The Lagrangian density of a Poincare invariant theory should not depend explicitly on the space-time coordinates. Does this mean $$ \partial_\mu \mathcal{L}=0~? $$

If this is the case doesn't the above equation imply that $\mathcal{L}$ is a constant? I know that $\mathcal{L}$ is a functional not a function but at the end it is still a function of space-time.


I) OP wrote (v2):

I know that $\mathcal{L}$ is a functional not a function.

Actually for local theories the Lagrangian density $$ \mathcal{L}(\phi(x), \partial\phi(x), \partial^2\phi(x), \ldots, ;x) $$ is a function of $\phi(x)$, $\partial\phi(x)$, $\partial^2\phi(x)$, $\ldots$, and $x$. In contrast, the action $S[\phi]=\int \!d^dx~\mathcal{L}$ is a functional of $\phi$, cf. e.g. this Phys.SE post.

II) OP wrote (v2):

The Lagrangian density of a Poincare invariant theory should not depend explicitly on the space-time coordinates. Does this mean $$\tag{1} \partial_{\mu} \mathcal{L}~=~0~? $$

If $\partial_\mu$ in OP's notation denotes explicit space-time differentiation $ \frac{\partial}{\partial x^{\mu}}$, then the answer is Yes. Note however that there are also implicit dependence on the space-time coordinates via the fields $\phi$. The total space-time derivative becomes

$$ d_{\mu}\mathcal{L}~=~\partial_{\mu} \mathcal{L} + \frac{\partial\mathcal{L}}{\partial\phi}\partial_{\mu}\phi + \frac{\partial\mathcal{L}}{\partial(\partial_{\nu}\phi)} \partial_{\nu}\partial_{\mu}\phi + \ldots . $$

III) Translational invariance (1) implies via Noether's theorem that the corresponding energy-momentum 4-vector $P_{\mu}$ is conserved.

  • $\begingroup$ In general, as far as I know, fields are maps from manifold M to manifold N. What do you mean saying "function of $\phi(x)$"? Do you mean the image of this map in some chosen local coordinates on N? $\endgroup$ – xxxxx Sep 17 '14 at 14:00
  • 1
    $\begingroup$ @xxxxx: The above answer uses common physics jargon/notation available in most physics textbooks. E.g. if $L(x,v,t)=\frac{m}{2}v^2$, then $\frac{\partial L(x(t),\dot{x}(t),t)}{\partial \dot{x}(t)}=m\dot{x}(t)$. On the other hand, more generally, mathematically speaking, if $M$ is the target space of the fields $\phi:\Sigma\to M$, $\Sigma$ is spacetime, and $E=M\times \Sigma \to \Sigma$ is a bundle over $\Sigma$, then the Lagrangian density $\mathcal{L}$ is a map $\mathcal{L}:JE\to \mathbb{R}$ , where $JE$ denotes the jet bundle for $E$. $\endgroup$ – Qmechanic Sep 17 '14 at 14:32
  • $\begingroup$ Thank you. But doesn't $\partial_\mu \mathcal{L}=0$ implies that $\mathcal{L}$ is a constant ? $\endgroup$ – Axion Sep 17 '14 at 14:55
  • $\begingroup$ $d_\mu \mathcal{L}=0$ implies that $\mathcal{L}$ is constant, $\partial_\mu \mathcal{L}=0$ implies that $\mathcal{L}$ doesn't explicity depends of $x^\mu$ but it depends of $\phi(x^\mu)$ instead, so $\mathcal{L}(\phi,\partial_\mu \phi,x^\mu) = \mathcal{L}(\phi,\partial_\mu \phi)$. $\endgroup$ – Héctor Sep 17 '14 at 15:13

You could say that $\partial_{\mu}\mathcal{L}=0$ only if the fields it applies to are treated as independent variables and not as functions of $x^\mu$. I.e. the Lagrangian does not explicitly refer to coordinates, but if you were to consider the fields as $\phi = \phi(x^\mu)$, then obviously $\partial_{\mu}\mathcal{L}\neq 0$.

The more rigorous way to say this is by considering a transformation which translates all the relevant physics in space-time by $a^\mu$. The Lagrangian gets transformed to $\mathcal{L}(x^\mu) \to \mathcal{L}_a(x^\mu)$ - for now we do not assume any form of this transformation. The zero derivative requirement could be specified as $$\mathcal{D}_a \mathcal{L} \equiv \lim_{\epsilon \to 0} \frac{\mathcal{L}_{a \epsilon}(x^\mu + \epsilon a ^\mu)-\mathcal{L}(x^\mu)}{\epsilon} = 0$$ I.e. if you do an infinitesimal translation of all the physics and compare the result with the original untranslated one (you must compare at the back-translated point), you get the same result. But by requiring this and assuming some reasonable mathematics, we get that it is equivalent to the condition $$ \mathcal{L}_a (x^\mu) = \mathcal{L}(x^\mu-a^\mu)$$ I.e. translating physics and back-translating coordinates yields the same result even if not infinitesimal. The important part is obviously the "all relevant physics" translating with this $a$-translation. This is almost a tautological statement - the Lagrangian depends only on "all relevant physics" and it's value gets transported with them.

The latter condition is a more special case of the ones mentioned by Frederic Brünner and m1rohit.

  • $\begingroup$ Thanks for your reply. Actually I shouldn't have used Poincare invariance in my question. I suppose the Lagrangian density can't depend on space-time explicitly even if this dependence is Poincare invariant. For example $\partial_\mu \phi \partial^\mu \phi+(x_\mu-x_\nu)^2$ can't be a sensible Lagrangian. So my question is do we require that $\partial_\mu \mathcal{L}=0$? $\endgroup$ – Axion Sep 17 '14 at 10:52
  • $\begingroup$ Poincaré invariant also means translation invariant. In the example you give, would the Lagrangian be a function of two different points? This argument would boil down either to violation of locality, violation of translation invariance, or a redundant constant term which does not change the dynamics. We really need not require $\partial_\mu \mathcal{L}=0$, the "sensibility" of the Lagrangian can really be derived from Poincaré invariance. On the other hand, introducing a two point Lagrangian means introducing completely different physics, so here the usual requirements have to be modified. $\endgroup$ – Void Sep 17 '14 at 11:30

Poincare invariance simply means that the Lagrangian does not change under a Poincare transformation:


Note that I have only written down Lorentz transformations, for simplicity. We now say that $\mathcal{L}$ is invariant if


This does not mean that the Lagrangian does not depend on spacetime coordinates. After all, the fields it contains are supposed to be functions of spacetime. It just tells us something about the transformation behaviour of the function. An invariant Lagrangian can be constructed out of expressions that transform like scalars, i.e. $\phi(x)\rightarrow\phi(\Lambda^{-1}x).$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.