Why is the gravitational force always attractive? Is there another way to explain this without the curvature of space time?

PS: If the simple answer to this question is that mass makes space-time curve in a concave fashion, I can rephrase the question as why does mass make space-time always curve concavely?

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    $\begingroup$ Curving spacetime made of rubber is a very problematic pop-sci concept; don't try to think of it as an actual model. $\endgroup$
    – user68
    Jun 24, 2011 at 16:53

8 Answers 8


Gravity is mediated by a spin two particle. Electromagnetism by spin 1.

Here is a link that answers your question:

even and odd spin do differ in that they require a product of charges with different signs to get attraction or repulsion:

spin even:

  • $q_1 q_2 > 0$: attractive
  • $q_1 q_2 < 0$: repulsive

spin odd:

  • $q_1 q_2 < 0$: attractive
  • $q_1 q_2 > 0$: repulsive

In the case of gravity, mediated by spin 2 particles, charge is mass, which is always positive. Thus, $q_1 q_2$ is always greater than zero, and gravity is always attractive. For spin 0 force mediators, however, there is no restriction on the charges and you can very well have repulsive forces. A better rephrasing of the question is: "Why do particles of odd spin generate repulsive forces between like charges, while particles of even spin generate attractive forces between like charges?"

Goes on to derive this

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    $\begingroup$ If OP is just asking about Einstein gravity with no cosmological constant, I think this answer is spot on. $\endgroup$
    – Qmechanic
    Jun 24, 2011 at 21:35
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    $\begingroup$ Interesting read, this however does not preclude the existence of "negative mass" particles/objects that would act repulsively. Gravity is only attractive because a)Like "charges" attract and b)only one "mass charge" observably exists. (a) is explained by theory, and (b) is still (AFAIK) an open question $\endgroup$
    – crasic
    Jun 25, 2011 at 12:51
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    $\begingroup$ @crasic Yes, there are experiments trying to see whether antiprotons are repulsed by gravity. cdsweb.cern.ch/record/1037532/files/spsc-2007-017.pdf . I also asked a related question a while ago : physics.stackexchange.com/questions/5521/… $\endgroup$
    – anna v
    Jun 25, 2011 at 13:19
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    $\begingroup$ The OP asked a question about classical gravity, and the answer is that classical gravity need not be attractive; GR is completely agnostic about what matter fields you give it, and it allows matter that violates energy conditions. The quantum-mechanical answer depends on general QFT ideas that aren't embodied in any known theory of quantum gravity, and it doesn't answer the question of why mass should always be positive. In fact, we know that all energy conditions are violated, for classical and/or quantum-mechanical reasons: arxiv.org/abs/gr-qc/0205066 $\endgroup$
    – user4552
    Aug 6, 2011 at 18:33
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    $\begingroup$ Need to note that gravitons ("gravity mediators") are not detected yet so everything that is based on them has very unstable grounds. $\endgroup$ Jun 1, 2019 at 21:23

If gravity is entropic, as suggested recently by Verlinde and earlier by others, gravity might be expected to be mostly attractive.

My speculative imaginary view of this has been that if the evolution of quantum physical objects is not perfectly conservative, then there might be processes which convert energy between scales. One possibility is that fields at the scales that determine gravitational interactions would be converted to lower scale matter that is currently not detectable. The result would be both an inflow and an outflow from regions that contain large masses, but on different scales and with different effects, attractive for matter that interacts more with the inflow, repulsive for matter that interacts more with the outflow.

At the human scale I suppose we have to imagine that we are pressed down by a flow of something that isn't the same as matter, it's the gravitational field if you will, and of course not an aether, metaphorically being sucked into the earth as food for a fundamental entropic process. Whatever exhaust there is from this process interacts with us enough less to be essentially undetectable on anywhere close to human scales, but would be detectable on either very large or very small scales.

