<p>There is no line at which things transition from Newtonian dynamics to MOND, it's more of a gradual continuous transition depending on the nature of the interpolating function <span class="math-container">$\mu(x)$</span>:</p>
<p>From Wikipedia <a href="https://en.wikipedia.org/w/index.php?title=Modified_Newtonian_dynamics&amp;oldid=1068788718" rel="nofollow noreferrer">Modified Newtonian dynamics</a>:</p>
<blockquote>
<p>This law, the keystone of MOND, is chosen to reproduce the Newtonian
result at high acceleration but leads to different (&quot;deep-MOND&quot;)
behavior at low acceleration:</p>
<p><span class="math-container">$$F_N = m\mu\left(\frac{a}{a_0}\right)a$$</span></p>
<p>Here <span class="math-container">$F_N$</span> is the Newtonian force, <span class="math-container">$m$</span> is the object's (gravitational)
mass, <span class="math-container">$a$</span> is its acceleration, <span class="math-container">$μ(x)$</span> is an as-yet unspecified
function (called the interpolating function), and <span class="math-container">$a_0$</span> is a new
fundamental constant which marks the transition between the Newtonian
and deep-MOND regimes. Agreement with Newtonian mechanics requires</p>
<p><span class="math-container">$$\mu(x) \to 1 \hspace{1cm} \textrm{for }x\gg1$$</span></p>
<p>and consistency with astronomical observations requires</p>
<p><span class="math-container">$$\mu(x) \to x \hspace{1cm} \textrm{for }x\ll1$$</span></p>
<p>Beyond these limits, the interpolating function is not specified by
the hypothesis...</p>
</blockquote>

CC BY-SA 4.0
2022-02-24T17:51:42.317Z
It's definitely a smooth transition, but an important parameter to call out is $a_0$ which is, for lack of a better word, the "scale height" of the MOND force and does control how quickly or slowly that transition happens.
zephyr
CC BY-SA 4.0
2022-02-24T21:29:58.303Z
So the theory does not indicate when the transition begins and ends.  But observationally we can see that MOND is not in effect at the solar syatem level but is at the level of a galaxy.  So observationally we should be able to see data points along the transition if we find large enough black holes with distant enough companion stars or if we observe stars close enough to the galactic center.
Brooks Nelson
CC BY-SA 4.0
2022-02-23T01:41:50.067Z
2022-02-23T01:41:50.067Z
GrapefruitIsAwesome
<p>This question is vague because in physics there is no &quot;MOND takes over&quot;. Instead we say that deviations from Newtonian mechanics become large, where &quot;large&quot; depends on our specific problem.</p>
<p>Let's ignore MOND (because the interpolating function - see GrapefruitIsAwesome's answer - is not known) and instead use the relativistic and non-relativistic expressions for kinetic energy. The non-relativistic version is:</p>
<p><span class="math-container">$KE = \frac{1}{2} mv^2$</span></p>
<p>where <span class="math-container">$m$</span> is the mass and <span class="math-container">$v$</span> is the velocity. The relativistic version is:</p>
<p><span class="math-container">$KE = (\gamma-1) mc^2$</span></p>
<p>where <span class="math-container">$c$</span> is the speed of light, <span class="math-container">$m$</span> is the rest mass, and <span class="math-container">$\gamma$</span> is the Lorentz factor, defined as <span class="math-container">$\frac{1}{\sqrt{1-v^2/c^2}}$</span>.</p>
<p>Try plotting these two expressions. You end up with something like this (this quick and dirty Excel plot is for <span class="math-container">$m = 1kg$</span> and uses <span class="math-container">$c = 1$</span>; the horizontal axis is <span class="math-container">$v/c$</span> and the vertical axis is the kinetic energy).</p>
<p><a href="https://i.sstatic.net/PvI9d.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/PvI9d.png" alt="enter image description here" /></a></p>
<p>The orange line is the relativistic kinetic energy, while the blue line is the non-relativistic kinetic energy.</p>
<p>Note that the two &quot;overlap&quot; for small <span class="math-container">$v$</span>. This is to be expected, because in this regime classical physics is a good approximation. However, the two diverge at larger speeds.</p>
<p>Your question in this context becomes &quot;at what speed does relativistic kinetic energy take over&quot; and there is no answer to it, because there are always deviations between the two expressions once <span class="math-container">$v &gt; 0$</span>. For example, at <span class="math-container">$v/c = 0.01$</span>, the non-relativistic equation gives a kinetic energy of 0.00005, while the relativistic equation says it is 0.0000500038. Does this discrepancy matter to you? If for your application the answer is &quot;yes&quot;, then this is a speed at which you need to take relativity into account. If &quot;no&quot;, then we say relativistic effects are negligible.</p>
<p>If MOND is correct, then Kepler's laws already fail in our solar system. Accordingly, your question is not really answerable.</p>

