As an extreme example, suppose a black hole (BH) were moving with respect to us. Would we not only feel a gravitational pull to where the BH is directly located now but also where it was in the past and where it's going in the future?

Essentially, does gravitational attraction also occur through the time dimension?

  • 4
    $\begingroup$ I'd have to check, but I believe you feel a pull towards the retarded position of the body in question. If memory serves me correctly, there are also solutions with the advanced position that are thrown away on grounds of causality. Such solutions are interesting to examine.... $\endgroup$ – R. Rankin Jun 10 '17 at 0:37
  • 4
    $\begingroup$ In GR, there is no gravitational pull, only spacetime curvature. In Newtonian gravity, 'gravitational pull' is instantaneous, i.e., the pull now is in the direction of the attracting mass now (universal time). $\endgroup$ – Alfred Centauri Jun 10 '17 at 0:59
  • $\begingroup$ You really need retarded positions. Non linearity and/or a shorter geodesic(s) does not alter the fact that field changes propagate with only the speed of light. Whether the speed of light is smaller near high concentrations of energy (like BH) is immaterial to the fact that you need to consider retarded positions. A moving mass will also produce an effect similar to the magnetic field of a moving charge (called "gravitomagnetism"). $\endgroup$ – Mihai B. Jun 12 '17 at 18:27
  • $\begingroup$ So to explain a bit more for what I am getting at. Right now I am 1 meter away from a mass moving perpendicular to the direction I am looking and it is curving spacetime in its current position. At the same moment through, is it curving spacetime at a future position effecting my current spacetime? If you treated time as a distance using Planck time to Planck length conversion (factor of c) I believe you would find the spacetime curving is there however it would be very small considering this large multiplier. $\endgroup$ – Joe Jun 12 '17 at 22:37
  • 2
    $\begingroup$ Joe, no. Gravitational effects can certainly affect the future, as long as it's inside the future light cone of what is causing it. But, that future cannot affect your present position, it would have to travel back in time - Ie, the light cone (appropriately constructed in the spacetime) from that future effect can only happen if causality is possible backwards. No way that can happen. I will modify my answer below, but it would be acausal to do that. $\endgroup$ – Bob Bee Jun 12 '17 at 23:58


Good question, and not clear what the answer is. It is certainly in 'almost all cases' that it is NOT pointing to the future position of the attracting body. Agree with @Alfred. It's a general relativity question, there is no classical field theory like Maxwell's to go by. I don't know the answer but know a couple things you can try. Also, as R. Rankin states, it is more probably pointing in the direction of the retarded solution, i.e. where it was when you go back to it at the speed of light, in the spacetime's light light geodesic.

You can check if you're willing to dig (it might already be solved and shown either way! I have not seen it) or do some thinking and math. You will need to find out in which spatial direction is the acceleration 3 vector pointing by getting the metric solution, and then calculating the timelike geodesic for some orbits, and then getting the 3 vector. But another step: which way the 3 vector is pointing at the test particle location does not tell you at what spatial point it'll intersect the source's position --- you need to transport that vector back. Since it'd be the path of the gravitational effect's propagation, which presumably is at the speed of light through the spacetime, i.e. extend that vector back in the spatial direction as a lightlike geodesic.

But there's a couple catches. First, what is the metric? Well, it's a Schwarzschild or Kerr solution, and then to have that central body move you can make a boost type coordinate transformation, and you would have the metric that an observer at infinity sees. Of course the geodesic is also with respect to that observer, and the pointing also. I do not know if the answer is invariant regardless of observer or frame of reference, if it was you could just leave it in the frame in which the central body is fixed and just get the answer then using solutions and geodesic equations in textbooks. So if you did it in that frame of reference it should be easy, but not sure it is invariant. Since you are dealing with some lightlike extension of a 3 vector, it's not obvious it's invariant.

I do know one case where the answer is almost certainly that it does not point to the body. It is in the Kerr metric, where the central body has rotation. In the region near but outside the horizon there is strong frame dragging, and in fact the test particles are pulled along --- Ie, the geodesics are not directed towards the center. If no rotation it seems to me from symmetry it would have to, but again I am not sure the answer is frame of reference invariant.

