1
$\begingroup$

The BMS symmetry is the product of the Poincare group with an abelian set of translations. In addition supersymmetry gives the momentum operator according to anticommutation of supergenerators. The question here is then whether the BMS symmetry is a signature of supersymmetry or supergravity. The gravitational memory of BMS symmetry is a translation of masses which occur with the passage of a gravitational wave. This may return masses to a configuration different than their original configuration. Is this translation in some ways a manifestation of supersymmetry?

I will give a bit of a heuristic answer to my own question on this for why I think this might be the case.

$\endgroup$
1
  • $\begingroup$ "The question here is then whether the BMS symmetry is a signature of supersymmetry or supergravity" - You make it sound as if this should follow in some way from the two sentences before it, but I can't quite see how. Could you be a bit more explicit? What specific theory are you talking about here, anyway - if you talk about "supersymmetry", you must specify the field content and the number of supercharges. $\endgroup$
    – ACuriousMind
    May 28, 2017 at 12:05

2 Answers 2

1
$\begingroup$

This question-and-answer seems certain to mislead people who read about "supertranslations" and "superrotations" in connection with BMS symmetry, and naturally assume that this already refers to supersymmetry. It does not - it is one of the rare cases in physics where the prefix "super-" does not refer to supersymmetry. (Another is John Wheeler's superspace.) From what I can tell, a supertranslation here means any infinitesimal transformation of the metric which asymptotically looks like a translation; and something similar holds for a superrotation.

Meanwhile, the square of an actual supersymmetry transformation is a translation, so Lawrence's idea seems to be that maybe these translation-like "BMS supertranslations" are also the square of some supersymmetry transformation.

$\endgroup$
1
  • $\begingroup$ I was not thinking of words, and I was not thinking of BMS translations as supertranslations, which indeed they are called. I had forgotten they were called that. I largely have just thought of these as translations without the super. That you gather these are the square of SUSY transformations is pretty spot on. In an S matrix form that might be what is going on. Your comments are helpful. $\endgroup$ May 28, 2017 at 2:44
0
$\begingroup$

The BMS symmetry is the product of the Poincare group with an abelian set of translations. In addition supersymmetry gives the momentum operator according to anticommutation of supergenerators. The question here is then whether the BMS symmetry is a signature of supersymmetry or supergravity. The gravitational memory of BMS symmetry is a translation of masses which occur with the passage of a gravitational wave. This may return masses to a configuration different than their original configuration. Is this translation in some ways a manifestation of supersymmetry?

I will give a bit of a heuristic answer to my own question on this for why I think this might be the case.

The generators of supersymmetry obey the anticommutator relationships $$ \{Q_a,~\bar Q_{b'}\}~=~2\sigma_{ab'}p^\mu. $$ where $p^\mu$ is the momentum operator. The momentum operator is the generator of translations with $U(x,~x_0)~=~e^{-ip(x-x_0}$ acting on a wave function $\psi(x_0)$ as $U(x,~x_0)\psi(x_0)~=~\psi(x)$. We have as well the BMS symmetry which is the semi-direct product of the Poincare group with an abelian group of translations ${\cal B}~=~SO(3,1)\times\mathbb R^4\rtimes{\cal A}$ which respectively are the Lorentz group, translatons and an abelian group of translations. It then seems plausible that the abelian group of translations could have some connection to supersymmetry.

The simplest way to see possibility is that a supergenerator can be thought of as the product of operators for bosons $a,~a^\dagger$ and those for fermions $b,~b^\dagger$ as $$ Q~=~a^\dagger b, Q^\dagger~=~b^\dagger a. $$ This bare-bones form contains all the essential information, and it will obey the rule that $$ \{Q,~Q^\dagger\}~=~a^\dagger a\{b,~b^\dagger\}~=~H. $$ where $\{b,~b^\dagger\}~=~1$ is used. Then for the relativistic case $H~=~\sqrt{p^2~+~m^2}$ and with $m~=~0$ gives the basic anti-commutator result.

Now we turn the attention to the Bondi metric $$ ds^2~=~ -(1~-~2m_B/r)du^2~-~2dudr~+~2\gamma_{z\bar z}r^2dzd\bar z~+~rC_{zz}dz^2~+~rC_{\bar z\bar z}d\bar z^2~+~ D^zC_{zz}dudz~+~D^{\bar z}C_{\bar z\bar z}du\bar z~+~\dots . $$ The simplest part of the Bondi metric is $$ ds^2~=~-du^2~-~2dudr~+~2\gamma_{z\bar z}r^2dzd\bar z, $$ which is just the Minkowski metric for a two sphere with the $z,~\bar z$ coordinates. The term $m_B$ is the mass term, and the mass is the so called Bondi mass and source of mass-energy propagating out to ${\cal I}^+$, where the coordinate $u$ is defined. The terms $C_{zz}$ and $C_{\bar z\bar z}$ determine Weyl curvature terms for gravitational wave propagating out. An Einstein field equation is\cite{key-7} $$ D_{\bar z}^2C_{zz}~-~D_z^2C_{\bar z\bar z}~=~0, $$ which gives the simple solution $C_{zz}~=~-2D_z^2C(z,~\bar z)$. Here $C(z,~\bar z)$ is a scalar potential. The change in this potential is a change in Weyl curvature with the passage of a gravitational wave.

Now consider the S-matrix $S~=~1~+~2\pi iT$ with $SS^\dagger~=~1$. It is not hard to show that $TT^\dagger~+~2\pi i(T~-~T^\dagger)$ $=~0$. The S-matrix related initial an final states as $$ S_{fi}~=~\lim_{t\rightarrow\infty}\langle\psi_f|\psi(t)\rangle~=~\langle\psi_f|S|\psi_i\rangle. $$ The S-matrix can be expressed according to the unitary operator $U(t,~t')~=~exp(-i\int_t'^tH)$. There is then a connection between this form and the T-matrix form with respect to black holes. To consider BMS theory and gravitational memory the black hole emits gravitational radiation as a quantum transition. This is noted in the Penrose diagram here. enter image description here

The circle represents a quantum loop of $O(\hbar)$ and the loop at a radius $\rho~=~c^2/g$ has a circumference $2\pi/g$, $c~=~1$. The unitary matrix for the circle is then $U~=~exp(2\pi H/g)$, where the Euclidean time $it~\rightarrow~\tau$ $=~2\pi/g$. The analogue of $SS^\dagger$ with the T-matrix then becomes the evaluation of $\langle\psi|S$ for the Hamiltonian corresponding to the final state and $S|\psi\rangle$ corresponding to the initial state. In this case the generator of the unitary matrix for the two states correspond to different accelerations and Hamiltonians. This is because the horizon has shifted. In the diagram above the quantum loop transitions into two particles in the regions I and II with the horizon as the blue vertical hyperbola. We then have $$ H H'^\dagger/gg'~+~2\pi(H/g~-~H'^\dagger/g')~=~0. $$ We then recognize, when we firm this notation up with Dirac notation so that $\bar Q~=~\gamma^0Q^\dagger$ that the second term is variation in the $\{Q,~\bar Q\}$. The first term is also the square of the momentum operator $p^2~=~-i\partial_\mu\partial^\mu$. The operator will then on the BMS metric coefficients $C_{zz}$ and $C_{\bar z\bar z}$ will in a covariant form give the Einstein field equation $D^2_{\bar z}C_{zz}~-~D^2_zC_{\bar z\bar z}~=~0$.

$\endgroup$

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.