Non inertial reference frames: Lagrangian Formulation

I have searched for the equations of motion arrived at through the Lagrangian formulation and I have seen few satisfactory derivations. There is one reference to it in John R Taylor's book on classical mechanics as an extremely hard problem(3 stars). Hence it is a good topic for our consideration.
First we begin by asking how a vector changing with time looks in an inertial reference \(\left(\frac{d\vec{r}}{dt}\right)_o\) as opposed to an non-inertial reference frame \(\left(\frac{d\vec{r}}{dt}\right)_q\) . Now consider two reference frames with basis vectors \(e_i\) and \(e^{'}_i\). So  \( \vec{r} = \sum_1^n r_i e_i \) or \(\vec{r} = \sum_1^n r^{'}_i e^{'}_i \). Assume that the former co-ordinate basis is inertial and the second is not. This means that: $$\left(\frac{d\vec{r}}{dt}\right)_o = \sum_1^n \frac{dr^{'}_i}{dt}e^{'}_i + \sum_1^n r^{'}_i \frac{de^{'}_j}{dt} = \sum_1^n \frac{dr}{dt}e_i \hspace{20mm} \text{eq.1}$$

If you look at the first term in the middle equation in eq.1 you will see a term that changes with respect to \(e^ {'}_i\).  This is the speed with repesct to the rotating frame. Consequently, eq.1 can be written as
$$\left(\frac{d\vec{r}}{dt}\right)_o = \left(\frac{d\vec{r}}{dt}\right)_q + \sum_1^n r^{'}_i \frac{de^{'}_i}{dt} \hspace{20mm} \text{eq.2}$$

But what exactly is this term \(\frac{de^{'}_i}{dt}\). We know that the rotating reference is rotating with some angular velocity \(\frac{d\Omega}{dt}\). So this term must involve angular velocity. We can expand this term in terms of its basis vectors to get that:
$$ \left(\frac{de^{'}_i}{dt}\right)_i = \left(\frac{d\Omega}{dt}\right)_{ij}e^{'}_j \hspace{20 mm} \text{eq.3} $$  Here I have merely written the ith component of the \(\frac{de^{'}_i}{dt}\) vector.  Eq.3 you will notice is actually matrix multiplication. At this juncture we could introduce levi-civita symbols and define the \( \left(\frac{d\Omega}{dt}\right)_{ij}\) to be \( \left(\frac{d\Omega}{dt}\right)_{k}\) in this manner \( \left(\frac{d\Omega}{dt}\right)_{ij}\)=\(\epsilon_{ijk}\left(\frac{d\Omega}{dt}\right)_{k}\). Hence,
$$ \left(\frac{de^{'}_i}{dt}\right)_i = \epsilon_{ijk}\left(\frac{d\Omega}{dt}\right)_{k}e^{'}_j \hspace{20mm} \text{eq.4}  $$ This is merely:
$$ \left(\frac{de^{'}_j}{dt}\right)_j =  \left(\frac{d\Omega}{dt}\right) \times e^{'}_j \hspace{20mm} \text{eq.5}$$ Let's define \(\frac{d\Omega}{dt}\) to be \(\omega\).  Consequently, eq.2 now becomes $$ \left(\frac{d\vec{r}}{dt}\right)_o = \left(\frac{d\vec{r}}{dt}\right)_q + \omega \times \vec{r} \hspace{20mm} \text{eq.6} $$

