@Joe_Bakhos, I will take one more chance with you. My email inbox is bombarded by ~100 emails per year by people who are extremely sure that they can solve the dark matter and dark energy problems. Unlike most physicists, I try to actually respond to some of them. Most of these theories are completely bunk, and the email exchanges often end with them attacking my professional competence and honesty. So far, you are the only one to ever apologize.
Now, there are many ways to test whether a modified gravity theory can produce a flat rotation curve. Most of these involve numerical integration of many orbits. These tools exist and I can point them out to you, but they are computationally very expensive to run, and many of them you outright cannot use because you only have an approximate force law that is only valid for point particles.
No worries; let’s start with a way more elementary test: circular motion at the outskirts of galaxies. This test is not at all conclusive, but if your theory cannot pass this very rudimentary test, then you have a problem. The idea is very simple, and only require high-school level math and physics.
This is a test of whether your theory has a hope of producing a flat rotation curve for stars/clumps of gas orbiting at the outskirt of galaxies. Most of the light we see in a galaxy is concentrated near the center of the galaxy. Therefore, far in the outskirt, the gravitational acceleration on an orbiting star due to the luminous part of the galaxy can be well approximated by a point source. This is very useful, as you only have the equation
F = -k \frac{M_1 M_2}{r^2} \cos\Theta
which is only valid for a point source.
The idea is the following:
- For stars and clumps of gas orbiting in a somewhat circular orbit, they experience two types of accelerations: centripetal and gravity. As these stars are not being flung inwards/outwards from the galaxy, these two accelerations balance each other in the radial direction:
a_{\rm centripetal} = \vec{a}_{\rm gravity} \cdot \hat{r}
- The centripetal acceleration is simply (where V is the orbital tangential velocity and R the orbital radius):
a_{\rm centripetal} = \frac{V^2}{R}
- As you can see, by substituting a_{\rm centripetal} to the equation in point 1), you can solve for V(R), the orbital tangential velocity as a function of the orbital radius. If the rotation curve is flat, then V would be constant with respect to R:
V(R) = \rm{constant}
- If your theory can produce V(R)=\rm{constant} for this very stripped-down system, then you are in business. Otherwise this is a problem and you have to come up with explanations why your theory is still valid.
Let me do a couple of examples for you. Let’s take the Galaxy M87, upon which astronomical observations have shown that regions beyond 20 kiloparsec possess a flat rotation curve.
For Newtonian gravity without dark matter, the acceleration experienced by the star in the radial direction is well approximated by
\vec{a}_{\rm gravity} \cdot \hat{r} = \frac{G M }{R^2}
where M is the mass of the galaxy. Plugging in this equation to the equation in point 1),
\frac{V^2}{R} = \frac{GM}{R^2}
Clearly then,
V(R) \propto \frac{1}{\sqrt{R}}
This is a Keplerian rotation curve, not a flat rotation curve, which again, is V(R) = \rm{constant}. Thus, Newtonian gravity without dark matter fails this test.
Another example: MOND, where for M87 at distances beyond 20 kiloparsec, gives an acceleration experienced by the star in the radial direction to be
\vec{a}_{\rm gravity} \cdot \hat{r} = \sqrt{\frac{GM}{R^2}a_0}
where a_0 is some constant. Plugging in this equation to the equation in point 1),
\frac{V^2}{R} = \sqrt{\frac{GM}{R^2}a_0}
Clearly then,
V(R) = (a_0 G M)^{1/4} = \rm{constant}
thus MOND passes this test: it can produce a flat rotation curve in this simplified calculation.
Now is your turn: plug in your equation for M87 at a distance of 20 kpc and beyond, and show that it can also give a flat rotation curve: V(R)=\rm{constant}. If it passed this test, then we can proceed to building more complicated models.
Don’t hesitate to ask me questions, but again, I am taking one last chance with you. If you attack my professional competence and honesty again, I will cease communication with you.
Edit: An additional caveat, @Joe_Bakhos, note that you do not need to produce V(R)=\rm{constant} exactly, but just to the extend of the error bars of astronomical measurements.
