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While looking at the metrics of different spacetimes, i came across the "Ellis wormhole", with the following metric:

$$c^2d\tau^2=c^2dt^2-d\sigma^2$$

where

$$d\sigma^2=d\rho^2+(\rho^2+n^2)d\Omega^2$$

I note that the temporal term has a constant coefficient. The Wikipedia article mentions:

There being no gravity in force, an inertial observer (test particle) can sit forever at rest at any point in space, but if set in motion by some disturbance will follow a geodesic of an equatorial cross section at constant speed, as would also a photon. This phenomenon shows that in space-time the curvature of space has nothing to do with gravity (the 'curvature of time’, one could say).

So this metric would not result in any "gravitational effects".

Looking at the Schwarzschild metric:

$$c^2d\tau^2=(1-\frac{r_s}{r})c^2dt^2-(1-\frac{r_s}{r})^{-1}dr^2-r^2(d\theta^2+\sin^2\theta d\phi^2)$$

Here we have a non-constant coeffcient for the first component. And this metric clearly has an attractive effect on particles, e.g. it's geodesics have the tendency towards $r\rightarrow0$.

  1. Does that mean the gravitational effect comes primarily from a "curvature of time" and not from spatial curvature? I assume part of the answer has to do with the motion through time being dominant for all but the fastest particles?

  2. Is the spatial curvature the primary cause of the visual distortion, e.g. the bending of light paths, in these metrics?

  3. I'm getting the picture that temporal curvature primarily affects objects moving fast through time (static and slow objects), and spatial curvature primarily affects objects moving fast through space (photons). Is this a good picture or completely wrong?

  4. If the spacetime around an "Ellis wormhole" is purely spatial, does that mean the faster i move (through space), the more i would feel the attraction and also second order effects like tidal forces?

  5. Are there physical metrics, e.g. valid solutions for the EFE which only have temporal curvature but no spatial curvature? Would such an object behave like a source of gravity, without the gravitational lensing?

  6. If such objects would be valid, would that mean you could pass them unharmed or even unnoticed at high speeds (moving fast through space), but would be ripped to pieces if you are moving slowly (moving fast through time)?

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  • $\begingroup$ I fear splitting them will cause much dupication, since they build on top of each other. However all questions are related so a good answer might answer them all. $\endgroup$ – Gotbread Aug 17 '16 at 20:19
  • $\begingroup$ That's fair enough, I can see your point $\endgroup$ – user108787 Aug 17 '16 at 20:21
  • $\begingroup$ @John Rennie wrote a very nice answer and his example of the FLRW metric in differenct coordinates illustrates the problem with interpreting spatial and temporal coordinates/curvatures. To realy look at them separately and to distinguish between time and space and temporal and spatial curvature I think one needs to look at the problem in the ADM/(3+1) formalism of GR. There you realy have a separation between spatial (intrinsic) and temporal (extrinsic) curvature. $\endgroup$ – N0va Aug 18 '16 at 12:49
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You need to be cautious about treating a time curvature and spatial curvature separately because this split is not observer independent. In some cases a metric can be written in coordinates where the $dt^2$ term is $c^2$ (or unity in geometric units) but this is just a choice of coordinates.

If you take, for example, the FLRW metric then we usually write it as:

$$ ds^2 = -dt^2 + a(t)\left(dx^2 + dy^2 + dz^2\right) $$

where $t$, $x$, $y$ and $z$ are the comoving coordinates. However it can also be written using conformal coordinates as:

$$ ds^2=a(\eta)^2(-d\eta^2+dx^2+dy^2+dz^2) $$

It's the same metric, describing the same spacetime geometry, but in one case the time coordinate looks as though it is curved while in the other case it looks as if it is flat. Both metrics are perfectly good descriptions of the geometry and we choose whichever version happens to be most convenient for our purposes.

But back to your question: the trajectory of a freely falling particle, i.e. its geodesic, is given by the geodesic equation:

$$ \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} = -\Gamma^\alpha_{\,\,\mu\nu}U^\mu U^\nu \tag{1} $$

In this equation $\mathbf x$ is the position $(t,x,y,z)$ of the particle in spacetime, $\mathbf U$ is the four velocity and the symbols $\Gamma^\alpha_{\,\,\mu\nu}$ are the Christoffel symbols that describe the curvature of the spacetime. You can think of this as a kind of equivalent to Newton's second law in that it relates the second derivative of position to the curvature.

Suppose we consider a stationary particle (stationary in our coordinates that is). Since the particle is stationary in space the components of the four velocity $U^x = U^y = U^z = 0$ and only $U^t$ is non-zero. In that case the geodesic equation (1) simplifies to:

$$ \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} = -\Gamma^\alpha_{\,\,tt}U^t U^t \tag{2} $$

Calculating the Christoffel symbols is a huge pain unless you have a copy of Mathematica to hand, but you can usually find them by Googling as indeed is the case for the Ellis wormhole (NB that link is a PDF) and the only non-zero Christoffel symbols are (I'll list them all in case the link above breaks):

$$\begin{align} \Gamma^\rho_{\theta\theta} &= -\rho \\ \Gamma^\rho_{\phi\phi} &= -\rho\sin^2\theta \\ \Gamma^\theta_{\theta \rho} = \Gamma^\theta_{\rho\theta} &= \frac{\rho}{n^2+\rho^2} \\ \Gamma^\theta_{\phi\phi} &= -\sin\theta\cos\theta \\ \Gamma^\phi_{\phi \rho} = \Gamma^\phi_{\rho\phi} &= \frac{\rho}{n^2+\rho^2} \\ \Gamma^\phi_{\phi\theta} &= \Gamma^\phi_{\theta\phi} = cot \theta \end{align}$$

Note that all the symbols $\Gamma^\alpha_{tt}$ are zero, so our geodesic equation (2) becomes:

$$ \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} = 0 $$

Or in other words in the Ellis wormhole a stationary particle remains stationary.