We are now finally in a position to formulate the equations of motion for non-inertial reference frames. Remember L = T - U where T is kinetic energy and U is potential energy. The Lagrangian in the non-inertial reference frame can then be written as
\begin{equation}
 L=\frac{m}{2}\dot{r}_o ^2 - U\hspace{10mm} \text{where}\hspace{10mm} \dot{r_o} = \dot{r_q} + \omega \times r
\end{equation}
\begin{eqnarray}
 \dot{r_o}\cdot \dot{r_o} =& \left(\dot{r_q} + \omega \times r\right) \cdot   \left(\dot{r_q} + \omega \times r\right) \\
  =&\dot{r_q}^2 + 2\dot{r_q}\cdot \left(\omega \times r \right) + \left(\omega \times r \right) \cdot \left(\omega \times r \right)
\end{eqnarray}
\begin{eqnarray}
\left(\omega \times r\right)_i \cdot \left(\omega \times r \right)_i =& \epsilon_{ijk} \omega_j r_k \epsilon_{ilm}\omega_l r_m\\
 =&\epsilon_{jkl}\epsilon_{ilm}  \omega_j r_k \omega_l r_m\\
 =&\left( \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl}\right) \omega_j r_k \omega_l r_m\\
  = &\delta_{jl}\delta_{km}\omega_j r_k \omega_l r_m - \delta_{im}\delta_{kl} \omega_{i}r_k \omega_l r_m\\
=& \omega_j\omega_j r_k r_k - \omega_j r_j r_k\omega_k \hspace{20mm} \text{eq.7}\\
\end{eqnarray}
\begin{equation}
\dot{r_o}\cdot \dot{r_o} = \dot{r_q}^2 + 2\dot{r_q}\cdot \left(\omega \times r \right) + \left(\omega r\right)^2 - \left(\omega \cdot r \right)^2 \hspace{20mm} \text{eq.8}
\end{equation}

Armed with eq.8 we can now go to the Euler-Lagrange Equation and deal with the derivative.
\begin{equation}
 \frac{\partial L}{\partial \dot{r}_q} =  m \left(\dot{r}_q + \omega \times r \right)
\end{equation}

To get the Lagrangian with respect to \(r_q\) is a bit more tricky. The hardest terms to deal with are clearly ( at least for me ) the last two terms in eq.8. So instead what I do is look at eq.7 and take the derivative with respect to \(r_k\) . This gives  \(2\omega_j\left(\omega_j r_k - r_j \omega_k\right)\). If you are familiar with levi-civita symbols then you can see it is simply a component of \(-\omega \times \left(\omega \times r_q \right)\). To be more explicit:
\begin{eqnarray}
 2 \omega_j\left(\omega_j r_k - r_j \omega_k\right) =& 2\omega_j \epsilon_{ijk}\omega_j r_k \\
 =& 2 \epsilon_{kji} \omega_j \left( \omega \times r \right)_i \\
 =& -2 \omega \times \left(\omega \times r \right)
\end{eqnarray}
 We finally have that
\begin{equation}
\frac{\partial L}{\partial r_q} = m \left(-\omega \times \dot{r}  - \omega \times
\left( \omega \times r \right) \right) - \nabla U .
\end{equation}

The next post will be looking at the angular velocity matrix that we used in eq.3. It turns out that it is anti symmetric. I will try to find out what this means about angular velocity and the relationship between non-inertial and inertial reference frames.  

Agenda for next few posts, Non-inertial reference frames

The purpose of the last few posts was mostly to understand normal co-ordinates and how to arrive at them without asking for a miracle, understanding convolutions and relationship to Laplace transform and lastly understanding Green's Function. By using a physical problem we can find a nice intuitional meanings of these mathematical objects. The next few posts will be to try to understand movement in non-inertial reference frames. The basis of the posts will be John R Taylor's book on classical mechanics although I take a peek at Landau's first volume. The summa will be Euler's rigid body equations.