Like I say, speculative, and also only a very small part of a whole entropic process. I can take a few or even a lot of downvotes, but pretend for a moment we're shooting the breeze at coffee. I haven't followed the Verlinde and other entropic gravity literature at all closely, so I don't know whether something like this of model has been suggested in a mathematical form, which would be required for it to be publishable (although being published definitely wouldn't be enough to make this not speculative).

If you ask for explanations for the established mathematics of General Relativity, I think the only currently possible response is speculation. GR is more grounded in empirical principles than in models that could be taken to be explanatory (which I say without prejudice insofar as I take empirically supported principled theories to be preferable to more-or-less ad-hoc models, except for the elusive question of how to imagine effective new empirically supported principles). I note that lurscher doesn't address your request for explanation.

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    $\begingroup$ Verlinde didn't add anything about entropic gravity that Ted Jacobson didn't said back in 1995 arxiv.org/abs/gr-qc/9504004, although compared with Jacobson exposition of the idea, Verlinde is almost completely incoherent, a true pain to read $\endgroup$
    – lurscher
    Jun 24, 2011 at 19:18
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    $\begingroup$ as Lubos has explained in his blog motls.blogspot.com/2010/01/… there are good constraints that a underlying connection between gravity and thermodynamics have to satisfy. Of course there is more to it than fluctuations breaking equivalence principle. Maybe fluctuations cannot contribute is some naive ways. In any case, there is no doubt that the relationship is a very intriguing one $\endgroup$
    – lurscher
    Jun 24, 2011 at 19:31
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    $\begingroup$ Fair enough, lurscher. The Wikipedia page says what you said slightly less trenchantly. Because Verlinde engages with the question at the Newtonian level instead of in terms of GR, it's arguably more accessible despite being less coherent. Of course anything serious has to make contact with GR in the sense of correspondence. $\endgroup$ Jun 24, 2011 at 19:32
  • $\begingroup$ no intention to be trenchant, sorry if it turned out like that. I just don't like the guy taking credit for an idea brought upon 15 years earlier (to which he doesn't seem to have added anything) $\endgroup$
    – lurscher
    Jun 24, 2011 at 19:34
  • $\begingroup$ @lurscher, Trenchant is too strong a word. No worries. One feature is that entropy is, like temperature to which it is thermodynamically dual, fundamentally a phase space observable. Quantum fluctuations, in contrast, are Lorentz invariant (by definition, because the vacuum state is defined to be Lorentz invariant). To me, gravity is unlikely to be about temperature and entropy, more likely to be about quantum fluctuations and the Lorentz invariant thermodynamic dual of quantum fluctuations (whatever that is). I think (local) Lorentz invariance is critical for GR, but no-one seems to get this. $\endgroup$ Jun 24, 2011 at 21:36

When I was a schoolboy, our teacher of physics asked once one of our brilliant students (a girl), something like: "What is the nature of gravity?". She thought for a moment and answered: "I do not know. And what is it? Our teacher replied: "If I had known..."

As long as gravity is not derived from other nature features, it should be considered as a fundamental law. We may study its properties but we may not ask such questions - just by definition of a fundamental law. It is as an axiom - it is given as such.

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    $\begingroup$ I think that asking questions like this can lead to new discoveries and insight into the nature of our universe. Why should we neglect a question on the grounds that there isn't a good answer yet? $\endgroup$ Oct 16, 2014 at 4:37
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    $\begingroup$ That sounds more like religion than science. $\endgroup$ Oct 4, 2015 at 23:38
  • $\begingroup$ @HalfdanFaber : if it is religion of the observations and the experiments, all is fine. Here, it is an observation, not a credo , particularly in Newton mechanics $\endgroup$
    – user46925
    Dec 29, 2015 at 3:14
  • $\begingroup$ So lets not ask why does electrons create an interference pattern. It is just how mother nature works. Asking radical and ridiculous questions is what makes science improve. Do not block it. $\endgroup$
    – user147133
    Mar 9, 2017 at 15:09

It is not true at all that gravitation is always attractive. In fact, if it were always attractive, the universe would not be expanding at an accelerated rate right now and an inflationary period would not have occurred.

