# Why boundary terms make the variational principle ill-defined?

Let me start with the definitions I'm used to. Let $$I[\Phi^i]$$ be the action for some collection of fields. A variation of the fields about the field configuration $$\Phi^i_0(x)$$ is a one-parameter family of field configurations $$\Phi^i(\lambda,x)$$ such that $$\Phi^i(0,x)=\Phi^i_0(x)$$ where $$\lambda\in (-\epsilon,\epsilon)$$. We take the map $$\lambda\mapsto \Phi^i(\lambda,x)$$ to be differentiable. In that case the first variation is defined by $$\delta \Phi^i(x) \equiv \dfrac{\partial}{\partial \lambda}\bigg|_{\lambda =0}\Phi^i(\lambda,x)\tag{1}.$$

Likewise the first variation of the action is defined to be $$\delta I[\Phi^i]\equiv\dfrac{d}{d\lambda}I[\Phi^i_\lambda],\quad \Phi^i_\lambda\equiv \Phi^i(\lambda,\cdot)\tag{2}.$$

Now, as I understand, the variational principle is the statement that the physical classical field configuration should be $$\Phi^i$$ such that $$\delta I[\Phi^i]=0$$ for any first variation $$\delta \Phi^i$$.

It so happens that most of the time $$I[\Phi^i]$$ is the integral over spacetime of some Lagrangian density $$d$$-form $$\mathcal{L}[\Phi^i]$$. Then if $$M$$ has some sort of boundary $$\partial M$$ it may happen that $$\delta I[\Phi^i]$$ has boundary terms contributing to it.

Now, in this paper the authors say that such boundary terms make the variational principle ill-defined (c.f. page 61):

As stated by Regge and Teitelboim, the action must posssess well defined functional derivatives: this must be of the form $$\delta I[\phi]=\int(\text{something})\delta \phi$$ with no extra boundary terms spoiling the derivative. The action must be differentiable in order for the extremum principle to make sense.

This is also alluded to in the WP page about the Gibbons-Hawking-York term in gravity:

The Einstein–Hilbert action is the basis for the most elementary variational principle from which the field equations of general relativity can be defined. However, the use of the Einstein–Hilbert action is appropriate only when the underlying spacetime manifold $${\mathcal {M}}$$ is closed, i.e., a manifold which is both compact and without boundary. In the event that the manifold has a boundary $$\partial\mathcal{M}$$, the action should be supplemented by a boundary term so that the variational principle is well-defined.

The boundary term alluded to above is introduced exactly to cancel one boundary term appearing when one varies the Einstein-Hilbert action. So again I take this as saying that if the variation of the EH action had such boundary term the variational principle wouldn't be well-defined.

Now, although this seems such a basic thing I must confess I still didn't get it:

1. Regarding the discussion in the linked paper, by repeated application of the Liebnitz rule, the variation of the Lagrangian density $$\cal L$$ always may be written as $${\delta \cal L} = E_i\delta \Phi^i +d\Theta\tag{3},$$ where $$E_i$$ are the equations of motion and $$\Theta$$ is the presympletic potential. The action thus is of the form $$\delta I[\Phi^i]=\int_M E_i \delta \Phi^i + \int_{\partial M}\Theta\tag{4},$$ I don't see how the presence of $$\Theta$$ stops us from defining $$E_i$$ as the functional derivatives.

Moreover, for me the most reasonable notion of differentiability for the action is to say that $$\lambda\mapsto I[\Phi^i_\lambda]$$ is a differentiable mapping. I don't see how boundary terms affect this.

So why boundary terms in $$\delta I[\Phi^i]$$ yields ill-defined functional derivatives? And in what sense this makes $$I$$ not differentiable?

2. More importantly, both the paper and the WP page on the GHY term allude to the variational principle being ill-defined if $$\delta I[\Phi^i]$$ contains boundary terms. We have a mapping $$\lambda\mapsto I[\Phi^i_\lambda]$$ and we seek an extremum of such map. I don't see how the fact that $$\delta I[\Phi^i]$$ has boundary terms would make this optimization problem ill-defined.