The other thing you can do is try a perturbation approach. And there the answer, if you are far enough away, is clearer. The linearized gravitational radiation goes as 1/r at infinity, and it is propagating radially. So any gravitational effect is radial, and I think you could prove and convince yourself that if the central body is in motion, or alternatively your frame of reference is, you will see it pointing radially at the retarded position. Since you are solving the linearized equations around a spherical geometry unperturbed solution, its equivalent to Maxwell's equations in a spherical spacetime. Not certain, but I'd think the answer would be that it should point to the retarded position, if you are far enough away. [there may still be a slight catch here: it's not really the quadrupolar radiation that is propagating, but really the monopole and if needed the Kerr dipole moment, that is moving with the central body. You'd have to prove that it travels radially, and far enough it does, but closer in a dipole pattern may need to be dealt with. But we already know from the Kerr solution, with a dipole moment, that close in the acceleration acquires a circular or axial component, and is not purely radial)

But General Relativity equations are very nonlinear, and the answer in general may just not be as one would get in classical electromagnetism in empty space, i.e., the retarded position and time. With the Kerr solution obviously you can find counterexamples because of frame dragging, not clear what intricacies may come up in spherical geometries. Somebody may be able to get, or already has, the answer with some elegant and simple reasoning.

Added in Edit on 6/12/17 US East Coast

The force points towards the retarded position for any source that is changing speed, ie. for any radiative propagation. For static (or maybe stationary) gravitational fields, it points to the current position. That is what Wikipedia says at https://en.wikipedia.org/wiki/Speed_of_gravity.

1)For a static solution, according to that article.

It is well known that the gravitational field points to the source if both source and (test) particle are in the same coordinate system. If the particle is moved with respect to the source, Wikipedia argues, in the article referenced above, that it’ll still point to the source, as it argues that the metric and thus equivalent force of gravity doesn’t change. That is actually wrong, because in a Lorentz boosted reference frame the 3 acceleration will be Lorentz transformed, and not (generally) the same as the un-boosted reference frame. In fact, one can find the answers for the 3 acceleration in another Wiki article at https://en.wikipedia.org/wiki/Acceleration_(special_relativity). The equation from there is

Equations for Lorentz boosted 3 acceleration, ie, Newtonian force

With a and u the 3 acceleration and velocity, and v the relative velocity from which the Lorentz factor is calculated. It does turn out from the equations above that if the relative velocity v of the two frames is perpendicular to a, then yes, a’ and a are collinear. Just not in generally Yes, in some approximations it can be close, but clearly not the same.

2) For a radiative type propagation due to accelerated, or more exactly, gravitational wave propagation, or anything due to dynamic changes in a non-stationary space time, the first Wikipedia article argues that the speed of gravity is c, with some measurements but still some controversy.

3) A better argument is derived from the known causal structure of spacetime, as elucidated by Penrose, Ellis and Hawking, and others.

For Minkowski spacetime it is clear that nothing can affect something in it’s past, and only inside it’s future light cone in the future. That’s depicted in the Wikipedia article https://en.wikipedia.org/wiki/Minkowski_space#Causal_structure

For general pseudo-Riemannian spacetime in general relativity the same is true. The best way to see this is using Penrose diagrams, which are no more than diagrams where the lightlike geodesics are plotted at plus/minus 45 degrees. It thus looks similar to Minkowski spacetime, expect other geodesics may be quite different. They are useful to analyze black holes, and see how once inside the horizon all orbits go into the singularity.


A Penrose diagram from is shown

from https://en.wikipedia.org/wiki/Penrose_diagram with the diagonal 45 degree lines the lightlike geodesics at all points.

From those diagrams in any spacetime one can see that except for singularity effects, causal orbits stay inside or on the surface of the light cones. Another picture from https://en.wikipedia.org/wiki/Minkowski_space#Causal_structure

A particle outside the future lightcone of an event

Shows an event E, and a particle P, and P’s apparent horizon where E can never affect P. The yellow area below E, in the past light cone of E, would include all events that can affect E. E cannot affect any of those. E cannot affect P in its present position and time, since P is outside it’s future light cone.

