Non-inertial reference frames (example calculations)

In this section we shall do the usual example of free-fall motion but the distinction will be the use of spherical co-ordinates to ease the use of mental geometric gymnastics and also a classical introduction to Larmor precession will be introduced. Because the earth is for the most part a sphere, spherical co-ordinates should be used with only very minor modifications. Gravity points in the -\(\hat{r}\) direction the \(\hat{\theta}\) is going from north pole to south pole and \(\hat{\phi}\) is going along a latitude. Ultimately we want to describe the motion as a function of latitude so the simple transformation of \(\frac{\pi}{2}- \theta = \lambda\) will be introduced ; \(\lambda\) is the latitude defined as being zero at the equator and goes to \(45^{\circ}\) in either direction.

First a note on unit vectors:
The \(\hat{r}\) is the one commonly introduced in text books and is reproduced below along with \(\hat{\theta}\) and \(\hat{\phi}\)

\begin{equation}
\hat{r}= \cos \phi \sin \theta \hat{x}+ \sin\phi \sin \theta \hat{y} + \cos \theta \hat{z}\\
\hat{\theta}= \cos \phi \cos \theta \hat{x}+ \sin\phi \cos \theta \hat{y} - \sin \theta \hat{z}\\
\hat{\phi}= -\sin \theta \hat{x}+  \cos \theta \hat{y}
\end{equation}

From these three equations one can easily find the inverse transformations that give \(\hat{x}\), \(\hat{y}\) and  \(\hat{z}\) in terms of the spherical unit vectors in this manner. To find what  \(\hat{x}\) is simply collect the terms next to the  \(\hat{x}\) in the above equations and group them together.
So  \(\hat{x} = \cos \phi \sin \theta \hat{r} + \cos \phi \cos \theta  \hat{\theta} -\sin \theta \hat{\phi}\). Same applies for the rest.
Also, if you know what the \(\hat{r }\) unit vector looks like in Cartesian unit vectors then \( \hat{\theta} \)is not very far away. First note that it is perpendicular to the \(\hat{r}\) unit vector and therefore while we had to project the \(\hat{r}\) onto the x-y axis by taking the \(\sin \theta \) in order to project the \(\hat{\theta}\) we use \(\cos \theta\). Since \(\hat{\theta}\) is going in the -\(\hat{z}\) direction we take the  -\(\sin \theta\) .

So why is all this important? Well, our \(\Omega\) or angular velocity will be in the \(\hat{z}\) direction and we will need to put it into spherical unit vectors. But from the above discussion, rather than playing with right triangles and twisting and contorting your head in unseemly manners. Simply collect the \(\hat{z}\) in the \(\hat{r}\) and  \(\hat{\theta}\) equations. So we have the following:
\begin{equation}
 \mathbf{\Omega \hat{z}} = \Omega \cos \theta \hat{r} - \Omega \sin \theta \hat{\theta}
\end{equation}

Imagine if you were on a test, you just saved yourself five minutes of agony. Now since \(\hat{\theta}\) moves in the -\(\hat{z}\)  we will need to re-write \(\mathbf{\Omega}\) in the following manner:

\begin{equation}
 \mathbf{\Omega \hat{z}} = \Omega \cos \theta \hat{r} + \Omega \sin \theta \hat{\theta}
\end{equation}

Also because we will put the origin at the surface of the earth and not the center we shall use the effective gravitational acceleration that already includes the effects of the centrifugal force. Thus our expression for equation of motion will have no expression involving position and we will not have to bite our lips for this time-saving step.

