How Can Energy Density Be Calculated from a Metric in General Relativity?

[utesfan100]

I have never been formally trained in GR, and have a question regarding the basics of how to calculate the energy density from a metric.

This question arises from thought experiments involving a field with a negative energy density. This is important only because I expect the energy density of the metric in question to be negative, but still conform to the EFE.

Suppose one has the following metric, dependent on a length scale of $r_p$ (the Prather radius of course :) ) :

$c^2 d\tau^2=(1-\frac{r_p}{2r})^4 c^2dt^2 - (1-\frac{r_p}{2r})^{-4} dr^2 -r^2d\Omega^2$

Based on how I cooked up this metric, I suspect this metric would be generated by an energy density of:

$\eta(r)=\frac{-c^4r_p^2}{8\pi Gr^4}(1-\frac{r_p}{2r})^{-2}$

1) How would I show this is correct?
2) If this is not correct, how would I find the actual metric for the suggested energy distribution?

 

[pervect]

The way it works is you specify the metric, then calculate the the Riemann tensor, (hard, and recommended to use automation), use the Riemann to find the Ricci tensor and the Ricci scalar via contraction (easier than the above, but will be handled by the same automation so let the automation do it), use both of those two to compute the Einstein tensor (ditto), and that specifies your energy density and (if it's not a dust solution) the pressure.

But this will give you the coordinate energy density. That may or may not be what you want.

An alternate technique involves specifying the metric via an orthonormal basis, which yields a more physical energy denisty.

Are you looking to find the energy density in some volume specified by dr, dtheta, dphi, or are you looking to find the energy density as measured by local clocks and rulers?

Note that the usual volume element you're used to (r^2 sin^2 theta dr dtheta dphi) is NOT going to be the volume element of your metric, and that in addition the energy density is going to be affected by your choice of time units, local clocks being different than coordinate clocks.I suspect there may be some communication difficulties on this point, so some better idea of what you intend to do with your energy density once you get it might prevent some conceptual problems.

 

[utesfan100]

Very good clarifying questions, getting to the heart of my uncertainty. I suspect that answers to these questions will help me find what I am looking for.

The way it works is you specify the metric, then calculate the the Riemann tensor, (hard, and recommended to use automation), use the Riemann to find the Ricci tensor and the Ricci scalar via contraction (easier than the above, but will be handled by the same automation so let the automation do it), use both of those two to compute the Einstein tensor (ditto), and that specifies your energy density and (if it's not a dust solution) the pressure.

Well, from above the metric appears to be:

$$g_{uv}=\begin{pmatrix} (1-\frac{r_p}{2r})^4c^2 & 0 & 0 & 0 \\ 0 & -(1-\frac{r_p}{2r})^{-4} & 0 & 0 \\ 0 & 0 & -r^2 & 0 \\ 0 & 0 & 0 & -r^2sin^2(\theta) \end{pmatrix}$$

Since I don't have automation, I guess I am stuck with the hard way :)

I have found http://scholarworks.umass.edu/cgi/v...r=1#search="Calculating Ricci tensor metric"" that suggests that for a spherically symmetric diagonal metric, such as this, $R_{uv}$ will also be diagonal. 60% done!

Further, $4R_{uu}=(\partial_u ln|g_{uu}|-2\partial_u)\partial_u ln|\frac{g}{g_{uu}}|-\sum \left[ (\partial_u ln|g_{\sigma\sigma}|)^2+(\partial_\sigma ln\frac{|g|}{g_{uu}^2}+2\partial_\sigma)g^{\sigma \sigma}\partial_\sigma g_{uu}\right]$, where the sum is over $\sigma\neq u$. Unfortunately, this looks like alien to me, which is why I am here.

Once I get this, and contract(?) to get R, it appears I should be done with the Tensor arithmetic for question 1.

Either this will match the expected value, or it won't. I expect the normal stress in the radial direction to equal the energy density. I don't see why my thought experiment should expect any lateral stresses, but I would be willing to accept them.

But this will give you the coordinate energy density. That may or may not be what you want

An alternate technique involves specifying the metric via an orthonormal basis, which yields a more physical energy denisty.

Are you looking to find the energy density in some volume specified by dr, dtheta, dphi, or are you looking to find the energy density as measured by local clocks and rulers?

Note that the usual volume element you're used to (r^2 sin^2 theta dr dtheta dphi) is NOT going to be the volume element of your metric, and that in addition the energy density is going to be affected by your choice of time units, local clocks being different than coordinate clocks..

I specifically want the energy density relative to this chart. I understand that the local energy density of a local observer would not match this energy density unless they first calibrated their rulers and clocks to the chart standard references, including the annoyance of the anisotropic length scales.

I suspect there may be some communication difficulties on this point, so some better idea of what you intend to do with your energy density once you get it might prevent some conceptual problems.

I can't tell you what I intend to do with my energy density in this calculation until I actually have a negative energy density field to talk about, now can I? :)

For question 2, it appears to be much harder to get to the metric from the energy distribution. Is this born out in experience?