Green's Function and coupled oscillators

          The purpose of this post is to get an physical intuition of what a Green's function. Since we are examining couple oscillators and hence coupled differential equations we will not be able to get the solution but we shall understand what the Green function is doing. Here again are the couple ODEs were are considering:

\begin{eqnarray}
m_1 \ddot{x}= -k_1x - k_{12}(x-y)\\
m_2\ddot{y} = -k_{2}y + k_{12}(x-y)
\end{eqnarray}

What we shall do is consider \(m_1\) as the system we are considering and then the driving force is given by \(m_2\). Now, at a specific moment in time ,t, the system fills an outside force called the impulse, this impulse could be described by a function over time or it could be one hit at a specific time.
            In fact when we usually solve differential equations we can interpret the differential equation as a system, now if it is homogeneous equation then we can interpret that as hitting the system for a split second and observing it. System then moves according to what the differential equation says. If on the other hand the differential equation is in-homogeneous, we can imagine the resulting movement of the system as a linear superposition of these constant split second hits over a time period. In general these split second hits will not have the same force. To get the solution to the in-homogeneous equation all we need to do is to add the response of the system to these split second hits. The response to these split second hits is described by the Green's Function. These split second hits could be actual mechanical hits or charges which provide electric impulse or masses which provide gravitational impulse. So what we need to do first is solve the homogeneous equation in other words find the Green's Function.

It is interesting to note that all we have described can be understood in terms of convolutions in the following manner:
Let D be differential operator acting upon the system,y, and a driving force f(t). So Dy = f.
Now we know from the properties of delta function that \( \int_{-\infty}^{\infty}\delta(t'-t)f(t') dt' = f(t)\) but this is merely \(\delta \star f\).  So we can re-write Dy = f as Dy =\( \delta \star f\). Let's suppose we can write 'y' as a convolution of two functions then D(\(A \star B\))= \(\delta \star f \). Using derivative properties of convolution we have \((DA) \star B \) = \(\delta \star f\). From this we see that B = f and DA = \(\delta\) . A here is the green's function which we will call G. Hence we have DG = \(\delta(t'-t)\), which is consistent with what I said about the green's function being the response of the system to a split second hit of the system. The split second hit here is given by the delta function and the magnitude of this hit is \( \int_{-\infty}^{\infty}\delta(t'-t)f(t') dt' = f(t)\) . From the above analysis we get that  \( y = G \star f\).

At last going back to our coupled ODEs, we will look at the fir st one which we will write in this form
$$ m_1 \ddot{x} +k_1x + k_{12}x = ky \hspace{20mm }\text{eq.1} $$. Remembering that  DG = \(\delta(t'-t)\) we shall write the following equation $$ m_1 \ddot{G_o} +k_1G + k_{12}G_o = \delta (t-t') \hspace{20mm} \text{eq.2} $$. This simplifies to $$\ddot{G_o} = - G_o\frac{k_1 + k_{12}}{m_1} \hspace{20mm}\text{eq.3} $$ we shall define \(\frac{k_1 + k_{12}}{m_1}\) as \(\omega_o ^2\)
The same procedure can be done with second ODE:
$$ m_2 \ddot{y} +k_2y + k_{12}y = kx \hspace{20mm }\text{eq.4} $$ . We arrive at  $$\ddot{G_1} = - G_1\frac{k_2 + k_{12}}{m_2} \hspace{20mm} \text{eq.5} $$ and define  \(\frac{k_2 + k_{12}}{m_2}\) as \(\omega_1 ^2\) .

