# Hamiltonian for a Lagrangian with coupling

I am dealing with the following Lagrangian density $$\mathscr{L}_{em}= -\frac{1}{2}\rho\omega^2 u^2 +\frac{1}{2}\nabla u:\Sigma :\nabla u-\frac{1}{2}\nabla\phi\cdot\epsilon\cdot\nabla\phi+\nabla\phi\cdot P:\nabla u$$ where $\rho,\omega\in\mathbb{R}^+$, $\Sigma_{ij,kl}=\Sigma_{ji,kl}=\Sigma_{ij,lk}=\Sigma_{kl,ij}$, $\epsilon_{ij}=\epsilon_{ji}$, $P_{ijk}=P_{ikj}$, $\phi$ is a scalar field and $u \in\mathbb{R}^3$.

I need to compute the associated Hamiltonian density.

If it was just

$$\mathscr{L}_m=-\frac{1}{2}\rho\omega^2 u^2 +\frac{1}{2}\nabla u:\Sigma :\nabla u$$

by defining the momentum $\sigma_m=\Sigma:\nabla u$, and using the Legendre transform

$$\mathscr{H}=p\cdot\nabla q(q,p)-\mathscr{L}(q,p),$$

where $q$ are the field variables and $p$ the momentum, I obtain

$$\mathscr{H}_m=\frac{1}{2}\rho\omega^2 u^2+\frac{1}{2}\sigma_m:\Sigma^{-1} :\sigma_m.$$

Also for

$$\mathscr{L}_e=-\frac{1}{2}\nabla\phi\cdot\epsilon\cdot\nabla\phi$$

I can obtain

$$\mathscr{H}_e=-\frac{1}{2} d_e\cdot\epsilon^{-1}\cdot d_e$$

with $d_e=-\varepsilon\cdot\nabla\phi$.

But now, what about the Hamiltonian density for $\mathscr{L}_{em}$? Can I write something like

$$\mathscr{H}_{em}=\frac{1}{2}\rho\omega^2 u^2 +\frac{1}{2}\sigma_m:\Sigma^{-1}:\sigma_m -\frac{1}{2}d_e\cdot\epsilon^{-1}\cdot d_e\pm d_e\cdot Q:\sigma_m~?$$

Or must I rely on the introduction of the momentum

$$\sigma_{em}=\Sigma:\nabla u+P^T\cdot\nabla\phi.$$

$$d_{em}=-\varepsilon\cdot\nabla\phi+P:\nabla u~ ?$$

Who is the matrix $Q$?

Is something related to this Phys.SE post: Lagrangian and hamiltonian of interaction ?

I am new in the argument, but every suggestion is appreciated.

• What does your $:$-notation mean? – Danu Dec 20 '14 at 14:05
• dot product is the inner product, i.e. $u\cdot u=u_i u_i$, double dot product is double inner product. The lagrangian can also written in einstein notation as $\mathscr{L}=-\frac{1}{2}\rho\omega^2 u_i u_i+\frac{1}{2}\partial_i u_j \Sigma_{ijhk}\partial_h u_k-\frac{1}{2}\partial_i\phi\epsilon_{ij}\partial_j\phi+\partial_i\phi P_{ijk}\partial_j u_k$. – Fabio Dec 20 '14 at 14:38
• And the variables are? – Danu Dec 20 '14 at 14:43
• $u$ and $\phi$ are the variables, – Fabio Dec 20 '14 at 16:14

## 2 Answers

In order to perform the (possibly singular) Legendre transformation, it is necessary to have information about pertinent rank conditions of the structure constants $\rho$, $\omega$, $\Sigma_{ij,k\ell}$, $\epsilon_{ij}$ and $P_{ijk}$.

In this answer, we will sketch how the (possibly singular) Legendre transformation is performed in principle:

1. We will use DeWitt's condensed notation to hide all spatial derivatives for simplicity.

2. Assume that the Lagrangian density $$\tag{1}\mathcal{L}~=~\mathcal{L}_2+\mathcal{L}_1+\mathcal{L}_0$$ is a quadratic function of the velocities $\dot{\Phi}^A$ (=temporal derivatives of the fields).

