Violations of Energy Conditions for Metric in Relativist's Toolkit

In summary, the metric $$ds^2 = -dt^2+dl^2+r^2(l)d\Omega^2$$ violates all energy conditions at ##l=0## due to the nonzero components of the Ricci tensor and the mixed Einstein tensor. In order for the energy conditions to hold, ##r_0 r’’_0## must be less than or equal to zero, which is not the case at ##l=0## according to the setup of the problem.
  • #1
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TL;DR Summary
I was working through problem #1 under section 2.6 (pg. 54) in “A Relativist’s Toolkit” by Poisson, and I basically just want to make sure I didn’t make a mistake in any of the hundreds of places that I could have. I really need to start making myself comfortable with the computer programs that can do tensor analysis for me, but until then, I’ll have to settle for someone who has the answers to tell me if I’m right or wrong.
Here’s the metric: $$ds^2 = -dt^2+dl^2+r^2(l)d\Omega^2$$where ##r(l)## is minimum at ##l=0## with ##r(0)=r_0## and ##r## approaching ##|l|## asymptotically as ##l## approaches ##\pm \infty##

Part a of the problem seemed pretty straightforward and intuitive, but part b asks which energy conditions are violated at ##l=0##, which required me to fill out a few pages of my notebook, but I finally ended up with the answer that WEC, NEC, and SEC are all violated since $$\rho+p_2 = \rho+p_3 = -\frac{1+r_0 r’’_0}{8\pi r^2_0} <0$$where indices 2 and 3 indicate the angular directions and ##r’’_0 \equiv \frac{d^2 r}{dl^2}## at ##l=0##. Finally, I have that the dominant energy condition is not necessarily violated as long as ##r_0 r’’_0 \geq 1## so that ##\rho \geq |p_i|##.

If someone has the answers and can give me a “look’s good” or a “doesn’t look good”, that’d be much appreciated. Or, you know, if someone wants to work through the problem themselves, I’d appreciate that even more!
 
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  • #2
*ahem*
Pencilvester said:
Summary: I was working through problem #1 under section 2.6 (pg. 54) in “A Relativist’s Toolkit” by Poisson, and I basically just want to make sure I didn’t make a mistake in any of the hundreds of places that I could have. I really need to start making myself comfortable with the computer programs that can do tensor analysis for me, but until then, I’ll have to settle for someone who has the answers to tell me if I’m right or wrong.

Here’s the metric: $$ds^2 = -dt^2+dl^2+r^2(l)d\Omega^2$$where ##r(l)## is minimum at ##l=0## with ##r(0)=r_0## and ##r## approaching ##|l|## asymptotically as ##l## approaches ##\pm \infty##

Part a of the problem seemed pretty straightforward and intuitive, but part b asks which energy conditions are violated at ##l=0##, which required me to fill out a few pages of my notebook, but I finally ended up with the answer that WEC, NEC, and SEC are all violated since $$\rho+p_2 = \rho+p_3 = -\frac{1+r_0 r’’_0}{8\pi r^2_0} <0$$where indices 2 and 3 indicate the angular directions and ##r’’_0 \equiv \frac{d^2 r}{dl^2}## at ##l=0##. Finally, I have that the dominant energy condition is not necessarily violated as long as ##r_0 r’’_0 \geq 1## so that ##\rho \geq |p_i|##.

If someone has the answers and can give me a “look’s good” or a “doesn’t look good”, that’d be much appreciated. Or, you know, if someone wants to work through the problem themselves, I’d appreciate that even more!
 
  • #3
Hm. I fed this into Maxima and got $$G_t{}^t={{2r_0r_0''-1}\over{r_0^2}}$$and$$G_2{}^2={{r_0''}\over{r_0}}$$That almost gives what you've quoted, except I was under the impression we needed to raise both indices on the stress-energy tensor to get ##\rho## and ##p## as the diagonal elements of the stress-energy tensor. And that gives something quite different.

Anyway, it's possible I've made a stupid mistake. It's been a long day and my phone is not the best platform for this. I'll try to recheck and post intermediate steps later, but more likely tomorrow (assuming someone else doesn't chip in first).
 
  • #4
From Maxima - non-zero Christoffel symbols, writing ##d\Omega^2=d\theta^2+\sin^2\theta d\phi^2##:
$$\begin{eqnarray*}
\Gamma^\theta_{l\theta}&=&{{r'}\over{r}}\\
\Gamma^\phi_{l\phi}&=&{{r'}\over{r}}\\
\Gamma^l_{\theta\theta}&=&-rr'\\
\Gamma^\phi_{\theta\phi}&=&{{\cos \theta}\over{\sin \theta}}\\
\Gamma^l_{\phi\phi}&=&-rr'\sin ^2\theta\\
\Gamma^\theta_{\phi\phi}&=&-\cos \theta\sin \theta
\end{eqnarray*}$$...plus ##\Gamma^i_{kj}=\Gamma^i_{jk}##. As per your notation, ##r'=dr/dl##. Non-zero components of the Ricci tensor:
$$\begin{eqnarray*}
R_{ll}&=&-{{2r''}\over{r}}\\
R_{\theta\theta}&=&-rr''-\left(r'\right)^2+1\\
R_{\phi\phi}&=&-rr''\sin ^2\theta-\left(r'\right)^2\sin ^2 \theta+\sin ^2\theta
\end{eqnarray*}$$where ##r''=d^2r/dl^2##. And finally the mixed Einstein tensor:$$\begin{eqnarray*}G_t{}^t&=&{{2rr''+\left(r'\right)^2-1}\over{r^2}}\\
G_l{}^l&=&{{\left(r'\right)^2-1}\over{r^2}}\\
G_\theta{}^\theta&=&{{r''}\over{r}}\\
G_\phi{}^\phi&=&{{r''}\over{r}}\end{eqnarray*}$$
 