 

[pervect]

Well, I crank it through and I get for the metric

$$g_{ab}= \left[ \begin {array}{cccc} 1-2\,{<br /> \frac {p}{r}}+3/2\,{\frac {{p}^{2}}{{r}^{2}}}-1/2\,{\frac {{p}^{3}}{{r<br /> }^{3}}}+1/16\,{\frac {{p}^{4}}{{r}^{4}}}&0&0&0\\0&-<br /> \left( 1-1/2\,{\frac {p}{r}} \right) ^{-4}&0&0\\0&0<br /> &-{r}^{2}&0\\0&0&0&-{r}^{2} \left( \sin \left( <br /> \theta \right) \right) ^{2}\end {array} \right]$$

the result for the Einstein tensor G_ab, which is R_ab - (R/2) g_ab, R being the Ricci scalar.

G_ab and is proportional to the stress-energy tensor - I hope you know what that is, or can look it up.

$$G_{ab} = \left[ \begin {array}<br /> {cccc} {\frac {1}{256}}\,{\frac {{p}^{2} \left( 384\,{r}^{6}-1024\,{r}<br /> ^{5}p+1136\,{r}^{4}{p}^{2}-672\,{r}^{3}{p}^{3}+224\,{p}^{4}{r}^{2}-40<br /> \,{p}^{5}r+3\,{p}^{6} \right) }{{r}^{10}}}&0&0&0\\0&<br /> -{\frac {{p}^{2} \left( 24\,{r}^{2}-16\,pr+3\,{p}^{2} \right) }{{r}^{2<br /> } \left( 2\,r-p \right) ^{4}}}&0&0\\0&0&3/8\,{\frac <br /> {{p}^{2} \left( 4\,{r}^{2}-4\,pr+{p}^{2} \right) }{{r}^{4}}}&0<br /> \\0&0&0&3/8\,{\frac { \left( \sin \left( \theta<br /> \right) \right) ^{2}{p}^{2} \left( 4\,{r}^{2}-4\,pr+{p}^{2} \right) <br /> }{{r}^{4}}}\end {array} \right]$$

It's got pressure terms as well as energy terms (the energy term is G_00) and I have no idea if the energy density part of the stress energy tensor is positive or negative. I made c=1, because I'm not used to dealing with it otherwise, and I'm more likely to make an obscure mistake by trying to include it than omit it. And I used p instead of r_p because it was easier to type.

 

[utesfan100]

Well, I crank it through and I get for the metric

...

G_ab and is proportional to the stress-energy tensor - I hope you know what that is, or can look it up.

...

It's got pressure terms as well as energy terms (the energy term is G_00) and I have no idea if the energy density part of the stress energy tensor is positive or negative. I made c=1, because I'm not used to dealing with it otherwise, and I'm more likely to make an obscure mistake by trying to include it than omit it. And I used p instead of r_p because it was easier to type.

Thanks! (Even though the results disprove my hypothisis.) [STRIKE]Are there any open source tools I can download for this type of analysis?[/STRIKE] Is Cadabra a good tool for this type of analysis?

That energy you got looks positive to me. I can appreciate the elegance of geometrized units. It is interesting that the transverse stresses are opposite in magnitude from the normal stress. This suggests a situation like the outward stress normal to the loading force experience by a rod being compressed.

If it is not too much trouble, what happens if the fourth powers of the binomial in the metric are made second powers?

 

[K^2]

Mathematica does the trick rather nicely if you have a set of functions for this sort of stuff.

With second power in place of fourth power.

$$G_{ab} = \left( \begin {array}<br /> {cccc}\frac{(p-2r)^2(p^2-8r^2)}{16 r^6}&0&0&0\\<br /> 0&\frac{-p^2+8r^2}{(p-2r)^2 r^2}&0&0\\<br /> 0&0&-\frac{p^2}{4r^2}&0\\<br /> 0&0&0&-\frac{p^2 sin^2\theta}{4r^2}\end {array} \right)$$

 

[utesfan100]

Mathematica does the trick rather nicely if you have a set of functions for this sort of stuff.

Unfortunately Mathematica is out of my budget range. I have just down loaded Maxima for my windows laptop, and it appears sufficient for this task. (Judging by the fact I was able to verify the results by pervect in a few minutes of playing with it)

It is nice to know that my getting bogged down trying to do tensor arithmetic by hand was due to the fact they are hard to do by hand :)

Question 2) is still unanswered, though I admit it has not been properly framed.

Suppose I have the Einstein tensor:

$$G_{ab} = \left( \begin {array}<br /> {cccc}\frac{p^2}{r^4(1-\frac{p}{2r})}&0&0&0\\<br /> 0&\frac{p^2}{r^4(1-\frac{p}{2r})}&0&0\\<br /> 0&0&0&0\\<br /> 0&0&0&0\end {array} \right)$$

How would I go about getting the metric for this distribution? I don't see where this feature is included in the computer algebra system toolbox. (Then again, I am very new to the toolbox, and have a blind spot front and center)

 

[K^2]

You have to solve a system of coupled non-linear differential equations. You are not very likely to get an analytic solution. There are many ways in which you can obtain a numerical solution. For a tensor like this, I bet Mathematica will do nicely. Sometimes, the only way to get a solution is to write your own solver following some ideas of what you think the solution should be like.