- #36
arivero
Gold Member
- 3,498
- 175
setting the record
Just to set the record straight, it is true that we must always keep the scent of two masses [tex]m_1,m_2[/tex]. In fact in the first calculations from Orion the [tex]m[/tex] in the angular moment is not the same [tex]m[/tex] that under the square roots, but that is only a abuse of notation and the result is the right one.
Still, one does not need to go deep to Plank mass and radius too soon. For the area swept by the particle 1 around the center of mass, it suffices to substitute
[tex]m_2 \to {m_2^3 \over (m_1+m_2)^2}[/tex]
and reciprocally for the particle 2. The analysis in terms of compton length is still possible.
I am unsure of which area to quantize in this case, if one or another or the sum, nor to speak of which reference frame to use. So lacking of a fundamental theory, it seems we aren't going to get some additional benefit from this additional freedom.
Perhaps a minor difference with the Keplerian case is that we have two masses to add for the total angular momentum. So the additional condition [tex]L=\hbar[/tex] does not imply directly [tex]m=m_P[/tex] now.
Just to set the record straight, it is true that we must always keep the scent of two masses [tex]m_1,m_2[/tex]. In fact in the first calculations from Orion the [tex]m[/tex] in the angular moment is not the same [tex]m[/tex] that under the square roots, but that is only a abuse of notation and the result is the right one.
Still, one does not need to go deep to Plank mass and radius too soon. For the area swept by the particle 1 around the center of mass, it suffices to substitute
[tex]m_2 \to {m_2^3 \over (m_1+m_2)^2}[/tex]
and reciprocally for the particle 2. The analysis in terms of compton length is still possible.
I am unsure of which area to quantize in this case, if one or another or the sum, nor to speak of which reference frame to use. So lacking of a fundamental theory, it seems we aren't going to get some additional benefit from this additional freedom.
Perhaps a minor difference with the Keplerian case is that we have two masses to add for the total angular momentum. So the additional condition [tex]L=\hbar[/tex] does not imply directly [tex]m=m_P[/tex] now.