For the next part of our analysis we shall continue with the first ODE and try to find the solution in terms of the force function.
\(G_o\) we know to be \(A\sin \omega t + B\cos\omega t\)  as a solution so then we apply the boundary condition that the system beings from rest at t = 0. Which means that B = 0. Thus our Green's function G = \(A\sin \omega_o t\). Now for t' < 0  the driving force is 0 and for t' > 0  the driving force is non-zero. This means that the full definition of our Green's function is $$G(t,t')= \begin{cases} 0 & t < t' \\ A\sin \omega_o t & t > t' \end{cases}$$ The remaining task is to find the value of our constant A. For that we go back to the original ODE in the following manner:
\begin{eqnarray}
\int_{t'-\epsilon}^ {t'+\epsilon}\left( m_1\ddot{G} + \omega_o G\right) dt = \int_{t'-\epsilon}^ {t'+\epsilon}\delta(t) dt\\
 m_1\frac{dG}{dt}|_{t'-\epsilon}^{ t'+\epsilon} + \omega_o \int_{t'-\epsilon}^ {t+\epsilon} G dt = 1\\
m_1\frac{dG}{dt}|_{t'+\epsilon}- m_1\frac{dG}{dt}|_{t'- \epsilon} = 1 \hspace{20mm} \text{eq.6}
\end{eqnarray}
Here \( \epsilon\) is some positive infinitesimal real number and we are doing the integrals in the limit as \(\epsilon\) goes to 0. So technically I should have had the limit signs above. Plugging G into eq.6 and remembering that G = 0 for t' < 0, we get that \(A_o\)= \(\frac{1}{m_1\omega_o}\)
Thus our solution is $$ x(t) = k_{12}\int_0^t y(t)A_o\sin(\omega_o (t-t')) dt' \hspace{20mm} \text{eq.7}$$ or if we had started with the second ODE $$ y(t) = k_{12}\int_0^t x(t)A_1\sin(\omega_1 (t-t')) dt' \hspace{20mm} \text{eq.8}$$

If this was not a coupled equation we would have f(t) instead of the green's function and we would integrate in order to find the solution. What eq.7 and eq.8 tell us is that if we apply isotropic conditions i.e \(k_1= k_2 \)and \(m_1 = m_2\) the green function is the same for both ODE and the solution to the system would be the convolutions of each other. Eq. 7 and Eq.8 show us  what Green's function is doing, we have a system feeling a force caused by the other mass coupled to our system ,  Green's function acts as a weighting function over time for the force we feel from the other system. Lastly, the fact that the argument in Green's function is t-t' shows causality. The solution up until time t is got by averaging the split second reactions of the system to the driving force up until that point. It matters where we are in time.

Laplace Transform and coupled oscillators

It is amazing that using Laplace transforms to solve ordinary differential equations is not taught to physics students that often and yet using these beasts reduces the problem to a simple algebra problem, trivializing what could have been a nasty problem. Let begin with the tools will need, namely, what action the Laplace transform has on derivatives. Assume that $$ \int_0^\infty e^{-(st)} x(t) = x(s)= \mathbf{L[x(t)]}$$