4) Gravitational Waves or Dynamic Gravitational Effects

So it is known that linearized gravitational on top of a spacetime travels at locally at speed c. Thus at any localized position it travels on it’s light cone, ie, on the Penrose diagram plus/minus 45 degree lines.


a) the gravitational effect of the source, if it is radiative (ie, dynamic), will only affect events inside or on its future light cone. So certainly it can affect the gravitational field in the future, on or inside that light cone.

b) But it cannot affect the test particle, which is then at any time in the past light cone of any future gravitational effects. The future cannot affect the past.

c)Can it affect, ie, cause an acceleration on the test particle at that instant when there is some dynamic change in the source? No, that dynamic change travels at the speed of light, inside the past light cone of the source at that instant. Thus it cannot point to the current position of the source, but closer to its retarded position. Since the gravitational radiation equations in Minkowski spacetime have the same causal structure as the Maxwell’s equations for radiation, in Minkowki spacetime it’ll be exactly the retarded position. I still cannot prove that it points to the retarded position, mathematically, but I think it should be possible to do so using Penrose and Hawking’s methods. It is clear that it cannot point to a future position nor the current position, for dynamic spacetimes with some gravitational radiation.

  • $\begingroup$ Are there any calculation or a simple example of a BH moving? $\endgroup$ – Brethlosze Jun 10 '17 at 11:10
  • $\begingroup$ There may be, I do not recall. I would have referenced it if I had. Maybe others have seen them, if not as I indicated you'll have to dig. $\endgroup$ – Bob Bee Jun 10 '17 at 19:33
  • $\begingroup$ I dont know where to dig... the sea is too broad for a pair of googgles $\endgroup$ – Brethlosze Jun 10 '17 at 19:35
  • 1
    $\begingroup$ The bottom line is that, in an inertial reference frame with some black hole (resp. charge) moving through it at constant velocity, if you are anywhere on a given constant $t$ plane with velocity $0$, then the force vector points towards the black hole's (resp. charge's) current position coordinatewise on that constant $t$ plane. The constant velocity case has no time lag! You are right that there are velocity dependent forces, though, in a transformed frame, which screws things up when both particles are travelling near the speed of light (suppressed by a factor of $uv/c^2$). $\endgroup$ – user12029 Jun 15 '17 at 0:57
  • 1
    $\begingroup$ It doesn't necessarily point to the retarded position. First, it's not clear to me what "point towards" would quite mean in this case outside of flat spacetime. But even in the sense you are imagining, the radiation of the black hole will scatter off itself and off of spacetime. This nonlinear influence can jostle the test particle around in all sorts of ways. $\endgroup$ – AGML Jun 19 '17 at 1:09

It points towards the present position

If the black hole has been moving at a constant velocity for forever, the force vector points towards its current position. This is the same situation that occurs in electromagnetism, which is elaborated on at length in many books (including Griffiths Electrodynamics and the linked Feynman lectures page). This reeks of "instantaneous communication", but it's a consequence of special relativity which is totally in agreement with special relativity. There is no instantaneous communication.

If you stopped and started the black hole, or wiggled it back and forth, there would be a speed of light delay for that new force/information to reach you. The key here is that the black hole has been moving at a constant velocity for an infinite amount of time in the past, so it has had an infinite amount of time to communicate all the information to you about where it will be for all time.

In such a restricted situation, it's not surprising at all! It's precisely as surprising as the fact that, in the situation of a black hole which has been stationary at the origin for an infinite amount of time, the force vector points towards the origin. There is no causality violation interaction here, and there is none in the above phenomenon. Like AGML says in another answer, "We are influenced by, and only by, the entire trajectory of the black hole in our past lightcone."


(This is my response to extended discussion in the comments! A tl;dr of this section is that I assume the usual definitions of "force" and "simultaneous" and all of the assumptions needed to define them.)

