Recent content by Sergio Rodriguez

Thank you! It's the first chapter. Section 1.20 Matrix Elements. Example 1.9, Self-test 1.9, page 33.

I have do this: As I don't know if f is an eigenfunction of Ω, I expand it as ## f = \sum_n c_nf_n## where ##f_n## are orthonormal eigenfunctions of Ω, so: ##\left<Ω\right> = \left<f\left|Ω\right|f\right> = \left<f\left|1*Ω*1\right|f\right> ## ##\sum_m\sum_n...

Homework Statement Show that if ##\left( \Omega f\right) ^* = -\Omega f ^* ## then ##\left< \Omega \right> = 0 ## for any real function f. where ##\Omega## is an operator Homework Equations It's a self test of the completeness relation --Molecular quantum mechanics (Atkins)-- so the equation...

Finally I got it! To simplify my expression I try the two ways you said me. First way: Δr = \frac {4πr^3f}{Gm_sM_e + 4πfr^2} Δr = \frac {\frac {4πr^3f}{r^2}}{\frac{Gm_sM_e}{r^2} + \frac {4πfr^2}{r^2}} Δr = \frac {4πrf}{\frac{Gm_sM_e}{r^2} + 4πf} Δr = \frac {4πrf}{GF + 4πf} and considering...

I have corrected my expression and now is dimensionally correct. The units is metres, the meters the radius has decrease in that lap when the radius is r. Now I try to find dr/dt 2πrf = \frac {-Gm_sM_e} {2r} + \frac {Gm_sM_e} {2(r - Δr} 2r(r - Δr) 2πrf = - (r-Δr) Gm_sM_e + rGm_sM_e...

Checking the dimensions of the two expressions, now I see that the book's one is speed units, and mine is dimensionally wrong, so I try to correct it but also try to find the book solution. Thanks!

Thanks, haruspex! I ' ll post them. And also check the units.

That expression of ME already has the KE inside.

Thanks! I'll check it.

Homework Statement A satellite of mass m_s orbits the Earth in a circular orbit of radius r_0. If the satellite orbits at the upper part of the atmosphere and the friction force f is constant, it would trace an spiral and fall to the earth. but if we suppouse that the friction force is small...

Yes, the KE is always positive. The GPE could be zero at the lowest level and positive at the initial position or begin with zero and negative later. But the SPE is positive because the spring is stretching and gaining energy.

Why do you use two different velocities for two tied bodies? And the system is set in motion by the gravity, and there is no GPE in the equations.

I don't know if you are looking after this, where the kinetic energy appears by calculation of the work done by a force: $$w= \int_A^B F dx =\int_A^B m \frac {dv} {dt} dx \, = m\int_A^B dv \frac {dx} {dt} =m\int_A^B vdv$$ $$w= m\left[ \frac {v^2} 2 \right] _A^B = m \frac 1 2 v_B^2 -...

I have the proof for the 8) As the second derivative of P is too long to do it, It's easier to proof that slope of the curve p(x) with x < b/2 is negative and the slope of the curve with x > b/2 is positive. I have do it and there is a minimum in x = b/2