For an asymptotically flat space-time, is the asymptotic symmetry group in presence of radiation the same as standard BMS group? If, instead, it is the larger symmetry group (see definition below), is it possible to construct charges for this group? What would the additional charges of this larger symmetry group will physically correspond to?


It is known that in the framework of geometric approach, one can define asymptotic symmetry group as group of conformal motion of the boundary $\mathscr{I}$. The isometry motion of physical space-time $(\mathcal{M},g)$ near $r\to\infty$ manifests as conformal motion in unphysical space-time $(\hat{\mathcal{M}},\hat{g})$, where $\hat{g}_{ab}=\Omega^2g_{ab}$ with $\Omega>0$ in $\mathcal{M}$ and $\Omega|_{\mathscr{I}}=0$. The generators of this motion satisfy (see the derivation here) \begin{equation} (\mathscr{L}_{\chi}\hat{g}_{ab}-2\Omega^{-1}\chi^c\partial_c\Omega \hat{g}_{ab})|_{\mathscr{I}}=0 \end{equation} For asymptotically flat space-time, under suitable choice of gauge (see Bondi-Sachs formalism), it is always possible to express \begin{equation} \hat{g}_{ab}|_{\mathscr{I}}=\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & q_{AB} \end{pmatrix} \end{equation} Only for this simple expression for metric $\hat{g}_{ab}$, the asymptotic form of $\chi$ agrees with BMS symmetry.

$\textbf{Q1}$ Is it always possible to express $\hat{g}_{ab}|_{\mathscr{I}}$ in the above form, in presence of radiation?

Consider the example of Robinson-Trautman space-time, which admits outgoing gravitational radiation and for any initial data, the system will asymptotically approach Schwarzschild solution: \begin{equation} ds^2=\left(\Delta \log P-2r\partial_u\log P-\frac{2m(u)}{r}\right)du^2+2dudr-\frac{2r^2}{P^2}d\zeta d\bar{\zeta} \end{equation} If we naively choose $\Omega=1/r$ and look at $d\hat{s}^2=\Omega^2ds^2$, we find (ignoring the overall minus sign): \begin{equation} \hat{g}_{ab}|_{\mathscr{I}}=\begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & P^{-2} \\ 0 & 0 & P^{-2} & 0 \end{pmatrix} \end{equation} For this metric, we see that gravitational field $\psi_4|_{\mathscr{I}}=-\partial_{\bar{\zeta}}[P^2\partial_{\bar{\zeta}}\partial_u\log P]$. For this non-zero radiative mode, it seems that we can't make any transformation to bring the angular part of $\hat{g}_{ab}$ down to unit sphere metric $q_{AB}$. Thus if we use the geometric approach, we see that $\chi$ does not correspond to the usual BMS generator but can be written as: \begin{equation} \chi^u=f+\mathcal{O}(l^2) \end{equation} \begin{equation} \chi^{l}=\left(\frac{1}{2}\mathscr{D}_Af^A+\frac{1}{2}f\partial_u\log\sqrt{|H|}\right)\;l+\mathcal{O}(l^2) \end{equation} \begin{equation} \chi^A=f^A-WH^{AB}\mathscr{D}_Bf\;l+\mathcal{O}(l^2) \end{equation} where $f$ and $f^A$ satisfies: \begin{equation} \partial_uf=\frac{1}{2}\mathscr{D}_Af^A+\frac{1}{2}f\partial_u\log\sqrt{|H|}-f\partial_u\log|W|-f^A\partial_A\log|W| \end{equation} \begin{equation} \mathscr{D}_{(A}f_{B)}-\frac{1}{2}\mathscr{D}_Cf^CH_{AB}+\frac{1}{2}f(\partial_uH_{AB}-H_{AB}\partial_u\log\sqrt{|H|})=0 \end{equation} For Robinson-Trautman metric, $W=1$ and $H_{AB}$ is the angular part. Note that the solution space is much larger than usual BMS group, as both $(f,f^A)$ are infinite-dimensional.

$\textbf{Q2}:$ Do we need to modify the expression for conserved charge in presence of radiation?

To get the context, imagine I have an isolated system which emits radiation from $u=u_1$ to $u=u_2$. For $u<u_1$ I can choose unit sphere cross-sections, which are also the good cuts ($\sigma|_{\mathscr{I}}=0$) and define BMS vector fields. During radiation these cross-sections will deform, consequently we will have $q_{AB}\to H_{AB}(u,x^C)$. If we stick to these cross-sections for $u>u_2$, the corresponding expression for $\chi$ will no longer be the usual BMS vector field. I suspect that there should be additional charges for this enlarged symmetry group, which is not addressed in the standard treatment (see Wald Zoupas construction or this for twistorial description). In the usual approach, we choose a different set of good cuts after radiation to define BMS charges. I'm not sure if this is justified, as we are missing out a large number of additional charges in this process