Finally we come to the general equation of motion:
\begin{equation}
 F_{rot} =  mg_{eff}+ F_{cor}\\
 F_{rot} = mg_{eff} + v_{rot} \times 2\Omega
\end{equation}

In the above expressions the subscript "rot" stands for rotational reference frame. Next we take the determinant to get the expression below:
\begin{equation}
v \times \mathbf{\Omega} = \hat{r} \left(v_{\phi}\Omega \sin \theta \right) + \hat{\theta}\left(v_{\phi}\Omega \cos \theta \right) + \hat{\phi}\left( v_{r}\Omega \sin \theta - v_{\theta}\Omega \cos \theta \right)
\end{equation}

We know have our equations of motion:
\begin{equation}
\ddot{r} = v_{\phi} \Omega \sin \theta - g_{eff} \\
\ddot{\theta} = v_{\phi}\Omega \cos \theta \\
\ddot{\phi} = \left( v_{r} \Omega \sin \theta - v_{\phi} \Omega \cos \theta \right)
\end{equation}

The procedure of getting the solution is by using a bit of perturbation theory. First solve the equations by assuming \(\Omega\) is zero to get the following equations:
\begin{equation}
\theta = 0\\
\phi = 0 \\
 r = h - \frac{g t^2}{2}\\
v_{\theta} = 0\\
v_{\phi} = 0\\
v_{r} = -gt
\end{equation}

Next assume that the angular velocity is non-zero but very small so that our zero order approximation are fairly accurate enough and plug the results in the equations of motions. The only equation that survives is \(\ddot{\phi} = -gt \Omega \sin \theta\). This gives the deviation in the \hat{\phi} direction. To get how this deviation varies according to latitudes simply put in the transformation mentioned at the beginning of this section.

The next example introduces the phenomenon of Larmor precession. Imagine a negative electric charge q is moving in an elliptical orbit about a positive charge q. What are the equations of motion in the rotating reference frame?
\begin{equation}
F_{rot} = 2m\left(v_{rot} \times \Omega \right)  + q\left(v_{rot} + \Omega \times r \right) \times \vec{B} + \left(\Omega \times r \right) \times \Omega   + \frac{q^2}{4\pi \epsilon r^2} \hat{r} \\
F_{rot} = 2m\left(v_{rot} \times \Omega \right)  + qv_{rot} \times \vec{B} + q\left(\Omega \times r \right) \times \vec{B}+ \left(\Omega \times r \right) \times \Omega   + \frac{q^2}{4\pi \epsilon r^2}  \hat{r}\\
F_{rot} = v_{rot} \times \left(2m \Omega - q\vec{B}\right) - q\left(\Omega \times r \right) \times \vec{B}+\left(\Omega \times r \right) \times \Omega   + \frac{q^2}{4\pi \epsilon r^2} \hat{r}
\end{equation}
Now in order not to have linear velocity in the rotating reference frame the first term in the last equation has to disappear and therefore \Omega has to be \(\frac{qB}{2m}\). This is the Larmor frequency at which the precession occurs. The particle has no translational velocity. We are then left with the following expression:
\begin{equation}
F_{rot} = -\frac{q^2}{2m}\left(\vec{B} \times r \right) \times \vec{B} + \frac{q^2}{2m}\left(\vec{B}\times r \right) \times \vec{B} - \frac{q^2}{4\pi \epsilon r^2} \hat{r}
\end{equation}

