avoiding Black-Holes by "Gluon Field Collapse" ?

[Widdekind]

Nucleons are ~10^-15m (=1 fm) across. The gluon-field between the three quarks, in nucleons, accounts for 99% of the mass-energy of the nucleon; "bare naked" quarks only mass ~3-4 MeV. And, the gluon field can be described by a potential energy, which increases linearly, with inter-quark separation distance ®, approximately U = +(1GeV / fm) x r. Ipso facto, if neutronized matter, in a relativistically compact object, were to collapse beyond the neutron-star stage; then gravity would compress the quarks, within the neutrons, together, reducing their separation distances, and so weakening their gluon fields. In effect, gravity would "take over" for the strong force, confining quarks without need for any Strong-Force-gluon-field. Compressing neutronized matter, by a factor of ~100, would reduce the mass-energy of nucleons, by the same factor, i.e. eliminate the gluon-field, reducing total effective mass-energy, down to the raw "bare naked" quarks themselves, whose total combined raw mass-energy is ~10MeV (=0.01GeV).

 

As a crude semi-Classical calculation, to estimate the effects, of such "gluon field failure", during gravitational collapse, assume hydrostatic equilibrium (one-zone approx.), in a spherically-symmetric object, wherein pressure derived from the quantum-energy (Fermi energy) of compressed particles:

[math]E_F \approx c \Delta p \approx \frac{\hbar c}{\Delta x}[/math]

 

[math]\frac{dP}{dr} = -\rho g[/math]

 

[math]\frac{P}{R} \approx \frac{M}{V} \frac{G M}{R^2}[/math]

 

[math]P = \frac{2}{3} \frac{E}{V}[/math]

 

[math]E = \frac{3 G M^2}{2 R}[/math]

 

[math]E = N_q \frac{\hbar c}{\Delta x} = 3 N_n \frac{\hbar c}{\Delta x}[/math]

 

[math]\frac{4 \pi \Delta x^3}{3} N_n \approx \frac{4 \pi R^3}{3}[/math]

 

[math]\Delta x = R N_n^{-1/3}[/math]

 

[math]E = \frac{3 \hbar c N_n^{4/3}}{R} = \frac{3 G M^2}{2 R}[/math]

 

[math]\hbar c N_n^{4/3} = \frac{G M^2}{2}[/math]

 

[math]M \equiv M_0 \times \left(\frac{R}{R_0}\right)[/math]

 

[math]M_0 = N_n m_H[/math]

 

[math]\hbar c = \frac{G m_H^2}{2} N_n^{2/3} \left(\frac{R}{R_0}\right)^2[/math]

where we have assumed that compression causes gluon-field failure, so reducing mass. Now, the maximum amount of mass eliminable via compression is 99%; the compression factor on the RHS could be as low as ~0.01. The LHS embodies quantum energy resistance to collapse; the RHS embodies gravity force causing collapse. For some number of neutrons N_n, the RHS would still exceed the LHS, even with maximum mass-loss:

[math]\frac{2 \hbar c}{G m_H^2} < N_n^{2/3} \times 10^{-4}[/math]

 

[math]\left( 10^4 \frac{2 \hbar c}{G m_H^2} \right)^{3/2} < N_n[/math]

 

[math]m_H N_n > 4 \times 10^6 M_{\odot}[/math]

Ipso speculato, galaxy-class Super-Massive Black-Holes (SMBH), of millions to billions of solar masses, might be "more compact" than star-class black-holes, which might merely be "super-neutron-stars", whose neutrons have compressed by up to 100x, reducing their masses by the same factor, due to "gluon field failure". Star-class BH might actually collapse, into giant-planet-massed "neutron-Jupiters" (~10-20 M_J). Galaxy-class SMBH might require even more intense physics to explain how they might resist total collapse, to actual singularity. Even so, these speculations suggest, that the original mass of material plunging into galaxy SMBH may have been a hundred times more than their surviving present masses imply, e.g. the original total mass of material, which collapsed into our Milky Way galaxy's central SMBH, may have been a half billion solar masses.

Edited October 12, 2012 by Widdekind

[Widdekind]

oops --

 

Gravity is generated by energy, not (only) by mass; Newton's equation is really:

 

[math]F_g = \frac{G E_1 E_2}{c^4 R^2}[/math]

Relativistic particles, by definition, have total (Kinetic) energies, far exceeding their rest-mass-energies; relativistic particles are quantum-mechanically "light" particles, having effectively no rest-mass. As such, a relativistically-compact neutron-star (say) is, effectively, a "neutrino-star", composed of effectively-massless particles, whose Kinetic energies are "doing most of their gravitating". While "gluon field failure" may occur, and would then eliminate ~99% of the neutrons' rest-mass(-energies), the relativistic neutrons "would barely notice"; again, their energies-of-motion were already generating the bulk of their gravity effects.

 

Classical limit (quantum pressure non-relevant):

[math]\frac{2 KE}{3 V R} \approx \frac{G M^2}{V R^2}[/math]

 

[math]KE \approx \frac{3 G M^2}{2 R}[/math]

 

Relativistic limit ([math]E \propto \frac{\hbar c}{\Delta x} \sim R^{-1}[/math]:

[math]E \approx \frac{3 G E^2}{2 c^4 R}[/math]

Thus, in the relativistic limit, the RHS (representing gravity) increases, w.r.t. the LHS (representing pressure), as R^-2; as R --> R_S, the RHS exceeds the LHS, and the relativistic, "light", particles implode. Hypothetical "gluon field failure" may occur, but even then, is probably not relevant.

[EquisDeXD]

I think you need more scientific evidence, because if gravity is powerful enough to surpass the strong force, what exactly do we have left over? We have no evidence that quarks break down into any other particle, and supposedly a singularity is infinitely dense, and furthermore a gluon field is something that each individual quark has, and the exchange of gluons occurs between quarks, and I don't know if there's any evidence to support that the exchange of gluons stops after a certain point. We can create degenerate matter here on Earth, but I still haven't heard of what your saying, and we most certainly don't know enough about the composition of a singularity to be able to apply conventional physics to it. And then, light particles aren't completely physical objects, they can't implode in on themselves. You can reflect them and create interference, stretch or shrink their frequency or stretch or shrink their wavelength, but implode doesn't have a meaning in wave mechanics.

