I am trying to derive my way through the paper https://arxiv.org/abs/1603.07706 and I am struggling to understand Eq.(3), where they expand the Bondi metric around Scri+.

Firstly, flat Minkowski space can be written as: $ds^2 = -dt^2 + dx^idx^i$. If we then introduce retarded Bondi coordinates: $$ \begin{cases} u = t - r \\ r^2 = x^ix^i \\ x^i = r\hat{x}^i(z,\bar{z}), \end{cases} $$ where $z,\bar{z}$ are coordinates on the two-sphere, we can derive the Bondi metric: $$ ds^2 = -du^2 - 2dudr + 2r^2\gamma_{z\bar{z}}dz\bar{z}, $$ where $\gamma_{z\bar{z}}$ is the metric of the two-sphere.

I have two questions:

  • How do we derive the Bondi metric using the new coordinates? I seem to always get a few ugly extra derivatives of the unit vector $\hat{x}$.

  • Once we have the metric, how do we perform the expansion around Scri+. I assume is should be of 1/r, but there are no such terms in the metric.

Attempt as solution for Question 1:

  • Using the above transformation we find that $dt^2 = du^2 + dr^2 + 2dudr$.

  • From $x^i = r\hat{x}(z,\bar{z})$ we find that: $dx^i = dr\hat{x}^i + r\left(\frac{\partial\hat{x}^i}{\partial z}dz + \frac{\partial\hat{x}^i}{\partial{ \bar{z}}}d\bar{z}\right)$, and hence: $$dx^idx^i = dr^2 + r^2\left(\left(\frac{\partial\hat{x}^i}{\partial z}\right)^2dz^2 + \left(\frac{\partial\hat{x}^i}{\partial \bar{z}}\right)^2d\bar{z}^2\right) + 2r^2\frac{\partial\hat{x}^i}{\partial z}\frac{\partial\hat{x}^i}{\partial \bar{z}}dzd\bar{z} + 2rdr\hat{x}^i\left(\frac{\partial\hat{x}^i}{\partial z}dz + \frac{\partial\hat{x}^i}{\partial \bar{z}}d\bar{z}\right)$$. Here we can identify $\frac{\partial\hat{x}^i}{\partial z}\frac{\partial\hat{x}^i}{\partial \bar{z}}$ as the metric of the 2-sphere (I think!), which we call $\gamma_{z\bar{z}}$

So now we are left with two extra terms, and I can't think of a way to make them vanish. I hope that we can say that second derivatives of $\hat{x}^i$ vanish, since it is a function parametrising a 2D surface.

Thanks for any help or hints!


1 Answer 1


My answer just completes your attempt for Question 1 (you were on the right track).

We have the following coordinate transformation : $$t = u+r$$ $$x_i = r f_i(z,\overline{z})$$ But the $f^i$'s are constrained by the following. $$r^2 = \sum_{i=1}^3 (x_i)^2 \Rightarrow \sum_{i=1}^3 (f_i)^2 =1 \tag{1}$$

Let's do the grunt work. $$dt = du+dr \Rightarrow dt^2= du^2 + dr^2 + 2dudr$$ $$dx_i = f_i dr + r {{\partial f_i}\over{\partial z}} dz + r{{\partial f_i}\over{\partial \overline{z}}} d\overline{z}$$ $$\Rightarrow \sum_{i=1}^3 {dx_i}^2 = \sum_{i=1}^3 [{f_i}^2 dr^2 + (r {{\partial f_i}\over{\partial z}})^2 dz^2 + (r{{\partial f_i}\over{\partial \overline{z}}})^2 d\overline{z}^2 + 2r^2 {{\partial f_i}\over{\partial z}}{{\partial f_i}\over{\partial \overline{z}}}dz d\overline{z} + 2f^i({{\partial f_i}\over{\partial z}}dz+{{\partial f_i}\over{\partial \overline{z}}}d\overline{z})rdr] \tag{2}$$ When the first term of $(2)$ is summed over $i$, it simply becomes $dr^2$ (using $(1)$).

Likewise when the last term is summed over $i$, we get (using $(1)$), $$\sum_{i=1}^32f_i({{\partial f_i}\over{\partial z}}dz+{{\partial f_i}\over{\partial \overline{z}}}d\overline{z})rdr = \sum_{i=1}^3 (2f_i df_i) rdr= (\sum_{i=1}^3 d({f_i}^2)) rdr = 0$$

Put this all together and we have, $$ds^2 = -dt^2 + \sum_{i=1}^3{dx_i}^2 = -du^2 - dr^2 - 2dudr + dr^2 + r^2\underbrace{\sum_{i=1}^3[( {{\partial f_i}\over{\partial z}})^2 dz^2 + ({{\partial f_i}\over{\partial \overline{z}}})^2 d\overline{z}^2 + 2 {{\partial f_i}\over{\partial z}}{{\partial f_i}\over{\partial \overline{z}}}dz d\overline{z}]}_{\text{This is the round metric on $\mathbb{S}^2$}}$$ If we label the coordinates $(z,\overline{z})$ as $(q^1,q^2)$, we can write the answer more neatly as $$ \Rightarrow ds^2 = -du^2 -2dudr + \sum_{i=1}^2\sum_{j=1}^2 (\gamma_{ij}dq^i dq^j)$$

EDIT : I don't understand Question 2. The metric you wrote in the question is the metric for the Minkowski space, but the expansion about $\mathcal{I}^+$ is asked for Asymptotically Minkowskian spaces.


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