Currently the only known source of expansion components of gravity is the cosmological constant, which, incidentally, is precisely the physical quantity that our theories fail to predict by the largest amount: 120 or 60 orders of magnitude, depending on whom you ask.

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    $\begingroup$ I don't think your argument holds, the repulsive force driving the expansion can simply be greater than the attractive gravitational force. Similarly, if space itself is undergoing inflation then there is no need for the gravitational force to be anything other than positive. $\endgroup$ Jun 25, 2011 at 11:11
  • $\begingroup$ --------exactly $\endgroup$
    – lurscher
    Jun 25, 2011 at 11:57

In general relativity, in general but we can consider the most simple case of a spherically symmetric gravitational field, gravity is always attractive as long as you are at rest with respect to the gravitational field. However, if you are moving with respect to the gravitational field, gravity can sometimes be considered to be "repulsive". If you are falling in radially, from initially being at rest at high altitude, you initially accelerate towards the central mass. However, you will reach a point, if the central mass is compact enough like a black hole, where you start to decelerate and when you are infinitely close to the "Schwarzschild radius" you will move infinitely slow. This is all if your movement is observed from a distant observer.

In the classical sense of the meaning, gravity will become "repulsive" as soon as you start to decelerate. The reason for this deceleration is that in contrast to "classical Newtonian mechanics", where the force only increases the momenta by increasing the velocity, and accelerating particles in an accelerator where the electromagnetic force increase the momenta by increasing the "lorentz factor" setting the speed of light to be the speed limit, in general relativity you have the third effect that the speed of light (again as measured by a distant observer) around a spherically symmetric mass distribution decreases with the radial distance.

The gravitational force will always be "attractive" if you are moving in towards a black hole in the sense that your velocity as a fraction of the local speed of light constantly increases but your velocity, as observed by a distant observer, will if you get close enought to a black hole inevitably start do decrease and it that sense gravitation will become "repulsive".

Also, in practice, Nasa/JPL is using the force on the right side of this expression to mimic relativistic effects in the weak fields of our solar system:

$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\left(1-\frac{4GM}{rc^2}+\frac{v^2}{c^2}\right)\hat{r} +\frac{4GM}{r^2}\left(\hat{r}\cdot \hat{v}\right)\frac{v^2}{c^2}\hat{v}$$

You can see that this becomes repulsive, even if you are not moving with respect to the central mass at a radial distance of $r=4GM/c^2$ which is twice the Schwarzschild radius in Schwarzschild coordinates.

If your moving radially inwards this becomes repulsive, if I am not doing any mistake, at: $$v=\frac{c}{\sqrt{3}}\sqrt{1-\frac{4GM}{rc^2}}$$

Notice that JPL is only using the expression above in the weak fields of our solar system, it gives the right value of the so called "anomalous precession of perihelion" but it is not supposed to work in the strong field limit. The expression is provided as number 4-61 on page 4-42 in the official documentation, Formulation for Observed and Computed Values of Deep Space Network Data Types for Navigation.

  • $\begingroup$ Do you happen to have a source where the computation is done indicating that this gives the right perihelion precession? $\endgroup$
    – jawheele
    Apr 25, 2021 at 4:31
  • $\begingroup$ @jawheele well no I have not seen an analytical solution but you can check the Nasa equation if you have a simple numerical integrator and it will give the right perihelion precession. $\endgroup$
    – Agerhell
    Apr 22, 2022 at 20:02
  • $\begingroup$ Right, I actually ended up checking this numerically myself not long after leaving that comment. Thanks for getting back to me! $\endgroup$
    – jawheele
    Apr 22, 2022 at 20:46

The inverse square is apparently a consequence of conservation of momentum. For two particles in orbit, Newton showed that the orbit is planar, and Bertrand https://en.wikipedia.org/wiki/Bertrand%27s_theorem showed that the forces between the two have to be either of the inverse square k/r^2 or space spring/Hook's law. So Newton's law of gravity and that of Coulomb have conservation of momentum as the origin.

