When we study exotic solutions of Einstein Field Equations (EFE), such as Traversable Wormholes, the elementary modus operandi to extract some analysis is to deal with EFE in the following form:

$$T_{ab} = \frac{1}{8\pi }G_{ab} \tag{1}$$

In other words, we want to search the type of matter which is defined via geometry; we want to know the matter that are generating a particular gravitational field. Therefore, calculate Einstein tensors are just $50\%$ of the job.

For diagonal metric tensors, the conclusions are quite direct. For example, suppose the interior solution of Morris-Thorne wormhole in a local coordinate chart given by $(t,r,\theta,\phi) \equiv (x^{0}, x^{1},x^{2},x^{3})$:

$$ g = -e^{2\Phi(r)} \mathrm{d}t \otimes \mathrm{d}t + \frac{1}{1-\frac{b(r)}{r}} \mathrm{d}r \otimes \mathrm{d}r + r^{2}\mathrm{d}\theta \otimes \mathrm{d}\theta + r^{2}sin^{2}(\theta)\mathrm{d}\phi \otimes \mathrm{d}\phi \tag{2}$$ or, $$ g\Big(\frac{\partial}{\partial x^{a}},\frac{\partial}{\partial x^{b}}\Big):= ds^{2} = -e^{2\Phi(r)} dt^{2} + \frac{1}{1-\frac{b(r)}{r}}dr^{2} + r^{2}d\theta ^{2} + r^{2}sin^{2}(\theta)d\phi ^{2} \tag{3}$$

where $\frac{\partial}{\partial x^{k}}$ are tangent basis vectors.

Now, in almost every analysis we want to simplify the calculations (and physically measure things). To do this we must to translate the tensor quantities into a tetrad frame (a local laboratory) like:

$$\begin{cases}\hat{\textbf{e}}_{(0)} = \frac{1}{\sqrt{-g_{00}}}\frac{\partial}{\partial x^{0}}\\ \hat{\textbf{e}}_{(1)} = \frac{1}{\sqrt{g_{11}}}\frac{\partial}{\partial x^{1}}\\\hat{\textbf{e}}_{(2)} = \frac{1}{\sqrt{g_{22}}}\frac{\partial}{\partial x^{2}}\\\hat{\textbf{e}}_{(3)} = \frac{1}{\sqrt{g_{33}}}\frac{\partial}{\partial x^{3}}\end{cases} \tag{4}$$

and read off the tetrad transformation matrix:

$$ e_{\hat{a}}^{i} = \begin{pmatrix} e^{-\Phi} & 0 & 0 &\\ 0 & (1-b(r)/r)^{1/2} & 0 & 0 \\ 0 & 0 & r^{-1} & 0 \\ 0 & 0 & 0 & (r sin(\theta))^{-1} \end{pmatrix} \tag{5}$$

Then, to transform tensors into the tetrad frame we perform the transformation like:

$$T_{\hat{a}\hat{b}} = e_{\hat{a}}\hspace{0.1mm}^{a}e_{\hat{b}}\hspace{0.1mm}^{b}T_{ab} \tag{6}$$

Knowing that, it is well known that the Einstein tensors of the Morris-Thorne metric are given by:

$$\begin{cases} G_{\hat{0}\hat{0}} = \frac{b'(r)}{r^{2}}\\ G_{\hat{1}\hat{1}} = -\frac{b(r)}{r^{3}} +2\Bigg(1-\frac{b(r)}{r}\Bigg)\frac{\Phi'(r)}{r}\\ G_{\hat{2}\hat{2}} = \Bigg(1-\frac{b(r)}{r}\Bigg)\Bigg[ \Phi''(r) + (\Phi'(r))^{2} - \frac{b'(r)r-b(r)}{2r(r-b(r))}\Phi'(r)-\frac{b'(r)r-b(r)}{2r^{2}(r-b(r))}+\frac{\Phi'(r)}{r}\Bigg] \\ G_{\hat{3}\hat{3}} = G_{\hat{2}\hat{2}} \end{cases} \tag{7}$$

where the prime $f' = \frac{\partial f}{\partial x^{1}}$

With Einstein tensors calculated it's direct to say that for a perfect fluid energy-momentum tensor (EMT):

$$ T_{ab}= (\rho + p)u_{a}u_{b} - pg_{ab} \tag{8}$$ , the exotic matter is given by:

$$\begin{cases} T_{\hat{0}\hat{0}} = \frac{1}{8\pi}\Biggr\{\frac{b'(r)}{r^{2}}\Biggr\}\\ T_{\hat{1}\hat{1}} = \frac{1}{8\pi}\Biggr\{-\frac{b(r)}{r^{3}} +2\Bigg(1-\frac{b(r)}{r}\Bigg)\frac{\Phi'(r)}{r}\Biggr\}\\ T_{\hat{2}\hat{2}} = \frac{1}{8\pi}\Biggr\{\Bigg(1-\frac{b(r)}{r}\Bigg)\Bigg[ \Phi''(r) + (\Phi'(r))^{2} - \frac{b'(r)r-b(r)}{2r(r-b(r))}\Phi'(r)-\frac{b'(r)r-b(r)}{2r^{2}(r-b(r))}+\frac{\Phi'(r)}{r}\Bigg]\Biggr\} \\ T_{\hat{3}\hat{3}} = \frac{1}{8\pi}\Biggr\{G_{\hat{2}\hat{2}}\Biggr\} \end{cases} \tag{9}$$

