0
$\begingroup$

I will try and be as concise and clear as possible, but I'm still trying to understand my own question.

Timelike observers can move freely through space, which allows them to set up experiments such as a red shift/blue shift experiment in a Schwarzschild spacetime. If the observers are at different radii stationary from the event horizon, they will experience different relative rates of time given by $$d\tau^2 = -\left(1- \frac{2M}{r}\right)dt^2 + \left(1- \frac{2M}{r}\right)^{-1}dr^2 + r^2 d\Omega^2$$ This is because $g_{00}$ is a function of $r$.

It is not possible, however, for two observers to remain stationary at two different points in time. It is also not possible for a single observer to send a signal from one point in time and receive that signal at the same point in space but a different point in time. This is baked into the causal structure of spacetime, it is unavoidable.

So in the FRW metric, for example, we can set $g_{00} = 1$ because it is possible to normalize away any $g_{00} = f(t)$ such that $g_{00} = 1$. Also known as a lapse function.

I turned to the Landau-Lifschitz pseudotensor for some possible insight into the gravitational potential across time and computed the $t^{00}$ value for the FRW spacetime as $$-6 \frac{\dot{a}^2}{a^2}$$

I'm not sure what to make of this.

To summarize: even when spatial separation does not exist, is there still a gravitational potential? (only across temporal separation, such as past to future)

Improve this question
$\endgroup$
4
  • $\begingroup$ "I'm still trying to understand my own question" - that could be a problem. "It is not possible, however, for two observers to remain stationary at two different points in time" - can you explain what you mean here by "stationary" ? "It is also not possible for a single observer to send a signal from one point in time and receive that signal at the same point in space but a different point in time" - sure it is - the signal could be reflected from some distant object, or it could travel around a loop of wire or fibreoptic cable. $\endgroup$
    – gandalf61
    Commented Aug 13, 2023 at 14:10
  • $\begingroup$ I used "stationary" in the sense that it is not possible for a timelike observer to remain stationary at a spacetime event. They must move along the coordinate time direction, and consequently their proper time will correspondingly move forward. Good point about the loop. I guess I meant something quite simple, just that we can't move freely across time in order to set up an experiment. But I think all of this is just coming back to gravity is the curvature of spacetime, not one or the other. Objects move through spacetime, and even the definition curvature requires >=2 dimensions. $\endgroup$
    – perchlorious
    Commented Aug 13, 2023 at 22:30
  • $\begingroup$ I think you got a factor of 2 wrong: the pseudotensor's tᵗᵗ component should be -3ȧ²/(a²8πG)=-3H²/(8πG), see here and here, which is simply the negative of the regular contravariant energy tensor component Tᵗᵗ (as it should be, since it was constructed to cancel that in order to allow zero energy conservation). $\endgroup$
    – Yukterez
    Commented Aug 14, 2023 at 1:49
  • $\begingroup$ Maybe you got it double because if you regard Λ as a geometric term instead of an energy density you would add Λ gμν to the energy tensor, but since the +Λ gμν term is for -+++ signature where gtt is negative the dark energy would be substracted and Tμν would be 0 in a flat dark energy only universe, so maybe you used De Sitter with constant H instead of general FLRW with a time dependend one and made the ± mistake to get -6H² instead of -3H² $\endgroup$
    – Yukterez
    Commented Aug 14, 2023 at 1:56

1 Answer 1

Reset to default
1
$\begingroup$

The short answer is no, but it depends a bit on how we understand your question.

The way I understand it, you say that if we have some mass in point A, a test mass in point B will feel its gravitational pull. So if this is true when the separation between A and B is spacelike why wouldn't be true when the separation is timelike?

The problem is that to discuss a gravitational force, or a gravitational potential, one has to depart from the covariant formulation of general relativity. In GR itself there is no force, and no potential. Only after you make some choices regarding a coordinate system can the prediction of GR be presented this way. But then you cannot simply go back and claim that time and space are just different directions in spacetime (you cannot truly do that in GR either but that is a different story).

Improve this answer
$\endgroup$
2
  • $\begingroup$ Yes, that was well summarized. Is there no pull from mass A at time A to mass A at time B? I actually mean does mass A get pulled through time by mass A. $\endgroup$
    – perchlorious
    Commented Aug 13, 2023 at 22:36
  • $\begingroup$ I would say no, since "pull" is a non-relativistic concept. If you put a mass in a position in space, you expect it to stay in that position, right? but if you put it in a moment in time, you wouldn't expect it to stay in that moment, right? Moreover the "speed" is constant, i.e. the moment of the mass change by 1 second each second so how would one see any pull? There is some caveat: a clock depends on the metric and in principle you might say that a gravity slowed down time and interpret this as a pull in time. Personally I think this is a wrong. $\endgroup$
    – yaron kedem
    Commented Aug 14, 2023 at 16:14

Not the answer you're looking for? Browse other questions tagged or ask your own question.