CC BY-SA 4.0
2022-02-25T03:04:50.350Z
2022-02-25T03:13:52.627Z
2022-02-25T03:13:52.627Z
Allure
<p>At what distance does MOND Modified Newtonian Dynamics take effect?  I understand MOND is described in relation to very slow accelerations, which is related to distance.  It is mentioned it has an affect within galaxies and between galaxies but not within solar systems.  But there obviously should be a 'line' at which Kepler's laws no longer apply and MOND takes over.  The smallest galaxies, near galactic centers, very large black holes, etc.</p>
<p>I'm also wondering how a body at the 'border' would behave.</p>

CC BY-SA 4.0
2022-02-24T18:58:20.473Z
The effect of MOND does **NOT** depend on distance at all. It depends solely on the field strength itself. This means that the distance from a gravitating object at which MOND effects would become important if MOND were correct depend on the mass of the object. You can work that out by simply setting the field strength equal to the critical value $a_0$: $d=\sqrt{GM/a_0}$. This is true **only** if there are no other sources of gravity that may affect the local field strength.
Walter
CC BY-SA 4.0
2022-02-24T19:00:07.353Z
Mond makes no precise predictions about the transition from Newtonian to MONDian regimes. So behaviour at the border is unclear.
Walter
CC BY-SA 4.0
2022-02-24T21:17:03.133Z
So if star A is x distance from its galactic center and star B is x distance from its galactic center their accelerations would be different under MOND if the second galactic center was say 10 times more massive? @Walter
Brooks Nelson
<black-hole><dark-matter><galactic-dynamics><newtonian-gravity><keplers-laws>
At what distance does MOND Modified Newtonian Dynamics take effect?
Brooks Nelson
# At what distance does MOND Modified Newtonian Dynamics take effect?

(Asked by: username_0 on 2022-02-23)

At what distance does MOND Modified Newtonian Dynamics take effect? I understand MOND is described in relation to very slow accelerations, which is related to distance. It is mentioned it has an affect within galaxies and between galaxies but not within solar systems. But there obviously should be a 'line' at which Kepler's laws no longer apply and MOND takes over. The smallest galaxies, near galactic centers, very large black holes, etc.

I'm also wondering how a body at the 'border' would behave.

**Tags:** black-hole, dark-matter, galactic-dynamics, newtonian-gravity, keplers-laws
> **Comments:**
> **username_1**: The effect of MOND does **NOT** depend on distance at all. It depends solely on the field strength itself. This means that the distance from a gravitating object at which MOND effects would become important if MOND were correct depend on the mass of the object. You can work that out by simply setting the field strength equal to the critical value $a_0$: $d=\sqrt{GM/a_0}$. This is true **only** if there are no other sources of gravity that may affect the local field strength.  
> **username_1**: Mond makes no precise predictions about the transition from Newtonian to MONDian regimes. So behaviour at the border is unclear.  
> **username_0**: So if star A is x distance from its galactic center and star B is x distance from its galactic center their accelerations would be different under MOND if the second galactic center was say 10 times more massive? @Walter  

