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Matt Addicoat

@maddicoat.bsky.social

Computational materials chemist. Special affection for combinatorial problems. New line in profile to distinguish me, the real human, from the spammy imposter.

125 Followers  |  90 Following  |  6 Posts  |  Joined: 09.01.2025
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Posts by Matt Addicoat (@maddicoat.bsky.social)

I wouldn't feel particularly bad. The messages that the imposter did send to a few colleagues were really generic low-grade spammy nonsense. 😒

06.01.2026 12:36 — 👍 1    🔁 0    💬 2    📌 0

Students did the calculations and wrote most of the paper in workshop sessions. Dealing with Reviewer 2 was, unfortunately, my job.

06.01.2026 10:56 — 👍 0    🔁 0    💬 0    📌 0
Preview
Associative vs. dissociative binding of CO2 on M4 transition metal clusters Density functional theory calculations were performed to determine reaction paths for the reaction of CO2 with M4 transition metal clusters (M = Nb, Mo, Ru, Rh, Pd, Ag, Pt). Geometries incorporating a...

Here's a cool thing that needs a human to do: These two papers doi.org/10.1039/D5CP... and doi.org/10.1039/D5CP... were published at the end of last year in @pccp.rsc.org. They are the result of the assessment of my masters Comp Chem class.

06.01.2026 10:56 — 👍 2    🔁 1    💬 1    📌 0

Whelp. It appears I have an impostor. To be clear, this is actually me, a real human compuational materials chemist <pinches self>. The imposter sems to have copied my profile, but is made of spam.

06.01.2026 10:23 — 👍 1    🔁 0    💬 2    📌 0
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So glad I got to present Maryam's work at the 8th #UKPorMat (@rsc-pormat.bsky.social) meeting in Birmingham today. We're hiring to continue this work: PDRA: vacancies.ntu.ac.uk/Job/JobDetai... PhD: pgrdashboard.ntu.ac.uk/ntu/?streamI...

04.06.2025 13:27 — 👍 4    🔁 3    💬 0    📌 0
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So glad I got to present Maryam's work at the 8th UKPorMat (@rsc-pormat.bsky.social) meeting in Birmingham today. We're hiring to continue this work: PDRA: vacancies.ntu.ac.uk/Job/JobDetai... PhD: pgrdashboard.ntu.ac.uk/ntu/?streamI...

04.06.2025 12:44 — 👍 9    🔁 3    💬 0    📌 0