What the heck is a periodic arrangement of lattices?
Crystals are a periodic repetition of some motiv. Due to periodicity, there is some flexibility in defining where one motiv ends and the next one begins, hence the different types of elementary cells. However, if you chose one point of the...
Here some pictures:
1. Prehnite, Norway
2. Essexite, Kaiserstuhl Volcanoe, Germany
3. Nummulith, Bad Adelholzen, Germany
4. Nummulith, Glauconitic filling after treatment with HCl. Bad Adelholzen
5. Limburgite, Limberg, Germany
6. Chondrite, North Africa
I just realized that the photos are not...
Andy, maybe you want to describe your setup more in detail. I would be very interested. These luminars are rather for macrophotography than for microscopy. How did you adapt them to your mic?
As far as I understand it, these sections are fossiliferous. These sections are usually thicker than petrographic sections, which makes identification of the mineral content far more difficult.
You should try to determine the thickness of your sample. To this end, use a rather highly magnifying...
Of course outliers are an issue! But first one has to define what an outlier is. An outlier may violate ##E(\epsilon_i)=0##. OLS is sensitive to this, it is not a robust method. A single outlier of this kind may lead to a slope estimate arbitrary far away from the true one.
The second kind of...
One of my most basic, yet important lessons in statistics is that "goodness" of an estimator is a multidimensional quantity. It should be consistent better asymptotically unbiased even better finite sample unbiased. Furthermore efficient and robust. Usually you cannot have all. There is even a...
The meaning of unbiasedness means that the expectation value of Y_i is equal to the population value y_i, E(Y_i)=y_i. Given that X_i= x_i, this is true for OLS.
Using ##Y_i=\beta_1 x_i + \beta_0 +\epsilon_i##,
$$ \hat{\beta}_1=\frac{\sum_i (x_i-\bar{x})(Y_i-\bar{Y})}{\sum_i (x_i-\bar{x})^2}.$$...
You do not have to assume that the error terms in linear regression are normally distributed. If they are not, the estimator is still consistent, but it may no longer be efficient. As long as the x values can be measured without error, it makes no difference whether one assumes the x values to...
An operator ##a^\dagger_i## generates a one particle state corresponding to a ##p_z## orbital centered at ##R_i##, ##p_z(r-R_i)=<r| a_i^\dagger|0>##. All one-particle state creation operators are linear superpositions of the ##a^\dagger_i##, so it is easy to get a wavefunction.
The question about the covalent bond is quite subtle. Usually, it is a bond between neutral atoms, so it is not immediately clear, why electromagnetism should lead to an attractive force. As first pointed out by Hellmann, the possibility to of delocalization of the electrons in a bond over two...
Personally, I like the phenomenological approach of Caratheodory: https://eclass.uoa.gr/modules/document/file.php/CHEM105/Άρθρα/Caratheodory-1909.pdf
This has been brought to modern mathematical language by Lieb and Yngvason...
I think I solved my problem: Writing ##R_z(\alpha)R_y(\beta)R_z(\gamma)## as ## R_z(\alpha)R_y(\beta)R_z(-\alpha)R_z(\gamma+\alpha)=R_{y'(\alpha)}(\beta) R_z (\gamma+\alpha)##, I can plot ##\alpha+\gamma## as a function of the orientation of the new rotation axis ##y'(\alpha)## and the rotation...
Dear Forum,
say I am projecting an ellipsoid along the z-axis to the xy-Plane. The resulting ellipsis is rotated around the z-axis by the angle gamma until the principal axes coincide with the x- and y axis.
Now before projecting, I rotate the ellipsoid first around the z- and then around the...
You are loosing also 50% more of the photons emitted by the sensor which otherwise would have been reflected back onto the sensor or onto some part in thermal equilibrium with the detector.
1. The main point is, that there is no difference in efficiency to detect individual photons. The number of photons reaching the detector from a patch of the moon at 301 K is simply higher than the number of photons from a 300 K patch, see the Planck formula.
2. Then you have to take into...
Photovoltaic mode, as you say, means, that there is no bias voltage, i.e., the diode is operated in short circuit (with a current meter). A dark current can also flow in zero bias mode. As there is no voltage difference, any dark current, due to the thermal generation of electron hole pairs...
In fact, the oxidation states of Ga and As remain +III and -III, respectively, Si has a higher electronegativity than Ga, but a lower one than As. As both Ga and As form 4 polar bonds with Si, the resulting oxidation states are +III and -III. We may compare this to the reaction of ammonia with...
