Dimensional analysis: Energy Transfer

[HAYAO]

Homework Statement

There is a paper in 1973:
T. Kushida, "Energy Transfer and Cooperative Optical Transitions in Rare-Earth Doped Inorganic Materials I. Transition Probability Calculation", J. Phys. Soc. Jpn. 1973, 34, 1318-1326. DOI: http://dx.doi.org/10.1143/JPSJ.34.1318
that explains the multipole-multipole energy transfer probability.

I need help with dimensional analysis of the following equation in this paper that looks EXACTLY like this:

$\bar{P}_{AB}^{(dd)} = \frac{1}{(2J_{a}+1)(2J_{b}+1)}\left ( \frac{2}{3} \right )\left ( \frac{2\pi }{\hbar} \right )\left ( \frac{e^{2}}{R^{3}} \right )^{2}\left [ \sum_{\lambda }^{ } \Omega _{A\lambda }\left \langle J_{a}\left \| U^{(\lambda )} \right \| J_{a}' \right \rangle^{2}\right ]\left [ \sum_{\lambda }^{ } \Omega _{B\lambda }\left \langle J_{b}\left \| U^{(\lambda )} \right \| J_{b}' \right \rangle^{2}\right ]S$

here,

$\bar{P}_{AB}^{(dd)}$: Dipole-dipole energy transfer probability. Unit: $[s^{-1}]$
$J_{a}$: Total angular momentum quantum number at state a of specie A. No unit.
$J_{b}$: Total angular momentum quantum number at state b of specie B. No unit.
$\hbar$: Reduced Planck constant. Unit: $[J s]$
$e$: Elementary charge. Unit: $[C]$
$R$: Distance between specie A and B. Unit: $[m]$
$\Omega _{A\lambda }$: Scaling parameter for specie A. $\lambda$ denotes tensor rank. Unit: $[m^{2}]$
$\left \langle J_{a}\left \| U^{(\lambda )} \right \| J_{a}' \right \rangle$: Reduced matrix element of $J_{a}\rightarrow J_{a}'$ transition of specie A. $\lambda$ denotes tensor rank. Unit: $[-]$
$\Omega _{B\lambda }$: Scaling parameter for specie B. $\lambda$ denotes tensor rank. Unit: $[m^{2}]$
$\left \langle J_{b}\left \| U^{(\lambda )} \right \| J_{b}' \right \rangle$: Reduced matrix element of $J_{b}\rightarrow J_{b}'$ transition of specie B. $\lambda$ denotes tensor rank. Unit: $[-]$
$S$: Spectral overlap integral of A and B. Unit: $[m]$

After dimensional analysis of the right side of the equation, it did not match with the unit on the left side of the equation.

Homework Equations

Dimensional analysis:
$\frac{1}{J\cdot s} \cdot \left ( \frac{C^{2}}{m^{3}} \right )^{2}\cdot m^{2}\cdot m^{2}\cdot m$

The Attempt at a Solution

$\frac{1}{J\cdot s} \cdot \left ( \frac{C^{2}}{m^{3}} \right )^{2}\cdot m^{2}\cdot m^{2}\cdot m$
$= \frac{C^{4}}{J\cdot s\cdot m}$
$= \frac{A^{4}\cdot s^{4}}{kg\cdot m^{2}\cdot s^{-2}\cdot s\cdot m}$
$= \frac{A^{4}\cdot s^{5}}{kg\cdot m^{3}}$

I broke them all down into SI units, but I have no idea how this is going to be $s^{-1}$. I think I am making a careless or fundamental mistake here, but I just can't figure it out. What do you guys think?

Thank you

 

[Charles Link]

I think the units could be cgs so that $e^2/r$ has units of energy. Meanwhile, it looks a lot like an application of Fermi's golden rule which needs a factor $\rho_f$, which is the density of final states per unit energy interval. Suggestion would be to look at Fermi's Golden Rule (google it) and see that it is dimensionally correct=then try to compare it to what this author has computed. $\\$ Editing: Also might his $P_{AB}$ refer to energy/unit time?

 

[HAYAO]

I think the units could be cgs so that $e^2/r$ has units of energy. Meanwhile, it looks a lot like an application of Fermi's golden rule which needs a factor $\rho_f$, which is the density of final states per unit energy interval. Suggestion would be to look at Fermi's Golden Rule (google it) and see that it is dimensionally correct=then try to compare it to what this author has computed. $\\$ Editing: Also might his $P_{AB}$ refer to energy/unit time?

Thank you Charles.

I actually already tried cgs just in case, but this was what I got:

$\frac{1}{J\cdot s} \cdot \left ( \frac{\left (cm^{\frac{3}{2}}\cdot g^{\frac{1}{2}}\cdot s^{-1} \right )^{2}}{m^{3}} \right )^{2}\cdot m^{2}\cdot m^{2}\cdot m$
$= \frac{\left ( cm^{3}\cdot g\cdot s^{-2} \right )^{2}}{J\cdot s\cdot m}$
$= \frac{cm^{6}\cdot g^{2}\cdot s^{-4}}{kg\cdot m^{2}\cdot s^{-2}\cdot s\cdot m}$
$= \frac{10^{-12}\cdot m^{6}\cdot 10^{-6}\cdot kg^{2}\cdot s^{-3}}{kg\cdot m^{3}}$
$= 10^{-18}\cdot m^{3}\cdot kg\cdot s^{-3}$

It didn't work.

