# Energy and momentum as partial derivatives of on-shell action in field theory

According to L&L, if we fix the initial position of a particle at a given time and consider the on-shell action as a function of the final coordinates and time, $S(q_1, \ldots, q_n, t)$, then...

$$E = -\frac{\partial S}{\partial t}$$

$$p_i = \frac{\partial S}{\partial q_i}$$

Is there a straightforward generalization of this to field theory? Something that would give the energy and momentum densities by differentiating the on-shell action (with respect to... something)?

Yes, this is e.g considered in Ref. 1. In field theory, the starting point is the off-shell action$^1$

$$\tag{1} I[\phi; t_f,t_i] ~:=~\int_{t_i}^{t_f} \! dt~\int_{\Sigma} d^3x~ {\cal L}(\phi(x,t),\dot{\phi}(x,t), \partial_x \phi(x,t);x,t) ,$$

where $t_i$ and $t_f$ denote initial and final times, respectively. We now impose appropriate boundary conditions (B.C.), e.g. Dirichlet B.C.

$$\tag{2} \phi^{\alpha}(x,t_i)~=~\phi^{\alpha}_i(x) \qquad \text{and}\qquad \phi^{\alpha}(x,t_f)~=~\phi^{\alpha}_f(x) .$$

We assume that for given B.C. (2) there exists a unique solution $\phi_{\rm cl}$ to the Euler-Lagrange equations. OP is interested in the (Dirichlet) on-shell action defined as

$$\tag{3} S[\phi_f,t_f; \phi_i,t_i] ~:=~ I[\phi_{\rm cl}; t_f,t_i].$$

Next define (Lagrangian) momentum field

$$\tag{4} \pi_{\alpha}(x,t) ~:=~\frac{\partial {\cal L}(\phi(x,t),\dot{\phi}(x,t), \partial_x \phi(x,t);x,t)}{\partial \dot{\phi}^{\alpha}(x,t)},$$

and energy

$$\tag{5} h(t)~:=~\int_{\Sigma} d^3x~\left(\sum_{\alpha}\pi_{\alpha}(x,t)\dot{\phi}^{\alpha}(x,t) -{\cal L}(\phi(x,t),\dot{\phi}(x,t), \partial_x \phi(x,t);x,t)\right).$$

Then one may show field-theoretically$^{2}$ that

$$\tag{6} \frac{\delta S}{\delta \phi^{\alpha}_f(x)}~=~ \pi_{\alpha}(x,t_f), \qquad \frac{\delta S}{\delta \phi^{\alpha}_i(x)}~=~ -\pi_{\alpha}(x,t_i) ,$$

and

$$\tag{7} \frac{\partial S}{\partial t_f}~=~-h(t_f), \qquad \frac{\partial S}{\partial t_i}~=~h(t_i).$$

Example: A free field Lagrangian density ${\cal L} = \frac{1}{2}\phi^2$ leads to

$$\tag{8} S(\phi_f,t_f; \phi_i,t_i) ~=~ \frac{1}{2(t_f-t_i)} \int_{\Sigma} d^3x~(\phi_f(x)-\phi_i(x))^2 .$$

References:

1. MTW; Section 21.1 and Section 21.2.

2. L.D. Landau & E.M. Lifshitz, Mechanics, vol. 1, 1976; $\S$ 43.

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$^1$ For actions in point mechanics, see e.g. this Phys.SE post.

$^2$ For a proof in point mechanics, see e.g. Ref. 2 and my Phys.SE answer here.

• Is there any way to get the linear momentum though? Because people always say position and linear momentum are conjugate variables, and apparently the situation does not change in field theory. But I no longer see the connection in that case. – Brian Bi Sep 4 '14 at 21:13
• The momentum density inside the stress-energy-momentum tensor (and its integrated version) do not appear to be directly related to the (Dirichlet) on-shell action in a simple manner, if that is your question. – Qmechanic Sep 5 '14 at 20:31

Yes indeed there is!

Firstly, the normal Hamilton-Jacobi equation goes through, so still the energy is given by the time derivative of the on-shell action.

But the relevant local object that is most natural in field theory is the stress-energy-momentum tensor, which contains densities and fluxes of energy and momentum. The question of what to vary to get this is perhaps unclear at first: the answer is to vary the background geometry on which the theory is defined.

More specifically, one varies the metric, which defines the local notions of distances and angles. In fact, in the end this turns out to be the best way to define the stress-energy tensor: it's (up to constants) the derivative of the action with respect to the background metric.

Incidentally, in gravitational theories such as GR the metric itself is a dynamical field, so this on-shell variation of the action with respect to the metric is by definition zero: one may define a "matter" stress-energy by just including part of the action, but there is no good local definition of total energy density and related things in such theories.

• I don't get how Hamilton--Jacobi comes through here. Any chance I can get you to do a simple example? – Brian Bi Aug 31 '14 at 0:54
• If you interpret the generalized coordinates $q_i$ of a system to be the fields, with the labels $i$ including the spatial coordinates $x$ (replacing any sums by integrals over $x$ where necessary), your favourite discussion of H-J will be in essence unchanged (except you might need to append the word 'density' to the phrase 'canonical momentum', use functional derivatives in place of ordinary derivatives, and perform similar generalisations). – Holographer Aug 31 '14 at 15:59
• I'd love to illustrate with a simple example, but I can't think of a situation in field theory where the problem can be solved in a simple way with general initial and final conditions, and the action evaluated explicitly on-shell... would be nice to see if anyone else can! – Holographer Aug 31 '14 at 16:13
• Yeah, I can totally see how the math might be intractable. I'm still not sure how to even write down H-J with a field though. What does the H-J equation look like for the electromagnetic field in vacuum, for example? – Brian Bi Sep 3 '14 at 20:17