Now the following follows:
 \begin{eqnarray}
 \mathbf{L[\dot{x}]}=&  [x(t)e^{-(st)}]_0^{\infty} + \int_0^{\infty}se^{-(st)}x(t) dt = -x(0) + sx(s) \hspace{20mm} (1)\\
\mathbf{L[\ddot{x}]} =& [\dot{x}e^{-(st)}]_0^{\infty} + \int_0^{\infty}se^{-(st)}\dot{x}dt \\
\mathbf{L[\ddot{x}]} =& -\dot{x}+ [sx(t)e^{-(st)}]_0^\infty + \int_0^{\infty}s^2e^{-(st)}x(t) dt\\
\mathbf{L[\ddot{x}]}=& s^2x(s) -sx(0)-\dot{x}(0) \hspace{20mm} (2)
\end{eqnarray}
I shall now refer to x(0) and y(0) as \(x_o\) and \(y_o\) respectively.
There is one more powerful result that relates laplace transforms to convolutions that would not hurt to prove. Remember, we are going to take the laplace transform of our  differential equations and at the end we will need to take the inverse laplace transform. The result we are about to prove will help us later and prevent us from doing nasty partial fractions. Let f*g represent the following integral\(\int_0^t f(\tau)g(t-\tau) d\tau\). Now let the laplace transform act upon it and let \(\bar{f}\bar{g}\)= \(\mathbf{L[f(t)*g(t)]}\) then \(\mathbf{L[f*g]}= \int_0^{\infty}e^{-(st)}\int_0^t f(\tau)g(t-\tau) d\tau dt\).
The above integral involves integrating vertically in the \(\tau,t\) plane so we can change that so that we integrate horizontally by doing the following:
$$
 \mathbf{L[f\star g]} = \int_0^{\infty}\int_{\tau}^{\infty} e^{-(st)}f(\tau)g(t-\tau) dt d\tau\\
 \mathbf{L[f\star g]}=\int_0^{\infty}f(\tau)[\int_{\tau}^{\infty} e^{-(st)}g(t-\tau)dt ]d\tau
$$
Changing variables to   \( u= t-\tau\) and doing the integral in the brackets in the second equation gives us the following
\begin{eqnarray}
 \mathbf{L[f \star g]} =& \int_0^{\infty}f(\tau)e^{-(s\tau)}\bar{g}(s) d\tau \\
 \mathbf{L[f \star g]} = &\bar{g}(s)\bar{f}(s)\\
\mathbf{L[f \star g]} = &\bar{f}(s)\bar{g}(s) \hspace{30mm} eq.3
\end{eqnarray}
From eq.3 we arrive at the powerful result, $$f(t) \star g(t) = \mathbf{L^{-1}[\bar{f}\bar{g}]} \hspace{30 mm} eq. 4 $$
With eq.1 ,eq.2  and eq.4 in our bag let's look back at the couple ordinary differential equations we originally derived from considering Newton's second law. They are the reproduce here:
\begin{eqnarray}
m_1\frac{d^2x_1}{dt^2}= -k_1x_1 - k_{12}x_1 + k_{12} x_2  \\
 m_2\frac{d^2x_2}{dt^2}= -k_2x_2 - k_{12}x_2 + k_{12} x_1
\end{eqnarray}
Let \(x_1\) = x and \(x_2\) = y for ease of writing and clarity.
Re-writing both equations, we get the following:
\begin{eqnarray}
m_1 \ddot{x}= -k_1x - k_{12}(x-y)\\
m_2\ddot{y} = -k_{2}y + k_{12}(x-y)
\end{eqnarray} Next apply laplace transform and assume oscillators begin from rest to get:
\begin{eqnarray}
(m_1s^2 + k_1+k_12)x - k_12y = m_1sx_o\\
 - k_12x+ (m_2s^2 + k_2+k_12)y = m_2sy_o
\end{eqnarray}
Hence we get the following matrix equation:
\begin{align}
\begin{pmatrix}
m_1s^2 + k_1 + k_{12} & -k_{12}\\
-k_{12} & m_2s^2 + k_2 + k_{12}
\end{pmatrix}\begin{pmatrix}
 x\\ y
\end{pmatrix} =
\begin{pmatrix}
 m_1sx_o\\
m_2sy_o
\end{pmatrix}
\end{align}