One has to be careful in describing anything in general relativity. On short scales close to a black hole, you have to focus on every nonlinearity involved, and talk about geodesics instead of "gravitational force", as well as worry about how every last detail is defined and how you measure it. On large scales between galaxies, the expansion of the universe becomes relevant. In between these scales, however, it is appropriate to treat the universe as flat like in special relativity. With "flat" space, you get lots of very lovely baggage: forces and velocities make more sense, you can define whether an observer thinks an event is in the future or the past, and you can easily define a black hole moving with constant velocity relative to an observer.

In certain situations, these assumptions aren't good to make.

In situations such as black holes and other point-like masses travelling at constant velocity through a galaxy or solar system, these assumptions are perfect and give you the unequivocal answer that the force vector points towards the "present" position.

"present" means something with a fixed $t$ coordinate. ie, something on the plane of simultaneity in your frame of reference. This is all a fancy way of saying that in a coordinate system $(t,x,y,z)$, the "present" is all events of the form $(t_0,x,y,z)$ for fixed $t_0$. It's a pretty simple definition of "present", and the complications when you switch reference frames constitute an important fact of special relativity.

So the metric $g_{\mu \nu}$ is equal to the metric of special relativity $\eta_{\mu \nu}$ plus a small deviation $h_{\mu \nu}$. We can apply whatever Lorentz transformations we want to this metric, and this ends up being (to lowest approximation) Newtonian gravity inside special relativity.

Working with that terminology and under those assumptions, the force of attraction points towards the black hole's present position.

The proof

I'll work in two dimensions and just apply Newtonian gravity and special relativity (equivalently: the Coulomb force of electromagnetism). We have to use the concept of the four-velocity and four-force in order to transform forces correctly. Note that I assume $c=1$ through the calculations.

You have a body of mass $M$ that moves forwards with velocity $u$. At time $t$ this has spacetime vector ${\bf x}_M=(t,u t,0)$ (a time, x position, y position vector). Denote $\gamma=\frac{1}{\sqrt{1-u^2}}$ (where I've set $c=1$).

We have a body of mass $m$ at position ${\bf x}_m=(t,x,y)$ and we want to figure out the force that acts upon it. It is stationary and non-moving in this frame (to simplify things!), and so it has four-velocity ${\bf v}_m=(1,0,0)$.

Lorentz transform to the new frame: ${\bf x}_m'=(\gamma t-\gamma u x,-\gamma u t+\gamma x,y)$, ${\bf v}_m'=(\gamma,-\gamma u)$. Note that under this transformation, ${\bf x}_M'=(\gamma t-\gamma u^2 t,-\gamma u t+\gamma u t,0)=(t/\gamma,0,0)$. But we know what the force acting on the particle should be in this frame: it's just the Newtonian force pointing to the origin. This gives the force vector $$\vec{f}'=\frac{-G M m}{(x'^2+y'^2)^{3/2}}(x',y')$$

The four-force in this frame, which I'll denote $\bf F'$, is written ${\bf F'}=(\gamma \vec{f}'\cdot \vec{v}',\gamma \vec{f}')$. We find:

$${\bf F'}=-\frac{G M m \gamma}{(x'^2+y'^2)^{3/2}}(-u x',x',y')$$

Now Lorentz transform back:

\begin{align*} {\bf F'}&=-\frac{G M m \gamma}{(x'^2+y'^2)^{3/2}}(0,\gamma x'+\gamma u(-u x'),y')\\ &=-\frac{G M m \gamma}{(x'^2+y'^2)^{3/2}}(0,x'/\gamma,y') \end{align*}

We just care about the direction of this. So the regular old vector force is in the direction of $(x'/\gamma,y')$, so in the direction of $(\gamma (x-u t)/\gamma,y)$, so in the direction of $(x-ut,y)$.

The force is in the direction of the current position of the mass.

What else do we get out of that giant computation?

If we didn't assume that the velocity of the particle is zero, we'd get a "magnetic" force! I'm lazy and don't like doing relativistic calculations by hand, so I wrote a mathematica script to find and simplify things, attached at end of post. I am probably just using a bulldozer to crack a walnut. I find that, if the velocity is not zero, the vector force is: $$\vec{f}=-\frac{G M m}{((x-tu)^2 \gamma^2+y^2)^{3/2}}\left(x-tu+u v_y y,y-u v_x y\right)$$

Note that if we throw the speed of light back in, the velocity dependent forces go like $\frac{u v_x}{c^2}$. So since this is suppressed by a factor of $c^2$, we expect it to be very small, hard to measure, and only relevant for microscopic objects which we can accelerate to relativistic speeds easily.

