4
$\begingroup$

I've taken to viewing tetrads as a linear transformations of the Minkowskian metric to some curved space. Really though I'm just using this as a device to picture their action (as it could be viewed in the passive or active diffeomorphic sense).

In trying to view it this way, I find it important to consider the metric $g_{\mu\nu}$ to be using the same coordinates as the Minkowskian one $\eta_{ab}$. This is just to ensure that their isn't a hidden coordinate transformation within the tetrad obscuring it's actual action upon the metric.

For example, we typically write the metric tensor using tetrads as:

$$g_{\mu\nu}=e_{\mu}^{a}\eta_{ab}e_{\nu}^{b}$$ Suppose we want to represent the Schwarschild vacuum solution (centered around the origin), this would be typically expressed as:

$$\left[\begin{array}{cccc} -(1-r_{s}/r) & 0 & 0 & 0\\ 0 & (1-r_{s}/r)^{-1} & 0 & 0\\ 0 & 0 & r^{2} & 0\\ 0 & 0 & 0 & r^{2}sin^{2}(\theta) \end{array}\right]=e(\overrightarrow{r})\left[\begin{array}{cccc} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right]e'(\overrightarrow{r})$$

Where $e(\overrightarrow{r})$ is our 4x4 tetrad. There is clearly a cartesian to spherical coordinate transformation lurking within our tetrads. So just for clarity, I'm opting to absorb the coordinate change into our minkowski metric (the reason will be clear soon) as:

$$\left[\begin{array}{cccc} -(1-r_{s}/r) & 0 & 0 & 0\\ 0 & (1-r_{s}/r)^{-1} & 0 & 0\\ 0 & 0 & r^{2} & 0\\ 0 & 0 & 0 & r^{2}sin^{2}(\theta) \end{array}\right]=e(\overrightarrow{r})\left[\begin{array}{cccc} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & r^{2} & 0\\ 0 & 0 & 0 & r^{2}sin^{2}(\theta) \end{array}\right]e'(\overrightarrow{r})$$

Now if we have another identical source centered at the point $\overrightarrow{a}$, which if isolated would be an identical Schwarschild solution I'm tempted to write:

$$g=e(\overrightarrow{r}-\overrightarrow{a})e(\overrightarrow{r})\left[\begin{array}{cccc} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & r^{2} & 0\\ 0 & 0 & 0 & r^{2}sin^{2}(\theta) \end{array}\right]e'(\overrightarrow{r})e'(\overrightarrow{r}-\overrightarrow{a})$$

In another words, I'm just applying a linear transformation (translated to be centered around the point represented by $\overrightarrow{a})$ once again to the metric. (It should be clear now why I had to remove the coordinate change).Then the total tetrad is given by the composition of the two:

$$e=e_{1}e_{2}$$

Can you actually do this? Basically I'm taking a page from quantum mechanics for multiparticle states. Would this work as an approximation does anyone know. I'm sure there's some coupling terms missing? This idea comes about pretty naturally in the linear transformation type of view of tetrads.

Anyway, I thought it could potentially be useful and I hadn't seen tetrads composed like that before. The indices seem a little strange here, but I'm more interested in how physical the metric is.

Basically: can we compose tetrads together to get many body metric solutions?

Addendum:

So if we compose two the two tetrads as mentioned above, we get:

$$g=\begin{array}{cccc} -(1-r_{s}/|\overrightarrow{r}|)(1-r_{s}/|\overrightarrow{r}-\overrightarrow{a})| & 0 & 0 & 0\\ 0 & (1-r_{s}/|\overrightarrow{r}|)^{-1}(1-r_{s}/|\overrightarrow{r}-\overrightarrow{a}|)^{-1} & 0 & 0\\ 0 & 0 & r^{2} & 0\\ 0 & 0 & 0 & r^{2}sin^{2}(\theta) \end{array}$$

Which admittedly looks strange, but when we graph say the $g_{00}$component vs $r$ for instance, we get: g_00 vs r Which is for $r_{s}=0.3$ and $a=1$. The red is for the two tetrads composed, while the green was obtained by simply superimposing two Schwarschild metrics (as Slereah mentioned in the comments). I'd be very curious to know just how good of an approximation this is.