Edited October 13, 2012 by EquisDeXD

[Widdekind]

Recall, throughout,

 

[math]\Delta p \approx \frac{\hbar}{\Delta x}[/math]

 

[math]\frac{2}{3} E_P \approx \frac{G}{c^4} \frac{E_g^2}{R}[/math]

 

[math]\alpha \equiv \frac{3 G m^2 N^{2/3}}{\hbar c}[/math]

 

 

 

 

---

Classical regime (no gluon field weakening):

 

[math]E_P \approx N \left( \frac{\Delta p^2}{2 m} \right)[/math]

 

[math]=\frac{N}{2} mc^2 \left( \frac{\lambda_C}{\Delta x} \right)^2[/math]

 

 

[math]E_g^2 \approx E^2 = N^2 \left( \left(m c^2 \right)^2 + \left( c \Delta p \right)^2 \right) [/math]

 

[math] = N^2 \left(m c^2 \right)^2 \left(1 + \left( \frac{\lambda_C}{\Delta x} \right)^2 \right) [/math]

Let [math]r \equiv \frac{\Delta x}{\lambda_C}[/math], [math]R = N^{1/3} \Delta x = N^{1/3} \lambda_C r[/math]. Then

 

[math]\therefore \frac{1}{r^2} \approx \frac{3 G N m c^2}{c^4 R} \left(1 + \left( \frac{\lambda_C}{\Delta x} \right)^2 \right)[/math]

 

[math]= \frac{3 G N^{2/3} m}{c^2 \lambda_C r} \left(1 + \left( \frac{1}{r} \right)^2 \right)[/math]

 

[math]= \frac{3 G N^{2/3} m^2}{\hbar c r} \left(1 + \left( \frac{1}{r} \right)^2 \right)[/math]

 

[math]\therefore r \approx \left( \frac{3 G m^2 N^{2/3}}{\hbar c} \right) \left( r^2 + 1 \right)[/math]

 

[math]\boxed{r \approx \alpha \left( r^2 + 1 \right)}[/math]

Roots:

 

[math]r = \frac{1 + \sqrt{1 - 4 \alpha^2}}{2}[/math]

Collapsing from [math]r \gg 1[/math], the larger radius would be reached first (and would be more likely to remain within the Classical regime). Ergo, only the larger root is considered. So, in the Classical regime (with no gluon field weakening), quantum pressure (LHS) can forestall gravity (RHS), for some compression factor [math]\frac{1}{2} < r < 1[/math]. Solutions exist, for increasing initial mass [math]N[/math], at decreasing final compaction ratio [math]r[/math], until:

 

[math]1 - 4 \alpha^2 = 0[/math]

 

[math]\alpha = \frac{1}{2} = \frac{3 G m^2 N^{2/3}}{\hbar c}[/math]

 

[math]m N = m \left( \frac{1}{2} \times \frac{\hbar c}{3 G m^2} \right)^{3/2} \approx \frac{1}{2} M_{\odot}[/math]

Checking assumptions, for the maximum compression ratio [math]r = \frac{1}{2}[/math], the quantum energy is twice the rest-mass energy, i.e. verging into the Relativistic regime. Relativistic corrections could apply.

 

 

 

 

---

Classical regime (with gluon field weakening):

With gluon field weakening, [math]m \approx m_0 \times \frac{R}{R_0}[/math]. Redefining [math]r \equiv \frac{\Delta x}{\lambda_0}[/math], then [math]m \approx m_0 \frac{N^{1/3} \lambda_0 r}{R_0} \equiv m_0 \frac{r}{r_0}[/math], where [math]r_0 \equiv \frac{R_0}{N^{1/3} \lambda_0}[/math] is the initial compression factor. Now,

 

[math]E_P \approx N \left( \frac{\Delta p^2}{2 m} \right)[/math]

 

[math]= \frac{N}{2} \left( \frac{r_0}{r} \right) m_0 c^2 \left( \frac{\lambda_0}{\Delta x} \right)^2[/math]

 

[math]= \frac{N m_0 c^2 r_0}{2} \frac{1}{r^3}[/math]

 

[math]E_g^2 \approx E^2 = N^2 \left( \left(m c^2 \right)^2 + \left( c \Delta p \right)^2 \right) [/math]

 

[math]= N^2 \left(m_0 c^2 \right)^2 \left( \left( \frac{r}{r_0} \right)^2 + \left( \frac{\lambda_0}{\Delta x} \right)^2 \right) [/math]

 

[math]= \left(N m_0 c^2 \right)^2 \left( \left( \frac{r}{r_0} \right)^2 + \left( \frac{1}{r} \right)^2 \right) [/math]

Thus

 

[math]\therefore \frac{N m_0 c^2 r_0}{3} \frac{1}{r^3} \approx \frac{G}{c^4} \frac{1}{N^{1/3} \lambda_0} \frac{1}{r} \left(N m_0 c^2 \right)^2 \left( \left( \frac{r}{r_0} \right)^2 + \left( \frac{1}{r} \right)^2 \right) [/math]

 

[math]r_0 \approx \frac{3 G m_0^2 N^{2/3}}{\hbar c} \left( r^2 \left( \frac{r}{r_0} \right)^2 + 1 \right) [/math]

 

[math]\boxed{r_0 \approx \alpha \left( r^2 \left( \frac{r}{r_0} \right)^2 + 1 \right)}[/math]

 

[math]r_0 - \alpha \approx \frac{r^4}{r_0^2}[/math]

Solutions exist, for increasing initial mass [math]N[/math], at decreasing final compaction ratio [math]r[/math], until [math]\alpha = r_0[/math]:

 

[math]r_0 = \alpha = \frac{3 G m_0^2 N^{2/3}}{\hbar c}[/math]

 

[math]m N = m \left( r_0 \times \frac{\hbar c}{3 G m_0^2} \right)^{3/2} \approx 1.4 r_0^{3/2} M_{\odot}[/math]

For neutron-stars, the initial compaction ratio [math]r_0 \approx 10[/math], since nuclear density matter has a spacing of 1 nucleon per fm, whereas the Compton wavelength of 1GeV nucleons is 10% of that value. Thus, gluon field weakening permits compact objects of initial mass up to [math]\approx 40 M_{\odot}[/math]. However, at those initial masses, the final compaction ratio [math]r \rightarrow 0[/math], implying diverging quantum (Fermi/Heisenberg) energies, thus violating the assumption of Classicality. Relativistic corrections may apply.

