# Extra term when calculating variation in Lagrangian density under infinitesimal Lorentz transform

Consider an (active) infinitesimal Lorentz transformation:

$$x^\mu \rightarrow x^\mu + {\omega^\mu}_\nu x^\nu,$$

so that any scalar field is transformed as

$$\phi(x) \rightarrow \phi'(x) = \phi(x) - {\omega^\mu}_\nu x^\nu \partial_\mu \phi(x) + O(\omega^2).$$

Now consider a Lagrangian density function $$\mathcal{L(\phi, \partial\phi)}$$ (with no explicit spacetime dependence). Every scalar field is associated to a Lagrangian density field $$\mathcal{L}[\phi](x) := \mathcal{L}(\phi(x), \partial\phi(x))$$, which is itself a scalar field. Therefore, it transforms with variation:

$$\delta \mathcal{L} = -{\omega^\mu}_\nu x^\nu \partial_\mu \mathcal{L}[\phi] = -\partial_\mu ({\omega^\mu}_\nu x^\nu \mathcal{L}[\phi]), \tag{1}$$

where the second equality arises because $$\omega$$ is antisymmetric. Since the Lagrangian only varies by a four-divergence, the action is unchanged. This makes perfect sense: all we've done is shift around spacetime by an orthogonal transformation, moving around the $$d^4 x\ \mathcal{L}$$ terms in the action integral, so the total action integrated over the whole of spacetime isn't going to change. So far so good.

The problem arises when I try to calculate $$\delta \mathcal{L}$$ a different way. I think we should be able to calculate the variation using:

\begin{align} \delta \mathcal{L} & = \mathcal{L}[\phi'] - \mathcal{L}[\phi] \\ & = \frac{\partial\mathcal{L}}{\partial\phi}\delta\phi + \frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)} \delta(\partial_\mu \phi) \\ & = - \left [ \frac{\partial\mathcal{L}}{\partial\phi} {\omega^\alpha}_\beta x^\beta \partial_\alpha \phi + \frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)} \partial_\mu ({\omega^\alpha}_\beta x^\beta \partial_\alpha \phi) \right ] \\ & = - {\omega^\alpha}_\beta x^\beta \left[ \frac{\partial\mathcal{L}}{\partial\phi}\partial_\alpha \phi + \frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)} \partial_\mu \partial_\alpha \phi \right] - \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)}{\omega^\alpha}_\mu \partial_\alpha \phi \\ & = -\partial_\alpha({\omega^\alpha}_\beta x^\beta \mathcal{L}) - \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)}{\omega^\alpha}_\mu \partial_\alpha \phi.\tag{2} \end{align}

As you can see, there's a second term that's appeared out of nowhere! Where have I gone astray?

I've tried double-checking that $$\delta (\partial_\mu \phi) = \partial_\mu (\delta \phi)$$ and it seems to work out, so I don't think that's the problem. I found this very old post but I find the argument that

$$\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \propto \partial^\mu \phi \tag{3}$$

unconvincing. Say, for example, you had a $$\frac{1}{2} \partial_\alpha \partial_\beta A^{\alpha\beta}$$ term - that's fine, because differentiating w.r.t. $$\partial_\mu \phi$$ symmetrises $$A$$, so it works out that the extra term is zero, but the proportionality doesn't hold. So is it possible, in complete generality, to prove that $$\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \partial_\nu \phi$$ is always symmetric in $$\mu$$ and $$\nu$$?

I think that the symmetry of the last term is a requirement for the Lagrange density to be a Lorentz scalar. Suppose that we use rotations rather than Lorentz's, so we do not have to worry about upstairs and downstairs indices. Then the change in the action integral under an infinitesimal rotation can come from two terms: 1) the change in the limits of integration --- this is the total divergence; 2) the change in the integrand at each point in the integration region. Consider the change at the point we rotate about (i.e $$x=0$$). Then the argument of the function does not change and we have simply $$\delta(\partial_\mu\phi)= \omega_{\mu\nu}\partial_\nu \phi$$. If the integrand is not to change at that point we need $$0 = \frac{\partial L}{\partial (\partial_\mu \phi)} \delta ({\partial_\mu \phi}) = \frac{\partial L}{\partial (\partial_\mu \phi)}\omega_{\mu\nu}\partial_\nu\phi.$$ Of course we could rotate about any point so $$x=0$$ is not special and so the same result must apply at all points.

After 10 months of further study, I think I am now able to answer my own question, so I'll do it for future reference.

The Lagrangian density field $$\mathcal{L}[\phi]$$ associated (functionally) to the field $$\phi$$ is not necessarily a Lorentz scalar field. Therefore, equation (1) may simply not hold. Of course, any example of practical significance will be a Lorentz scalar field, and imposing this condition, by asserting that (1) holds, combines with (2) to create a condition for the Lagrangian to be a Lorentz scalar field, which is that

$$\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} {\omega^\nu}_\mu \partial_\nu \phi = 0$$

for all (antisymmetric) $$\omega$$.

I haven't checked this by hand, but I suspect the example that I give in equation (3) is exactly a case where the Lagrangian density field is not a Lorentz scalar. This makes sense, since $$A$$ is a distinguished tensor and thus can provide a preferred basis.

This works if you use the fact that $$\phi$$ also satisfies the Euler-Lagrange equation for the density $$\mathcal{L}$$. Instead of what you wrote, consider

$$\begin{eqnarray} \delta \mathcal{L} & = & \mathcal{L}[\phi + \delta \phi] \\ & = & \frac{\partial \mathcal{L}}{\partial \phi}\delta \phi + \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \partial_{\mu} (\delta\phi) \\ & = & \frac{\partial \mathcal{L}}{\partial \phi}\delta \phi + \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \delta \phi \right) - \left( \partial_{\mu} \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \right) \delta\phi \\ & = & \frac{\partial \mathcal{L}}{\partial \phi}\delta \phi + \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \delta \phi \right) - \frac{\partial \mathcal{L}}{\partial \phi}\delta \phi \\ & = & \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \delta \phi \right) \end{eqnarray}$$

Going from the second line to the third, used the product rule. Going from the third line to the fourth used the Euler-Lagrange equation for $$\phi$$. This results in a total divergence again, so no change in action.

This doesn't get you back to exactly the same form of the divergence that you had before, but I'm not sure that matters as long as the action is unchanged.

• Hi there Brick. What you have shown is that a solution to the EL equations will leave the action invariant, but we already knew that, because it's a solution to the EL equations, for which the starting point is that the action be left invariant. My question applies not just to the solution space, but to the whole configuration space. Jun 28, 2021 at 19:43
• What assumption (if any) do you have about $\phi$, $\mathcal{L}$, and the EL equations pre-transform? (I.e. about $\mathcal{L}[\phi]$ not $\mathcal{L}[\phi']$ in the notation of the question.) Jun 28, 2021 at 20:05
• Or maybe from the other way around, it's not clear why you think your Eq. 1 should necessarily hold when $\mathcal{L}$ depends on derivatives of $\phi$ in addition to $\phi$ itself, and if it doesn't depend on the derivatives then your "extra" term, of course, vanishes already. Jun 28, 2021 at 20:23
• The field is arbitrary (so subject to smoothness conditions, boundary conditions, yada yada yada). And indeed, Equation 1 should not hold for arbitrary (local) Lagrangian functionals -- see my new answer :) Jun 28, 2021 at 20:45