Response to Comment on “Reversible disorder-order transitions in atomic crystal nucleation”

Author:

Jeon Sungho1,Hwang Sang-Yeon2,Ciston Jim3,Bustillo Karen C.3,Reed Bryan W.4,Hong Sukjoon1,Zettl Alex567,Kim Woo Youn2,Ercius Peter3,Park Jungwon89,Lee Won Chul1

Affiliation:

1. Department of Mechanical Engineering, BK21FOUR ERICA-ACE Center, Hanyang University, Ansan, Gyeonggi 15588, Republic of Korea.

2. Department of Chemistry, Korea Advanced Institute of Science and Technology, Daejeon 34141, Republic of Korea.

3. National Center for Electron Microscopy, Molecular Foundry, Lawrence Berkeley National Laboratory, Berkeley, CA 94720, USA.

4. Integrated Dynamic Electron Solutions Inc., Pleasanton, CA 94588, USA.

5. Department of Physics, University of California, Berkeley, CA 94720, USA.

6. Materials Sciences Division, Lawrence Berkeley National Laboratory, Berkeley, CA 94720, USA.

7. Kavli Energy NanoSciences Institute, Berkeley, CA 94720, USA.

8. School of Chemical and Biological Engineering and Institute of Chemical Process, Seoul National University, Seoul 08826, Republic of Korea.

9. Center for Nanoparticle Research, Institute for Basic Science, Seoul 08826, Republic of Korea.

Abstract

Yu et al . suggested calculating precisely the size ranges of the three parts of our figure 3A, adjusting the free-energy levels in figure 3B, and considering the shape effect in the first-principles calculation. The first and second suggestions raise strong concerns for misinterpretation and overinterpretation of our experiments. The original calculation is sufficient to support our claim about crystalline-to-disordered transformations.

Publisher

American Association for the Advancement of Science (AAAS)

Subject

Multidisciplinary

Reference15 articles.

1. Comment on “Reversible disorder-order transitions in atomic crystal nucleation”

2. Reversible disorder-order transitions in atomic crystal nucleation

3. The simplest instance of Simpson’s paradox is that the average of averages is not the average. For example the average of {1 1 1} is 1 ( n = 3) and the average of {5} is 5 ( n = 1). The average of the two averages is 6/2 = 3 but the average of all numbers is 8/4 = 2. What Yu et al. performed is to average variable numbers of data points and then to apply curve fitting to the averages which has the mathematically equivalent problem as this example has.

4. The interpretation of interaction in contingency tables;Simpson E. H.;J. R. Stat. Soc. Ser. B,1951

5. S. Stevens Introduction to Statistics: Think & Do (Worldwide Center of Mathematics ver. 4.1 2019) chap. 2.5.

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