Help with Problem on Photons and Atoms | Cohen-Tannoudji et al.

  • Thread starter Jimmy Snyder
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In summary, the charge density and current are given by: \rho(r,t) = \Sigma_{\alpha} q_{\alpha}\delta[r - r_{\alpha}(t)] and j(r,t) = \Sigma_{\alpha} q_{\alpha} v_{\alpha} \delta[r - r_{\alpha}(t)]. Using the product and chain rules, these equations become: \rho(r,t) + \nabla \cdot \j(r,t) = 0 and \frac{\partial}{\partial t}\rho(r,t) = -\Sigma_{\alpha} q
  • #1
Jimmy Snyder
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This is from page 1 of "Photons and Atoms" by Cohen-Tannoudji et. al.

Given a set of particles of charge [itex]q_{\alpha}[/itex] and position [itex]r_{\alpha}(t)[/itex] the charge density [itex]\rho[/itex] and current j are given by
[tex]\rho(r,t) = \Sigma_{\alpha} q_{\alpha}\delta[r - r_{\alpha}(t)][/tex]
[tex]j(r,t) = \Sigma_{\alpha} q_{\alpha} v_{\alpha} \delta[r - r_{\alpha}(t)][/tex]

Show that:

[tex]\frac{\partial}{\partial t}\rho(r,t) + \nabla \cdot \j(r,t) = 0[/tex]

I don't know how to get started. The only thing I know of that looks like it might help is:

[tex]x\delta'(x) = -\delta(x)[/tex]

But I don't see how to apply it here.
 
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  • #2
Just use the following (product and chain) rules :

[tex]\vec \nabla \cdot (f\vec v)=(\vec \nabla f)\vec v+f(\vec \nabla \cdot \vec v)[/tex]

[tex]\frac{\partial}{\partial t} f(\vec r(t))=(\vec \nabla f)\cdot\frac{\partial \vec r}{\partial t}[/tex]
And it's be a one or two line proof.

Only thing is you get things like [tex]\vec \nabla \delta(\vec r-\vec r_\alpha(t))[/tex]. I`m not really sure myself what that would be, but just leave it like that and you're fine.
 
  • #3
Galileo said:
Just use the following (product and chain) rules :

[tex]\vec \nabla \cdot (f\vec v)=(\vec \nabla f)\vec v+f(\vec \nabla \cdot \vec v)[/tex]

[tex]\frac{\partial}{\partial t} f(\vec r(t))=(\vec \nabla f)\cdot\frac{\partial \vec r}{\partial t}[/tex]
And it's be a one or two line proof.

Only thing is you get things like [tex]\vec \nabla \delta(\vec r-\vec r_\alpha(t))[/tex]. I`m not really sure myself what that would be, but just leave it like that and you're fine.

Thanks for looking at this Galileo. I made a mistake when I posted it. The positions of the particles are a function of time [itex]r_{\alpha}(t)[/itex]. I edited the original and I also repeat the corrected formulas here:

[tex]\rho(r,t) = \Sigma_{\alpha} q_{\alpha}\delta[r - r_{\alpha}(t)][/tex]
[tex]j(r,t) = \Sigma_{\alpha} q_{\alpha} v_{\alpha} \delta[r - r_{\alpha}(t)][/tex]

If I apply the partials to the formulas and use your suggestions, I get the following:

[tex]\frac{\partial}{\partial t}\rho(r,t) = \Sigma_{\alpha} q_{\alpha} \frac{\partial}{\partial t}\delta[r - r_{\alpha}(t)] = -\Sigma_{\alpha} q_{\alpha} (v_{\alpha} \nabla \delta(r - r_{\alpha}(t)))[/tex]
[tex]\nabla j(r,t) = \Sigma_{\alpha} q_{\alpha} (\nabla v_{\alpha} \delta(r - r_{\alpha}(t)) + v_{\alpha} \nabla \delta(r - r_{\alpha}(t)))][/tex]

And so:
[tex]\frac{\partial}{\partial t}\rho(r,t) + \nabla \cdot \j(r,t) = \Sigma_{\alpha} q_{\alpha} \nabla v_{\alpha} \delta(r - r_{\alpha}(t))[/tex]

I'm not there yet, but this is a great stride forward.
 
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  • #4
jimmysnyder said:
And so:
[tex]\frac{\partial}{\partial t}\rho(r,t) + \nabla \cdot \j(r,t) = \Sigma_{\alpha} q_{\alpha} \nabla v_{\alpha} \delta(r - r_{\alpha}(t))[/tex]

Now, why is the right-hand side equal to zero?
 
  • #5
George Jones said:
Now, why is the right-hand side equal to zero?
I figured that out on my way home from work. Thanks to Galileo and to George. I remember your help with GR earlier.
 

1. What is the relationship between photons and atoms?

The relationship between photons and atoms is that photons are the fundamental particles of light, while atoms are made up of protons, neutrons, and electrons. Photons are the carriers of electromagnetic radiation and interact with atoms through absorption, emission, and scattering processes.

2. Can you explain the concept of quantization in relation to photons and atoms?

Quantization is the idea that certain properties of particles, such as energy and momentum, can only exist in discrete values. In the context of photons and atoms, this means that the energy levels of an atom can only change in specific increments, corresponding to the absorption or emission of a photon with a specific energy.

3. How do photons and atoms interact in the process of atomic excitation and de-excitation?

When an atom absorbs a photon, the photon's energy is transferred to an electron in the atom, causing it to jump to a higher energy level. This is known as atomic excitation. Conversely, when an electron in an excited state drops down to a lower energy level, it emits a photon, a process known as de-excitation.

4. What is the significance of the Rabi frequency in the study of photons and atoms?

The Rabi frequency is a measure of the strength of the interaction between photons and atoms. It is proportional to the amplitude of the electric field of the photon and the dipole moment of the atom. It is an important parameter in understanding the dynamics of atomic systems and is used in various applications, such as quantum computing and spectroscopy.

5. Can you provide an example of an application of the principles discussed in "Help with Problem on Photons and Atoms | Cohen-Tannoudji et al."?

One example of an application of the principles discussed in this book is in atomic clocks. The precise energy levels of atoms can be used as a reference for accurate timekeeping. By using lasers to manipulate the energy levels of atoms and measuring the frequency of the emitted photons, atomic clocks can keep time with incredible accuracy, making them essential in various technologies such as GPS and telecommunications.

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