# Diagonalized Lagrangian

1. Oct 23, 2008

### shehry1

My Graduate Mechanics text (Walecka and Fetter) says in the chapter on Small Oscillations, that the Modal Matrix Diagonalizes the Lagrangian

$$L = \sum_{\sigma=1}^\infty (\dot{\zeta_\sigma}^2 - \omega_\sigma \zeta_\sigma^2)$$

where $$\zeta$$ are the normal coordinates related to the original coordinates $$\eta$$ via the modal matrix.

But as far as I understand, the Lagrangian is simply a number, not a matrix. So what does the diagonalization of the Lagrangian mean.

2. Oct 24, 2008

### dx

The Lagrangian is not a number, it's a function. In the case of small oscillations, it happens to be a quadratic form, and it can be written as

$$L = Q^{T}AQ$$,

Where $$Q = (q_1, q_2, ..., q_s, \dot{q}_1, \dot{q}_2, ..., \dot{q}_s)$$. When the $$q_i$$ are normal coordinates, $$A$$ becomes a diagonal matrix.

3. Oct 24, 2008

### shehry1

Sorry for saying that its a number but I still didn't get it. And I think I basically want to know how can Lagrangian the function be transformed into Lagrangian the matrix. e.g. we have $$2L = \dot{\eta}^T m \dot{\eta} - \eta^T v \eta$$. But the $$\eta$$ are essentially column vectors, giving me a function in the form of a 1x1 matrix after multiplication. So what does it mean to write the Lagrangian in the form of a matrix

4. Oct 24, 2008

### dx

Well, its not exactly a matrix. But matrices are not the only things that can be diagonalized. Quadratic forms for example can be diagonalized. If you have some random set of generalized coordinates and velocities $$X = (x_i, \dot{x}_i)$$, then the Lagrangian for small oscillations is of the form

$$L = \frac{1}{2}\sum (m_{ik} \dot{x_i} \dot{x_k} - k_{ik} {x_i} {x_k})$$.

This can be written in matrix notation as $$L = X^{T}AX$$, where A is in general not diagonal. This is the modern way of representing quadratic forms. It turns out that there is a specific set of coordinates $$Q_i$$ in terms of which the Lagrangian simplifies into the form

$$L = \frac{1}{2}\sum_{\alpha} (\dot{Q_{\alpha}^2} - {\omega_{\alpha}^2} {Q_{\alpha}^2})$$.

When you write this in the form $$Q^{T}AQ$$, A will be a diagonal matrix.

Last edited: Oct 24, 2008