If you tack on a third dimension you can write the formula as a cross product. If you define $$\vec{E}=-\frac{G M}{((x-tu)^2 \gamma^2+y^2)^{3/2}}\left(x-tu,y,0\right)$$ $$\vec{B}=-\frac{G M}{((x-tu)^2 \gamma^2+y^2)^{3/2}}\left(0,0,uy\right)$$ then the force law that I wrote down is $$\vec{f}=m\vec{E}+\vec{v}\times \vec{B}$$ which is recognizable as an analogue of the Lorentz force law

Source code

Finding the vector force in the case where the velocity isn't zero:

(* lorentz transforms from unprimed to prime. invlorentz transforms back. *)
lorentz={{gamma,-u gamma,0},{-u gamma, gamma, 0},{0,0,1}};
invlorentz={{gamma,u gamma,0},{u gamma, gamma, 0},{0,0,1}};

(* unprimed frame. nu= 1/Sqrt[1-vx^2-vy^2] *)

(* primed frame. alpha = 1/Sqrt[1-vx'^2-vy'^2] *)
fourvelocityprime=lorentz.fourvelocity;(* the first component of the four velocity is the lorentz factor *)
(* divide through by alpha to get the two-dimensional vector velocity *)

(* Calculate the vector force using Newton's law in the primed frame *)
fprime=-G M m/(xprime^2+yprime^2)^(3/2){xprime,yprime};
(* Put the 2d vector force and vector velocity together into the four-force in the primed frame. *)
fourforceprime=Simplify[Join[{alpha fprime.vprime},alpha fprime]];
(* Lorentz transform back into the unprimed frame. Replace gamma and simplify! *)
(* Components 2 and 3 of the four-force are nu times the vector force. *)
  • $\begingroup$ This is the correct answer. Static fields behave this way universally, whether we're talking Gravity, EM, etc. for any field source moving at constant velocity. $\endgroup$ – JPattarini Jun 12 '17 at 17:16
  • 2
    $\begingroup$ "It points toward the present position". Is there no problem with the word "present"? $\endgroup$ – Philip Roe Jun 12 '17 at 17:24
  • 1
    $\begingroup$ @PhilipRoe I assume present to mean "fixed $t$ hyperplane in minkowski space". I'll add that to the answer. $\endgroup$ – user12029 Jun 12 '17 at 17:26
  • $\begingroup$ The present position in what coordinates? Why do you assume the velocity to be "constant"? $\endgroup$ – AGML Jun 18 '17 at 21:05
  • 1
    $\begingroup$ Right. I'm inclined to agree with you if the problem is to describe the influence of a real, very distant black hole upon observers on Earth. But it seems to me that OP has in mind the strong field, since the question is under the GR tag, in which case these things do matter (they are even of quantitative significance during calculations of e.g. inspiral trajectories into supermassive black holes) $\endgroup$ – AGML Jun 19 '17 at 3:24

I'd suggest the following formulation of your question:

Start with some "background" spacetime with metric $g_{ab}$. A pointlike mass $m$ moves through this metric along some arbitrary worldline $\gamma$. The presence of $m$ distorts the spacetime, resulting in some "true" solution $g'_{ab}$. We assume the background spacetime has a mass scale $M>>m$, so that the difference between $g'_{ab}$ and $g_{ab}$ is in some sense "small".

A third mass, $\mu$, moves along a geodesic $y$ of $g'_{ab}$. $\mu$ is small enough that its own gravitational influence may be completely ignored. At constant time in some coordinate system, what is the relationship between $y$ and $\gamma$?