Last edited:
  • #5
Ibix said:
$$\begin{eqnarray*}
\Gamma^l_{\theta\theta}&=&-rr'\\
\Gamma^l_{\phi\phi}&=&-rr'\sin ^2\theta\\
\end{eqnarray*}$$
Wow, I missed those two negative signs, and I’m a little surprised a mistake that early didn’t make my stress-energy components completely unrecognizable from what they should have been. But they were wrong enough that I think I was subsequently wrong about at least the dominant energy condition.

Using the correct tensor components, and a basis where ##\left( {\mathbf e_{\alpha}} \right)^\alpha =\sqrt{|g^{\alpha \alpha}|}## (not summed) and##\left( {\mathbf e_{\alpha}} \right)^\beta = 0## thereby satisfying ##g_{\alpha \beta} \left( {\mathbf e_{\mu}} \right)^\alpha \left( {\mathbf e_{\nu}} \right)^\beta = \eta_{\mu \nu}## I get $$\rho + p_1 = -\frac{2r’’_0}{r_0}$$ And in order for ##\rho \geq |p_1|##, ##r_0 r’’_0## must be less than or equal to zero. From the setup of the problem, ##r’’_0## must be positive, and I’m pretty sure ##r_0## is implicitly positive, and so all energy conditions are violated at ##l=0##.

Is this correct? This is the first energy condition problem I’ve worked on, so I just want to make sure I’m doing this right.
 
  • #6
Pencilvester said:
Summary: I was working through problem #1 under section 2.6 (pg. 54) in “A Relativist’s Toolkit” by Poisson, and I basically just want to make sure I didn’t make a mistake in any of the hundreds of places that I could have. I really need to start making myself comfortable with the computer programs that can do tensor analysis for me, but until then, I’ll have to settle for someone who has the answers to tell me if I’m right or wrong.

Here’s the metric: $$ds^2 = -dt^2+dl^2+r^2(l)d\Omega^2$$where ##r(l)## is minimum at ##l=0## with ##r(0)=r_0## and ##r## approaching ##|l|## asymptotically as ##l## approaches ##\pm \infty##

I think the intent here is that the coordinates are ##t, l, \theta, \phi## and that r(l) is an auxiliary function (and not a coordinate)? So that if we write out the line element in full, we have

$$ds^2 = -dt^2 + dl^2 + r^2(l)(d\theta^2 + \sin^2 \theta d\phi^2)$$

We then have the obvious orthonormal basis of one-forms

$$dt \quad dl \quad r(l) d\theta \quad r(l) \sin \theta \, d\phi$$

Feeding the above into my automated program (the input to the program is the list of the coordinates, the basis vectors, and the basis inner product which is diag(-1,1,1,1)) we get:

$$8 \pi \rho = G_{\hat{t}\hat{t}} = \frac{ 1 - 2 \frac{\partial^2 r } {\partial l^2} r - \left( \frac{ \partial r} {\partial l } \right) ^2 } {r^2}$$

$$8 \pi p_{l} = G_{\hat{l}\hat{l}} = \frac{ \left( \frac{\partial r } {\partial l} \right)^2 - 1 }{r^2}$$

$$8 \pi p_{\theta} = 8 \pi p_{\phi} = G_{\hat{\theta}{\theta}} = G_{\hat{\phi}\hat{\phi}} = \frac{ \frac{ \partial^2 r} {\partial l^2 } } {r} $$

Here the "hats" indicate the use of the orthonormal basis, which we defined by giving its dual (the cobasis), and I've used symbolic indices for the components of the Einstein tensor, rather than numeric indices.

For comparison purposes with other posters with different notation, it might be useful to note that

$$G_{\hat{t}\hat{t}} = -G^\hat{t}{}_\hat{t} = -G^t{}_t \quad G_{\hat{l}\hat{l} } = G^\hat{l}{}_\hat{l} = G^l{}_l \quad G_{\hat{\theta}\hat{\theta}} = G^{\hat{\theta}}{}_{\hat{\theta}} = G^{\theta}{}_{\theta}$$

The components of the rank(1,1) Einstein tensors (one upper index, one lower index) are the same in the coordinate basis as the orthonormal basis except for sign. So if there is some confusion about the (very useful) tool of the orthonormal co-basis, one can just look at the rank(1,1) tensor components which have the same magnitude in the coordinate basis as the orthonormal basis, and flip the sign of ##G^t{}_t##.
 
  • #7
pervect said:
Feeding the above into my automated program (the input to the program is the list of the coordinates, the basis vectors, and the basis inner product which is diag(-1,1,1,1)) we get...
So I’m correct in saying all energy conditions are violated at ##l=0## since
Pencilvester said:
I get
$$\rho + p_1 = -\frac{2r’’_0}{r_0}$$And in order for ##\rho \geq |p_1|##, ##r_0 r’’_0## must be less than or equal to zero. From the setup of the problem, ##r’’_0## must be positive, and I’m pretty sure ##r_0## is implicitly positive, and so all energy conditions are violated at ##l=0##.
Yes?
 

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