3. We may for later convenience redefine the fields $$\tag{2}\Phi^A~\longrightarrow ~\Phi^{\prime A}~=~R^A{}_B~\Phi^B.$$

4. After possible redefinition (2) of the fields $$\tag{3}\Phi^A~=~\{\phi^{\alpha}, \ldots \},$$ we may assume that $\mathcal{L}_2$ is of the form $$\tag{4} \mathcal{L}_2~=~\frac{1}{2}\dot{\phi}^{\alpha}m_{\alpha\beta}\dot{\phi}^{\beta},$$ where the symmetric matrix $m_{\alpha\beta}$ is invertible.

5. Let us for simplicity assume that $\mathcal{L}_1$ is quadratic in the variables $\Phi^B$.

6. After possible redefinition (2) of the fields $$\tag{5}\Phi^A~=~\{\phi^{\alpha};z^I; \lambda^a\},$$ we may assume that $\mathcal{L}_1$ is of the form $$\tag{6}\mathcal{L}_1~=~A_{\alpha}\dot{\phi}^{\alpha} +\frac{1}{2}z^I\omega_{IJ}\dot{z}^J,$$ where the matrix $\omega_{IJ}$ is invertible, and $A_{\alpha}$ depends linearly on the fields $\Phi^B$.

7. Define momenta $$\tag{7}\pi_{\alpha} ~:=~\frac{\partial\mathcal{L}}{\partial \dot{\phi}^{\alpha}} ~=~m_{\alpha\beta}\dot{\phi}^{\beta}+A_{\alpha}.$$

8. By possibly redefining the fields $$\tag{8}z^I~\longrightarrow ~z^{\prime I}~=~r^I{}_J~z^J,$$ with $$\tag{9}z^I~=~\{q^i;p_j\},$$ and possibly throwing away total time derivative terms, we may assume that $$\tag{10}\frac{1}{2}z^I\omega_{IJ}\dot{z}^J~=~p_i \dot{q}^i.$$

9. Introduce momenta $\rho_b$ to the auxiliary variables $\lambda^a$.

10. The Hamiltonian density becomes $$\tag{11} {\cal H}~=~\frac{1}{2}(\pi_{\alpha}-A_{\alpha})(m^{-1})^{\alpha\beta}(\pi_{\beta}-A_{\beta})-\mathcal{L}_0,$$ cf. the Faddeev-Jackiw method. See also e.g. this Phys.SE post.

11. In the Hamiltonian formulation, $\{\phi^{\alpha};\pi_{\beta}\}$, $z^I =\{q^i;p_j\}$ and $\{\lambda^a;\rho_b\}$ are canonical variables.

• That you explained is exactly what I am trying to do, and I can not explain/compute in a simple way. The essence is in the inversion of the matrix $m$ that in my case is (reading row-wise) $m=[\Sigma,P^T,P,-\epsilon]$. I have no other term, except $\sim u^2$ that in your notation should end in $\mathcal{L}_0$. What I would like to do is "to inherit" the free Lagrangian/Hamiltonian results and to write the Hamiltonian in the case of interaction by using this two terms and a part due to interaction, i.e. what I sketched as $d_e\cdot Q:\sigma_m$. – Fabio Dec 21 '14 at 7:46
• I updated the answer. Tl;dr: Do not compute. More input needed. – Qmechanic Dec 21 '14 at 13:54

Thanks to the procedure suggested by Qmechanic, I have clarified myself. I need just to invert the matrix $m$, since it has for the momenta

$\sigma_{em}=\Sigma :\nabla u+P^T\nabla \phi$

$d_{em}=P:\nabla u-\epsilon\cdot\nabla\phi$

or

$\left(\begin{array}{c}\sigma_{em}\\d\end{array}\right)= \left(\begin{array}{cc}\Sigma & P^T\\ P & -\epsilon\end{array}\right) \cdot\left(\begin{array}{c}\nabla u\\\nabla\phi\end{array}\right)$.

In my case, the matrix $m$, has tensor entries, but considering the symmetries for $\Sigma$, $\epsilon$ and $P$, it can be reduced to a 9x9 matrix $m'$ with scalar entries and the vector for the "velocities" has now just 9 entries. Then is matter to do the computation.

The procedure suggested by Qmechanic is quite general and I've very appreciated his/her suggestion, Thanks!