Notice that if all the m's are the same and all the k's are the same then the matrix on the left becomes symmetric. Thus one of the reasons for symmetric matrices is that there is a translational in-variance in  the problem. In other words one can't tell which is the right or left oscillator. Pick one as your left oscillator, the moment you turn around, I shall turn the whole system around and now your left oscillator will be on your right and you will not know the difference.  Now, what I shall do next is to solve for x using cramer's rule( it turns out that cramer's rule can be useful especially if the coefficients are functions). So x(s)=
\begin{equation} \frac{m_1s x_o[m_2s^2+k_2+ k_{12}] - k_{12}m_2sy_o}{(m_1s^2+k_{12}+k)(m_2s^2+k_2+k_{12}) - k_{12}^2}
\end{equation}
We shall now make a simplifying assumption namely that \(m_1 = m_ 2 = m\) , \(k_1 = k_{12} = k_{2} = k\) and lastly \(\frac{k}{m}= \omega^2\). With that in mind x(s) becomes
\begin{equation}
\frac{m^2s^3x_o - kmsy_o -2msk}{(ms^2+3k)(ms^2+k)}\\
= \frac{3sx_o}{2(s^2+ 3\omega^2)}-\frac{sx_o}{2(s^2+\omega^2)} - \frac{\omega^2sy_o}{(s^2+3\omega^2)(s^2+\omega^2)} - \frac{2s\omega}{(s^2+ 3\omega^2)(s^2+ \omega^2)}
\end{equation}
We are now in position to use eq. 4 derived earlier. Remember the result says that if you recognize a product of laplace transforms \(\bar{f}\bar{g}\), then do this \( \int_0 ^ t f(\tau)g(t-\tau) d\tau\) which will give you \(\mathbf{L^{-1}[\bar{f}\bar{g}}]\). Since I recognize sines and cosines in the equation, I can do the following convolutions:
\begin{equation}
x(t ) = x_o\left(\frac{3 \cos(\sqrt{3}t\omega)}{2}- \frac{\cos(t \omega)}{2}\right) - \omega y_o \left(\cos(\omega \sqrt{3}t) \star \sin \omega t \right) - 2 \left(\cos (\sqrt{3}\omega t) \star \sin \omega t \right)
\end{equation}
Then using wolfram alpha for the actual integrals, one gets:
\begin{equation}
 x(t) = x_o\left(\frac{3 \cos(\sqrt{3}t\omega)}{2}- \frac{\cos(t \omega)}{2}\right) - \frac{y_0(\cos(t\omega)- \cos(\sqrt{3}t\omega))}{2} - \frac{\cos(t\omega)- cos(\sqrt{3}t\omega)}{\omega}
\end{equation}.

Let's assume that both oscillator began from equilibrium, then \(x_o\) and \(y_o\) are zero. To get y(t) it is easier to go back to eq.1 and solve for y getting \(\frac{\ddot{x}}{\omega^2} + 2x = y\) and plugging the appropriate equations for \(\ddot{x}\) and \( x\), one gets the following:
\begin{equation}
y(t) =  \frac{-\cos(\omega t) - \cos(\sqrt{3}t \omega)}{\omega}
\end{equation}

The mystery of normal co-ordinates

I must begin with a confession, namely that I have never understood the point of normal co-ordinates mostly because we can only arrive at them after we have completely solved the problem and found all the solutions. The point of them to decouple our differential equations, but they decouple our differential equation after we have found all the solutions. Nevertheless, they are quite interesting and understanding them will give you an easier time understanding basis vectors and operators in quantum mechanics.
Now if we have two vectors that span \(\mathbf{R}^2\) then we can write any vector \(\vec{x}\) in \(\mathbf{R}^2\) as a linear combination of them i.e \(\vec{x} = \sum_{i=1}^N a_i\mathbf{e}_i\) for constants \(a_i\) and basis vectors \(\mathbf{e_i}\).  But supposing we needed to describe this vector in another set of co-ordinates because the vector looked simpler, then we need a transformation operation to move from on set of co-ordinate basis to another. This transformation operation is usually called a standard matrix in linear algebra. So quick example before we move forward.
Let's begin with the standard Cartesian co-ordinates \(\hat{x}\) = (1, 0) and \(\hat{y}\) = (0, 1). Supposing we needed to rotate this co-ordinates vectors  \(45^{\circ}\) counterclockwise. Then that means that \(\hat{x}\) moves to \(\hat{x}^{'}\) = (1,1) and \(\hat{y}\) moves to \(\hat{y}^{'}\) = (-1,1). Thus, our standard matrix does this: \(T(\hat{x})\) = (1, 1) and \(T(\hat{y})\) = (-1,1).We then make the standard matrix, T, from how it operates on our original basis vectors in the following manner:
\begin{equation}
  \begin{pmatrix}
    T(\hat{x}) & T(\hat{y}) \\
     1              &     -1\\
     1              &       1
  \end{pmatrix}
\end{equation}

