# Legendre Transformation of Lagrangian density ?

1. May 2, 2015

### DOTDO

Hi

I began to study the basics of QED.

Now I am studying Lagrangian and Hamiltonian densities of Dirac Equation.

I'll call them L density and H density for convenience :)

Anyway, the derivation of the H density from L density using Legendre transformation confuses me :(

I thought because parameters of them are space-time components, it should be

But I found that this is related to the De Doner - Weyl Theory,

and the H density used in textbook is

where the dot represents time derivative.

So, my question is,

why we consider Legendre transformation on only time derivative of phi ?

Is it just 'defined' to consider energy of the system?

Then what does the covariant H density defined in the De Donder - Weyl theory mean?

2. May 3, 2015

3. May 3, 2015

### stevendaryl

Staff Emeritus
From the lagrangian density $\mathcal{L}$, you can define the canonical stress-energy tensor $\mathcal{T}$ as follows:

$\mathcal{T}^{\mu \nu} = \dfrac{\partial \mathcal{L}}{\partial \partial_\mu \phi_j} \partial^\nu \phi_j - g^{\mu \nu} \mathcal{L}$

This is a conserved current in the first index:

$\partial_\mu \mathcal{T}^{\mu \nu} = 0$

Then you can define a hamiltonian density $\mathcal{H}$ in terms of $\mathcal{T}$:

$\mathcal{H} = \mathcal{T}^{tt} =\dfrac{\partial \mathcal{L}}{\partial \partial_t \phi_j} \partial^t \phi_j - g^{tt} \mathcal{L}$

This is the same as the expression in the textbook, if you're using the metric where $g^{tt} = +1$ and defining $\dot{\phi_j} = \partial_t \phi_j$

The hamiltonian is the integral of the hamiltonian density over all space:

$H = \int d^3 x \mathcal{H}$

It's the hamiltonian, not the hamiltonian density, that is constant:

$\dfrac{d}{dt} H = 0$