Dynamic light scattering laplace inversion

[Steve Drake]

Hi,

Those of you familar with dynamic light scattering (DLS), will know that a common method used to obtain a particle size distribution is via a laplace inversion of the autocorrelation function.

What I want to know is why? What does Laplace space have to do with DLS (I've only learned basics Laplace transforms and Laplace inversions from simple elec eng...). My understanding is that the scattered light depends on how fast the particles are diffusing. And that the propagating light from the sample to the detector undergoes an optical Fourier transform.

I also know that you can take the Fourier transform of the ACF to obtain the power spectrum...

But where does laplace come into this? All papers I read just say that the spectrum can be described by the equation

g^{(1)}=\int G(\Gamma )e^{(-\text{$\Gamma $t})}

Thanks

 

[M Quack]

Hint: How is G[Gamma] related to g(t)?

Gamma = q^2 D

where D is the diffusion coefficient. I guess that the particle size can be derived from D.

Then G(Gamma) gives you a measure for the number of particles with that size.

 

[Steve Drake]

Hint: How is G[Gamma] related to g(t)?

Gamma = q^2 D

where D is the diffusion coefficient. I guess that the particle size can be derived from D.

Then G(Gamma) gives you a measure for the number of particles with that size.

Yes, I do know that \Gamma is Dq²,

for example the simplest case is just an exponential decay exp(-Dq²\tau), where we obtain the diffusion coefficient D, and intern the radius (stokes Einstein).

But what I don't understand is why does the Laplace inversion describe g⁽¹⁾?

 

[Andy Resnick]

Hi,

Those of you familar with dynamic light scattering (DLS), will know that a common method used to obtain a particle size distribution is via a laplace inversion of the autocorrelation function.

What I want to know is why? What does Laplace space have to do with DLS (I've only learned basics Laplace transforms and Laplace inversions from simple elec eng...). My understanding is that the scattered light depends on how fast the particles are diffusing. And that the propagating light from the sample to the detector undergoes an optical Fourier transform.

I also know that you can take the Fourier transform of the ACF to obtain the power spectrum...

But where does laplace come into this? All papers I read just say that the spectrum can be described by the equation

g^{(1)}=\int G(\Gamma )e^{(-\text{$\Gamma $t})}

Thanks

From my perspective, the Laplace transform is very similar to a Fourier transform- the Laplace transform can be used when dissipative processes occur. If you like, think of exp(iwt), but allow the frequency w to be imaginary (or complex, if you want to think more generally). Because DLS probes diffusive transport, the processes are dissipative.

Does that help?

 

[M Quack]

Laplace inversion is nothing else than inverting the equation you cite.

G(Gamma) = Laplace(g(t))

If there is only one contribution then G(Gamma) = delta(Gamma-Dq^2)

If there are more then G(Gamma) will be a distribution. The width of the distribution will tell you about polydispersity, the peak position about avg. size, etc.

 

[Steve Drake]

Thanks guys,

So basically its inverting that equation for each delay time (t) and trying to get the best fit?

But what I still don't understand is why you can't just put a trial distribution in for G (say Gaussian), and then just do numerical integration to solve for \Gamma?

 

[M Quack]

You can do that, but you need to have a guess about the trial function first. You can also assume a Gaussian and then fit the parameters (FHWM, peak position, ...) until the (numerical) integral matches the measured spectrum.

A Laplace inversion does not require a model.

 

[Steve Drake]

Thanks, I think I am beginning to understand it more now. I think my problem is the lack of knowledge of laplace stuff, i never can understand it. Fourier makes sense but not laplace. So the laplace inversion can take into account the possibility of bimodal solutions or solutions with a broad or narrow distribution. But if you do it numerically you have to put in say a Gaussian of a certain FWHM and peak position, do the numerical integration, change the FWHM or peak, do it again etc etc till the best fit, and then it may not even be a Gaussian shaped distribution.

So if for instance you put in particles of a certain size and you KNOW all aspects of the distribution (say gaussian), would the laplace inversion give the same answer as a gaussian + numerical (in a perfect world), or a close one...?

Thanks a lot again

 

[M Quack]

I guess yes.

It should be possible to do a numerical Laplace inversion without any prior assumtions about the data, similar to a numerical Fourier transform.

 

[Steve Drake]

Ok thanks a lot for this,

so in general, The numerical methods of laplace inversion are a 'quick way' to fully describe the data points rather than just doing numerical integration with heaps of different parameters?