Ballentine Equation 5.13 on conservation of momentum

In summary, Ballentine uses geometrical arguments to obtain the initial magnitude of a hydrogen atom's bound electron momentum. equation (5.13) obtains this momentum by multiplying together the components of ##\mathbf{P}_a##, ##\mathbf{P}_b##, and ##\mathbf{P}_o##.
  • #1
112
13
In Chapter 5.3, Ballentine uses geometrical arguments to obtain the initial magnitude of a hydrogen atom's bound electron momentum. How does equation (5.13) obtain? I tried to naively compute
$$p_e^2 \equiv \textbf{p}_e\cdot \textbf{p}_e = p_a^2+p_b^2+p_o^2 + 2\textbf{p}_a\cdot \textbf{p}_b - 2\textbf{p}_a\cdot \textbf{p}_0 - 2\textbf{p}_0\cdot \textbf{p}_b $$ $$= p_a^2+p_b^2+p_o^2 + 2p_ap_b\cos(\pi - \phi) - 2p_ap_0\cos \theta - 2p_bp_0\cos \theta$$
but then could not go any further. Am I misunderstanding the geometrical relationships of the vectors in Figure 5.1?

Screen Shot 2023-03-29 at 10.27.43 AM.png
Screen Shot 2023-03-29 at 10.27.55 AM.png
 
Physics news on Phys.org
  • #2
EE18 said:
$$p_e^2 \equiv \textbf{p}_e\cdot \textbf{p}_e = p_a^2+p_b^2+p_o^2 + 2\textbf{p}_a\cdot \textbf{p}_b - 2\textbf{p}_a\cdot \textbf{p}_0 - 2\textbf{p}_0\cdot \textbf{p}_b $$ $$= p_a^2+p_b^2+p_o^2 + 2p_ap_b\cos(\pi - \phi) - 2p_ap_0\cos \theta - 2p_bp_0\cos \theta$$
Am I misunderstanding the geometrical relationships of the vectors in Figure 5.1?
The angle between ##\mathbf{P}_a## and ##\mathbf{P}_b## is not ##\pi - \phi##.

##\mathbf{P}_a## lies in the yellow plane that makes angle ##\phi/2## to the horizontal gray plane. You might try finding expressions for the x, y, and z components of ##\mathbf{P}_a## (shown in blue).

1680112505407.png
 
  • Like
Likes Ishika_96_sparkles, malawi_glenn and PeroK
  • #3
TSny said:
The angle between ##\mathbf{P}_a## and ##\mathbf{P}_b## is not ##\pi - \phi##.

##\mathbf{P}_a## lies in the yellow plane that makes angle ##\phi/2## to the horizontal gray plane. You might try finding expressions for the x, y, and z components of ##\mathbf{P}_a## (shown in blue).

View attachment 324201
Thank you so much for that diagram, it helps me tremendously.

It seems like I have, by symmetry, that ##\textbf{p}_a \cdot \textbf{p}_b = p_{ax}^2 -p_{ay}^2 + p_{az}^2##. It then remains to find these components in terms of the given angles and ##p_a##. Now clearly ##p_{az} = \tan(\phi/2)p_{ay}##, ##p_{az} = p_a \cos \theta##, and ##p_a^2 = p_{ax}^2 +p_{ay}^2 + p_{az}^2## so that at least in theory I have three equations with which I can substitute away ##p_{ax}^2 -p_{ay}^2 + p_{az}^2## in the above in terms of the angles and ##p_a##. However it seems very ugly -- is there a cleaner way to do it or is it necessarily ugly?
 
  • #4
EE18 said:
Now clearly ##p_{az} = \tan(\phi/2)p_{ay}##, ##p_{az} = p_a \cos \theta##
I think you meant the second equation to represent ##p_{ax}##.

Consider writing ##\mathbf{p}_a## in unit vector notation $$\mathbf{p}_a =p_{ax} \mathbf{i} +p_{ay} \mathbf{j} +p_{az} \mathbf{k}$$ Each of the components can be expressed in terms of the magnitude ##p_a## and the angles ##\theta## and ##\phi/2##. For example, you know ##p_{ax} = p_a \cos \theta##.

Do the same for ##\mathbf{p}_b##.

For ##\mathbf{p}_o## we have simply ##\mathbf{p}_o = p_0 \mathbf{i}##. Then use equation (5.12) to find the component expression for ##\mathbf{p}_e##.
 
  • Like
Likes EE18
Back
Top