Response to #34:
Hi Jon,
Apologises for the length of this post, but wanted to try and table some equations for the dust ball based on your description. I don’t necessarily expect a response, but would welcome any correction of my assumptions as I find the model an extremely useful way of trying to understand some of the wider, albeit speculative, ideas within cosmology.
Thanks for the pointer to Newton’s Shells, this does indeed suggest the standard Newtonian equation is not applicable within the overall perspective of the dust ball model and should be replaced by the one you presented, i.e.
[1] F = \frac{4\pi Gm\rho r}{3}
In a spatially flat model, each galaxy's speed away from the center is equal to the Newtonian escape velocity of the total matter within the spherical ball defined by that galaxy's radius from the center.
Later you make the statement above, which led me to wonder if we also need to revise the standard definition of the escape velocity for the dust ball universe. Normally, the
‘classical’ escape velocity is derived from the balance of kinetic energy (outward expansion) against potential energy (inward gravitation). However, on the basis that gravitational potential is the integral of force, given in [1], would the minimum dust ball escape velocity be given by:
[2] E_k = \frac {mv^2}{2} = \frac{4\pi Gm\rho}{3} * \frac {r^2}{2}= E_p
[3] v = r \sqrt {\frac{4\pi G\rho}{3}}
I recognise that the premise of this derivation is weak, as the expression for kinetic energy is only valid for non-relativistic velocity, but suggests that:
[4] v = r*X but similar in form to Hubble v = Hr
However, if we set [k=0] and [\Lambda=0] in Friedmann’s equation we appear to get
[5] H^2 =\frac {v^2}{r^2} = \frac{8\pi G\rho}{3}
[6] v = r \sqrt {\frac{8\pi G\rho}{3}}
Is the inference that recession velocity of the dust ball is twice the required escape velocity, i.e. comparing [3] to [6]? On the assumption that we can still equate equation [1] to [ma] to derive an expression for [a=-g]:
[7] a=\frac{4}{3}\pi G\rho r=-g
Which suggests that not only does gravity not fall off via the inverse square law, but that it actually increases with radius [r]. However, this also seems to have some obvious similarities to Friedmann’s Acceleration equation,
[8] \left(\frac {\ddot a}{a}\right )= - \frac{4}{3}\pi G (\rho+3P/c^2)
However, on the assumption that the expansion of the dust ball is actually accelerating, does this suggest that [3P/c^2 > -2 \rho], i.e. that \ddot a > g
Finally, if possible, I wanted to check a couple of issues associated with expansion of the dust ball:
- Given that the dust ball expands under the pressure of dark energy, there was an implication that the dark energy per unit volume does not get
‘diluted’, but rather remains constant. If so, wouldn’t the effective mass associated with this energy increase and therefore cause the mass-density [\rho] of the dust ball to increase with the expansion over time? If so, does the increase of [g] with radius [r] remain linear?
- Overall, the logic of your thought experiment seems sound. However, I have a attached a diagram showing 3 points in space (2-D only), which defines a triangle. Each point of the triangle is subject to different levels of expansive acceleration and gravitational deceleration, which is proportional to [r], subject to the caveat above. Is it the assumption of a spatially flat, homogeneous model that this triangle remains geometrically similar under expansion?
Again, apologises for the length, but wanted to table any equations and issues for future reference. Anyway, thanks again for helping me with this model.
