# Derivation of Bondi metric

I was trying to understand the BMS formalism from the beginning. The whole formalism depends on the expression of the Bondi metric. I don't understand how this expression of the metric was derived. Even in the first paper by Bondi (1960), its derivation has not been given. Is there a way to derive it?

This is the reference article.

https://doi.org/10.4249/scholarpedia.33528

The easy way to answer your question is to say that in $$D$$ dimensions there are $$D$$ coordinate choices to be made and the Bondi metric has $$D$$ gauge conditions $$g_{rr} = 0 , \qquad g_{ra} = 0 , \qquad \partial_r \det ( r^{-2} g_{ab} ) = 0 ,$$ where $$a=1,\cdots,D-2$$. Therefore, once can always bring any metric into Bondi form.

To be honest, the above answer is sufficient. However, I'm guessing that what you're really looking for is the physical meaning behind the coordinates $$u$$, $$r$$ and $$x^a$$ so lets do that as well.

Let $$(M,g)$$ be a Lorentzian $$D$$ dimensional spacetime. Let $$x^\mu$$ be generalized coordinates on a local region of $$M$$.

We foliate this spacetime by a family null hypersurfaces $$\Gamma_c$$ described by $$u(x) = c$$. $$u(x)$$ is defined such that it increases towards the future. The outward-pointing null normal on this family of surfaces is $$n_\mu = - \partial_\mu u , \qquad n^\mu n_\mu = 0 . \tag{1}$$ Each of null hypersurfaces $$\Gamma_c$$ are generated by null geodesics and $$n^\mu$$ is the tangent vector to these null geodesics. Let the coordinates $$x^a$$ label the null generators (i.e. $$x^a$$ is fixed along any given null geodesic). Further, let $$r$$ be a parameter along these null geodesics such that $$n^\mu = e^{-2\beta} \frac{\partial x^\mu}{\partial r} . \tag{2}$$ Until this point, the definitions were completely general and are applicable in any spacetime. Note that $$r$$ is an arbitrary choice of parameterization so we can choose it however we want. For instance, one convenient choice might be to set $$\beta = 0$$ since in that case $$r$$ would the affine parameter along the geodesics. This choice was made by Newman and Unti when they studied the BMS group (see here). More generally, the affine parameter is defined by $$d\lambda = e^{2\beta} d r.$$

BMS made a different choice which made reference to the fact that the spacetime is asymptotically flat as follows. We note that expansion of the null geodesics is given by $$\theta = \nabla_\mu n^\mu = \frac{d}{d\lambda} \log A .$$ $$\theta$$ captures the change of the area $$A$$ of the transverse section of $$\Gamma_u$$ as a function of the affine parameter $$\lambda$$.

Now, in flat spacetime the transverse section $$\Gamma_u$$ is simply a sphere of radius $$r$$, so its area is $$A = C r^{D-2}$$ The affine parameter along the null geodesics is $$r$$ so $$\theta = \frac{D-2}{r}.$$ BMS made a coordinate choice that generalizes the equation above to more general metrics, $$\theta = e^{-2\beta} \frac{D-2}{r} . \tag{3}$$

We are now ready to write down the metric in Bondi coordinates $$x^\mu = (u,r,x^a)$$. In this coordinate system, equations (1) and (2) read $$n_\mu = ( - 1 , 0 , 0_a ) , \qquad n^\mu = ( 0 , e^{-2\beta} , 0 , 0^a ) .$$ However, we have $$n^\mu = g^{\mu\nu} n_\nu \implies (0,1,0,0^a) = ( - g^{uu} , - g^{ur} , - g^{ua} )$$ It follows that $$g^{uu} = 0$$, $$g^{ur} = e^{-2\beta}$$ and $$g^{ua} = 0$$. We define $$g^{rr} = U$$ and $$g^{ra} = U^a$$ so the inverse metric in these coordinates takes the form $$g^{\mu\nu} = \begin{pmatrix} 0 & - e^{-2\beta} & 0^b \\ - e^{-2\beta} & U & U^b \\ 0^a & U^a & g^{ab} \end{pmatrix}$$ Inverting this metric, we find that the line element is $$ds^2 = - U du^2 - 2 e^{2\beta} du dr + g_{ab} ( dx^a + U^a du ) ( dx^b + U^b du ) .$$ This is Bondi form of the metric.

Finally, we must also simplify equation (3) in these coordinates. First, we note the property $$\det(g_{\mu\nu}) = e^{2\beta} \det (g_{ab} ) .$$ Then, \begin{align} \theta &= \nabla_\mu n^\mu \\ &= \frac{1}{\sqrt{ - \det(g_{\mu\nu})}} \partial_\mu \left( \sqrt{ - \det(g_{\mu\nu})} n^\mu \right) \\ &= \frac{1}{2} e^{-2\beta} \partial_r \log \det(g_{ab}) \end{align} Equation (3) then implies $$\frac{1}{2} \partial_r \log \det(g_{ab}) = \frac{D-2}{r} \implies \partial_r \det ( r^{-2} g_{ab} ) = 0.$$