2
$\begingroup$

Ellis Wormholes are a enticing metric due to their lack of serious mass, however, I can find almost no information about them other than a few basic values.

What are their travel times? Travel stresses? Flat space requirements? I will appreciate anyone who can link me to more information about these metrics (that are not locked behind a paywall). Thank you.

| cite | improve this question | | | | |
$\endgroup$
2
$\begingroup$

A recent (2019) Arxiv submission by Hyat Huang and Jinbo Yang, https://arxiv.org/abs/1909.04603, generalizes the Ellis Wormhole to one with charge. AFAI can see, it is a proper treatment and there are many references.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ Thank you! Nice and too the point, from this the equations for travel time should be doable. $\endgroup$ – MrKred Apr 7 at 11:16
0
$\begingroup$

I'll base my answer on the first question: your title.

Could Ellis Wormholes be used for Interstellar Travel?

The short answer is no, due to the physical content of standard (well-stablished) General Relativity. But let's see more deeply about this (boring) fact.

I) General Relativity in a (planck-lenght) nutshell:

So, gravity is a natural interaction which receives a well-stablish, non-quantum, description in a form of a tensor equation known as Einstein Field Equations (EFE): $$ G_{\mu\nu} =: R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu} = \frac{8\pi G}{c^{4}} T_{\mu\nu} \tag{1}$$

This equation tell's you all about the relationship between the geometry (The left-hand side (LHS) $G_{\mu\nu}$) and energy content ((The right-hand side (RHS) $T_{\mu\nu}$) of a given spacetime, i.e., given a matter content this equation can give you: Dynamical Equations.

II) Metric tensors and the LHS of EFE

Now, the important fact is to note that the LHS of EFE is all about geometry, because $G_{\mu\nu}$ (the Einstein Tensor) is constructed upon the componentes of a quantity called Riemann tensor, given by:

$$R^{\mu}_{\alpha \beta \gamma } = \frac{\partial \Gamma^{\mu}_{\alpha \gamma}}{\partial x^{\beta}} - \frac{\partial \Gamma^{\mu}_{\alpha \beta}}{\partial x^{\gamma}} + \Gamma^{\mu}_{\sigma \beta} \Gamma^{\sigma}_{\gamma \alpha} - \Gamma^{\mu}_{\sigma \gamma} \Gamma^{\sigma}_{\beta \alpha} \tag{2}$$

This tensor calculates the curvature of a given manifold (spacetime); this tensor tell's you with your spacetime is flat or curved.

Now, the very nature of this tensor is given by the symbols $\Gamma$'s, called Christoffel Symbols or Levi-Civita Connection Coefficients, and they are given by:

$$\Gamma^{\nu}_{\delta \xi} =: \frac{1}{2}g^{\nu \zeta}\Bigg\{ \frac{\partial g_{\nu \delta }}{\partial x^{\xi}}+\frac{\partial g_{\nu \xi }}{\partial x^{\delta}}-\frac{\partial g_{\delta \xi }}{\partial x^{\zeta}} \Bigg\} \tag{3}$$

Well, you can see that all the geometrical content of EFE is given in terms of the $g_{\mu\nu}$, the components the metric tensor. This, mathematical, fact is based upon the metric tensor defined on the Manifold (the spacetime essentially).

So, as you can see, the tensors which defines LHS of EFE are the very metric tensor, which carries the physical notion of "gravitational potential" and the Ricci Tensor $R_{\mu\nu}$, which is a contraction of Riemann tensor $[1]$, hence carries the information of the curvature of spacetime. To be precise, the Einstein tensor $G_{\mu\nu}$ isn't too straightfoward to obtain,you have to construct it under specific conditions $[2]$.

As you can, also, see these geometrical tensors gives you the precise notion of the interpretation of gravity as curvature of spacetime.

III) Energy Conditions

So, knowing about the arbitrarity of metric tensors, you can in fact propose anyone. You can do that because metric tensors are the solutions of EFE. You are only constrained by Riemmanian geometry, Levi-Civita connections, Lorentizian signature and metric tensor properties. Then wormholes are particular solutions which have no problem, a priori, with EFE.

The fact is; you can propose an metric tensor (then calculate the Einstein Tensors and so on...) but when you propose a energy-momentum tensor $T_{\mu\nu}$ (i.e. the source of energy which are producing the curvature) you have to be consistent with obeservable and reasonable (known) matter fields.

This ideia of "reasonable matter" lies on constrains imposed on energy tensor $T_{\mu\nu}$, called Energy Conditions $[3]$. For example: The Weak Energy Condition (WEC):

$$ \rho= T_{\mu\nu}v^{\mu}v^{\nu}\geq 0 \tag{4}$$

states that the matter density observed is always positive.

Wormholes tends to have "difficulties" on these type of analysis, for example, in Ellis wormhole the energy density is given by:

$$\rho = -\frac{c^{8}n^2}{8\pi G (n^2+\rho^2)} \tag{5}$$

which violates the WEC.

IV) Conclusion

Spacetimes which have "energy condition pathologies" can be interpreted (IN STANDARD GENERAL RELATIVITY) as not suitable for real gravitational physics. That's the fate for Ellis Wormhole: we need an Exotic Matter to sustain the geometry, therefore we cannot use for interestellar travel.

$$* * *$$

$[1]$ https://mathworld.wolfram.com/RicciCurvatureTensor.html

$[2]$ https://www.einsteinrelativelyeasy.com/index.php/general-relativity/80-einstein-s-equations

$[3]$ https://en.wikipedia.org/wiki/Energy_condition

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ Well, I was implying that if Exotic Matter or Energy could be obtained, could an Ellis Wormhole be used for Interstellar Travel. I am too much interested in wormhole physics to not be aware of that downside. $\endgroup$ – MrKred Apr 7 at 11:11

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.