This direct conclusion occurs due to the diagonal character of metric tensor. Therefore, we can always use (I think) perfect fluid energy-momentum tensors, because they "match" Einstein tensors without off-diagonal terms. Furthermore, in every tetrad the physical interpretation of $T_{\hat{a}\hat{b}}$ is:

$$ T_{\hat{a}\hat{b}} = \mathrm{Diag}[\rho, -\chi, p,p] \implies \tag{10}$$

$$ \begin{cases} T_{\hat{0}\hat{0}} = \rho \hspace{20mm} \mathrm{Energy} \hspace{1mm} \mathrm{Density} \\ T_{\hat{1}\hat{1}} = -\chi \hspace{16mm} \mathrm{Tension} \\ T_{\hat{2}\hat{2}} = p \hspace{20mm} \mathrm{Pressure}\\ T_{\hat{3}\hat{3}} = p \hspace{20mm} \mathrm{Pressure} \end{cases}$$

The great conclusion of this analysis are the so called wormhole structure equations:

$$\begin{cases} \rho = \frac{1}{8\pi}\Biggr\{\frac{b'(r)}{r^{2}}\Biggr\}\\ \chi = -\frac{1}{8\pi}\Biggr\{-\frac{b(r)}{r^{3}} +2\Bigg(1-\frac{b(r)}{r}\Bigg)\frac{\Phi'(r)}{r}\Biggr\}\\ p = \frac{1}{8\pi}\Biggr\{\Bigg(1-\frac{b(r)}{r}\Bigg)\Bigg[ \Phi''(r) + (\Phi'(r))^{2} - \frac{b'(r)r-b(r)}{2r(r-b(r))}\Phi'(r)-\frac{b'(r)r-b(r)}{2r^{2}(r-b(r))}+\frac{\Phi'(r)}{r}\Bigg]\Biggr\} \end{cases} \tag{11}$$

Note that for every given shape function $b(r)$, and redshift function $\Phi(r)$ we have a model of wormhole.


I wish to replicate this methodology for a non-diagonal metric tensor called rotating wormhole. The symbolic equations (in a suitable tetrad frame) are given by:

$$\begin{cases} T_{\hat{0}\hat{0}} = \frac{1}{8\pi}G_{\hat{0}\hat{0}}\\ T_{\hat{1}\hat{1}} = \frac{1}{8\pi}G_{\hat{1}\hat{1}}\\ T_{\hat{2}\hat{2}} = \frac{1}{8\pi}G_{\hat{2}\hat{2}} \\ T_{\hat{3}\hat{3}}= \frac{1}{8\pi}G_{\hat{3}\hat{3}} \\ T_{\hat{0}\hat{3}}= \frac{1}{8\pi}G_{\hat{0}\hat{3}} \end{cases} \tag{12}$$

The problem is therefore the off-diagonal term $G_{\hat{0}\hat{3}}$ which implies a off-diagonal matter term $T_{\hat{0}\hat{3}}$.

I wish to ask finally: which kind of matter tensor (in a tetrad frame) has the form of:

$$T_{\hat{a}\hat{b}} = \begin{pmatrix} T_{\hat{0}\hat{0}} &0&0& T_{\hat{0}\hat{3}}\\0&T_{\hat{1}\hat{1}}&0&0\\0&0&T_{\hat{2}\hat{2}}&0\\0&0&0&T_{\hat{3}\hat{3}}\end{pmatrix} \hspace{10mm}?\tag{13}$$

  • 1
    $\begingroup$ This will never happen. The stress tensor is always symmetric. Furthermore, the systematic method for generating stress tensors is as the functional derivatives of the matter part of the action (see Carrol's book for example). Otherwise you're just writing down whatever you like anyway. $\endgroup$ Commented Jan 27, 2021 at 20:08
  • $\begingroup$ You are saying that there isn't a energy-momentum tensor like $(13)$? Furthermore, how can we manage the fact that EFE are asking for a $T_{\hat{0} \hat{3}}$ term. And yes, I see the problem since T_{ab} = T_{ba}, but again, EFE are asking that term since $G_{\hat{0} \hat{3}}$ isn't zero. $\endgroup$
    – M.N.Raia
    Commented Jan 27, 2021 at 20:27
  • $\begingroup$ If it isn't symmetric, it can't sit on the RHS of the EFE. Furthermore, $G$ is, by definition, symmetric. Depending on your preferred derivation of the field equations, this is a key defining feature. If nothing else, you can write it out in terms of the Ricci tensor and observe it's symmetric by symmetry of the Ricci tensor and metric, both of which are definitional. $\endgroup$ Commented Jan 27, 2021 at 20:49
  • $\begingroup$ However, I will also point out that if whatever source you're working with only wrote (12), they almost certainly meant for the (3,0) component to be implied since symmetry is not optional. $\endgroup$ Commented Jan 27, 2021 at 20:51
  • $\begingroup$ @RichardMyers I will correct. $\endgroup$
    – M.N.Raia
    Commented Jan 27, 2021 at 20:52

1 Answer 1


I think you are making some assumptions which in general are not true. Let's break it down to elementary facts/assumptions:

I (fact): $T_{ab}$ is symmetric, therefore one can always find (at each event) basis vectors such that the components of $T_{ab}$ are diagonal;

II (fact): $g_{ab}$ is symmetric, therefore one can always find (at each event) basis vectors such that the components of $g_{ab}$ are diagonal;

III (assumption, not true in general): the basis vectors which realize II also realize I;

IV (assumption, not true in general): The set of basis which realizes II always contains coordinate basis.

Try to reformulate your argument eliminating assumptions III and IV and see if the doubt persists.


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