As I understand it, Milgrom's MOND model can be interpreted in one way by saying that Newton's Law of Gravitation on its own is insufficient to predict the gravitational acceleration when acceleration is very small.

A second interpretation is that Newton's Second Law (force = time rate of change in momentum, in the absence of other unbalanced forces) breaks down at very small accelerations. I have not found an understandable account of this interpretation, so I will focus on the first interpretation.

Milgrom, 1983 suggested a new fundamental constant with dimensions of acceleration and whose suggested magnitude (from fitting to observed astronomical data) is $|a_0| \approx 2*10e-8 cm/s^2$. For example this is the magnitude of the Newtonian gravitational acceleration $a=GM/r^2$ caused by a 3Kg cannonball, acting on a pea at a distance of 1m.

For a given source mass and reacting massive test particle, three regimes, based on the appropriate formula for acceleration $a_r$ at distance $r$, can be defined:-

(1) High acceleration Newtonian regime where $|a_r|>>|a_0$| and acceleration is well-described by Newton's Law: $ a_r = F_N/m = -GM/r^2$. The $-$ sign indicates that the acceleration is directed in the opposite direction to that in which distance $r$ from source to receiver increases.

(2) Transitional regime where some kind of transitional formula applies. Within this regime lies the transition distance $r_{t}=\sqrt{GM/a_0}$ where, for given $M$, the formulae for regimes (1) and (3) give the same acceleration : $ -GM/r^2 = a_r = -\sqrt{a_0 GM}/r $

(3) Low acceleration Deep MOND regime where where $|a_r|<<|a_0|$ and acceleration is given by:- $F_N/m = -GM/r^2 = a_r|a_r/a_0|$ and thus $a_r = -\sqrt{a_0 GM}/r.$

This seems fairly straightforward to me in the case of an isolated test particle some distance from a static concentrated source mass. For given $GM$ the appropriate acceleration regime is a variable function of source:target distance $r$. As the test particle moves away from the source, acceleration falls of initially as $\propto 1/r^2$ and eventually as $\propto 1/r$, with a gradual transition between the two regimes. And using the circular orbit centripetal force relation ($-V_R^2/r = a$) it is not hard to see how a fixed central mass $M$ can generate the sort of flat $V_R$ vs $r$ rotation curve interpreted (from observed doppler shifts) in the outer parts of galactic disks:-

$$\frac{V_R^2}{r} =-a_r = \frac{\sqrt{a_0 GM}}{r} \implies V_R^2 = \sqrt{a_0 GM} \implies V_R = constant $$

Also, over a range of galactic masses $M_i$, we derive $ {V_{iR}}^4 \propto M_i $ and, if luminosity $L_i \propto M_i$ then the observed Tully-Fisher relation $ {V_{iR}}^4 \propto L_i $ is obtained (as MOND was designed to do).

But I have a problem with this. While the model seems clear in the case of an isolated test particle far from the gravitational source I do not understand how it applies to a test particle in the interior of a large massive object such as a star, planet, moon, or asteroid. At any point inside a spherically symmetrical example of such an object (apart from a very small region at the centre) the net gravitational acceleration from all the object mass closer to the object centre, will be much larger than the given value of $a_0$. (The effect of object mass further out from the centre will net to zero in accord with Newton's Shell Theorem).

One way that a test particle might respond differently to gravitons from the object and to gravitons from a far distance source might be if some characteristic of the gravitons of either or both sources acted as an indicator of distance (or age, or divergence which both depend on distance). But Milgrom and others, e.g. Sanders & McGaugh, 2002, have emphasized that a distance dependency is not supported by the data. Moreover, how would the reaction mechanism "know" to respond differently to gravitons coming from one distant source e.g. a star of 2 solar masses compared to to gravitons coming from two distant (but close to each other) sources of same total mass (e.g. a binary pair of stars with each star having 1 solar mass).

I know that Milgram and other MOND supporters do not strongly defend their model in terms of any particular physical mechanism but rather point to its claimed success in predicting galactic rotation curves and the potential insights that might be gleaned from that. Maybe modern (astro)physics simply does not reject a model simply because of lack of even an intuitive mechanism. Maybe there is a plausible explanation of how massive particles in a star can react differently to gravitons from remote objects. Or maybe I am missing something obvious in the MOND model.

What do you think?