Consider just how easy it is to manipulate a metric like this. We could apply the tetrad to the Minkowski metric

$$g_{scharschild}=e\eta e'$$

Then say translate it (T) away from the origin:

$$\Longrightarrow Te\eta e'T'$$

Lorentz boost it $\Lambda$ in some direction:

$$\Longrightarrow\Lambda Te\eta e'T'\Lambda'$$

add another identical Schwarschild source:$$\Longrightarrow e\Lambda Te\eta e'T'\Lambda e'$$ then decide to return to the origin and original Lorentz frame:

$$g=\left(T^{-1}\Lambda^{-1}e\Lambda T\right)e\eta e'\left(T'\Lambda e\Lambda^{-1}T^{-1}\right)$$

Now our metric has a Schwarschild source at the origin, and another identical source speeding by it in some arbitrary direction determined by our $\Lambda$.

Now all of our transformations are Lie group elements, so if we write them in exponential form, I'd expect higher order terms like:

$$e^{X}e^{Y}=e^{X+Y+\frac{1}{2}[X,Y]+etcetera}$$

Which would probably be the source of any error here? This seems very suited to computer modeling such systems, is this how you guys go about it?

$\endgroup$
  • $\begingroup$ It would be more of a sum to represent multiple sources, and of course that won't work in GR since it is nonlinear. $\endgroup$ – Slereah Dec 18 '18 at 11:06
  • $\begingroup$ @Slereah Honestly, I'm curious enough that I may try and work it out for a simple setup like the above and see what the stress energy tensor corresponds to. Or if we viewed the tetrad as an element of Diff(M) we might get our nonlinear terms appearing as terms in the campbell-baker-hausdorff formula when we compose two elements/tetrads? $\endgroup$ – R. Rankin Dec 18 '18 at 11:12
  • $\begingroup$ @Slereah Preliminary graphing of the resulting metric tensor is very similar to a direct adding of the two sources, but with extra coupling showing up. I'll try and post something later on. $\endgroup$ – R. Rankin Dec 18 '18 at 12:11
  • $\begingroup$ I guess the first thing you should do is plug it into the field equation and find out what you get for stress energy tensor. $\endgroup$ – Andrew Steane Dec 18 '18 at 19:20
  • $\begingroup$ @AndrewSteane Haha yeah I know. I'll try and get to it and post my results unless someone beats me to it. thanks $\endgroup$ – R. Rankin Dec 18 '18 at 19:21
2
$\begingroup$

This Ansatz composition will generally be accurate only at first post-Minkowski order, i.e. for regions of space-times where we can find coordinates such that $$g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu} \,,$$ where any component of the metric deviation $h_{\mu\nu} \ll 1$ and we thus neglect any $\mathcal{O}(h^2)$ terms. An important feature of that theory is that if we have two separate solutions of the linearized Einstein's equation for the metric deviation $h^{(1)}_{\mu\nu}, \, h^{(2)}_{\mu\nu}$, then $h^{(1)}_{\mu\nu} + h^{(2)}_{\mu\nu}$ is also a solution (with sources superposed).

Now consider the fact that any orthonormal tetrad for $g^{(i)}_{\mu\nu} = \eta_{\mu\nu} + h^{(i)}_{\mu\nu}$ ($i=1,2$ labels the two solutions) must be in the form $$e^{(i)a}_\mu = \Lambda^a_b (\delta^b_\mu + \chi^{(i)b}_\mu)$$ where it is easy to show that $2\delta^a_\nu\chi^{(i)b}_\mu\eta_{ab} = h^{(i)}_{\mu\nu}$ and we can simply write $\chi^{(i)b}_\mu = h^{(i)b}_{\;\;\;\;\mu}$. It is now easy to see that $$e^{(1)\mu}_\kappa e^{(2)a}_\mu \eta_{ab} e^{(2)b}_\nu e^{(1)\nu}_\lambda = \eta_{\lambda \kappa} + h^{(1)}_{\lambda \kappa} + h^{(2)}_{\lambda \kappa} + \mathcal{O}(h^2)\,.$$ However, this is as far as it goes.