 

 

 

 

---

Relativistic regime (no gluon field weakening):

 

[math]E_P \approx N \left( \Delta p c \right)[/math]

 

[math]= N mc^2 \left( \frac{\lambda_C}{\Delta x} \right)[/math]

 

[math]= N mc^2 \frac{1}{r}[/math]

 

 

[math]E_g^2 \approx E^2 = N^2 \left( \left(m c^2 \right)^2 + \left( c \Delta p \right)^2 \right) [/math]

 

[math]= \left(N m c^2 \right)^2 \left(1 + \left( \frac{\lambda_C}{\Delta x} \right)^2 \right)[/math]

 

[math]= \left(N m c^2 \right)^2 \left(1 + \left( \frac{1}{r} \right)^2 \right) [/math]

Thus

 

[math]\frac{2 N mc^2}{3 r} \approx \frac{G}{c^4 R} \left(N m c^2 \right)^2 \left(1 + \left( \frac{1}{r} \right)^2 \right) [/math]

 

[math]2 = \frac{3 G N^{2/3} m}{c^2 \lambda_C} \left(1 + \left( \frac{1}{r} \right)^2 \right)[/math]

 

[math]= \frac{3 G N^{2/3} m^2}{\hbar c} \left(1 + \left( \frac{1}{r} \right)^2 \right)[/math]

 

[math]\therefore 2 r^2 \approx \left( \frac{3 G m^2 N^{2/3}}{\hbar c} \right) \left( r^2 + 1 \right)[/math]

 

[math]\boxed{2 r^2 \approx \alpha \left( r^2 + 1 \right)}[/math]

 

Classical

:

[math]\therefore r \approx \alpha \left( r^2 + 1 \right)[/math], [math]\alpha \longrightarrow \frac{1}{2}[/math], [math]r \longrightarrow 1[/math]

Relativistic

:

[math]\therefore 2 r^2 \approx \alpha \left( r^2 + 1 \right)[/math], [math]\alpha \longrightarrow 2[/math], [math]r \longrightarrow \infty[/math]

 

[math]\left( 2 - \alpha \right) r^2 \approx \alpha[/math]

Solutions exist, for increasing initial mass [math]\alpha, N[/math], at increasing [math]r[/math], until [math]\alpha = 2[/math].

 

[math]\frac{3 G N^{2/3} m^2}{\hbar c} = 2[/math]

 

[math]m N = m \left( 2 \times \frac{\hbar c}{3 G m^2} \right)^{3/2} \approx 4 M_{\odot}[/math]

In the relativistic regime, Pressure can resist up to [math]2^3 \times[/math] as much mass, as compared to the Classical case. However, at those masses, the final compaction ratio diverges, implying uncompacted, i.e. non-Relativistic regimes. For a final compaction ratio of one, the limit of the Classical regime, the Relativistic equations imply [math]\alpha = 1[/math], implying an initial mass of [math]\approx 1.4 M_{\odot}[/math], nearly thrice as much as the Classical equations imply, for the same final compaction ratio.

 

 

 

 

---

Relativistic regime (with gluon field weakening):

 

[math]E_P \approx N \left( \Delta p c \right)[/math]

 

[math]= N m_0c^2 \left( \frac{\lambda_0}{\Delta x} \right)[/math]

 

[math]= N m_0c^2 \frac{1}{r}[/math]

 

 

[math]E_g^2 \approx E^2 = N^2 \left( \left(m c^2 \right)^2 + \left( c \Delta p \right)^2 \right) [/math]

 

[math]= \left(N m_0 c^2 \right)^2 \left( \left( \frac{r}{r_0} \right)^2 + \left( \frac{\lambda_0}{\Delta x} \right)^2 \right)[/math]

 

[math]= \left(N m_0 c^2 \right)^2 \left( \left( \frac{r}{r_0} \right)^2 + \left( \frac{1}{r} \right)^2 \right) [/math]

Thus

 

[math]\frac{2 N m_0c^2}{3 r} \approx \frac{G}{c^4 R} \left(N m_0 c^2 \right)^2 \left( \left( \frac{r}{r_0} \right)^2 + \left( \frac{1}{r} \right)^2 \right) [/math]

 

[math]2 \approx \frac{3 G N^{2/3} m_0}{2 c^2 \lambda_0} \left( \left( \frac{r}{r_0} \right)^2 + \left( \frac{1}{r} \right)^2 \right)[/math]

 

[math]\approx \frac{3 G m_0^2 N^{2/3}}{\hbar c} \left( \left( \frac{r}{r_0} \right)^2 + \left( \frac{1}{r} \right)^2 \right)[/math]

 

[math]\boxed{2 r^2 \approx \alpha \left( r^2 \left( \frac{r}{r_0} \right)^2 + 1 \right)}[/math]

Roots:

 

[math]r^2 = \frac{2 \pm \sqrt{ 4 - 4 \alpha^2 r_0^{-2}}}{2 \alpha r_0^{-2}}[/math]

 

[math]r^2 = r_0^2 \left( \frac{1 \pm \sqrt{1 - \alpha^2 r_0^{-2}} }{\alpha} \right)[/math]

Solutions exist, for increasing initial mass [math]N[/math], at decreasing final compaction ratio [math]r[/math], until

 

[math]1 = \alpha^2 r_0^{-2}[/math]

 

[math]\alpha = r_0[/math]

 

[math]= \frac{3 G N^{2/3} m^2}{\hbar c}[/math]

 

[math]m N = m \left( r_0 \frac{\hbar c}{3 G m^2} r_0 \right)^{3/2} \approx 1.4 r_0^{3/2} M_{\odot}[/math]

For neutron-stars, beginning collapse at nuclear density (1 nucleon per fm³), [math]r_0 \approx 10[/math], since [math]\lambda_0 \approx 10^{-16}m[/math]. Thus, gluon field weakening allows pressure to overcome gravity, for larger initial masses, up to about [math]40 M_{\odot}[/math]. At that critical threshold mass, the compression ratio:

 

[math]r^2 = \frac{r_0^2}{\alpha} = r_0 \approx 10[/math]

 

[math]r \approx 3[/math]

At that compression ratio, nucleons would be squeezed down to a third of "normal" size, and so would have been reduced to a third of their "normal" mass, i.e. only [math]\approx 15 M_{\odot}[/math] of mass would remain. Meanwhile, the nucleons' quantum (Fermi/Heisenberg) energies would only be about a third of their initial rest-mass-energies; yet, that would now equal their reduce-to-a-third-initial-rest-masses. So, the total mass-energy residing in the resulting compact object, equaling the sum of remaining mass-energy, and quantum energy, would be [math]\approx 30 M_{\odot}[/math]. Notably, the Classical & Relativistic equations both point to the same threshold initial mass of [math]\approx 40 M_{\odot}[/math].

Edited October 14, 2012 by Widdekind

[EquisDeXD]

Recall, throughout,

 

[math]\Delta p \approx \frac{\hbar}{\Delta x}[/math]

 

[math]\frac{2}{3} E_P \approx \frac{G}{c^4} \frac{E_g^2}{R}[/math]

 

---

Classical regime (no gluon field weakening):

 

[math]E_P \approx N \left( \frac{\Delta p^2}{2 m} \right)[/math]

 

[math]=\frac{N}{2} mc^2 \left( \frac{\lambda_C}{\Delta x} \right)^2[/math]

 

[math]E_g^2 \approx E^2 = N^2 \left( \left(m c^2 \right)^2 + \left( c \Delta p \right)^2 \right) [/math]

 

[math] = N^2 \left(m c^2 \right)^2 \left(1 + \left( \frac{\lambda_C}{\Delta x} \right)^2 \right) [/math]

Let [math]r \equiv \frac{\Delta x}{\lambda_C}[/math], [math]R = N^{1/3} \Delta x = N^{1/3} \lambda_C r[/math]. Then

