# Invariant interval and gravity

Gravity is inversely proportional to the distance between objects.

Do we use Euclidean distance or the invariant interval for that distance?

Using the invariant interval makes everything a bit more complex: An object that is one lightyear away, is also one year "younger" when we observe it. Does that temporal distance contribute, so that when we calculate trajectories and physical phenomena on a distance - they are different from what's observed locally?

Does this affect observed things such as the trajectory of two equally distant (from us) objects and rotational velocity of distant objects?

An example:

When we look towards the center of the milky way, we're looking about 27 000 years back in time; i.e. the temporal distance is 27000 years, and spatial distance is 27 kly

When we look at the center of the andromeda galaxy, it's about 25 Mly away. When we look at another object in the andromeda galaxy, which is also 25 Mly away - the temporal distance between those objects, relative to us is 0.

So; is the relative temporal distance between objects and the observer incorporated in Einsteins field equations?

• Are you talking about gravity in the context of Newtonian or Einsteinian (i.e. general relativistic) physics? – Danu Sep 2 '14 at 21:29
• @Danu I assume that if you use invariant interval, it de facto becomes relativistic physics? – frodeborli Sep 2 '14 at 21:31
• Are you aware of the distinction between special and general relativity? Special relativity says nothing about gravitation specifically, while general relativity completely redefines it. – Danu Sep 2 '14 at 21:32
• Pedant alert: Gravity has an inversely proportional potential; force scales by $r^{-2}$ – user121330 Sep 2 '14 at 21:34
• I saw this comment thread grow while I wrote my (short) answer. Advice to the OP: Do not think about gravity when trying to understand SR. Gravity is the region of GR, try thinking about electromagnetic phenomena (mainly light being emitted/travelling/whatever) when thinking about SR. – ACuriousMind Sep 2 '14 at 21:38

Gravity (the potential that is, not the force) is only inversely proportional to actual, spatial distance in the Newtonian approximation of GR, also known as the weak field limit, i.e. for bodies (like the earth) that are very far from becoming a black hole. One obtains it e.g. in the Schwarzschild solution of GR for $\frac{r_S}{r}\ll 1$.

But, in general, gravity is not proportional to anything - the presence of matter determines the geometry of spacetime through the Einstein field equations, and "gravity" is the observation that that geometry differs - in whatever form - from the gravity-free flat space that is Minkowski space.

About the temporal distance being included or not included - the Einstein field equations are, as a tensor equation, invariant under arbitrary coordinate transformations, just as every good SR equation is invariant under Lorentz transformations. Both invariances mean that the term temporal distance (which is not invariant) cannot really play a role in these, because, from another frame, that temporal distance will be different (or non-existent, or have a different sign). You can, invariantly, talk about spacetime distances being time-like, light-like or space-like depending on their sign. But, there is nothing to include here - no matter whether SR or GR, we have given up on assigning any special treatment to time: It is just another coordinate on the spacetime manifold (well, in SR, it's the one with the odd sign in the metric, but that's really it), and our equations and techniques apply to everything on that manifold equally.

• I think OP wants you to talk about Newtonian gravity, 'enhanced' by a restriction on the speed of light ;) – Danu Sep 2 '14 at 21:39
• I've tried to reformulate my question; my question is not about gravity - it's about observed effects. – frodeborli Sep 2 '14 at 21:58
• @frodeborli: You've lost me. Yes, what we "observe" is not always what we "calculate" (see e.g. Terrell rotation for the actual observation of the Lorentz contraction), but I don't get what you are asking about the trajectories - what do you mean by different, and why do you think they should be? – ACuriousMind Sep 2 '14 at 22:03
• @ACuriousMind I believe distant objects that have a temporal distance, as observed by us to be 0 should display different behaviour than objects that have a temporal distance of for example 27000 years. I added an example. – frodeborli Sep 2 '14 at 22:14
• @frodeborli: You still haven't told us what difference in behaviour you would expect and why. And it is really not obvious (to me). – ACuriousMind Sep 2 '14 at 22:21

Hopefully this should clarify:

Let's say you're looking at a ball of negative charge 1 lightyear away, but that it disappeared 6 months ago (i.e., you still see it). You still feel the negative charge because the electric field lines from the ball travel precisely at the speed of light.

Edit: Since gravitational field lines also travel at the speed of light, the same effect holds true for the effect of one massive object on another in space.

• I don't feel that this is an answer to my question, and it'll be hard to measure anyway. I'd place a (small) bet on the propagation of gravity occuring at the speed of light, but once the field has been set up, a change in direction and distance may be instantaneous. I guess this could be tested by observing two objects travelling in parallel in space. – frodeborli Sep 2 '14 at 22:47
• I don't feel that this is an answer to my question,[...] It really is an answer to your question. If you believed that the gravitational force depended on the invariant interval, then for an event 1 ly away and 1 year in the past, the interval would be zero. Would the gravitational force then be infinite? Clearly not. – user4552 Sep 3 '14 at 1:17
• The question is then do "gravitational field lines" travel at the speed of light, or are they instantaneous? I have read differing views on the subject but most have suggested the former - but I am open to correction. This is not controversial. Gravitational effects propagate at c. – user4552 Sep 3 '14 at 1:18
• @BenCrowell Thanks, for some reason I couldn't find anywhere that stated that clearly. – Señor O Sep 3 '14 at 14:30