One symptomatic issue that arises when one superposes two gravitating sources is the fact that the Einstein equations "know" whether the resulting system is equilibrium or not. If it is not, Einstein equations make them in equilibrium by topological defects in the space-time usually called struts or strings that "support" the sources. It is a feature that arises only by the "interaction" of the two original solutions and is certainly not present in any one of them individually. These effects arise already at quadratic post-Minkowski order, and since the "tetrad composition" you propose does not produce these defects, it cannot be valid beyond first post-Minkowski order.

If you want to take a look at some space-times where something like a superposition of sources is possible (usually on the level of some potential), then I recommend to check out the Weyl metrics, or the Majumdar-Papapetrou solutions, especially how they were examined in Hartle & Hawking (1972).

$\endgroup$
  • $\begingroup$ Could you steer me towards literature on the struts and strings you mention? All I can seem to find are references to QFT based topological phase transitions, which are outside the current scope of what I'm looking at. $\endgroup$ – R. Rankin Dec 19 '18 at 3:20
1
$\begingroup$

This is not a direct answer to your question of two sources, but is a comment on your idea of Schwarzschild being a linear transformation of the Minkowski metric. The idea of a gravitational source causing a linear transformation also seems to work for gravitational waves (GWs) which appear to do a linear transformation (strain) to the Minkowski metric. $$ g_{\mu\nu}=e_{\mu}^{a}\eta_{ab}e_{\nu}^{b} $$ $$ \left[\begin{array}{cccc} -1 & 0 & 0 & 0\\ 0 & 1+2h_{+} & 2h_{X} & 0\\ 0 & 2h_{X} & 1-2h_{+} & 0\\ 0 & 0 & 0 & 1 \end{array}\right] = \left[\begin{array}{cccc} -1 & 0 & 0 & 0\\ 0 & 1+h_{+} & h_{X} & 0\\ 0 & h_{X} & 1-h_{+} & 0\\ 0 & 0 & 0 & 1 \end{array}\right] \left[\begin{array}{cccc} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right] \left[\begin{array}{cccc} -1 & 0 & 0 & 0\\ 0 & 1+h_{+} & h_{X} & 0\\ 0 & h_{X} & 1-h_{+} & 0\\ 0 & 0 & 0 & 1 \end{array}\right] $$

Inherent in your idea that the transformation can be active or passive (ie: transform the metric or the far away observer by the inverse transformation), is that the transformation acts on all objects in the effected region (eg: the coordinate vectors $dx^{\nu}$ and photon momentum vectors $k^{\mu}$). The phase change along a photon’s path is $d\phi = g_{\mu\nu} k^{\mu} dx^{\nu}$ transforms like a scaler under the GW transformation and therefore is unchanged by the GW. If $dx^{\nu}$ is the coordinate distance between two mirrors of either arm of a Michelson interferometer, then the final interference pattern will be unchanged by a GW.…contrary to the GW detections claimed by LIGO.

Notice that if the GW transformation was regarded as passive and the inverse done to the far away observer, of course we wouldn’t expect the observer to see the interference pattern to change!

Your neat idea that gravity does a linear Lie Group transformation that acts on all objects in the region of the GW and can be active or passive, I think conflicts with the idea that LIGO can detect GWs.

$\endgroup$
  • $\begingroup$ Thanks for the input Gary! I think I figured out the nonlinear extension of this notion (mentioned in my comment to Void above) Basically making the differential form of the infinitesimal generators covariant derivatives rather than partial ones. Using the Lie group properties, some interesting solutions might be found. $\endgroup$ – R. Rankin Dec 19 '18 at 2:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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