Yes, the equation is derived from Fermi's golden rule. However, the $\rho_f$ is included as the spectral overlap integral $S$, so I do not believe we have to worry about that.

The rate is supposed to be events/unit time. The author implicitly says so, and so does other papers (O.L. Malta, J. Non-Cryst. Solids 2008, 354, 4770-4776).

 

[haruspex]

I think the units could be cgs so that $e^2/r$ has units of energy.

Ithink it fairly clear there's an ε₀ missing. Are you saying that the cgs units are so arranged that this takes the value 1? If so, you may be right, I am not familiar with it.
Either way, you do get to turn $e^2/r$ into dimensions of energy. Working that through, I get, for the whole expression, energy x distance / time. So even interpreting P_AB as energy/time there is a spare length dimension.

I tried looking up spectral overlap integral, but cannot find anything online which makes it clear what dimension that should have. I found an integral with λ⁴.dλ in it, and nothing obvious to cancel all those extra length dimensions, so I gave up.

 

[Charles Link]

Thank you Charles.

I actually already tried cgs just in case, but this was what I got:

$\frac{1}{J\cdot s} \cdot \left ( \frac{\left (cm^{\frac{3}{2}}\cdot g^{\frac{1}{2}}\cdot s^{-1} \right )^{2}}{m^{3}} \right )^{2}\cdot m^{2}\cdot m^{2}\cdot m$
$= \frac{\left ( cm^{3}\cdot g\cdot s^{-2} \right )^{2}}{J\cdot s\cdot m}$
$= \frac{cm^{6}\cdot g^{2}\cdot s^{-4}}{kg\cdot m^{2}\cdot s^{-2}\cdot s\cdot m}$
$= \frac{10^{-12}\cdot m^{6}\cdot 10^{-6}\cdot kg^{2}\cdot s^{-3}}{kg\cdot m^{3}}$
$= 10^{-18}\cdot m^{3}\cdot kg\cdot s^{-3}$

It didn't work.

Yes, the equation is derived from Fermi's golden rule. However, the $\rho_f$ is included as the spectral overlap integral $S$, so I do not believe we have to worry about that.

The rate is supposed to be events/unit time. The author implicitly says so, and so does other papers (O.L. Malta, J. Non-Cryst. Solids 2008, 354, 4770-4776).

I tried to look up the O.L. Malta paper you cited, but unfortunately they are wanting $35.95 for the pdf.

 

[HAYAO]

Ithink it fairly clear there's an ε₀ missing. Are you saying that the cgs units are so arranged that this takes the value 1? If so, you may be right, I am not familiar with it.
Either way, you do get to turn $e^2/r$ into dimensions of energy. Working that through, I get, for the whole expression, energy x distance / time. So even interpreting P_AB as energy/time there is a spare length dimension.

Yes, which is also what I got above. I don't understand...

I tried looking up spectral overlap integral, but cannot find anything online which makes it clear what dimension that should have. I found an integral with λ⁴.dλ in it, and nothing obvious to cancel all those extra length dimensions, so I gave up.

Well the integral part in the above equation comes from the final density of states in the Fermi's Golden Rule. The unit for that is number of states/unit energy. Unit energy can be expressed in wavenumber [cm^-1], so that makes the density of state with an unit of [cm]. The spectral overlap you presented there comes from FRET, but there are a lot of things going on before that equation is derived. I am not sure if it is the same spectral overlap defined in the paper above. In the paper above, the spectral overlap is explicitly in form of inverse energy in wavenumbers so [cm] as well.

 

[haruspex]

Unit energy can be expressed in wavenumber

Sure, but that's the sort of thing that breaks dimensional analysis. The connection is via the factor hc, which has dimension.
If we introduce the missing ε₀ and stick with states/unit energy instead of converting to states*wavenumber I think you will find the dimensionality of the expression reduces to just 1/time, as you originally expected.

 

[HAYAO]

Okay guys, I've had some help from Charles and tried solving it all in CSG unit system. I think I've got it. Here it is (I've abbreviated scaling such as 10² in changing from m to cm):

$\frac{1}{cm^{2}\cdot g\cdot s^{-2}\cdot s}\cdot \left ( \frac{\left ( cm^{\frac{3}{2}}\cdot g^{\frac{1}{2}}\cdot s^{-1} \right )^{2}}{cm^{3}} \right )^{2}\cdot cm^{2}\cdot cm^{2}\cdot \frac{1}{cm^{2}\cdot g\cdot s^{-2}}$
$= \frac{1}{cm^{2}\cdot g\cdot s^{-2}\cdot s}\cdot \left ( \frac{ cm^{3}\cdot g\cdot s^{-2}}{cm} \right )^{2}\cdot \frac{1}{cm^{4}}\cdot cm^{2}\cdot cm^{2}\cdot \frac{1}{cm^{2}\cdot g\cdot s^{-2}}$
$= \frac{1}{cm^{2}\cdot g\cdot s^{-2}\cdot s}\cdot \left ( cm^{2}\cdot g\cdot s^{-2}\right )^{2}\cdot \frac{1}{cm^{2}\cdot g\cdot s^{-2}}$
$= \frac{\left ( cm^{2}\cdot g\cdot s^{-2}\right )^{2}}{\left ( cm^{2}\cdot g\cdot s^{-2}\right )^{2} s}$
$= s^{-1}$

Looks like it works! Thank you guys!