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Differences between revisions 1 and 5 (spanning 4 versions)
 ⇤ ← Revision 1 as of 2015-12-14 08:19:48 → Size: 3917 Editor: KeithLofstrom Comment: ← Revision 5 as of 2015-12-14 18:54:59 → ⇥ Size: 8969 Editor: KeithLofstrom Comment: Deletions are marked like this. Additions are marked like this. Line 1: Line 1: #format jsmath Line 2: Line 3: [[ Rws | Really Wild Stuff ]] Line 5: Line 8: I have a 0.67 acre (0.2711 hectare) suburban lot, and I theoretically own the mineral rights all the way to the center of the Earth. The earth's surface is 510 million km², or 51 billion hectares - my lot is 5.32 e-12 of the surface of the earth, and also of the total mass of the earth (5.6736e24 kg), so my "share" of that is 3e13 kg. It is above continental granite, but relatively thin - the Juan de Fuca plate, mostly basalt without beneficiated ore bodies, is subducting down into the mantle only a few kilometers beneath me. There are a few dormant volcanos nearby. I have a 0.67 acre ( 2711 m², 0.2711 hectare) suburban lot, and I theoretically own the mineral rights all the way to the center of the Earth. The earth's surface is 510 million km², or 51 billion hectares - my lot is 5.32 e-12 of the surface of the earth, and also of the total mass of the earth (5.6736e24 kg), so my "share" of that is 3e13 kg. The property is above continental granite, but relatively thin - the Juan de Fuca plate, mostly basalt without beneficiated ore bodies, is subducting down into the mantle only a few kilometers beneath me. There are a few dormant volcanos nearby. Line 7: Line 10: So - assuming some magic way to keep the sides from collapsing, how much energy would it cost to cut a hole all the way to a point in the center? How much material is that, and how much is it worth, purified to commercial grade metals? But let's assume this volume is "average" for the Earth's composition, and make the same kind of calculations that are made for materials in asteroids. Assuming some magic way to keep the sides from collapsing, how much energy would it cost to cut a hole all the way to a point in the center and lift everything out? How much material is that, and (assuming market prices did not collapse) how much is it worth, purified to commercial grade metals? Line 9: Line 12: ||<-7> Continental Crust 0.47%, my share is 1.43e11 kg, est. 20 km thick, lift energy 1.4e16 J |||| SiO₂ || 60.6% || 8.67e10 kg || Glass || 8.67e10 kg || $1.00/kg ||$8.6e10 |||| Al₂O₃ || 15.9% || 2.27e10 kg || Aluminum || 1.20e10 kg || $1.80/kg ||$2.2e10 || ||<-7> '''Continental Crust 0.47%, my share is 1.43e11 kg''' ||||!SiO₂ || 60.6% || 8.67e10 kg || Glass || 8.67e10 kg || $1.00/kg ||$8.6e10 ||||!Al₂O₃ || 15.9% || 2.27e10 kg || Aluminum || 1.20e10 kg || $1.80/kg ||$2.2e10 || Line 17: Line 20: || TiO₂  || 0.7% || 1.00e09 kg || Titanium || 6.00e08 kg || $4.30/kg ||$2.6e09 |||| P₂O₅  || 0.1% || 1.43e08 kg || Phosphorus || 6.24e07 kg || $76/kg ||$4.7e09 || || !TiO₂ || 0.7% || 1.00e09 kg || Titanium || 6.00e08 kg || $4.30/kg ||$2.6e09 |||| !P₂O₅ || 0.1% || 1.43e08 kg || Phosphorus || 6.24e07 kg || $76/kg ||$4.7e09 || Line 23: Line 26: || U || 1.8 ppm || 2.57e05 kg || Uranium || 2.57e05 kg || $140/kg ||$3.5e07 || || U || 1.8 ppm || 2.57e05 kg || Uranium || 2.57e05 kg || $140/kg ||$3.5e07 || Line 26: Line 29: ||<-7> Mantle, 67.3%, my share is 2.02e13 kg, 35 to 2900 km, lift energy xxxxxx J |||| Mg || 22.2% || 4.48e12 kg || Magnesium || 4.48e12 kg || $4.00/kg ||$1.8e13 || ||<-7> '''Mantle, 67.3%, my share is 2.02e13 kg, 35 to 2900 km'''              |||| Mg || 22.2% || 4.48e12 kg || Magnesium || 4.48e12 kg || $4.00/kg ||$1.8e13 || Line 29: Line 32: || Fe || 6.3% || 1.27e12 kg || Cast Iron || 1.27e12 kg || $0.20/kg ||$2.5e11 || || Fe || 6.3% || 1.27e12 kg || Cast Iron || 1.27e12 kg || $0.20/kg ||$2.5e11 || Line 32: Line 35: || P || 260 ppm || 5.25e09 kg || Phosphorus || 5.25e09 kg || $76/kg ||$4.0e11 || || P || 260 ppm || 5.25e09 kg || Phosphorus || 5.25e09 kg || $76/kg ||$4.0e11 || Line 36: Line 39: ||<-7> Core 32.2%, my share is 9.8e12 kg, 2900 to 6370 km, lift energy xxxxxx J |||| Fe || 85% || 8.33e12 kg || Cast Iron || 8.33e12 kg || $0.20/kg ||$1.7e12 || ||<-7> '''Core 32.2%, my share is 9.8e12 kg, 2900 to 6370 km'''              |||| Fe || 85% || 8.33e12 kg || Cast Iron || 8.33e12 kg || $0.20/kg ||$1.7e12 || Line 39: Line 42: ||<-6> '''Estimated market value of my minerals, 34 trillion dollars'''    || $3.4e13 || ||<-6> '''Estimated market value of my minerals, 34 trillion dollars''' || '''$3.4e13 ''' || Line 41: Line 44: -------- Line 42: Line 46: === Lift Energy compared to Asteroid Re-orbit Energy ===The material is in a tall, skinny, inverted pyramid. We will optimistically assume uniform density to make the calculation easier, though this is hugely less optimistic than the absurd idea that we can actually do anything like this.Assume a perfect sphere with a surface gravity of $g$, radius $R_e$, and density $\rho$. Assume the surface area is $A$. How much energy $d E$ does it take to lift a slab from depth $R$ with thickness $d R$ to the surface?The area of a slab at depth $R$ is $A_r = A ~ ( R / R_e )^2$. The mass of this slab is $d m = \rho ~ A_r ~ dR = \rho ~ A ~ ( R / R_e )^2 ~ d R$Gravity inside a uniform sphere is linearly proportional to radius, so the gravity at radius $R_y$ is $g_y = g ~ R_y / R_e$ . The amount of energy to lift the slab form depth $R$ a distance of $d R_y$ is $d E_R ~ = ~ g_y ~ d m ~ d R = ( g ~ R_y / R_e ) ( \rho ~ A ~ ( R / R_e )^2 ) ~ d_R ~ d R_y ~ = ~ ( \rho ~ g ~ A ~ R^2 / {R_e}^3 ) ~ R_y d_R ~ d R_y$ Let's integrate that over $R_y$ from $R$ to the surface $R e$ :$E_R ~ = ~ ( \rho ~ g ~ A ~ R^2 / {R_e}^3 ) ( {R_e}^2 - R^2 / 2 ) = ( \rho ~ g ~ A / 2 ) ( ( R^2 / R_e ) - ( R^4 / {R_e}^3 ) )$The total energy for the entire column of material integrates over $R$ from 0 to $R_e$:$E ~ = ~ \rho ~ g ~ A ( R_e )^2 / 15$The volume of the inverted pyramid is $A R_e / 3$, so the assumed-uniform density $\rho$ is the total mass $M$ divided by volume:$\rho ~ = ~ 3 ~ M / ( A ~ R_e )$Producing a very simple result for the whole column:$E ~ = ~ g ~ M ~ R_e / 5$If $g$ = 9.8 m/s², $M$ = 3e13 kg, and $R_e$ = 6371 km, the total lift energy $E$ is 3.7e20 J, about 50 minutes of earth solar illumination, or 9 hours of 10% efficient global PV. The energy per kilogram is 12.5 MJ, or about 3.47 kilowatt hours, or a $\delta V$ equivalent of 5000 m/s. Assuming that the heat capacity of earth mantle material is 500 J/kg-K (POMA), a kilogram brought up from the 4000 K lower mantle boundary would add about 2 MJ/kg, a total of 6e19 J, about 8 minutes of total earth surface solar energy.=== Asteroid Delivery Energy === If the volume of material was shaped into a spherical asteroid, it would have a radius $R_A ~ = ~ ( A ~ R_e / 4 \pi )^{1/3} ~ = ~ 1.11 km. But since we need reaction mass, let's assume we are tossing the material off a much larger asteroid with a superhigh acceleration launcher.Moving a chunk of asteroid from the asteroid belt into low earth orbit is difficult - let's assume we will merely change its orbit from a circular 2 AU to an elliptical orbit between 2 AU and 1 AU, and we will use the atmosphere to bring the material down, and that the material somehow remains intact and impacts somewhere safe. We will assume it has the same composition as our hypothetical wedge of earth and is equally valuable. The 1 AU earth orbits at 30 km/s, and the 2 AU asteroid belt orbits$ \sqrt{2} $slower, 21.2 km/s. An elliptical orbit between 1 AU and 2 AU has an eccentricity$ e $= 1/3, a semimajor axis$ a $= 1.5 AU, a characteristic velocity$ v_0 $= 26 km/s, an aphelion velocity$ v_a $= 17.3 km/s, and a perihelion velocity$ v_p $= 34.6 km/s . So, assuming a relatively small escape velocity from the source asteroid, the asteroid launch velocity is 21.2 - 17.3 = 3.9 