So why boundary terms make the variational principle ill-defined? In other words, why a well-defined variational principle demands $$\delta I[\Phi^i]$$ to be of the form $$\delta I[\Phi^i]=\int({\text{something}})\delta \Phi^i$$ as the authors of the paper seem to claim?

If we have non-vanishing boundary terms, then the map $$\lambda \mapsto I[\Phi_\lambda^i]$$ is not differentiable in the following sense. Using somewhat less sophisticated notation, let

$$I[\Phi^i_\lambda:\eta] := \int_{\mathcal M} \mathcal L\left(\Phi^i_0(x)+\lambda\cdot \eta(x),\partial\Phi_0^i(x)+\lambda\cdot\partial\eta(x)\right) d^4x$$

for some arbitrary differentiable function $$\eta$$. This map is certainly differentiable, and we find that $$\left.\frac{d}{d\lambda}I[\Phi^i_\lambda:\eta]\right|_{\lambda=0} = \int_{\mathcal M}\left(\frac{\partial \mathcal L}{\partial \Phi_0^i}-\partial_\mu \left[\frac{\partial \mathcal L}{\partial(\partial_\mu \Phi_0^i)}\right]\right)\cdot \eta(x) \ d^4x+ \oint_{\partial\mathcal M} n_\mu\frac{\partial \mathcal L}{\partial (\partial_\mu \Phi_0^i)}\eta(x) \ dS$$

where $$n_\mu$$ are the components of the surface normal vector. This is differentiability in the sense of Gateaux. However, this Gateaux derivative generically depends on which $$\eta$$ we choose.

The ultimate goal is to demand that the variation in the action functional vanish regardless of our choice of $$\eta$$. Assuming that the boundary term vanishes, this implies that

$$\int_{\mathcal M}E[\Phi_0^i]\eta(x) d^4x = 0 \implies E[\Phi_0^i] = 0$$

However, in the presence of the boundary terms, no such implication is possible. For any particular field configuration, the variation in the action integral becomes

$$\left.\frac{d}{d\lambda}I[\Phi^i_\lambda:\eta]\right|_{\lambda=0} = \int_{\mathcal M} f(x) \eta(x) d^4x + \oint_{\partial \mathcal M} n_\mu g^\mu(x)\eta(x) dS$$

For this to vanish for arbitrary $$\eta$$, either both integrals need to vanish or they need to cancel each other. In the former case, the boundary terms are not present after all, while the latter case doesn't actually work. To see this, imagine that

$$\int_{\mathcal M} f(x) \eta(x) d^4x =- \oint_{\partial \mathcal M} n_\mu g^\mu(x)\eta(x) dS = C \neq 0$$

for some choice of $$\eta$$, and note that we can always add to $$\eta$$ a smooth function which vanishes on the boundary but has support at any region of the bulk that we choose. This would change the first integral but not the second, thus breaking the equality. Consequently, though the two integrals may cancel for some choices of $$\eta$$, they cannot possibly cancel for all choices of $$\eta$$ (again, unless they both vanish in the first place).

Even worse in a certain sense, the presence of the non-vanishing boundary terms implies, for reasons which follow immediately from those above, that the variation can be made to take any value in $$\mathbb R$$ by appropriate scaling of $$\eta$$.

One can think of this as rather analogous to multivariable calculus. The existence of partial (Gateaux) derivatives of some function (the action functional) along any particular direction (for arbitrary choice of $$\eta$$) is not sufficient to guarantee that the map is differentiable. In this case, with an eye toward our ultimate goal of having a vanishing functional derivative which independent of $$\eta$$, we define a functional as differentiable if its Frechet derivative can be put in the form

$$\left.\frac{d}{d\lambda}I[\Phi^i_\lambda:\eta]\right|_{\lambda=0} = \int_{\mathcal M} E[\Phi_0^i] \ \eta(x) d^4x$$

and define its functional derivative to be $$E[\Phi_0^i]$$.

I'd like to make a quick note on your statement

I don't see how the presence of $$\Theta$$ stops us from defining $$E_i$$ as the functional derivatives.

