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Following up on the questions raised here:

If you take all the matter and energy out of a significantly large volume of spacetime, what you'll be left with is a small chunk of spacetime that - in the assumptions of $\lambda CDM$ - will be Lorentzian. By Lorentzian, we mean that two particles with no mass and at rest with respect to each other and separated by a distance, $\zeta$, will stay separated by $\zeta$ for all time.

I can't reconcile this with the equations of GR. GR has this λ constant which exists on all scales. Whatever λ is - and we really don't know - this property is intrinsic to all spacetime. That is, it can't be removed. The net effect of this intrinsic property is that space grows exponentially with time.

Now put two test particles with no mass in this completely empty, 'local' space. Won't they accelerate away from each other? Reduce the scale and repeat the experiment. The acceleration is the same. How is this spacetime, even on the smallest scale, Lorentzian if test particles act like it's curved?

Qmechanic
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Quark Soup
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2 Answers2

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The point is that "locally flat", "locally Riemannian", "locally Lorentzian", do not mean that all geometric properties tend to become the corresponding ones in flat space when restricting to a sufficiently small neighborhood of a point. Roughly speaking, this is only valid for those properties that depend on the metric coefficients up to the first derivatives computed at the given point. Mathematically speaking this fact corresponds to the existence of normal neighbourhoods of points.

As soon as a property depends on higher derivatives it cannot be cancelled by restricting in an arbitrarily small neighborhood of a point. The typical example is the relative acceleration of "infinitesimally" close geodesics. Notice that it is computed with respect to the affine parameter so that it exists also on a sphere where that parameter has the dimension of space and not time. This acceleration depends on the Riemann tensor (made of second derivatives of the metric).

Finally, physically speaking, the curvature is the relative acceleration of free falling bodies in GR: the Riemann curvature tensor at an event can be completely measured by measuring that relative acceleration of congruences of timelike geodesic at that event.

Valter Moretti
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  • I believe this addresses my thought experiment, but I was wondering if you could be more specific. Could you explicitly state what happens to the two test particles in the thought experiment? If I take measurements of the acceleration between the two test particles in smaller and smaller segments of spacetime (smaller deltas of space, smaller deltas of time), I will see that the acceleration remains the same no matter the scale. How is spacetime, that is otherwise devoid of all matter and energy - locally Lorentzian if I still detect an acceleration? – Quark Soup Jan 24 '21 at 14:05
  • It holds $\frac{D^2\delta u X^a}{ds^2}= {R^a}_{bcd}T^bT^c \delta u X^d$ where $T$ is the tangent vector $T^a=\frac{dx^a}{ds}$ to one of the two geodesics $x=x(s)$. The other geodesic $x'=x'(s)$ is infinitesimally close to it: $\delta u X^a(a) = x'^a(s) - x^a(s)$. You see that you always see relative acceleration $\frac{D^2\delta u X^a}{ds^2}$ if the Riemann tensor $R$ does not vanish provided $\delta u \neq 0$. – Valter Moretti Jan 24 '21 at 14:35
  • However the acceleration does not remain the same. This is obvious, since everything is smooth if the acceleration remained the same for arbitrarily close material points (even with zero rest mass), taking the limit of coincidence you would obtain an acceleration of a point with respect to itself! That makes no sense. Nevertheless you have always relative acceleration for arbitrarily close free falling material points. – Valter Moretti Jan 24 '21 at 14:47
  • It seems to me that you are interested in the relative acceleration of free falling bodies comoving with the universal reference frame of FLRW spacetime. There the distance is measured in the spatial sections of the model whose geodesics are timelike, instead you are referring to light-like geodesics and they cannot be comoving. However the above geodesic deviation equation is valid in all cases. Referring to FLRW models: $X$ must be tangent to the spatial sections and $T$ parallel to the preferred timelike worldlines. – Valter Moretti Jan 24 '21 at 14:55
  • I appreciate the math, but please understand, I'm looking for the answer, in plain English, whether the local geometry of spacetime is Lorentzian everywhere. In the textbooks, this is given as an axiom that isn't worthy of critical analysis. As per your last comment, I'm not looking for comoving coordinates. In this example, there would be no acceleration in comoving coordinates (by definition). I believe I understand what you're saying with the formulas, but a plain answer to my thought experiment would be helpful. – Quark Soup Jan 24 '21 at 15:04
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    No: there is acceleration in comoving coordinates of FLRW spacetime, just because there is an accelerated expansion. However it is not possible to clearly understand these topics without mathematics, because these are technical issues and they must be handled technically. In plain English what one can says is that there could be acceleration between arbitrarily close free falling bodies depending on the fact that the spacetime is curved. Any locally flat notion cannot affect this result. – Valter Moretti Jan 24 '21 at 15:36
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The surface of a sphere is locally Euclidean.

Consider two points on the equator of the Earth maybe a meter apart. Start them on northward trajectories. For a very short time, they will appear to be at constant distance. But by the time they reach 5°N latitude, you will notice that the distance between them is smaller by about half a millimeter. But, crucially, you had to travel some 500 km to see this effect.

This is the difference between local and global properties. The idea of calculus, if you like, is that a curve is locally a straight line. Calculus fails with noisy structures, and you have to invent something called stochastic calculus to patch it, because noise when you zoom in does not look like a straight line. But calculus succeeds on a circle because if you zoom in at a point on the circle, it looks more and more like a straight line.

Differential geometry is just making it clear what sort of calculus you're doing, in the sense of asking, if I zoom in more and more on spacetime, what exactly does it look like? What is my “manifold” equivalent for a “straight line” here? For the sphere, it is a two-dimensional euclidean space, if I look very very closely at the points on the sphere I can pretend that the sphere is made up of a patchwork of these little 2D euclidean spaces, caveat being that I would have to go to an infinitely small mesh to truly cover the sphere with them.

I hope that clarifies a little bit. Just like in the circle we build a curve out of little straight lines, and in the sphere we build curves out of little flat planes, the idea of general relativity is that we can build any big space-time out of little hyperbolically-flat spacetimes.

CR Drost
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  • We all agree that a small enough section of a sphere (or a circle) is flat. Could you please address the problem of two massless test particles in otherwise empty spacetime. Will these particles accelerate away from each other or not? If they do, then how can his spacetime be considered flat (Lorentzian)? If not, then explain at what scale the $\lambda$ is no longer relevant. – Quark Soup Jan 23 '21 at 21:32
  • @GluonSoup It's a limit process. A circle doesn't turn into a polygon (and a sphere doesn't turn into a polyhedron) if you zoom in far enough. But as you zoom in, the small arc you're looking at gets flatter and flatter. It only becomes perfectly flat when you've zoomed in infinitely, and the size of the arc is zero. – PM 2Ring Jan 23 '21 at 23:00
  • @GluonSoup I think what you may be missing is that time is a part of the manifold. As such, asking if the test particles accelerate away from each other is asking a question about world-lines, and it is the same question about the two particles on the equator heading North. They have trajectories described by great circles that intersect at the North Pole so yes they do get closer, but by definition you have canceled out the first order change by wanting them to be moving parallel, and it is a second-order change caused by the curvature of the surface, that they get closer. – CR Drost Jan 23 '21 at 23:16
  • Is there any chance I can get you to answer my questions directly? Is there any limit you can find where these two test particles are not accelerating? – Quark Soup Jan 24 '21 at 18:01