 

[math]\therefore \frac{1}{r^2} \approx \frac{3 G N m c^2}{c^4 R} \left(1 + \left( \frac{\lambda_C}{\Delta x} \right)^2 \right)[/math]

 

[math]= \frac{3 G N^{2/3} m}{c^2 \lambda_C r} \left(1 + \left( \frac{1}{r} \right)^2 \right)[/math]

 

[math]= \frac{3 G N^{2/3} m^2}{\hbar c r} \left(1 + \left( \frac{1}{r} \right)^2 \right)[/math]

 

[math]\therefore r \approx \left( \frac{3 G m^2 N^{2/3}}{\hbar c} \right) \left( r^2 + 1 \right)[/math]

Roots:

 

[math]\alpha \equiv \frac{3 G m^2 N^{2/3}}{\hbar c}[/math]

 

[math]r = \frac{1 + \sqrt{1 - 4 \alpha^2}}{2}[/math]

Collapsing from [math]r \gg 1[/math], the larger radius would be reached first (and would be more likely to remain within the Classical regime). Ergo, only the larger root is considered. So, in the Classical regime (with no gluon field weakening), quantum pressure (LHS) can forestall gravity (RHS), for some compression factor [math]\frac{1}{2} < r < 1[/math], until the initial mass (Number of particles) exceeds:

 

[math]1 - 4 \alpha^2 = 0[/math]

 

[math]\alpha = \frac{1}{2} = \frac{3 G m^2 N^{2/3}}{\hbar c}[/math]

 

[math]m N = m \left( \frac{\hbar c}{6 G m^2} \right)^{3/2} \approx \frac{1}{2} M_{\odot}[/math]

Checking assumptions, for the maximum compression ratio [math]r = \frac{1}{2}[/math], the quantum energy is half of the rest-mass energy, i.e. remaining within the Classical regime. Trans-relativistic corrections could apply.

 

---

Classical regime (with gluon field weakening):

With gluon field weakening, [math]m \approx m_0 \times \frac{R}{R_0}[/math]. Redefining [math]r \equiv \frac{\Delta x}{\lambda_0}[/math], then [math]m \approx m_0 \frac{N^{1/3} \lambda_0 r}{R_0} \equiv m_0 \frac{r}{r_0} r[/math], where [math]r_0 \equiv \frac{R_0}{N^{1/3} \lambda_0}[/math] is the initial compression factor. Now,

 

[math]E_P \approx N \left( \frac{\Delta p^2}{2 m} \right)[/math]

 

[math]= \frac{N}{2} \left( \frac{r_0}{r} \right) m_0 c^2 \left( \frac{\lambda_0}{\Delta x} \right)^2[/math]

 

[math]= \frac{N m_0 c^2 r_0}{2} \frac{1}{r^3}[/math]

 

[math]E_g^2 \approx E^2 = N^2 \left( \left(m c^2 \right)^2 + \left( c \Delta p \right)^2 \right) [/math]

 

[math]= N^2 \left(m_0 c^2 \right)^2 \left( \left( r \right)^2 + \left( \frac{\lambda_0}{\Delta x} \right)^2 \right) [/math]

 

[math]= \left(N m_0 c^2 \right)^2 \left( r^2 + \left( \frac{1}{r} \right)^2 \right) [/math]

Thus

 

[math]\therefore \frac{N m_0 c^2 r_0}{3} \frac{1}{r^3} \approx \frac{G}{c^4} \frac{1}{N^{1/3} \lambda_0} \frac{1}{r} \left(N m_0 c^2 \right)^2 \left( r^2 + \left( \frac{1}{r} \right)^2 \right) [/math]

 

 

\frac{1}{r^2} \approx \frac{3 G N m c^2}{c^4 R} \left(1 + \left( \frac{\lambda_C}{\Delta x} \right)^2 \right)[/math]

 

[math]= \frac{3 G N^{2/3} m}{c^2 \lambda_C r} \left(1 + \left( \frac{1}{r} \right)^2 \right)[/math]

 

[math]= \frac{3 G N^{2/3} m^2}{\hbar c r} \left(1 + \left( \frac{1}{r} \right)^2 \right)[/math]

 

[math]\therefore r \approx \left( \frac{3 G m^2 N^{2/3}}{\hbar c} \right) \left( r^2 + 1 \right)[/math][/indent]

Roots:

 

[math]\alpha \equiv \frac{3 G m^2 N^{2/3}}{\hbar c}[/math]

 

[math]r = \frac{1 + \sqrt{1 - 4 \alpha^2}}{2}[/math]

Collapsing from [math]r \gg 1[/math], the larger radius would be reached first (and would be more likely to remain within the Classical regime). Ergo, only the larger root is considered. So, in the Classical regime (with no gluon field weakening), quantum pressure (LHS) can forestall gravity (RHS), for some compression factor [math]\frac{1}{2} < r < 1[/math], until the initial mass (Number of particles) exceeds:

 

[math]1 - 4 \alpha^2 = 0[/math]

 

[math]\alpha = \frac{1}{2} = \frac{3 G m^2 N^{2/3}}{\hbar c}[/math]

 

[math]m N = m \left( \frac{\hbar c}{6 G m^2} \right)^{3/2} \approx \frac{1}{2} M_{\odot}[/math]

But see, not only can we not use classical assertions but we don't have scientific evidence for what forms when the compression is too high, all we know is matter has to move to higher energy levels in order to not violate the Pauli Exclusion principal, not to mention the relativistic effects of time actually stopping after the event horizon to an outside observer.

Edited October 13, 2012 by EquisDeXD

[Widdekind]

But see, not only can we not use classical assertions but we don't have scientific evidence for what forms when the compression is too high, all we know is matter has to move to higher energy levels in order to not violate the Pauli Exclusion principal, not to mention the relativistic effects of time actually stopping after the event horizon to an outside observer.

 

Without speculated "gluon field weakening", the Classical-to-Relativistic equations imply maximum masses of [math]\approx \frac{1}{2} - \frac{3}{2} M_{\odot}[/math].

 

With those speculations, maximum (initial) masses of [math]\approx 40 M_{\odot}[/math] can survive gravitational collapse (by compressing and shedding gluon-generated mass-energy). Thus, the entire range of stellar-mass black-hole objects, up to dozens of solar masses (e.g. Cygnus X-1, SS 433), can be explained, by the "gluon field weakening" speculation. Galaxy-scale super-massive black-hole objects, may not be explicable. If so, then "star class" vs. "galaxy class" black hole objects may be qualitatively distinct.