There's a good bit of truth in what you say. Indeed, if all you want is the Euler-Lagrange equations for the field, then you could argue that the correct formal prescription is to vary the action, throw away any boundary terms, and then demand that the variation vanish. It seems a bit inelegant, but it would give you the equations you're looking for.

One runs into problems, however, when one moves to the Hamiltonian framework. Ambiguity in boundary terms leads to ambiguity when trying to define e.g. notions of total energy of a particular spacetime. In the absence of surface terms, the Hamiltonian vanishes for $$g_{ij}, \pi^{ij}$$ which obey the equations of motion; choosing a boundary term amounts to choosing a value for the integral of the Hamiltonian over all of spacetime, and the GHY term yields the ADM energy.

Such boundary terms are apparently also quite important for quantum gravity, but this is an area with which I am wholly unfamiliar, so I cannot possibly comment intelligently on it.

Let me ask something, you say "However, in the presence of the boundary terms, no such implication is possible". If we demand $$\delta I[\Phi_0^i]=0$$ wrt any variation, then in particular this would hold for compactly supported $$\eta(x)$$. This would not imply $$\int_{\mathcal M}E[\Phi_0^i] \eta(x) d^4x = 0$$ for all compactly supported $$\eta(x)$$ and in turn imply $$E[\Phi_0^i]=0$$ even in the presence of boundary terms? What goes wrong here?

It sounds like you are weakening the requirement that the action be stationary under arbitrary variation to the requirement that the action only be stationary under variations with compact support. If you do this, then you get the implication (and therefore the EL equations) back. However, this means that you are shrinking the space of "candidate" field configurations to those which are identical to the initial one at the boundary.

If you are not interested in any kind of time evolution at the boundary, then this is fine; in general, this is too restrictive. One could imagine, for instance, a combination of initial condition and evolution equations which would necessarily change the field at the boundary. Imposing fixed (Dirichlet) boundary conditions in addition to the evolution equations and this particular initial condition would lead to no solutions at all.

To make matters worse, in the particular case of gravity, the Lagrangian density actually contains second derivatives of the metric by way of a total derivative

$$\partial_\mu (h^{\mu\nu} \partial_\nu \Phi_0^i)$$ which is a possibility I did not consider in the work I did above. In this case it follows that the boundary term becomes

$$\oint_{\partial M} n_\mu \big[g^\mu(x) \eta(x) + h^{\mu \nu}(x)\partial_\nu \eta(x)\big] dS$$

In this case, it would not suffice to hold the variation fixed at the boundary - we would also need to hold its derivatives fixed as well. This is unacceptable, as the equations of motion are themselves second-order; fixing both $$\Phi_0^i$$ and $$\partial_\nu \Phi_0^i$$ at the boundary would generically overdetermine the system, except in those serendipitous cases in which $$n_\mu h^{\mu\nu} \rightarrow 0$$.

• Thanks for the great answer ! Let me ask something, you say "However, in the presence of the boundary terms, no such implication is possible". If we demand $\delta I [\Phi_0^i]=0$ wrt any variation, then in particular this would hold for compactly supported $\eta(x)$. This would not imply $$\int_M E[\Phi_0^i]\eta(x)d^4x = 0,$$ for all compactly supported $\eta(x)$ and in turn imply $E[\Phi_0^i] = 0$ even in the presence of boundary terms? What goes wrong here?
– Gold
Apr 8, 2020 at 13:55
Here is one comment. If we adapt OP's definition $$\delta I[\Phi^i]=\int_M E_i \delta \Phi^i + \int_{\partial M}\Theta_i \delta \Phi^i\tag{4},$$
then in order for the bulk-term $$E_i$$ and the boundary-term $$\Theta_i$$ to be uniquely defined, we must for starters impose that they are not differential operators of non-zero order (acting on $$\delta \Phi^i$$) but just functions (i.e. differential operators of zero order), because else we could use tricks a la integration by parts to redistribute what belongs to the bulk and what belongs to the boundary. It turns out for the EH action on a manifold with a boundary, that this is not possible without the GHY boundary term (because of higher spacetime derivatives in the EH action).