So, now our matrix equation to move from one basis co-ordinate to another is the following
\begin{align}
 \begin{pmatrix}
  \hat{x}\\ \hat{y}
 \end{pmatrix}
 =   \begin{pmatrix}
     1              &     -1\\
     1              &       1
  \end{pmatrix}
 \begin{pmatrix}
  \hat{x}^{'}\\
 \hat{y}^{'}
\end{pmatrix}
\hspace{20mm} Eq.1
\end{align}
Now  going back to our problem we found our solution in matrix notation to be the following:

\begin{align}
\begin{pmatrix}x_a(t) \\x_b(t)\end{pmatrix}= \begin{pmatrix}1 & 1 \\ 1 & - 1\end{pmatrix}\begin{pmatrix} \cos \omega_1 t \\ \cos \omega_2 t \end{pmatrix}
\hspace{20mm} Eq.2
\end{align}
Notice the similarity, we can think of the solution we found as living naturally in another co-ordinate space that has been rotated \(45^{\circ}\) clockwise. In order to reconstruct our solution the matrix in eq.2 will have to be an identity matrix on the right hand side of the equation(because then the first solution is simply in terms of \(\cos \omega_1 t\) and the second solutions is simply in terms of \(\cos \omega_2 t\)). And how in the world do we do this ? We simply multiply both sides by from the left by the inverse of the matrix. The inverse matrix acting on \(x_a(t)\) and \(x_b(t\) gives us how to construct our normal co-ordinates. Our inverse matrix is $$ \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\end{pmatrix}$$  If we multiply this by the column vector [\(x_a (t) , x_b (t)\)] we arrive at our normal co-ordinates. This then is the systematic way of getting at normal co-ordinates.  I have eschewed from discussing similar matrices because that is always done and rarely is any connection made with this problem. Similar matrices will have to wait when I write a discussion on the matrix formulation of quantum mechanics.

Introduction to coupled oscillators

Today, we begin with two coupled oscillators. The two oscillators \(m_1\) and  \(m_2\) are connected with spring constant \(k_{12}\). Spring constant \(k_1\) connects \(m_1\) to a fixed position and \(k_2\) connects \(m_2\) to a fixed position on the other end.
 The first is to figure out Newton's equations governing each mass. We do that by pulling on one of the masses(say \(m_2\)) to the right by distance \(x_2\). Now what effect will this have on \(m_1\)? Well, \(m_1\).will be displaced from its position by \(x_1\) distance. The force on spring constant \(k_1\) will want to return the spring back to equilibrium so it will be \(-k_1x_1\). Spring constant \(k_{12}\) will want to return to equilibrium so it will be \(-k_{12}x_{1}\). But there is a third force( the one causing all the stretching), this acts opposite to the two opposite forces and is equal to \(k_{12}x_2\). All this analysed from the perspective of \(m_1\). Thus far we have this for \(m_1\): \begin{equation} m_1\frac{d^2x_1}{dt^2}= -k_1x_1 - k_{12}x_1 + k_{12} x_2   (1)\end{equation}

The second step is to analyse the system from the perspective of \(m_2\).  So perturb the system by pulling \(m_1\) to the left by amount \(x_1\) and watching the springs holding \(m_2\). What do we see? First, we see \(k_2)\) pulling \(m_2\) back to equilibrium so the force is \(-k_2 x_2\) and so does \(k_{12}\) with force \(-k_{12}x_2\). Again there is the last force that is causing all the stretching and is acting in opposition to the first two forces, its amount is \(k_{12}x_1\). Thus we arrive at the last equation for the second mass, namely:   \begin{equation} m_2\frac{d^2x_2}{dt^2}= -k_2x_2 - k_{12}x_2 + k_{12} x_1 (2) \end{equation}