 

I think you need more scientific evidence, because if gravity is powerful enough to surpass the strong force, what exactly do we have left over? We have no evidence that quarks break down into any other particle, and supposedly a singularity is infinitely dense, and furthermore a gluon field is something that each individual quark has, and the exchange of gluons occurs between quarks, and I don't know if there's any evidence to support that the exchange of gluons stops after a certain point. We can create degenerate matter here on Earth, but I still haven't heard of what your saying, and we most certainly don't know enough about the composition of a singularity to be able to apply conventional physics to it. And then, light particles aren't completely physical objects, they can't implode in on themselves. You can reflect them and create interference, stretch or shrink their frequency or stretch or shrink their wavelength, but implode doesn't have a meaning in wave mechanics.

 

First, if gravity overcame the Strong force, then (inexpertly, simplistically) quarks would simply stop generating gluons, since gravity & pressure would keep them confined; and gluons are only generated, when quarks try to stray away from each other.

 

Second, relatedly, the Strong force is explicitly short-ranged; gluons do not propagate more than ~1 fm. Beyond that distance, the energy in the gluon field has already become so intense, that new quark-antiquark pair production occurs.

 

Third, you could (hypothetically) create a black-hole from photons of light, if you compressed their combined energy = mass x c², within their combined Schwarzschild radius R_S = G E / c⁴.

 

Fourth, i offer that you may have heard (implicitly) of these speculated phenomena. For, current physical theory, without "gluon field weakening", only explains low-mass compact objects, barely above one solar mass. Whereas, "gluon field weakening" allows objects dozens of times more massive, to survive gravitational collapse (by shedding gluon-generated mass). Thereby, this simple parsimonious picture explains all known "star-class" stellar-massed black holes, up to dozens of solar masses, e.g. Cygnus X-1, SS 433.

[EquisDeXD]

Without speculated "gluon field weakening", the Classical-to-Relativistic equations imply maximum masses of [math]\approx \frac{1}{2} - \frac{3}{2} M_{\odot}[/math].

 

With those speculations, maximum (initial) masses of [math]\approx 40 M_{\odot}[/math] can survive gravitational collapse (by compressing and shedding gluon-generated mass-energy). Thus, the entire range of stellar-mass black-hole objects, up to dozens of solar masses (e.g. Cygnus X-1, SS 433), can be explained, by the "gluon field weakening" speculation. Galaxy-scale super-massive black-hole objects, may not be explicable. If so, then "star class" vs. "galaxy class" black hole objects may be qualitatively distinct.

Actually, I guess that does kind of make sense because gluon fields do seem to get weaker as you get closer to the source, but at such a small scale you'd have to probably describe the singularity as one equation since you wouldn't be able to distinguish between many particles, but conventionally you don't need field weakinging to explain the black hole size, if it has more mass, it's bigger, gluon fields don't need to weaken. The more mass that's added, the larger the distortion in the fabric of space, which means there's a greater boundary for the event horizon.

 

 

 

First, if gravity overcame the Strong force, then (inexpertly, simplistically) quarks would simply stop generating gluons, since gravity & pressure would keep them confined; and gluons are only generated, when quarks try to stray away from each other.

No, the exchange of gluons is analogous and quarks can't be isolated, they can't "stray away" unless you put high amounts of energy on them, which when you do the energy get's converted into more quarks making it so that no matter what a quark is always paired with one other quark, and in order for that to happen there has to be an exchance of gluons.

 

Second, relatedly, the Strong force is explicitly short-ranged; gluons do not propagate more than ~1 fm. Beyond that distance, the energy in the gluon field has already become so intense, that new quark-antiquark pair production occurs.

Really? I thought it depended on the relativistic mass of a gauge boson.

 

Third, you could (hypothetically) create a black-hole from photons of light, if you compressed their combined energy = mass x c², within their combined Schwarzschild radius R_S = G E / c⁴.

Well energy does have a relative mass equivalence, but it's pretty hard to create a black hole out of light, even the strongest light beams in the world don't distort the fabric of space too much.

 

Fourth, i offer that you may have heard (implicitly) of these speculated phenomena. For, current physical theory, without "gluon field weakening", only explains low-mass compact objects, barely above one solar mass. Whereas, "gluon field weakening" allows objects dozens of times more massive, to survive gravitational collapse (by shedding gluon-generated mass). Thereby, this simple parsimonious picture explains all known "star-class" stellar-massed black holes, up to dozens of solar masses, e.g. Cygnus X-1, SS 433.

Or you could look at my theory which merely states that the wave functions combine, thus creating a more massive particle that has an unbelievably small uncertainty, which allows mass to constantly be added and a singularity to always become smaller.

Edited October 14, 2012 by EquisDeXD

[illuusio]

Nucleons are ~10^-15m (=1 fm) across. The gluon-field between the three quarks, in nucleons, accounts for 99% of the mass-energy of the nucleon; "bare naked" quarks only mass ~3-4 MeV. And, the gluon field can be described by a potential energy, which increases linearly, with inter-quark separation distance ®, approximately U = +(1GeV / fm) x r. Ipso facto, if neutronized matter, in a relativistically compact object, were to collapse beyond the neutron-star stage; then gravity would compress the quarks, within the neutrons, together, reducing their separation distances, and so weakening their gluon fields. In effect, gravity would "take over" for the strong force, confining quarks without need for any Strong-Force-gluon-field. Compressing neutronized matter, by a factor of ~100, would reduce the mass-energy of nucleons, by the same factor, i.e. eliminate the gluon-field, reducing total effective mass-energy, down to the raw "bare naked" quarks themselves, whose total combined raw mass-energy is ~10MeV (=0.01GeV).

 

 

 

Few pointers... If gravitation can take over strong force to me it sounds that they both are based on same phenomenon. Another thing is compressing quarks together. Because baryons do have rotation frequency, compressing them together (into contact), would cause them to annihilate before compressed together. Same applies to quarks (because quarks generate baryons rotation frequency). To some extend you can compress baryons together but there is a limit for sure.

[EquisDeXD]

Few pointers... If gravitation can take over strong force to me it sounds that they both are based on same phenomenon. Another thing is compressing quarks together. Because baryons do have rotation frequency, compressing them together (into contact), would cause them to annihilate before compressed together.

If they annihilate, what do the composite particles compress into? And what happens to the energy? How is it "compressed" into the singularity?

 

Same applies to quarks (because quarks generate baryons rotation frequency). To some extend you can compress baryons together but there is a limit for sure.

 

I was wondering what happened after the point of degeneracy, but have we actually observed this happening? Can we actually test this? I know degenerate matter can be made on Earth, but can we overcome the strong force?

Edited October 14, 2012 by EquisDeXD

[illuusio]

If they annihilate, what do the composite particles compress into? And what happens to the energy? How is it "compressed" into the singularity?

 

 

 

I was wondering what happened after the point of degeneracy, but have we actually observed this happening? Can we actually test this? I know degenerate matter can be made on Earth, but can we overcome the strong force?

 

In "slow" compression they compress mainly into photons and to smaller material (E***r particles) and energy is distributed to all over. Singularity, nah, that train went few billion years ago.

You can use particle colliders they show what happens with too much "compression".