Non-inertial frames: Matrix formulation

In the Lagrangian formulation  we came across this equation \( \left(\frac{de^{'}_i}{dt}\right)_i = \left(\frac{d\Omega}{dt}\right)_{ij}e^{'}_j  \) and it was mentioned that this matrix anti-symmetric. First we may give a heuristic argument as to why we may expect it to be. Notice that it arrives we take the derivative of a unit vector. Now, in what way can a unit vector change? It certainly can't change its length therefore the matrix in the equation can't have real eigenvalues. Remember a diagonalizable matrix acting upon a vector returns that same vector multiplied by a constant. As we have mentioned before this constant cannot be real for our matrix, it certainly can not be zero  so the only other option is that this eigenvalue is imaginary. But we also know that anti-symmetric matrices have imaginary eigenvalues hence we may expect our matrix to be anti-symmetric. This argument is not a proof because I am sure there are normal matrices with imaginary eigenvalues. But since we are in SO(3) we have some hope. Here then is a more rigorous argument to show that it is anti-symmetric. The argument goes as follows:
\begin{eqnarray}
 0 =& \frac{d}{dt}\left(e_i \cdot e_j \right)\\
 0 = &\frac{de_i}{dt}e_j + \frac{de_j}{dt}e_i\\
 0 =&\left(\frac{d\Omega}{dt}\right)_{ij}e_j e_j + \left(\frac{d\Omega}{dt}\right)_{ji}e_i e_i\\
 0 =  &\left(\frac{d\Omega}{dt}\right)_{ij} + \left(\frac{d\Omega}{dt}\right)_{ji}
\end{eqnarray}
 Looking at the last step we see that taking the transpose of the matrix and adding it to the matrix gives us the zero matrix. Hence we conclude it is an anti-symmetric matrix. QED
Since we start out by knowing that the unit vector cannot change its length in time we know that this change is actually related to a rotation. But at this point you should be disturbed, the equation we started with in the first paragraph has one specific direction namely, \( e_j\). This is where knowing a lot about matrices comes in. It turns out that all cross products can be represented as anti-symmetric matrices. The  equation we started off with is actually a  cross product! Re-writing \( \left(\frac{d\Omega}{dt}\right)\) as \(\omega \times \) where \(\omega = \frac{d\Omega}{dt}\) we have our connection to cross products.
Now that we know we have anti-symmetric matrix on our hand, let's reformulate newton's second law in non-inertial frames using matrices.
Lets suppose we have a vector  r, then we know that there is a standard matrix such that the vector's representation in another reference frame is given by Mr' where M is the standard matrix and r' is the vector in terms of another set of basis vectors. Since we are looking at orthogonal matrices then \(M^{-1}= M^{T}\) . That means that \(r' = M^{T}r\).Now look at the first derivative of r.
\begin{eqnarray}
 \dot{r} =& \dot{M}r' + M\dot{r^{'}}\\
  =&\dot{M}M^{T}r + M\dot{r^{'}}
\end{eqnarray}

Before we move any further lets take note that \(MM^{T} = I\)  and let's take the derivative of this expression to get \(\dot{M}M^{T} + M\dot{M^{T}} = 0\). It is important to notice that the matrix \(\dot{M}M^{T}\) is anti-symmetric based on the preceding equation. Taking its transpose and adding the two together gives us 0. Marvels of all marvels this is the same matrix we discussed earlier under a different guise. If you still have not witnessed the miracle, hold my statement with your grains of salt and read on. Let's define \(\dot{M}M^{T}\) to be the operator O. this will make our work later on easier.
Let's take the second derivative:
\begin{eqnarray}
\ddot{r} =& \dot{O}r + O\dot{r} + \dot{M}\dot{r}' + M \ddot{r}'\\
=& \dot{O}r + O\dot{r} - O^{T}\dot{r}+ O^{T}\dot{M}r' + M\ddot{r}'\\
=& \dot{O}r + 2 O\dot{r} + O^{T}\dot{M}r' + O\ddot{r}'\\
=& \dot{O}r + 2 O\dot{r} + O^{T}\dot{M}M^{T}r + M\ddot{r}'\\
=& \dot{O}r + 2 O\dot{r} - OOr + M\ddot{r}'\\
\end{eqnarray}
For the second step I went back to the first derivative of r to substitute what \(\dot{r}'\) was.

Finally, let's take a closer look at the last step we get when we calculate the second derivative. My claim was that O which is \(\dot{M}M^{T}\) is really \(\omega \times \) which is also the matrix \(\left(\frac{d\Omega}{dt}\right)_{ij}\). Let's substitute into the last step for the second derivative the expression \(\omega \times \) for O. We get the following:
\begin{equation}
\ddot{r} = \dot{\omega} \times r + 2 \omega \times \dot{r} - \omega \times \omega \times r + M\ddot{r}'
\end{equation}

What does all this mean? Well, mostly that matrices are unbelievably  awesome !