Some post involving diodes made me reconsider pn junctions, with I first learned almost 40 years ago. Coming from a chemistry background, something always felt strange, but I could not tell what. Now I realized:
While e.g. Gallium arsenide is a semiconductor itself and can be described roughly...
Thanks Jim for splitting this off!
I think it is easier to consider instead of a diode a pure semiconductor crystal, made of e.g. CdTe. A photon with an energy larger than the band gap may produce an electron-hole pair which may be detected by applying a voltage and measuring the current. Now...
By the Fluctuation-Dissipation theorem,
https://en.wikipedia.org/wiki/Fluctuation-dissipation_theorem
Both the diode and the resistor will constantly generate a fluctuating current. If the object you want to detect is in thermal equilibrium with your electronic device, you won't be able to...
The question is not so much about detection of electron hole pairs as about the statistics of this process. If the detector has the same temperature or a higher one than the object whose radiation you want to detect, you will not be able to distinguish the signal due to your star from the...
Good question. I would expect the solubility to be very low. Where does your question come from? Is this related to some chemical reaction or rather spectroscopy?
You are welcome!
This reasoning can be formalized. Pauling already introduced the electronegativity, which determines the ionicity of the bonds, in terms of the electron affinity and ionization energy.
According to Koopman's theorem, both are given basically in terms of the LUMO and HOMO...
I think one confusion with the definition of polarisation is that, while it can in deed be described as a density of hypothetical infinitesimal dipoles, it is in general not sufficient to only consider the real dipole moments of the atoms or molecules making up the substance, although sometimes...
I remember that even for localized charge distributions the polarization will involve higher order multipole moments. In principle, P is fixed by the condition that it's divergence equals the charge distribution up to the rotation of an arbitrary field, as div rot =0. The irrotational part may...
Interesting project. I would not use tallow but rather some cheap olive oil which has a more similar spectrum of fatty acids as compared to neatsfoot oil. This gives you a type of turkish red oil. Sulphonation definitely will work with concentrated sulphuric acid. It is extremely important to...
I suppose you can find data on HeH+, HeD+ and HeT+.
There are accurate calculations:
https://aip.scitation.org/doi/10.1063/1.4759077
These molecules are of great interest, as they form in the decay of TH, TD and T_2 which are intensively studied in the context of measuring the mass of neutrinos.
You can describe identical distinguishable particles in QFT using paraboson or parafermion creation and anihilation operators. This also allows to describe Boltzmann statistics as a limit in QFT.
Interesting, just some days ago I read this article
https://journals.plos.org/plosone/article?id=10.1371/journal.pone.0264092
which puts serious doubts on earlier arrival in the americas.
Yes, this is the point I wanted to make. Even if particles are distinguishable at a microscopic level (as in the DNA example or in case of isotopic substitution pattern), the information contained in these particle "labels" corresponds to a huge information entropy, which has to be taken into...
I like as an example to consider random DNA molecules. If the molecules are long enough, no two molecules will have the same sequence. Macroscopically, the DNA will behave as a pure uniform substance. You can't devise a macroscopic process which would allow you to generate work from mixing two...
Maybe you can think of it like this. This degeneracy can be lifted in several ways. For example, in a spin 1/2 problem, you can add either a sigma_z or a sigma_x term to lift the degeneracy. Assume you take the degenerate case as the limit of a small sigma_z potential tending to 0. Now if you...
This fascinating story about Tanis was first - and quite unusually - published in "The New Yorker", and not in a scientific journal. A following scientific publication in PNAS did not contain many of the claims made in the New Yorker. Paleontologists seem to be rather sceptical concerning...
This case is known as "degenerate perturbation theory" and is discussed in any textbook. Specifically, you first have to diagonalize the perturbation V in the set of degenerate zeroth order states.
Here another attempt of an answer:
I think the main question is how people did know let say how many grams of e.g. silver correspond to 1 mole without being able to count the number of atoms.
The point is that chemists observed already in the 18th century, the law of constant proportions...
The SI system is system of units geared towards engineering. In this respect, the mole makes sense, as, other than with the dozen, a chemist (and most physicists either) have no means to count the atoms or molecules inside the amount of substance they are working with, neither would they have...
I am not sure whether this statement about "not an eigenstate of the Hamiltonian" is correct. We are talking here about phases, which usually require the idealisation of systems of infinite extent. But in infinite systems, also broken symmetry states are eigenstates of a Hamiltonian.
Even if the frequency for t>0 is fixed, if you do a Fourier transform of the whole time dependence of the electric field, it will contain all frequencies. You simply don't have monochromatic photons, if the wavetrain is not infinitely long.