Convert Continuous Inverse Scale Parameter to Physically Relevant Units

Click For Summary
SUMMARY

The discussion focuses on converting a continuous inverse scale parameter (CISP) into physically relevant units, specifically in the context of diffusion equations. The user presents equations from two sources: an error function from APEC and a diffusion equation from P.G. Shewmon's "Diffusion in Solids." The key issue is determining the correct relationship between the concentration parameter (alpha) and the CISP (h) to derive the diffusion coefficient (D). The user proposes a normalization method to simplify calculations, ultimately arriving at a diffusion coefficient of D = 0.007 um/h.

PREREQUISITES
  • Understanding of diffusion equations in solid materials.
  • Familiarity with error functions and their applications in physics.
  • Knowledge of concentration parameters in diffusion studies.
  • Basic mathematical skills for manipulating equations and units.
NEXT STEPS
  • Research the derivation and applications of the error function in physical sciences.
  • Study the principles of diffusion in solids as outlined in P.G. Shewmon's "Diffusion in Solids."
  • Explore methods for normalizing data in experimental physics.
  • Investigate the implications of concentration parameters on diffusion coefficients.
USEFUL FOR

Researchers and students in materials science, physicists studying diffusion processes, and anyone involved in experimental data analysis related to concentration and diffusion parameters.

Salish99
Messages
28
Reaction score
0
How to convert a continuous inverse scale parameter into a physically relevant quantity:
1) What is a CISP, and why is called continuous and why inverse?

2) how do I deal with it:
Example:
On
http://www.apec.umn.edu/faculty/gpederso/documents/4501/risk45DistFunc.pdf
the error function is defined as
f(x)= h/sqrt(Pi()) x e^(-(hx)2)

Now, in P.G. shewmon. Diffusion in solids. McGrawHill NY, 1963, the function for the diffusion of a solid thin film into a bulk material is given as
c(x,t) = alpha/sqrt(4Pi()Dt) x exp (-x2/4Dt)

if I sub in one equation into the other, then for the first term the continuous inverse scale parameter
h = alpha/sqrt(4Dt)
but for the term in the exponential part of the equation
h = 1/sqrt(4Dt)

So, I MUST set alpha = 1 and that's not physically right for diffusion experiments.

Alpha is the concentration of the solute (i.e. the stuff in the thin film that we want to investigate the diffusion of), in terms of counts or intensity. That changes over distance.
It does not occur within the exponential term.

How do I convert h into physically meaningful data? Do I assume the concentration alpha remains outside the first term equation?

Let's say, concentration alpha is 6000 If I fit my data to the erf, I get my output fitting parameter h as 0.71. What is D now
is it option a:
D = alpha2/(h24t),
or option b
D = 1/(h24t), in which I don't take the initial surface concentration into account.

Thanks for your thoughts / help.
 
Last edited by a moderator:
Physics news on Phys.org


Update: I devised the following method:
I traced the maximum in my data, and divided the entire data by that maximum, thus normalizing it to 1. Thus, as a result now all my concentration is normalized to 1, and I can assume alpha = 1.
Once I calculate the D out of the equation D = 1/(h24t), in the case of 500 h and an h of 0.16283 this comes to D = 0.007406 1/h. I assume the unit of the inverse scale parameter is micrometer.
Then it would be:
D=0.007 um/h

Can I do that? - How can I find the units of my inverse scale parameter?
Or do I still have to multiply D by the former maximum concentration values?
What is a continuous inverse scale parameter, and why is called continuous and why is it inverse (and inverse to what?)?
These links doe not cover my questions:
https://www.physicsforums.com/showthread.php?t=269208
https://www.physicsforums.com/showthread.php?t=275747
https://www.physicsforums.com/showthread.php?t=232883
https://www.physicsforums.com/showthread.php?t=253505
https://www.physicsforums.com/showthread.php?t=178318
 


anyone who can kindly help me? :shy:
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 2 ·
Replies
2
Views
623
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
4K