This is a complex question, but the meat of it does seem to me to have a definitive answer:

  • The "effect of $m$ upon $\mu$" can, it seems to me, be viewed as the difference between the four-acceleration of $\mu$ along $y$ and the four-acceleration of the geodesic of $g_{ab}$ that $\mu$ would have travelled along if $m$ were not there. We need $m$ to be small in order for the above to make sense. Call that difference $a_\mathrm{pert}$.
  • You are asking: if we draw a geodesic in $g_{ab}$ initially tangent to $a_\mathrm{pert} (t)$ from $y(t)$ to $\gamma$, does that geodesic intersect $\gamma$ also at $t$ ("does the acceleration of the test mass point towards the location of the black hole in the present"?)? That question is not coordinate-invariant and has no answer.
  • We can still address the spirit of your question. The time-dependence of the Einstein equations can be viewed as a system of hyperbolic PDEs. The influence of a perturbation to spacetime at some event $e(t,x,y,z)$ is therefore confined to the characteristics of the PDEs at $e$, which in the case of GR means the lightcone.
  • In other words, the perturbation to the motion of $\mu$ at time $t$ can depend only on the portion of $\gamma$ which is in the past lightcone of $y(t)$.
  • This allows us to definitively answer part of your question:

NO, we do NOT feel attraction (or any influence) due to where the particle is going in the future.

  • However, we in general will be affected by the entire past history of the particle. This is called the "tail effect" and it is very important to black hole perturbation theory. It encloses a number of mechanisms that can be heuristically understood:

    1. Imagine you are a fish swimming around near a submarine (this is a useful analogy because the Navier-Stokes equations are also hyperbolic, but nonlinear). The submarine moves the water, and this will affect your own motion. This will happen only if you are in the wake of the submarine (i.e. only if the submarine is in your past soundcone). Analogously, only the motion of $m$ within $\mu$'s past lightcone can affect $\mu$.

    2. If the dynamics of the water were totally linear, then the form of the wake will be due to where the submarine was when the part of the wake striking you was emitted - the retarded time of the submarine. Similarly, if gravitation were linear like electrodynamics, we would mostly need to worry only about the retarded time of $m$ to figure out the perturbation to $\mu$'s motion, and not $m$'s entire history.

    3. But water, like gravity, does not behave linearly. The push of the submarine upon the water at its retarded time will in general interact with other pushes that the submarine made at other earlier times. These can form eddies and in general mess things up very badly. Even worse, these effects will also influence the motion of the submarine. To figure out the details you need to do a very complex, confusing calculation of what is called the "self-force". Analogously in gravity, the mass $m$ will emit gravitational radiation etc. This radiation will not simply travel linearly outwards, but will also scatter off itself and off the background spacetime. This will affect nonlinearly the motion of both $m$ and $\mu$. But while the details are very complex, their dependence is still bound to $m$'s past.

The definitive answer:

We are influenced by, and only by, the entire trajectory of the black hole in our past lightcone.

Edit: here is a simplified version of the basic argument.

The effect of the black hole's motion upon the gravitational field can be understood as a perturbation to spacetime. The Einstein equations are hyperbolic in their time-dependence, with characteristic curves being the lightlike geodesics. The influence of perturbations to hyperbolic PDE systems are confined to the characteristics of those systems. Thus the motion of the black hole at time $t$ influences spacetime only in the future lightcone of its position at $t$.

For similar reasons, the motion of the test mass can, too, only depend upon the spacetime in its past lightcone. This will intersect the perturbation due to the black hole at the black hole's retarded time. The perturbation at the retarded time will depend upon the entire history of the black hole at that retarded time. So the motion of the test mass depends upon the whole history of the black hole at its retarded time, but not in the future of such.

  • $\begingroup$ This is, by the way, more or less a reprocessing of the ideas in link.springer.com/article/10.12942/lrr-2011-7#Sec1 $\endgroup$ – AGML Jun 18 '17 at 21:04

Gravitational attraction takes place only in one direction. Assuming speed of gravity same as that of light, that direction would be towards the retarded position of the black hole. Just by simple mechanics, the resultant force has to be in a single direction. Curvature of space at one point of space and time, will only have a resultant slope in a single direction. Even if it felt the force towards past and future positions, they would cancel one another out, except the components that are towards current (retarded) position. But then you would have increased gravity due to the added components and would violate the gravitational formula. So, no, it is only towards the current (retarded) position of the black hole.


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.