions described by the new Lagrangian

K(Q,$\displaystyle\dot{Q}$,t) = L(Q,B$\displaystyle\dot{Q}$ + $\displaystyle\dot{B}$Q,t) = L(Q,B($\displaystyle\dot{Q}$ + $\displaystyle\Omega$Q),t) (41)

This formula illustrates the benefits of the Lagrange formalism when dealing with coordinate changes.

It is interesting to see how the Coriolis, centripetal and inertia forces can be derived from the Lagrangian formalism. Let L = T - U where T = (1/2)m($\dot{q}$)2. If B is orthogonal then $\langle$Bx,By$\rangle$ = $\langle$x,y$\rangle$ where $\langle$ $\cdot$ , $\cdot$ $\rangle$ is the dot product (so ($\dot{q}$)2 = $\langle$$\dot{q}$,$\dot{q}$$\rangle$ ). Hence, the new Lagrangian is:

K(Q,$\displaystyle\dot{Q}$,t) = $\displaystyle{\textstyle\frac{1}{2}}$m($\displaystyle\dot{Q}$)2 + m($\displaystyle\Omega$Q,$\displaystyle\dot{Q}$) + $\displaystyle{\textstyle\frac{1}{2}}$m($\displaystyle\Omega$Q,$\displaystyle\Omega$Q) - U(BQ,t).      (42)
The extra two terms translate into the additional terms in the equations of motion which were called Coriolis, centripetal and inertia forces.

5.3.1 Equations of motion

We find the generalized momentum:

$\displaystyle{\frac{\partial K}{\partial \dot Q}}$ = m$\displaystyle\dot{Q}$ + m$\displaystyle\Omega$Q (43)


$\displaystyle{\frac{\partial K}{\partial Q}}$ = - m$\displaystyle\Omega$$\displaystyle\dot{Q}$ - m$\displaystyle\Omega^{2}_{}$Q - $\displaystyle{\frac{\partial\tilde U}{\partial Q}}$. (44)

where $\tilde{U}$(Q,t) = U(BQ,t) is the new potential. After identifying

F(Q) = - $\displaystyle{\frac{\partial\tilde U}{\partial Q}}$ (45)

(we use B - 1 = B T here!) we obtain exactly the same result as in 40.

5.3.2 The Hamiltonian

Having calculated the generalized momentum in rotating coordinates

P = $\displaystyle{\frac{\partial K}{\partial \dot Q}}$ = m$\displaystyle\dot{Q}$ + m$\displaystyle\Omega$Q (46)

we may find the Hamiltonian in rotating coordinates:
H(P,Q) = P$\displaystyle\dot{Q}$ - K(Q,$\displaystyle\dot{Q}$)   
  = P(P/m - $\displaystyle\Omega$Q) - $\displaystyle{\textstyle\frac{1}{2}}$(P/m - $\displaystyle\Omega$Q)2 + $\displaystyle\tilde{U}$(Q,t)   
  = $\displaystyle{\textstyle\frac{1}{2}}$$\displaystyle{\frac{P^2}{m}}$ + $\displaystyle\langle$$\displaystyle\Omega$P,Q$\displaystyle\rangle$ + $\displaystyle\tilde{U}$(Q). (47)

The Hamiltonian equations of motion are

$\displaystyle\dot{Q}$ = P/m - $\displaystyle\Omega$Q,        
$\displaystyle\dot{P}$ = - $\displaystyle{\frac{\partial\tilde U}{\partial Q}}$ - $\displaystyle\Omega$P.      (48)

In three dimensions ( n = 3 ) these equations could also be written as

$\displaystyle\dot{Q}$ = P/m - $\displaystyle\omega$ x Q,   
$\displaystyle\dot{P}$ = - $\displaystyle{\frac{\partial\tilde U}{\partial Q}}$ - $\displaystyle\omega$ x P. (49)

5.4 A note on the direction of angular velocity

It is easy to make the sign mistake in determining the direction of $\omega$ . A simple rule eliminates this mistake: $\omega$ is opposite to the angular velocity of a point at rest in the original coordinate system expressed in the rotating coordinate system. For example, if we use a coordinate system rigidly attached to the earth then a point on the surface of the earth resting in non-rotating coordinates appears to be moving west for an observer in the rotating coordinate system. Thus $\omega$ is pointing towards the north pole, with length equal to

$\displaystyle{\frac{2\pi}{24\cdot 3600}}$ (50)

next up previous
Next: 6 Restricted circular three-body Up: Lagrangian and Hamiltonian mechanics Previous: 4 Hamiltonian mechanics
Marek Rychlik