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## 2 answers

### Answer by username_2 on 2022-02-23. Score: 4

There is no line at which things transition from Newtonian dynamics to MOND, it's more of a gradual continuous transition depending on the nature of the interpolating function $\mu(x)$:

From Wikipedia [Modified Newtonian dynamics](https://en.wikipedia.org/w/index.php?title=Modified_Newtonian_dynamics&oldid=1068788718):

> This law, the keystone of MOND, is chosen to reproduce the Newtonian
> result at high acceleration but leads to different ("deep-MOND")
> behavior at low acceleration:
>
> $$F\_N = m\mu\left(\frac{a}{a\_0}\right)a$$
>
> Here $F\_N$ is the Newtonian force, $m$ is the object's (gravitational)
> mass, $a$ is its acceleration, $μ(x)$ is an as-yet unspecified
> function (called the interpolating function), and $a\_0$ is a new
> fundamental constant which marks the transition between the Newtonian
> and deep-MOND regimes. Agreement with Newtonian mechanics requires
>
> $$\mu(x) \to 1 \hspace{1cm} \textrm{for }x\gg1$$
>
> and consistency with astronomical observations requires
>
> $$\mu(x) \to x \hspace{1cm} \textrm{for }x\ll1$$
>
> Beyond these limits, the interpolating function is not specified by
> the hypothesis...

> **Comments:**
> **username_3**: It's definitely a smooth transition, but an important parameter to call out is $a_0$ which is, for lack of a better word, the "scale height" of the MOND force and does control how quickly or slowly that transition happens.  
> **username_0**: So the theory does not indicate when the transition begins and ends.  But observationally we can see that MOND is not in effect at the solar syatem level but is at the level of a galaxy.  So observationally we should be able to see data points along the transition if we find large enough black holes with distant enough companion stars or if we observe stars close enough to the galactic center.  

---

### Answer by username_4 on 2022-02-25. Score: 0

This question is vague because in physics there is no "MOND takes over". Instead we say that deviations from Newtonian mechanics become large, where "large" depends on our specific problem.

Let's ignore MOND (because the interpolating function - see GrapefruitIsAwesome's answer - is not known) and instead use the relativistic and non-relativistic expressions for kinetic energy. The non-relativistic version is:

$KE = \frac{1}{2} mv^2$

where $m$ is the mass and $v$ is the velocity. The relativistic version is:

$KE = (\gamma-1) mc^2$

where $c$ is the speed of light, $m$ is the rest mass, and $\gamma$ is the Lorentz factor, defined as $\frac{1}{\sqrt{1-v^2/c^2}}$.

Try plotting these two expressions. You end up with something like this (this quick and dirty Excel plot is for $m = 1kg$ and uses $c = 1$; the horizontal axis is $v/c$ and the vertical axis is the kinetic energy).

[![enter image description here](https://i.sstatic.net/PvI9d.png)](https://i.sstatic.net/PvI9d.png)

The orange line is the relativistic kinetic energy, while the blue line is the non-relativistic kinetic energy.

Note that the two "overlap" for small $v$. This is to be expected, because in this regime classical physics is a good approximation. However, the two diverge at larger speeds.

Your question in this context becomes "at what speed does relativistic kinetic energy take over" and there is no answer to it, because there are always deviations between the two expressions once $v > 0$. For example, at $v/c = 0.01$, the non-relativistic equation gives a kinetic energy of 0.00005, while the relativistic equation says it is 0.0000500038. Does this discrepancy matter to you? If for your application the answer is "yes", then this is a speed at which you need to take relativity into account. If "no", then we say relativistic effects are negligible.

If MOND is correct, then Kepler's laws already fail in our solar system. Accordingly, your question is not really answerable.


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