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Is every spacetime metric physically realizable? I know that given any spacetime metric, you could work out a stress-energy tensor for each position that would result in that metric.

However, I also know that building a wormhole requires negative energy densities, which are probably not possible. If you had some kind of exotic matter with negative mass, then would every metric be possible, or could you get things like requiring energy to come from nowhere or pressurized mass not expanding?

It might help if I give context to this question. I've heard that the Alcubierre drive was just someone making a metric and solving for the stress-energy tensor. From what I've been able to decipher of the math, that certainly looks like it's all they did. I'm wondering if I could make my own crazy spacetime tensors like that, or if they did something I didn't notice where they showed that negative energy density was the only impossible thing.

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  • $\begingroup$ I don't have enough info to write up a full answer... For a wormhole, in essence the negative mass-energy is needed to create a repulsive flavor of gravity to avoid the hole collapsing under its own stresses. It could be used to construct otherwise impossible spacetime curvatures, but it would still have limits I'd assume. Since we can't play around with negative mass, only imagine it, I don't think we'll find an answer soon. $\endgroup$ – Xeren Narcy Mar 2 '15 at 3:37
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    $\begingroup$ We have Einstein's field equations. We don't need to actually test this stuff out. $\endgroup$ – DanielLC Mar 2 '15 at 5:50
  • $\begingroup$ This question might be related physics.stackexchange.com/q/4015 $\endgroup$ – MBN Mar 2 '15 at 9:26
  • $\begingroup$ Should this question read, "Is every space-time metric physically realizable?" $\endgroup$ – innisfree Jul 30 '15 at 11:41
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You never have to make energy come from nowhere, the fact that the Einstein tensor has zero covariant divergence means that you can write any spacetime and the corresponding stress-energy tensor will have zero covariant divergence.

So in short if you write down an arbitrary spacetime, and look at conservation of energy and momentum for the required stress-energy it will not need to violate any conservation laws that we expect normal matter to follow.

And zero divergence is related to the old result that changes in energy (or momentum) are determined only by energy (or momentum respectively) net flowing in or out of a region. So that's something you don't have to worry about.

But negative mass isn't an obvious fix. If $E=+\sqrt{m^2c^4+\vec p^2c^2}$ then a negative mass doesn't necessarily make the energy negative and it is energy density, not mass density, that is a source of curvature.

For exotic spacetimes you need exotic matter or a nontrivial topology. And exotic matter is a euphemism for exotic stress energy tensors. How exotic and how much is then the question.

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  • $\begingroup$ It is not true to say that the energy in a region of space only depends on the energy going in and out of it. An obvious counterexample would be cosmological spacetime. $\endgroup$ – John Davis Jun 28 '15 at 21:27
  • $\begingroup$ I'm not disagreeing with local conservation of energy, but that doesn't translate into conservation of energy for arbitrary regions of space over arbitrary periods of time. It means energy conservation for arbitrarily small regions of space over arbitrarily small periods of time. $\endgroup$ – John Davis Jun 28 '15 at 22:12
  • $\begingroup$ @Timaeus it's already invalid for a non-infinitesimal bounded region. Unlike in flat spacetime, there is no canonical way of adding up the four-momentum at different points into an overall total, since parallel transport is path-dependent. You can get partial results in some special cases but there is no general answer to "how much energy is in this extended region", whether it's bounded or not. $\endgroup$ – Robert Mastragostino Jul 30 '15 at 16:43
  • $\begingroup$ @RobertMastragostino I'm a little surprised to get this comment now when I already updated the answer 8 hours ago to just say that we expect the same degree of conservation for exotic matter as for regular matter, that they both have a zero covariant divergence for the stress-energy. I had already edited out any statement that energy is conserved and just say that a zero divergence is related to the old result of energy conservation. Were you objecting to comments I hadn't deleted yet or did you think I needed to be more explicit about how energy isn't conserved for normal nonexotic matter? $\endgroup$ – Timaeus Jul 30 '15 at 16:54
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The problem with general relativity is it is too general and allows solutions that it would be very difficult to believe are physical.

The first problem is that the metric is defined upon a topology. For example the physically extremely prohibitive condition that a spacetime is empty and with zero cosmological constant is not enough to uniquely define a spacetime, as it still allows a whole host of topologies- some of which have multiple distinct ways in which this condition can be realized.

The second problem is that some spacetimes exhibit extreme pathologies that it is hard to imagine they could represent actual physical situations. For example in some spacetimes that are non-time orientable there is no local distinction between past and future.

Thirdly spacetimes can represent physically unrealizable situations such as averaged negative energy density.

The upshot is that physicists impose a whole host of implicit and explicit conditions to restrict unphysical solutions.