[EquisDeXD]

In "slow" compression they compress mainly into photons and to smaller material (E***r particles) and energy is distributed to all over. Singularity, nah, that train went few billion years ago.

 

So I know that as a photon approaches the event horizon that it's wave-length stretches as it moves to a lower energy potential, but relative to the observer while inside the black hole this happens as it approaches the singularity doesn't it? So if matter get's compressed into light, but light's wave-length becomes larger, wouldn't the singularity essentially be composed of radio waves? How can photons be compressed if their uncertainty increases? It seems like a photon at the exact position of the singularity would have infinite uncertainty, assuming I'm interpreting it right, but I guess that would explain the thermo-dynamical effects of black holes and how they would evaporate if the uncertainty of the interior energy of the black hole can extend beyond the event horizon.

Also, does this mean that currently my theory that matter compressed belong the point of degeneracy can be treated as a single wave function is no good? Unless you can treat a compost of light waves trapped in a space as such...

 

You can use particle colliders they show what happens with too much "compression".

Are you sure those photons just aren't the byproduct of the energy put into accelerating the particles in the first place? Because last I've heard scientists made a "quark-gluon" plasma, nothing about annihilating quarks.

Edited October 14, 2012 by EquisDeXD

[illuusio]

So I know that as a photon approaches the event horizon that it's wave-length stretches as it moves to a lower energy potential, but relative to the observer while inside the black hole this happens as it approaches the singularity doesn't it? So if matter get's compressed into light, but light's wave-length becomes larger, wouldn't the singularity essentially be composed of radio waves? How can photons be compressed if their uncertainty increases? It seems like a photon at the exact position of the singularity would have infinite uncertainty, assuming I'm interpreting it right, but I guess that would explain the thermo-dynamical effects of black holes and how they would evaporate if the uncertainty of the interior energy of the black hole can extend beyond the event horizon.

Also, does this mean that currently my theory that matter compressed belong the point of degeneracy can be treated as a single wave function is no good? Unless you can treat a compost of light waves trapped in a space as such...

 

 

Are you sure those photons just aren't the byproduct of the energy put into accelerating the particles in the first place? Because last I've heard scientists made a "quark-gluon" plasma, nothing about annihilating quarks.

 

 

Within that generated plasma there is not enough pressure to create annihilation of quarks (repulsion still exists).

 

When photon approaches the event horizon of black hole it will gain energy (blue shifting).

 

 

[EquisDeXD]

When photon approaches the event horizon of black hole it will gain energy (blue shifting).

 

Wait, how does the photon "gain" energy by moving to a lower energy potential? And still, what is a singularity made out of and then if there's no increase in uncertainty how can heat leak out from a black hole? How do photons even get compressed?

Edited October 14, 2012 by EquisDeXD

[Klaynos]

!

Moderator Note

Off topic threads split to here;

http://www.scienceforums.net/topic/69743-illusio-on-ether-and-blueshift/

[Widdekind]

Recall, throughout,

 

[math]\Delta p \approx \frac{\hbar}{\Delta x}[/math]

 

[math]\frac{2}{3} E_P \approx \frac{G}{c^4} \frac{E_g^2}{R}[/math]

 

[math]\alpha \equiv \frac{3 G m^2 N^{2/3}}{\hbar c}[/math]

 

 

 

 

---

trans-Classical regime (with gluon field weakening):

Adding the fourth-order (trans-)relativistic correction terms:

 

[math]\rho \equiv \frac{p}{mc}[/math]

 

[math]E - mc^2 \approx mc^2 \left( \frac{1}{2} \rho^2 + \frac{3}{8} \rho^4 \right)[/math]

 

With gluon field weakening, [math]m \approx m_0 \times \frac{R}{R_0}[/math]. Redefining [math]r \equiv \frac{\Delta x}{\lambda_0}[/math], then [math]m \approx m_0 \frac{N^{1/3} \lambda_0 r}{R_0} \equiv m_0 \frac{r}{r_0}[/math], where [math]r_0 \equiv \frac{R_0}{N^{1/3} \lambda_0}[/math] is the initial compression factor. Now,

 

[math]\rho \equiv \frac{\Delta p}{m c} \approx \frac{h c}{m c^2} \frac{1}{\Delta x} = \frac{m_0}{m} \frac{\lambda_{C,0}}{\Delta x} \equiv \frac{r_0}{r} \frac{1}{r} = \frac{r_0}{r^2}[/math]

 

[math]\boxed{\rho = \frac{r_0}{r^2}}[/math]

 

 

[math]E_P \approx N mc^2 \left( \frac{1}{2} \rho^2 + \frac{3}{8} \rho^4 \right)[/math]

 

[math]= N m_0c^2 \frac{r}{r_0} \left( \frac{r_0^2}{2 r^4} \right) \left( 1+ \frac{3 r_0^2}{4 r^4} \right)[/math]

 

[math]\boxed{E_P \approx \left( N m_0c^2\right) \left( \frac{r_0}{2 r^3} \right) \left( 1+ \frac{3 r_0^2}{4 r^4} \right)}[/math]

 

 

[math]E_g^2 \approx N^2 \left( \left(m c^2 \right)^2 + \left( c \Delta p \right)^2 \right) [/math]

 

[math]= N^2 \left(m_0 c^2 \right)^2 \left( \left( \frac{r}{r_0} \right)^2 + \left( \frac{\lambda_{C,0}}{\Delta x} \right)^2 \right) [/math]

 

[math]\boxed{E_g^2 \approx \left(N m_0 c^2 \right)^2 \left( \left( \frac{r}{r_0} \right)^2 + \left( \frac{1}{r} \right)^2 \right)}[/math]

Thus

 

[math]\frac{2}{3} \left( N m_0c^2\right) \left( \frac{r_0}{2 r^3} \right) \left( 1+ \frac{3 r_0^2}{4 r^4} \right) \approx \frac{G}{c^4} \frac{1}{R} \left(N m_0 c^2 \right)^2 \left( \left( \frac{r}{r_0} \right)^2 + \left( \frac{1}{r} \right)^2 \right)[/math]

 

[math]\left( \frac{r_0}{r^3} \right) \left( 1+ \frac{3 r_0^2}{4 r^4} \right) \approx \frac{3 G}{c^4} \frac{1}{N^{1/3} \Delta x} \frac{\lambda_{C,0}}{\lambda_{C,0}} \left(N m_0 c^2 \right) \left( \left( \frac{r}{r_0} \right)^2 + \left( \frac{1}{r} \right)^2 \right)[/math]

 

[math]\left( \frac{r_0}{r^3} \right) \left( 1+ \frac{3 r_0^2}{4 r^4} \right) \approx \frac{3 G m_0^2 N^{2/3}}{\hbar c} \frac{1}{r} \left( \left( \frac{r}{r_0} \right)^2 + \left( \frac{1}{r} \right)^2 \right)[/math]