But even this result needs to be treated with some care because you have to understand your coordinates to interpret it. To show this consider the FLRW metric that I referred to above. I won't go through the details but you can do exactly the same calculation for the FLRW metric and reach the same conclusion:

$$ \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} = 0 $$

But remember that in the FLRW metric the coordinates are comoving coordinates, not the coordinates that you or I use when e.g. measuring the distances to distant galaxies, and the comoving coordinates are moving relative to everyday coordinates (which is why distant galaxies are moving and indeed accelerating relative to us). Even when we find that in a particular coordinate system a stationary particle remains stationary, this doesn't mean we would actually observe a stationary object to remain stationary.

(Though as it happens in the Ellis wormhole spacetime you and I would observe that a stationary object remains stationary.)

I think this addresses your questions 1 to 4. As for your questions 5 and 6, as it happens I asked exactly the same question in What makes a coordinate curved? and the answer is that at least two principal curvatures must be non-zero. So you cannot find a geometry/coordinate system where the curvature is only in the time coordinate.

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Consider a local Lorentz frame, $g_{ij} = \mathrm{diag}(1,-1,-1,-1)$. An observer (really a congruence of observers) at rest with respect to this frame has velocity vector $u^i = \delta^i_0$. It experiences no force (geodesic deviation) if it obeys the geodesic equation, which in this case becomes simply $\gamma^i{}_{00} \equiv 0$. From the compatibility of the connection with the metric we know that $\gamma^i{}_{00} \equiv 0$ if and only if $\gamma^0{}_{i0} \equiv 0$, and from the first Cartan equation $$ d\omega^i = \omega^j \wedge \gamma^i{}_j = \gamma^i{}_{jk}\omega^j\wedge\omega^k, $$ we know that this is certainly the case if $d\omega^0 \equiv 0$. Here $\gamma^i{}_j$ are the connection forms and $\gamma^i{}_{jk}$ are the components (Ricci rotation coefficients). Given a set of coordinates it is natural to consider an observer to be static if the velocity vector is given by $$ u^\mu = \frac{1}{\sqrt{g_{00}}}\delta^\mu{}_0 $$ (here I let $\mu,\nu,\ldots$ signify coordinate indices). Considering then the static case ($g_{0\mu} \equiv 0$ for all $\mu = 1,2,3$) we find that upon setting $e^\mu_0 = u^\mu$ we have $d\omega^0 \equiv 0$ whenever $g_{00}$ is constant. Such is the reasoning behind the statement that gravitational attraction arises from non-constant $g_{00}$. As you can see it is most definitely a simplification.

As is apparent from the above exposure, any observer not at rest with respect to our local Lorentz frame may experience a force (though the nature of its correspondence to high speeds depends on the exact form of the metric and/or connection forms).

As to your questions regarding the effects on light, it is important to recall that light follows null geodesics. Therefore they will always be affected by the nature of the rotation coefficients $\gamma^i{}_{00}$ but also by at least some other coefficients. It would require a speed greater than that of light (spacelike observer) to escape to effects of $\gamma^i{}_{00}$, but this is clearly unphysical.

Although, as John Rennie links to in his answer, it is meaningless to talk about curvature in one direction, in light of the above considerations we might ponder the case where $\gamma^i{}_{00} = \gamma^0{}_{i0}$ are the only non-zero rotation coefficients. This corresponds concretely to the simplest case of the greater the speed with respect to our frame, the smaller the "curvature effects" on the motion (though as noted above, one would require a speed greater than that of light to escape them altogether). Then $d\omega^i \equiv 0$ for all $i = 1,2,3$. By the second Cartan equation $$ d\gamma^i{}_j = \gamma^k{}_j\wedge\gamma^i{}_k + \frac{1}{2}R^i{}_{jk\ell}\omega^k\wedge\omega^\ell, $$ we immediately find $$ d\gamma^0{}_i = -\gamma^0{}_{i0|j} \omega^0 \wedge \omega^j = R^0{}_{i0j} \omega^0 \wedge \omega^j $$ to give the only (potentially) non-zero curvature components, up to symmetries. Note that we take $i,j \neq 0$, whence in particular it follows that the Ricci tensor is zero if and only if the Riemann tensor is. Therefore we can at least conclude that such solutions cannot be vacuum, and therefore cannot describe the exterior of any object.

EDIT: In fact, I was a bit lazy in concluding the above. Making the contractions, and ignoring any cosmological constant, we find that the Einstein field equations yield $T_{0i} = 0$ for all $i$, whence any (non-flat) solution must violate the dominant energy condition. Thus we can further conclude that such a solution is unphysical, since there are timelike observers that observe energy to flow faster than the speed of light, i.e. timelike vectors $v^i$ such that $T_i{}^jv^i$ is spacelike (namely all observers not at rest with respect to our frame).

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