It is also worth noting that Hook's law can be shown to come out of/limiting case of the inverse square in the case of 'crowdedness', where there are too many interactants and very small space to move. This can be shown easily by taking three particles along a line interacting under the inverse square and give the middle a nudge keeping the end particles fixed. If we take it that there is no charge with zero mass, Coulomb's law follows too.

Even without the help of Bertrand theorem, it is possible to derive the Maxwell equation from just charge conservation and its continuity equation. See this reg and quote: https://pdfs.semanticscholar.org/3251/31eadb62c8fdfdaaad7b21a308992ff3a4d2.pdf '' We show how the covariant form of Maxwell’s equations can be obtained from the continuity equation for the electric charge ''. Clearly the same can be done using mass conservation and we get the gravitomagnetic equations out of it.

In general, Maxwell/gravitomagnetic equations work for both attractive and repulsive forces. But if two masses are locked in an orbit, they must be under attraction. The facts that similar masses attract whereas similar electric charges repel has to rely on experimental knowledge.

New research from MIT on a new long-range attraction force connected to spin. https://www.youtube.com/watch?time_continue=10&v=1ZZcgBmS5W4


Gravity is not always attractive:

Gravity is thought of as a weak boring force. But if you get fast spinning black holes and strong gravitational waves interacting, then you can get 'repulsion and attraction' effects happening out of purely gravitational forces.

So repulsive gravity does not exist in quiet spacetimes, but if you look at what happens when you shine a gravitational wave onto a rapidly spinning black hole, you will find that adjusting the frequency only slightly from the ideal level allows you to push the hole away from you (when the bh absorbs a wave) or pull it closer (when the bh adds energy to the radiation beam via superradiance).

See: http://arxiv.org/pdf/1312.4529v2.pdf - figure 4. With a Gravitational Wave

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    $\begingroup$ Dear Tom Andersen. It is often frown upon to post nearly identical answers to similar posts. In such cases, it is often better to just flag/comment about duplicate questions, so they can get closed. $\endgroup$
    – Qmechanic
    Dec 29, 2015 at 1:38
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    $\begingroup$ And the above is not true. Gravitational wave carries energy and momentum, gravity is still attractive. When the gravitational wave hits something or gets absorbed by something sure it'll transfer momentum. An electromagnetic wave does the same, but it is by itself neither attractive nor repulsive. If you want to find repulsive mass you need to find mass witha negative pressure, or negative nergy density. Those were not it. $\endgroup$
    – Bob Bee
    Jul 6, 2016 at 4:19


dipole +- <--- some distance ---> +- dipole

Two dipoles are always attractive (or a dipole and another charge). If they are like this +- ... -+ or -+ ...+- the dipoles will rotate and the configuration become attractive +- ... +- or -+ ... -+ .

They obey a 1/r³ relation.

If you can consider that inside the baryons (neutron, proton) can exist a configuration of dipoles you have an answer. (read the book of Douglas Pinnow, 'Our Resonant Universe'. It is a monography of a model of particles where this happens).

How do we go from 1/r³ to 1/r²? Integrate along the path.
Why? Explore the concept of polarizable vacuum.

The consensus is that gravitation is not electromagnetism, but in that way it is always attractive. And I like it.

But the order of magnitude of gravitation is $10^{-35}$ (more or less, by memory) of EM, so how can it be EM?
Can you figure out two EM radiators in each dipole in opposition of phase (one the +, other the -) extremely near one of the other? Yes, the radiated EM field has to be extremely faint.

(The connection with EM is much more compelling, IMO, that a connection with thermodynamics or other...)


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