If you place the first two equations underneath each other you will recognize matrix multiplication. We can thus make the following matrix equation:

\begin{align}
\begin{pmatrix}m_1 x_1^{''}\\m_2 x_2^{''}\end{pmatrix}= \begin{pmatrix}-(k_1+k_{12})& k_{12}\\ k_{12}&-(k_2+k_{12})\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}
\end{align}.
It is important to notice that the matrix is symmetric. This implies that it has real eigenvalues and its eigenvectors are orthogonal, this will come into play later on. We then the make an ansatz , namely that the solutions are of the form \(x_1\) = \(A_1\)\(e^{i\omega t}\) and \(x_2\)= \(A_2\)\(e^{i\omega t}\) and plug into our matrix equation. The exponetial function will drop out living the following:


\begin{align}
\begin{pmatrix}-m_1 A_1\omega ^2\\-m_2 A_2\omega^2\end{pmatrix}= \begin{pmatrix}-(k_1+k_{12})& k_{12}\\ k_{12}&-(k_2+k_{12})\end{pmatrix}\begin{pmatrix}A_1\\A_2\end{pmatrix}
\end{align}.

The above equation can be re-arranged to give the following :

\begin{align}
  0= \begin{pmatrix}(k_1+k_{12}-m_1\omega^2) & k_{12}\\ k_{12}&(k_2+k_{12}-m_2\omega^2)\end{pmatrix}\begin{pmatrix}A_1\\A_2\end{pmatrix}
\end{align}.
These last few steps are usually the clear steps in text books. One can easily follow the math, but I decided to re derive for the sake of clarity and mellifluous flow of logic. Usually textbooks jump to saying a non trivial solution is got by making the determinate of the 2 by 2 matrix equal to zero but why is this? How does the determinate suddenly rear its head?
Consider the following matrix equation Ax=b. If A is a non-singular matrix then this equation has only one solution. Now make b=0. It still has one solution and due to the invertible matrix theorem the only solution is x being the zero vector. But that is the trivial solution to our problem. So instead we look for matrices that are singular and thus have more than just the zero vector as the solution. Now, how do we find these matrices? Simple, their defining characteristic is that their determinant is zero. Thus we set the determinat of our matrix equal to zero.

I shall now skip to the solution. You already know from the hundreds of materials out there that one gets two kinds of solutions. 1) Both oscillators have the same amplitude 2) Out of phase with each other. This means we can write the solution thusly:
\(x_a(t)\)= \(A_1\) \(\cos \omega_1 t\) +  \(A_1\) \(\cos \omega_2 t\)  and \(x_b(t )\)= \(A_1\) \(\cos \omega_1 t\) - \(A_1\) \(\cos \omega_2 t\). Again put one equation underneath the other and a matrix equation suddenly appears:
\begin{align}
\begin{pmatrix}x_a(t) \\x_b(t)\end{pmatrix}= \begin{pmatrix}1 & 1 \\ 1 & - 1\end{pmatrix}\begin{pmatrix} \cos \omega_1 t \\ \cos \omega_2 t \end{pmatrix}
\end{align}
In this form it becomes apparent that there must be some transformation we can apply so that we are not taking linear combination of cosines, this transformation will help us decouple the two cosines. Why is it apparent? The matrix with all the 1s is diagonalizable.
This is the stage where in texts all logic breaks down and the author brings up some transformation seemingly out of thin but that does the trick. The student is then left wondering what if I had 3 or 4 oscillators how could I figure out the transformation. That is the subject of the next post. What we want to understand is a general and systematic way of getting at these so called "normal co-ordinates".

The purpose

I have often found myself scouring the web looking for a clue to a proof, or why  a negative sign suddenly appears. If I am lucky, I more often than not, find the necessary clue at some blog where it is beautifully explained. This blog hopefully will be written in that spirit. It is meant for the student who is desperately searching the internet for an explanation that everyone has said is obvious. Also I shall try to explain little curiosities in physics that are never asked or answered in class, for example how come the symmetric matrices always appear in physics; what is this a reflection of ? Is it some coincidence or a reflection of some fundamental principle.
The first thorough examination will be on 2 coupled oscillators. The problem will be solved using matrices, Laplace transforms, Lagrangian mechanics  and Green's Functions.