meta user
network profile
| bio | website | |
|---|---|---|
| location | Zurich, Switzerland | |
| age | ||
| visits | member for | 2 years, 6 months |
| seen | 8 hours ago | |
| stats | profile views | 58 |
Mathematics student at ETH Zürich.
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Jun 27 |
awarded | Popular Question |
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Apr 2 |
accepted | Measuring extra-dimensions |
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Mar 18 |
asked | Measuring extra-dimensions |
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Feb 4 |
comment |
Vibrational motion of linear diatomic molecule Thanks a lot for your answer! |
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Feb 4 |
accepted | Vibrational motion of linear diatomic molecule |
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Feb 3 |
revised |
Vibrational motion of linear diatomic molecule added 38 characters in body |
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Feb 3 |
asked | Vibrational motion of linear diatomic molecule |
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Feb 2 |
accepted | Peaks on top of Bremsstrahlung |
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Feb 2 |
comment |
Peaks on top of Bremsstrahlung Thank you for answering! This makes sense. =) Also thanks for the link, @dmckee. |
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Feb 1 |
asked | Peaks on top of Bremsstrahlung |
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Jan 18 |
awarded | Popular Question |
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Jan 2 |
comment |
Confusion concerning the use of wavenumber in exercise about Fabry-Perot etalon Thank you very much, by the way. =) |
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Jan 1 |
comment |
Confusion concerning the use of wavenumber in exercise about Fabry-Perot etalon Ah, this clears everything up. Using one single term for two things which have the same units and come up in the same contexts is just plane vicious! :-( physicists... |
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Jan 1 |
accepted | Confusion concerning the use of wavenumber in exercise about Fabry-Perot etalon |
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Jan 1 |
comment |
How could something have negative mass? @Anamitra: If you set $i^2 = 1$ but $i \ne 1$, then you have simply given a new name to $-1$... Note that for such a "definition": $(i-1)(i+1) = i^2 - 1 = 0$. So $i=\pm 1$. Also $i$ is usually not defined by $i^4 = 1$ (or rather can not be defined in such a way), but to be a solution to $X^2 + 1 = 0$ (which itself has some degeneracy, but not in an essential way). |
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Dec 31 |
asked | Confusion concerning the use of wavenumber in exercise about Fabry-Perot etalon |
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Dec 23 |
accepted | Puzzled by $\mathrm p = -i\hbar \nabla = m \dot{\mathrm r}$? |
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Dec 23 |
comment |
Puzzled by $\mathrm p = -i\hbar \nabla = m \dot{\mathrm r}$? Thank you for answering and clearing this up! |
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Dec 22 |
asked | Puzzled by $\mathrm p = -i\hbar \nabla = m \dot{\mathrm r}$? |
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Dec 17 |
accepted | Counting the number of modes |
