<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jim</i>
<br />Dr Joe, How do you determine the gravity of a proton? ...I think the solar field becomes equal to the gravity of the universe at some distance near 50 or 100AU.
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Thanks, Jim, for the post. This "gravity of the universe" figure, is interesting! Do you recall the source?
Regarding the gravity of a proton, I sketched the calculation in other posts on this messageboard years ago, and also in my 2002 article in "Aircraft Engineering and Aerospace Technology", but here's a review:
1. Let's try to construct a proton of the smallest size possible, using college quantum mechanics to make the most compact possible radial Gaussian mass distribution allowed by the proton's mass, according to the Heisenberg uncertainty principle. Let's start with the formula E = sqrt(m^2*c^4 + p^2*c^2). I'm considering the proton's constituent deBroglie waves; in this view, there is no rest mass because all the energy is in the internal "p". Because of equipartition of energy, one part of the proton's mass is available for the deBroglie waves in each of the three directions, and a fourth part is available for "spin". Along, say, the x-axis, I have momentum p such that the root-mean-square value of p satisfies p^2*c^2 = 0.25 * m^2*c^4, i.e. p = 0.5 * m*c. A deBroglie wave of wavenumber k, has momentum p = hbar*k. So, the rms value of k (along the x-axis) is 0.5*m*c/hbar.
2. Heisenberg's uncertainty principle is really only a theorem in the first year graduate mathematical course called "real analysis". The theorem says, that in Gaussian wave packets, the bigger the rms k value (i.e. wavenumber) along, say, the x-axis, the smaller the packet's x dimension. The formula is k*x = 0.5, where the k & x are the rms values for the wave packet. (There's another theorem that says that Gaussian wave packets are the best you can do; for other kinds of wave packets, k*x > 0.5.)
3. Now I have a Gaussian wave packet, the smallest possible ( x = 0.5/k, where k is given in #1, so x = hbar/(m*c), the Compton radius for the proton), and it's radially symmetric. I used a statistical table of the normal distribution, but with modern computers, if one knows how to program BASIC or some other language, it's fairly easy, to have the computer add up the effects of the "shells" and find the gravitational force inside the proton at various radii, bearing in mind that the proton's mass is distributed in this Gaussian form. There is a radius at which the gravitational force is maximum.
4. Note that I call hbar/(m*c), the Compton radius. Usually, h/(m*c) is called the Compton wavelength. So my "Compton radius" is the Compton wavelength divided by 2*pi, recalling that hbar is defined as h/(2*pi). The gravitational force for the above Gaussian packet is maximum at 1.3688 Compton radii, and the mass inside this radius, is 0.40096 times the total proton mass. So, for this compressed proton, the maximum interior gravitational acceleration is G*m/(Compton radius)^2 * 0.40096/1.3688^2 = (using physical constants from my old Handbook of Chemistry & Physics) 5.398/10^5 cm/sec^2.
5. The gravitational acceleration due to the sun, at a distance of 1 AU, is omega^2 * r = 0.5930 cm/sec^2. This would seem to imply that the gravitational acceleration due to the sun at 104.8 AU, equals the maximum gravitational self-acceleration inside a proton.
6. Now let's recall the Copenhagen interpretation of quantum mechanics. If only one dimension, say, x, can be measured, then the energy available for localization along the x-axis isn't m*c^2/4, as I assumed above; it's simply m*c^2. The proton effectively can be sqrt(4) = 2x smaller, and its internal gravitational acceleration 4x greater. This corresponds to 104.8/2 = 52.4 AU.
