Assume we have the following Lagrangian field density where $x, x'$ both three dimensional real vectors are coordinates and $t$ represents time, field is given by $\phi$. Assume for the sake of concreteness that $\phi$ is a function that returns three-dimensional real vector.

$$ \mathscr{L} = f(x) (\partial_t\phi)^2 + \int g(x, x', \phi(x), \phi(x')) dx'. $$

Define canonical momentum to be:

$$ \pi = \frac{\partial \mathscr{L}}{\partial(\partial_t\phi)} = 2f(x)\partial_t\phi. $$


  1. Is it true that we can write equations of motion as: $$ \partial_t \pi_i = - \sum \limits_j \frac{\partial}{\partial x_j} P_{ij}? $$ Here, $P_{ij}$ is some function.

  2. Is there an equivalent law of conservation of energy? How does it look like?

  1. To keep the notation simple let us consider bi-local point-mechanics $$ S[q,v]~=~\int \! dt~dt^{\prime}~{\cal L}(t,t^{\prime}), \tag{A} $$ where we use the shorthand notation $$ {\cal L}(t,t^{\prime})~=~{\cal L}(q(t),q(t^{\prime}),v(t),v(t^{\prime}),t,t^{\prime}).\tag{B}$$ (It is in principle straightforward to generalize to multi-local field theory.)

  2. Define momentum $$p(t)~:=~\frac{\delta S[q,v]}{\delta v(t)} ~=~\int \! dt^{\prime}~\frac{\partial[{\cal L}(t,t^{\prime})+{\cal L}(t^{\prime},t)]}{\partial v(t)},\tag{C}$$ force $$F(t)~:=~\frac{\delta S[q,v]}{\delta q(t)}~=~\int \! dt^{\prime}~\frac{\partial[{\cal L}(t,t^{\prime})+{\cal L}(t^{\prime},t)]}{\partial q(t)},\tag{D}$$ Lagrangian $$ L(t)~:=~\int \! dt^{\prime}~[{\cal L}(t,t^{\prime})+{\cal L}(t^{\prime},t)], \tag{E}$$ and energy function $$ H(t)~:=~p(t)v(t)-L(t).\tag{F}$$

  3. Next we identify $$\dot{q}(t)~\approx~v(t),\tag{G}$$ and assume the stationary action principle (SAP). The Lagrangian EOM becomes $$\dot{p}(t)~\approx~F(t).\tag{H}$$

  4. One may show that if ${\cal L}(t,t^{\prime})$ has no explicit time dependence, then the energy function $H(t)$ is conserved on-shell.

  5. For more on non-local action functionals, see e.g. this Phys.SE post.

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  • $\begingroup$ Thanks! 1. Why do you define Lagrangian (and respective quantities) in such a symmetric way? 2. Why is Lagrangian defined not in a way that action is integral of Lagrangian (which I think is the usual convention)? 3. When you write in point 4 that $\mathcal{L}$ has no explicit time dependence do you mean that both partial derivatives are zero? $\endgroup$ – HydrodynamicsPlease Aug 6 '19 at 1:00
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    $\begingroup$ 1. The symmetric form comes from the SAP. 2. What you want to call the Lagrangian for non-local theories is not set in stone. The definition (E) is here chosen to have eq. (F). 3. Yes. $\endgroup$ – Qmechanic Aug 6 '19 at 12:56

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