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
Pencilvester
184
42
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?
 

1. What are energy conditions in relativity?

Energy conditions are a set of mathematical inequalities that describe the behavior of energy and matter in the context of Einstein's theory of general relativity. They are used to determine the physical properties of space-time and the distribution of energy and matter within it.

2. What are the violations of energy conditions?

Violations of energy conditions occur when the mathematical inequalities that describe the behavior of energy and matter in relativity are not satisfied. This can happen in certain situations, such as near black holes or in the early universe, where the energy and matter are extremely dense and the effects of gravity are strong.

3. Why are violations of energy conditions important?

Violations of energy conditions are important because they can provide insights into the fundamental laws of physics and the behavior of space-time. They can also help us understand the extreme conditions in the universe, such as the behavior of black holes and the early stages of the universe.

4. How do scientists detect violations of energy conditions?

Scientists can detect violations of energy conditions through various methods, such as studying the behavior of particles near black holes or observing the expansion of the universe. They can also use mathematical models and simulations to study the effects of extreme conditions on energy and matter.

5. Can energy conditions be violated in everyday life?

No, energy conditions are not violated in everyday life. They only occur in extreme situations, such as near black holes or in the early universe, where the effects of gravity are very strong. In our daily lives, the energy conditions are satisfied and the laws of physics hold true.

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