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    $\begingroup$ Another example is a binary star system at large galactic radius. The total field is strong because of the companion, therefore MOND gives a different acceleration for binaries than gas atoms, and different again for single stars. But, it is well observed that they travel together everywhere in a galaxy. $\endgroup$
    – eshaya
    Aug 16 at 17:30
  • $\begingroup$ I had been wondering whether stars and gas in an outer parts of a galaxy (where the rotation velocity is relatively flat) do indeed travel at the same rotational velocity. Can you possibly point me to any papers that might confirm this? $\endgroup$
    – steveOw
    Aug 27 at 17:56
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    $\begingroup$ You have to look at papers with optical rotation curves and then find the HI rotation curve to compare. Not likely anyone can publish a paper saying they are similar. If they differ though, that paper would be Nobel Prize worthy. $\endgroup$
    – eshaya
    Aug 28 at 14:07
  • $\begingroup$ Is MOND different from newtonian mechanics in having this problem? At least when applying his shell theorems. Inside a "solid" sphere the net force increases proportional to r. Until the "surface" is reached, at which point it begins decreasing proportional to r^2. The universe exists in between his idealized solid sphere and hollow spherical shell. $\endgroup$
    – Livid
    Oct 17 at 3:01
  • $\begingroup$ @Livid Newtonian mechanics doesn't have a problem with low gravitational accelerations nor does it predict low acceleration anywhere inside a star (except at the centre of mass). $\endgroup$
    – steveOw
    Oct 17 at 11:21

1 Answer 1


Hi bit late to the party but I think there are two issues here. 1) How to deal with superpositions of small Newtonian gravitational accelerations and their conversion to the Milgromian expectation and 2) What underlying physical mechanism could be responsible for Milgromian dynamics.

To deal with the superposition issue first. This is crucial for correctly understanding MOND/Milgromian dynamics. MOND acts on the total gravitational field not on decomposed vectors. MOND is not a linear theory so decomposing the gravitational field in individual contributions and applying the transition function between the Newtonian and the Milgromian regime to each decomposed vector will give a different (incorrect) result compared to applying the transition function to the full gravitational field. Inside a star the gravitational accelerations of the individual atoms on a test particle would have small accelerations compared to the transition acceleration $a_0$ but it is not correct to apply the transition function on the acceleration caused by each atom individually. You have to calculate the complete net Newtonian gravitational acceleration and apply MOND on that.

If you insist on decomposing your gravitational field because it simplifies your problem then you have to take into account the external field effect. This is somewhat of a misnomer, perhaps the term "total field effect" would be more descriptive. This is not particularly relevant for the internal dynamics of stars because they don't show any Milgromian behaviour whatsoever but it is relevant for diffuse objects (dwarf galaxies, molecular clouds, wide binaries, etc.) located in strong external gravitational fields (from host galaxies, galaxy clusters, spiral arms, etc). Here too it is the total field that counts.

To calculate the Milgromian result for a system that you have decomposed for ease of calculation:

$$g_{obs}=(g_{N int}+g_{N ext})\nu((g_{N int}+g_{N ext})/(a_0))-g_{N ext}\nu((g_{N ext})/(a_0)))$$

Compared to the Milgromian result for a system which you haven't decomposed (or is isolated such that the sum of all external accelerations is much smaller than the internal accelerations and much smaller than $a_0$):

$$g_{obs}=g_{N int}\nu(g_{N int})$$

Here $\nu$ is the transition function. See also this paper by Lelli et al. Note that in the above two cases I've given a one dimensional simplification. As long as you are using the above two equations along the direction of greatest change in acceleration this will work fine. If that is not an option you will have to solve the full field equation numerically:


As for the second issue you raise, what the underlying physical mechanism is like, this is currently unknown. Because Milgromian dynamics is non-linear it is probably necessary that gravitons undergo some sort of phase transition at appropriate densities.

  • $\begingroup$ Your second paragraph effectively restates the core problem I have with MOND e.g. it surely cannot apply inside a star. I found this problematic because I was under the impression that the flat or gently-rising sections of observed galactic rotation velocity curves (e.g. m33) applied to gas and stars rather than just just gas. $\endgroup$
    – steveOw
    Jul 21 at 23:19
  • $\begingroup$ You write "You have to calculate the complete net Newtonian gravitational acceleration and apply MOND on that." But that is effectively what I implied in my sentence:- "At any point inside a spherically symmetrical example of such an object (apart from a very small region at the centre) the net gravitational acceleration from all the object mass closer to the object centre, will be much larger than the given value of 𝑎0.". I appreciate your engagement but I don't think you have recognized the core issue of my (admittedly verbose) question. $\endgroup$
    – steveOw
    Jul 26 at 18:02
  • $\begingroup$ Often the break or choice of what is external vs internal is arbitrary. That requires the observed acceleration to be symmetric in internal and external sources. The formula here is not. $\endgroup$
    – eshaya
    Aug 16 at 20:27

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