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  • $\begingroup$ I'm talking about solutions that are very difficult to believe are physical. I'm talking about solutions that violate basic conservation laws. $\endgroup$ – DanielLC Mar 3 '15 at 7:25
  • $\begingroup$ My point is that physicists impose conditions on solutions to exclude those which they don't think are physical. There's multiple different conditions which can be imposed and which ones should be imposed will always be subject to debate. Taking the Alcubierre metric as example, it violates various energy conditions, but it doesn't violate the strongest commonly imposed causality condition. $\endgroup$ – John Davis Mar 3 '15 at 8:07
  • $\begingroup$ The method you describe will never lead to a stress-energy tensor that violates conservation laws (@Timaeus pointed this out in his/her answer.) If you have a metric, calculate its Einstein tensor $G_{\mu \nu}$, and set $T_{\mu \nu} = G_{\mu \nu}/(8 \pi G)$, then you'll automatically get $\nabla^\mu T_{\mu \nu} = \nabla^\mu G_{\mu \nu} = 0$ (by the Bianchi identity.) $\endgroup$ – Michael Seifert Jun 28 '15 at 16:38
  • $\begingroup$ @MichaelSeifert energy conditions are not the same as local conservation laws. $\endgroup$ – John Davis Jun 28 '15 at 21:14
  • $\begingroup$ Of course. My comment was responding to @DanielLC's comment about whether solutions generated this way could "violate basic conservation laws." $\endgroup$ – Michael Seifert Jun 29 '15 at 0:02
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It is an extremely complicated problem. Imposing restrictions on the stress energy tensor will give you different possible spacetimes (which spacetimes is also quite a complex problem), but if those restriction apply in all cases is quite hard to prove. The stress energy tensor still has to obey the field equations of the various matter fields that exist, but it is not easy to know what that means for the geometry.

Here's a sample of the various restrictions, and their implication on a stress energy tensor $diag(\rho, p)$ :

  • The Trace Energy Condition (TEC). It's an old one that is not really used anymore, due to being too generally violated. It states that the trace of the stress energy tensor is never negative : $T^\mu_\mu \geq 0$, or $p \leq \frac{\rho}{3}$. It was thought to always be valid until it was shown in the 60's that matter in neutron stars probably violated it (with $p = \rho$). It implies that the Einstein tensor's trace is also positive, of course, but I don't know about its implications, as it is quite old and hasn't been used in almost 50 years. So we can forget about it.
  • The Strong Energy Condition (SEC). If you have a timelike vector $X^\mu$, then $R_{\mu\nu} X^\mu X^\nu \geq 0$, or, equivalently, $T_{\mu\nu} + \frac{1}{2}g_{\mu\nu}T\geq 0$, meaning $\rho + 3p \geq 0$ and $\rho + p \geq 0$.
  • The Dominant Energy Condition (DEC). If you have a timelike vector $X^\mu$, then $T_{\mu\nu} X^\mu X^\nu \geq 0$, and $T_{\mu\nu} X^\mu$ is a future pointing causal vector, or $\rho \geq 0$ and $|p|\leq\rho$.
  • The Weak Energy Condition (WEC). If you have a timelike vector $X^\mu$, then $T_{\mu\nu} X^\mu X^\nu \geq 0$, or $\rho \geq 0$ and $p + \rho \geq 0$.
  • The Null Energy Condition (NEC). The most basic of "classical" energy conditions. If you have a null vector $k^\mu$, then $T_{\mu\nu} k^\mu k^\nu \geq 0$, or $p + \rho \geq 0$.

Many of those definitions are related, and you have the following implications :

$DEC \rightarrow WEC \rightarrow NEC$

$SEC \rightarrow NEC$

Unfortunately, a variety of classical and quantum effects violate them. Simple scalar field theories have been shown to violate the SEC, as well as interacting fermionic theories. The accelerating expansion of the universe also seems to violate it. The weak energy condition is violated by squeezed vacuum states. The NEC is violated by non-minimally coupled scalar fields, superpositions of free states, the Casimir effect and Hawking radiations.

To remedy this situation, averaged version of those theorems were created, that are more difficult to violate. They are of the form

$\int_\gamma T_{\mu\nu} k^\mu k^\nu d\tau \geq 0$

for the Averaged Null Energy Condition (ANEC), for instance. You integrate the values over an entire null geodesic with respect to its parameter. Similar definitions exist for the weak condition (AWEC) and strong condition (ASEC), but over a timelike geodesic. Also among those are quantum inequalities, based on the observation that instances of negative energy tend to be limited in space and time, of the form

$\int dt \langle T_{\mu\nu} X^\mu X^\nu \rangle_\omega g(t) \geq f(t)$

Where $\omega$ is a Hadamard quantum state and $g(t)$ a smooth compact support function. All of those energy conditions are also violated, mostly by Casimir-type effects, which have no variation of energy density.

There has been recently additional, non-linear energy conditions that have been attempted to solve this problem, the so called quantum energy conditions. These include the Flux Energy Condition (FEC), which requires that the flux of energy should be causal :

$(\langle T^{\mu\nu}\rangle V_\nu) (\langle T^{\rho\sigma}\rangle V_\tau) g^{\nu\tau} \geq 0$

the Quantum Flux Energy Condition (QFEC), which I will not write because god damn, which is that the flux shouldn't be too spacelike (weaker condition of the FEC). The Determinant Energy Condition (DETEC), where the determinant of the stress energy tensor is non-negative, or in the case of the quantum version (QDETEC), not too negative. The Trace-of-Square Energy Condition (TOSEC), $\langle T^{\mu\nu}\rangle\langle T_{\mu\nu}\rangle \geq 0$, and its quantum version (QTOSEC), $\geq$ to some bound.

The QTOSEC seems to hold to most weird quantum states so far, so that's about as strong a statement as you can do that will hold for all known phenomenon

A few references :

http://arxiv.org/pdf/gr-qc/0205066.pdf

http://arxiv.org/pdf/1405.0403.pdf

http://arxiv.org/pdf/1306.2076.pdf

http://arxiv.org/pdf/1208.5399v1.pdf

http://arxiv.org/pdf/gr-qc/9410043v1.pdf

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