 

[math]\left( \frac{r_0}{r^2} \right) \left( 1+ \frac{3 r_0^2}{4 r^4} \right) \approx \alpha \left( \left( \frac{r}{r_0} \right)^2 + \left( \frac{1}{r} \right)^2 \right)[/math]

 

[math]\boxed{r_0 \left( 1+ \frac{3 r_0^2}{4 r^4} \right) \approx \alpha \left( r^2 \left( \frac{r}{r_0} \right)^2 + 1 \right)}[/math]

 

[math]r_0^3 \left( r^4 + \frac{3 r_0^2}{4} \right) \approx \alpha \left( r^8 + r_0^2 r^4 \right)[/math]

 

[math]r^4 \left( r_0^3 - \alpha r_0^2 \right) \approx r^8 \alpha - \frac{3 r_0^5}{4}[/math]

Roots:

 

[math]r^4 = \frac{r_0^2 \left( r_0 - \alpha \right) \pm \sqrt{r_0^4 \left( r_0 - \alpha \right)^2 + 3 \alpha r_0^5}}{2 \alpha}[/math]

 

[math]r^4 = r_0^2 \frac{\left( r_0 - \alpha \right) \pm \sqrt{\left( r_0 - \alpha \right)^2 + 3 \alpha r_0}}{2 \alpha}[/math]

 

[math]r^4 = r_0^2 \frac{\left( r_0 - \alpha \right) \pm \sqrt{r_0^2 + \alpha^2 + \alpha r_0}}{2 \alpha}[/math]

Prima facie, stable solutions exist for any initial mass [math]\left( \propto \alpha \right)[/math], at decreasing compression ratio r.

 

 

 

 

---

estimated Pressures inside neutron-stars

[math]\frac{P}{R} \approx \frac{3 G M^2}{4 \pi R^5}[/math]

 

[math]P \approx \frac{3 G}{4 \pi M^2} \times \left( \frac{c^2}{2 G} \right)^4 \left( \frac{R_S}{R} \right)^4[/math]

 

[math]= \frac{c^8}{64 \pi G^3 M^2} \left( \frac{R_S}{R} \right)^4[/math]

 

[math]\approx \frac{3 \times 10^{35} Pa}{m^2 r^4}[/math]

 

[math]\approx \frac{3 \times 10^{30} bar}{m^2 r^4}[/math]

where m measures the neutron-star mass in solar masses, and r measures the neutron-star radius in Schwarzschild radii. For a one-and-a-half solar-mass neutron-star, at three Schwarzschild radii, that would be about 10³³ Pa = 10¹⁶ Tbar. By comparison, the maximum tension sustainable, by a hypothetical "gluon cable" (derived from slowly stretching a quark out of its original nucleon), would be about:

 

F / A = (1 GeV / fm) / fm

²

 

= 10

³⁵

Pa

 

= 10

¹⁸

Tbar

Thus, as neutron-stars compress down towards their Schwarzschild radii, internal pressures approach maximum theoretical nuclear pressures.

Edited October 15, 2012 by Widdekind

[illuusio]

 

 

---

estimated Pressures inside neutron-stars

[math]\frac{P}{R} \approx \frac{3 G M^2}{4 \pi R^5}[/math]

 

[math]P \approx \frac{3 G}{4 \pi M^2} \times \left( \frac{c^2}{2 G} \right)^4 \left( \frac{R_S}{R} \right)^4[/math]

 

[math]= \frac{c^8}{64 \pi G^3 M^2} \left( \frac{R_S}{R} \right)^4[/math]

 

[math]\approx \frac{3 \times 10^{35} Pa}{m^2 r^4}[/math]

 

[math]\approx \frac{3 \times 10^{30} bar}{m^2 r^4}[/math]

where m measures the neutron-star mass in solar masses, and r measures the neutron-star radius in Schwarzschild radii. For a one-and-a-half solar-mass neutron-star, at three Schwarzschild radii, that would be about 10³³ Pa = 10¹⁶ Tbar. By comparison, the maximum tension sustainable, by a hypothetical "gluon cable" (derived from slowly stretching a quark out of its original nucleon), would be about:

 

F / A = (1 GeV / fm) / fm

²

 

= 10

³⁵

Pa

 

= 10

¹⁸

Tbar

Thus, as neutron-stars compress down towards their Schwarzschild radii, internal pressures approach maximum theoretical nuclear pressures.

 

How much is the internal pressure inside of a black hole then?

[EquisDeXD]

Recall, throughout,

 

[math]\Delta p \approx \frac{\hbar}{\Delta x}[/math]

 

[math]\frac{2}{3} E_P \approx \frac{G}{c^4} \frac{E_g^2}{R}[/math]

 

[math]\alpha \equiv \frac{3 G m^2 N^{2/3}}{\hbar c}[/math]

 

 

 

 

---

trans-Classical regime (with gluon field weakening):

Adding the fourth-order (trans-)relativistic correction terms:

 

[math]\rho \equiv \frac{p}{mc}[/math]

 

[math]E - mc^2 \approx mc^2 \left( \frac{1}{2} \rho^2 + \frac{3}{8} \rho^4 \right)[/math]

 

With gluon field weakening, [math]m \approx m_0 \times \frac{R}{R_0}[/math]. Redefining [math]r \equiv \frac{\Delta x}{\lambda_0}[/math], then [math]m \approx m_0 \frac{N^{1/3} \lambda_0 r}{R_0} \equiv m_0 \frac{r}{r_0}[/math], where [math]r_0 \equiv \frac{R_0}{N^{1/3} \lambda_0}[/math] is the initial compression factor. Now,

 

[math]\rho \equiv \frac{\Delta p}{m c} \approx \frac{h c}{m c^2} \frac{1}{\Delta x} = \frac{m_0}{m} \frac{\lambda_{C,0}}{\Delta x} \equiv \frac{r_0}{r} \frac{1}{r} = \frac{r_0}{r^2}[/math]

 

[math]\boxed{\rho = \frac{r_0}{r^2}}[/math]

 

 

[math]E_P \approx N mc^2 \left( \frac{1}{2} \rho^2 + \frac{3}{8} \rho^4 \right)[/math]

 

[math]= N m_0c^2 \frac{r}{r_0} \left( \frac{r_0^2}{2 r^4} \right) \left( 1+ \frac{3 r_0^2}{4 r^4} \right)[/math]

 

[math]\boxed{E_P \approx \left( N m_0c^2\right) \left( \frac{r_0}{2 r^3} \right) \left( 1+ \frac{3 r_0^2}{4 r^4} \right)}[/math]

 

 

[math]E_g^2 \approx N^2 \left( \left(m c^2 \right)^2 + \left( c \Delta p \right)^2 \right) [/math]

 

[math]= N^2 \left(m_0 c^2 \right)^2 \left( \left( \frac{r}{r_0} \right)^2 + \left( \frac{\lambda_{C,0}}{\Delta x} \right)^2 \right) [/math]

