Hello, first of all this is great playlist thanks. I also want to ask what is required to make the eccentricity zero ? Constant “A” or maybe angular momentum(l0) ? It looks like it cant be l=0 but in QM, electron has a orbit even angular momentum eigenvalue (l) of zero. Hope you can help on that, thanks alot.
Thanks for watching, glad the videos are useful for you! The case of l0=0 corresponds to a radial "orbit" (purely 1D motion in which case the equation of motion is mu r-double-dot = F(r)). So no circle, ellipse, hyperbola or parabola... For a circular orbit, you need the final expression for r(phi) to equal a constant (c) for all angles phi. For that to happen, you need epsilon=0, which means A=0. In that case we have w(phi)=0 as well (which places a constraint on u(phi) in terms of gamma*mu/l0^2).
@@jamesbattat4672 I haven’t got any notifications and didn’t know that you replied but thank you very much for the reply and help. I guess my confusion is coming from that the electron which has n=0 l=0 has a spherical shaped probablity while l=0 corresponds to 1D motion. I know that the solution has angular part (which corresponds to the spherical harmonics) besides radial but still I have no idea why the full solution which is multipication of angular and radial components is shaped as sphere or how does radial solution effect the shape of the probability distribution. I know you are not explaning electron orbit exclusively but in my QM course, the radial solution of electron same with this.
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Thank you for these videos James. They are amazing and they help a lot to understand the concepts of these topics. Much appreciated, simply amazing.
Glad they are helpful!
Hello, first of all this is great playlist thanks. I also want to ask what is required to make the eccentricity zero ? Constant “A” or maybe angular momentum(l0) ? It looks like it cant be l=0 but in QM, electron has a orbit even angular momentum eigenvalue (l) of zero. Hope you can help on that, thanks alot.
Thanks for watching, glad the videos are useful for you! The case of l0=0 corresponds to a radial "orbit" (purely 1D motion in which case the equation of motion is mu r-double-dot = F(r)). So no circle, ellipse, hyperbola or parabola... For a circular orbit, you need the final expression for r(phi) to equal a constant (c) for all angles phi. For that to happen, you need epsilon=0, which means A=0. In that case we have w(phi)=0 as well (which places a constraint on u(phi) in terms of gamma*mu/l0^2).
@@jamesbattat4672 I haven’t got any notifications and didn’t know that you replied but thank you very much for the reply and help. I guess my confusion is coming from that the electron which has n=0 l=0 has a spherical shaped probablity while l=0 corresponds to 1D motion. I know that the solution has angular part (which corresponds to the spherical harmonics) besides radial but still I have no idea why the full solution which is multipication of angular and radial components is shaped as sphere or how does radial solution effect the shape of the probability distribution. I know you are not explaning electron orbit exclusively but in my QM course, the radial solution of electron same with this.