I'm facing a problem which states the following:

Suppose that it is discovered that Newton's law of gravitation is incorrect and that the force $F$ on test particle of mass $m$ due to a body of mass $M$ has an additional term that does not depend on $M$, as follows:

$$F = - \frac{GmM}{r^2} + \frac{A m r}{3}$$

Where $A$ is a positive constant. Assuming that Newton's sphere theorem continues to hold, derive the appropriate form fo the Friedmann equations in this case.

How I proceeded

Consider the motion of a point like particle of mass $m$ located on the surface of a sphere of homogeneous density $\rho$ and radius $R = a r$. The acceleration this particle is subjected is

$$\ddot{R} = r \ddot{a} = \frac{-GM}{r^2} + \frac{AR}{3} = -\frac{G}{R^2}\left(\frac{4}{3}\pi R^3 \rho\right) + \frac{AR}{3} = -\frac{4}{3}\pi G\rho R + \frac{AR}{3}$$


$$\ddot{R} = r\ddot{a} = \frac{1}{3}\left[A - 4\pi\rho G\right]ra$$

$$3\frac{\ddot{a}}{\dot{a}} = A - 4\pi\rho G$$

Adding the pressure term and arranging:

$$3\frac{\ddot{a}}{\dot{a}} = A - 4\pi\rho G(\rho + 3P)$$

Is this correct?

Thank you so much!

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    $\begingroup$ I've added the homework-and-exercises tag. In the future, please use this tag on this type of question. $\endgroup$ – user4552 Jan 6 '19 at 14:30
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    $\begingroup$ FWIW, $A$ is like a cosmological constant. $\endgroup$ – Qmechanic Jan 6 '19 at 16:00
  • $\begingroup$ @Qmechanic Ahh, that makes so sense now! $\endgroup$ – Les Adieux Jan 6 '19 at 16:35

If we want to derive the Friedmann Equation by using Newtonian Mechanics we can always use the Energy of the system.


And the potential energy related to this force would be, $$U=-\int Fdr$$ $$U=-\int(\frac{-MmG}{R^2}+\frac{AmR} {3})dr$$ $$U=-\frac {MmG} {R}-\frac {AmR^2} {6}$$

Total energy of the particle of mass m will be, $$E=T+U$$ $$E=\frac {mV^2} {2}-\frac {MmG} {R}-\frac {AmR^2} {6}$$

Multiplying both side with $\frac {2} {mR^2}$

$$\frac {2E} {mR^2} = \frac {V^2} {R^2}-\frac {2MG} {R^3}-\frac {A} {3}$$

but $V=\dot{R}=\dot{a}(t)r$ and we know that $R=a(t)r$ hence

$$\frac {2E} {mR^2} = \frac {\dot{a}(t)^2} {a(t)^2}-\frac {2MG} {R^3}-\frac {A} {3}$$

finally writing

$$\frac {\dot{a}(t)^2} {a(t)^2}=\frac {8\pi G\rho} {3}+\frac {A} {3}+\frac {2E} {ma(t)^2r^2}$$

if we wanted to write in relativistic form we would have,$2E/mr^2=-\kappa c^2/R^2$ and energy density $\epsilon=\rho c^2$ or $\rho=\epsilon /c^2$ where $\rho$ is the matter density.

so we have,

$$\frac {\dot{a}(t)^2} {a(t)^2}=\frac {8\pi G} {3c^2}\epsilon(t)+\frac {A} {3}-\frac {\kappa c^2} {a(t)^2R^2}$$

For the acceleration, from $$F=ma=m\ddot{R}=-k/R^2+AmR/3$$

$$\ddot{R}=-MG/R^2+AR/3$$ $$\frac {\ddot{R}} {R} =-MG/R^3+A/3$$

or $$\frac {\ddot{a}(t)} {a(t)}= -\frac {4\pi G\rho} {3} + \frac {A} {3}$$

In general the acceleration equation is written as $$\frac {\ddot{a}(t)} {a(t)}= -\frac {4\pi G}{3c^2}(\epsilon+3P)$$

But the equation of state for matter is $P=w\epsilon=0$ since $w=0$ for matter hence we have $$\frac {\ddot{a}(t)} {a(t)}= -\frac {4\pi G}{3c^2}(\epsilon) + \frac {A} {3}$$ (Since we have an additional term, our equation is a bit different)

or in terms of matter density,

$$\frac {\ddot{a}(t)} {a(t)}= -\frac {4\pi G\rho} {3} + \frac {A} {3}$$

Which is exactly what we derived using the $F=ma$

  • $\begingroup$ More or less. I want to find Friedmann equations for that model. Shalln't we take in consideration also the "pressure term"? $\endgroup$ – Les Adieux Jan 6 '19 at 15:06
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    $\begingroup$ Sorry, In this case its better to go from the Energy, I assumed you wanted to derive the acceleration equation. We do not need to consider the pressure term for matter, since pressure would be zero. And the question do not asks about the radiation or for other terms. I ll edit my answer according to that. $\endgroup$ – Reign Jan 6 '19 at 15:19
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    $\begingroup$ I did the editing. Hope this helps you better $\endgroup$ – Reign Jan 6 '19 at 15:47
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    $\begingroup$ I did a bit more editing :) $\endgroup$ – Reign Jan 6 '19 at 16:07
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    $\begingroup$ $\ddot{R} = \ldots$ after you derived the acceleration from Newton's law. A minus sign is missing? $\endgroup$ – Les Adieux Jan 6 '19 at 17:00

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