 

[math]\boxed{E_g^2 \approx \left(N m_0 c^2 \right)^2 \left( \left( \frac{r}{r_0} \right)^2 + \left( \frac{1}{r} \right)^2 \right)}[/math]

Thus

 

[math]\frac{2}{3} \left( N m_0c^2\right) \left( \frac{r_0}{2 r^3} \right) \left( 1+ \frac{3 r_0^2}{4 r^4} \right) \approx \frac{G}{c^4} \frac{1}{R} \left(N m_0 c^2 \right)^2 \left( \left( \frac{r}{r_0} \right)^2 + \left( \frac{1}{r} \right)^2 \right)[/math]

 

[math]\left( \frac{r_0}{r^3} \right) \left( 1+ \frac{3 r_0^2}{4 r^4} \right) \approx \frac{3 G}{c^4} \frac{1}{N^{1/3} \Delta x} \frac{\lambda_{C,0}}{\lambda_{C,0}} \left(N m_0 c^2 \right) \left( \left( \frac{r}{r_0} \right)^2 + \left( \frac{1}{r} \right)^2 \right)[/math]

 

[math]\left( \frac{r_0}{r^3} \right) \left( 1+ \frac{3 r_0^2}{4 r^4} \right) \approx \frac{3 G m_0^2 N^{2/3}}{\hbar c} \frac{1}{r} \left( \left( \frac{r}{r_0} \right)^2 + \left( \frac{1}{r} \right)^2 \right)[/math]

 

[math]\left( \frac{r_0}{r^2} \right) \left( 1+ \frac{3 r_0^2}{4 r^4} \right) \approx \alpha \left( \left( \frac{r}{r_0} \right)^2 + \left( \frac{1}{r} \right)^2 \right)[/math]

 

[math]\boxed{r_0 \left( 1+ \frac{3 r_0^2}{4 r^4} \right) \approx \alpha \left( r^2 \left( \frac{r}{r_0} \right)^2 + 1 \right)}[/math]

 

[math]r_0^3 \left( r^4 + \frac{3 r_0^2}{4} \right) \approx \alpha \left( r^8 + r_0^2 r^4 \right)[/math]

 

[math]r^4 \left( r_0^3 - \alpha r_0^2 \right) \approx r^8 \alpha - \frac{3 r_0^5}{4}[/math]

Roots:

 

[math]r^4 = \frac{r_0^2 \left( r_0 - \alpha \right) \pm \sqrt{r_0^4 \left( r_0 - \alpha \right)^2 + 3 \alpha r_0^5}}{2 \alpha}[/math]

 

[math]r^4 = r_0^2 \frac{\left( r_0 - \alpha \right) \pm \sqrt{\left( r_0 - \alpha \right)^2 + 3 \alpha r_0}}{2 \alpha}[/math]

 

[math]r^4 = r_0^2 \frac{\left( r_0 - \alpha \right) \pm \sqrt{r_0^2 + \alpha^2 + \alpha r_0}}{2 \alpha}[/math]

Prima facie, stable solutions exist for any initial mass [math]\left( \propto \alpha \right)[/math], at decreasing compression ratio r.

 

 

 

 

---

estimated Pressures inside neutron-stars

[math]\frac{P}{R} \approx \frac{3 G M^2}{4 \pi R^5}[/math]

 

[math]P \approx \frac{3 G}{4 \pi M^2} \times \left( \frac{c^2}{2 G} \right)^4 \left( \frac{R_S}{R} \right)^4[/math]

 

[math]= \frac{c^8}{64 \pi G^3 M^2} \left( \frac{R_S}{R} \right)^4[/math]

 

[math]\approx \frac{3 \times 10^{35} Pa}{m^2 r^4}[/math]

 

[math]\approx \frac{3 \times 10^{30} bar}{m^2 r^4}[/math]

where m measures the neutron-star mass in solar masses, and r measures the neutron-star radius in Schwarzschild radii. For a one-and-a-half solar-mass neutron-star, at three Schwarzschild radii, that would be about 10³³ Pa = 10¹⁶ Tbar. By comparison, the maximum tension sustainable, by a hypothetical "gluon cable" (derived from slowly stretching a quark out of its original nucleon), would be about:

 

F / A = (1 GeV / fm) / fm

²

 

= 10

³⁵

Pa

 

= 10

¹⁸

Tbar

Thus, as neutron-stars compress down towards their Schwarzschild radii, internal pressures approach maximum theoretical nuclear pressures.

 

But gluon fields have strength varying at distance, I think by the square of the distance in fact. If quarks were compressed to an even greater point beyond degeneracy wouldn't that make the gluon field between any two quarks ever stronger? And, if matter is compressed into photons, how does weakening a gluon field do anything? Photons have no color charge.

Edited October 15, 2012 by EquisDeXD

[Widdekind]

How much is the internal pressure inside of a black hole then?

Lots ?

 

(FYI, the close correspondence between the NS internal pressure, and nuclear energy density, is (on second thought) not accidental -- by definition, NS are those space objects compressed to nuclear mass(-energy) density. (And Pressure is proportional to energy density.) So the fact that all the numbers work out so (astrophysically) closely is not surprising. Nevertheless, those energy densities are order-of-magnitude 10³⁵ J m^-3.)

 

But gluon fields have strength varying at distance, I think by the square of the distance in fact. If quarks were compressed to an even greater point beyond degeneracy wouldn't that make the gluon field between any two quarks ever stronger? And, if matter is compressed into photons, how does weakening a gluon field do anything? Photons have no color charge.

No -- in my understanding, Gluon fields are not modeled with (1/r) potentials, like EM fields. They are modeled with linear ® potentials, so modeling a force constant with distance. The Strong (gluon) force only "kicks in" when quarks stray far from each other; at zero separation, there is zero interaction (called "asymptotic freedom").

 

Quarks are not photons. But, when relativistic, such that their energies >> rest-mass energies, then relativistic quarks move like photons. (They still keep their color charges, and still interact strongly)

[EquisDeXD]

No -- in my understanding, Gluon fields are not modeled with (1/r) potentials, like EM fields. They are modeled with linear ® potentials, so modeling a force constant with distance. The Strong (gluon) force only "kicks in" when quarks stray far from each other; at zero separation, there is zero interaction (called "asymptotic freedom").

Hmm, that seems highly contradictory to what I remember, do you have any links?

 

Quarks are not photons. But, when relativistic, such that their energies >> rest-mass energies, then relativistic quarks move like photons. (They still keep their color charges, and still interact strongly)

Quarks move like photons? I've never heard of anything like that, if matter gets compressed beyond the point of degeneracy, then a singularity has to either be a lot larger which you can't prove and doesn't seem to follow conventional particle physics since quarks only have a finite relative size for how small they can get, or there's no quarks.