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 !
\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 !
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