  • $\begingroup$ standard BMS group is precisely the asymptotic symmetry group of radiating asymptotically flat spacetimes, obtained for boundary conditions containing non-trivial news function. If the asymptotic symmetry group is different from BMS (maybe extending it) then it should correspond to a different (possibly more generic) boundary conditions. $\endgroup$
    – A.V.S.
    Jan 29, 2023 at 6:13
  • $\begingroup$ @A.V.S. I'm not sure if that's the case. I have seen discussion where we compare the change in BMS charge before and after radiation, but never while the system is radiating. This I mentioned in my comment above. If I have a radiating system it may not (?) satisfy all the boundary conditions which is needed to derive BMS group, like in the case of Robinson Traurman metric $\endgroup$
    – KP99
    Jan 29, 2023 at 8:04
  • $\begingroup$ @A.V.S. I have updated the question $\endgroup$
    – KP99
    Jan 29, 2023 at 12:52

1 Answer 1


For an asymptotically flat space-time, is the asymptotic symmetry group in presence of radiation the same as standard BMS group?

Asymptotic symmetry group (ASG) is defined as a quotient group of allowed nontrivial diffeomorphisms by trivial diffeomorphisms. Here “nontrivial” or “trivial” depends on how such diffeomorphisms act on boundary (asymptotic) data, while “allowed” means that there could be further restrictions to ensure e.g. finiteness of obtained quantities.

So, ASG is not a characteristic of a spacetime but of boundary conditions which a class of spacetimes may satisfy. Standard boundary conditions of the Bondi–Sachs formalism certainly allow for the description of gravitational radiation from an isolated system (and in fact, the search for such description was the motivation for the development of this formalism). So a radiative spacetime can have charges corresponding to BMS symmetries, and since null infinity is permeable to energy flux, those charges are not really “conserved”. This spacetime at the same time can satisfy some other boundary condition and thus can possess another set of quantities corresponding to the symmetries of this other ASG (of course those sets of charges are not independent of one another).

If, instead, it is the larger symmetry group (see definition below), is it possible to construct charges for this group?

This other ASG (corresponding to the new boundary conditions) can extend the BMS group. In literature there are two distinct proposals for such extensions:

  • the Barnich–Troessaert group $(\mathrm{Vir} × \mathrm{Vir}) ⋉ \mathcal{S}$, see this and this;

  • the Campiglia–Laddha group $\mathrm{Diff}(S^2) ⋉ \mathcal{S}$, see this and this.

… where $\mathcal{S}$ is the abelian subgroup of supertranslations.

Some of the charges constructed for those extended ASG turn out to be divergent, so a renormalization process is needed.

Is it always possible to express $\hat{g}_{ab}|_\mathscr{I}$ in the above form, in presence of radiation?

For the Robinson–Trautman space-times that used as example in the question it seems that some of spacetimes in this class do not satisfy standard (Bondi–Sachs) boundary conditions, so the answer is “no”. See e.g. this paper where a gravitational wave emission process of the RT spacetime is interpreted as endowing Schwarzschild black hole with “superboost hair” (superboost being generators of extended BMS group symmetries). In the impulsive limit of such spacetime the metric $q_{AB}$ would be the metric of unit sphere globally for $u<0$ (before emission) but only locally for $u>0$ everywhere except a number of isolated points.

  • $\begingroup$ Really interesting. It seems that Barnich-Troessart group is a special case of $\chi$ I have defined above: I need to impose: $W=1$, $\partial_uH_{AB}=H_{AB}\partial_u\log\sqrt{H}$ and $\partial_uf^A=0$ to get their result. I was thinking of writing a linearized RT space-time by implanting soft hairs for $\chi$ vector-field on Schwarzschild BH , but this has been done. I am wondering if expression for charge and renormalization can be addressed by using a different approach. Thank you for the response! $\endgroup$
    – KP99
    Feb 1, 2023 at 12:12
  • $\begingroup$ A follow up question: Is my reasoning for Q2 justified? My point was that the charge should be calculated on the same family of cross-sections before and after radiation. For the BMS case, both Wald-Zoupas and Twistor construction gives the same result, but I feel that the arguments in twistor construction is not self-consistent. Should I post a separate question for this? $\endgroup$
    – KP99
    Feb 2, 2023 at 11:51
  • $\begingroup$ @KP99: Re your Q2, it seems to me that “good cut” is needed only for things that are dependent not on full BMS symmetry group but on its specific subgroup (like angular momentum w.r.t. pure rotations). If we have a general “BMS charge” then it should not depend on the specific cross-sections (as long as news remains zero on the entirety of the cross-section). $\endgroup$
    – A.V.S.
    Feb 2, 2023 at 16:23
  • $\begingroup$ Ah, I see. I just checked that if $g_{AB}=H_{AB}+c_{AB}l+O(l^2)$, then asymptotic shear $\sigma|_{\mathscr{I}}=m^a\delta l_a=\frac{1}{2}c_{AB}m^Am^B$. So the good cut condition doesn't seem to depend on $H_{AB}$. Which means I can have unit sphere cross-sections (which is needed to define BMS $\chi$) having non-zero $\sigma$ after radiation. So my reasoning in Q2 was not correct $\endgroup$
    – KP99
    Feb 2, 2023 at 17:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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