km/s, only 61% of the energy of extraction of the wedge from the earth. Note that if both operations are solar powered, the Earth's surface gets about 2.8x more solar illumination.The asteroidal material arrives at the earth's orbit with a relative velocity of 4.6 km/s, an energy of 10.6 MJ/kg, which is added to the earth escape velocity ( 10.7 km/s ) energy of 57.3 MJ/kg, totalling 68 MJ/kg. That energy would be divided between impact energy and atmospheric heating, but would all become atmospheric heat after the impact energy dissipates. Total energy for 3e13 kg is 2e21 joules, equivalent to 4.5 hours of earth surface solar energy.=== Conclusion ===Overall, the two processes are within an order of magnitude of each other, both delivering gravitationally sorted but otherwise non-beneficiated rock of approximately equal (low) value to the earth's surface. The "core the earth" approach is obviously silly - besides access to a nickel-rich core, there many disadvantages compared to a mine 20 km deep and 67 acres in area, or a mine 200 meters deep and 6700 acres in area, which would require far less energy to remove. The product is the same - uninteresting rock, unless this was done around a concentrated ore body. '''Asteroid mining to provide raw materials to Earth is ridiculous.''' Backyard Minerals Space materials delivered to earth as manufactured goods do not make much sense. The materials are not beneficiated ores, and manufacturing requires a huge and interconnected network of factories. Let's compare it to something equally silly. I have a 0.67 acre ( 2711 m², 0.2711 hectare) suburban lot, and I theoretically own the mineral rights all the way to the center of the Earth. The earth's surface is 510 million km², or 51 billion hectares - my lot is 5.32 e-12 of the surface of the earth, and also of the total mass of the earth (5.6736e24 kg), so my "share" of that is 3e13 kg. The property is above continental granite, but relatively thin - the Juan de Fuca plate, mostly basalt without beneficiated ore bodies, is subducting down into the mantle only a few kilometers beneath me. There are a few dormant volcanos nearby. But let's assume this volume is "average" for the Earth's composition, and make the same kind of calculations that are made for materials in asteroids. Assuming some magic way to keep the sides from collapsing, how much energy would it cost to cut a hole all the way to a point in the center and lift everything out? How much material is that, and (assuming market prices did not collapse) how much is it worth, purified to commercial grade metals?  Continental Crust 0.47%, my share is 1.43e11 kg SiO₂ 60.6% 8.67e10 kg Glass 8.67e10 kg 1.00/kg || 8.6e10 Al₂O₃ 15.9% 2.27e10 kg Aluminum 1.20e10 kg 1.80/kg || 2.2e10 FeO 6.7% 9.58e09 kg Cast Iron 7.45e09 kg 0.20/kg || 1.5e09 CaO 6.4% 9.15e09 kg Calcium 6.54e09 kg 2.00/kg || 1.3e10 MgO 4.7% 6.72e09 kg Magnesium 5.39e09 kg 4.00/kg || 2.1e10 Na₂O 3.1% 4.43e09 kg Sodium 3.29e09 kg 3.30/kg || 1.1e10 K₂O 1.8% 2.57e09 kg Potassium 2.14e09 kg 22/kg || 4.7e10 TiO₂ 0.7% 1.00e09 kg Titanium 6.00e08 kg 4.30/kg || 2.6e09 P₂O₅ 0.1% 1.43e08 kg Phosphorus 6.24e07 kg 76/kg || 4.7e09 Ni 90 ppm 1.29e07 kg Nickel 1.29e07 kg 3.90/kg || 5.0e07 Cu 68 ppm 9.72e06 kg Copper 9.72e06 kg 2.00/kg || 1.9e07 Nd 33 ppm 4.72e06 kg Neodymium 4.72e06 kg 60/kg || 2.8e08 Ga 19 ppm 2.72e06 kg Gallium 2.72e06 kg 200/kg || 5.4e08 U 1.8 ppm 2.57e05 kg Uranium 2.57e05 kg 140/kg || 3.5e07 In 160 ppb 22900 kg Indium 22900 kg 700/kg || 1.6e07 Se 50 ppb 7150 kg Selenium 7150 kg 150/kg || 1.1e06 Mantle, 67.3%, my share is 2.02e13 kg, 35 to 2900 km Mg 22.2% 4.48e12 kg Magnesium 4.48e12 kg 4.00/kg || 1.8e13 Si 21.2% 4.28e12 kg Glass 9.17e12 kg 1.00/kg || 9.2e12 Fe 6.3% 1.27e12 kg Cast Iron 1.27e12 kg 0.20/kg || 2.5e11 Ca 2.6% 5.25e11 kg Calcium 5.25e11 kg 2.00/kg || 1.1e12 Al 2.4% 4.85e11 kg Aluminum 4.85e11 kg 1.80/kg || 8.7e11 P 260 ppm 5.25e09 kg Phosphorus 5.25e09 kg 76/kg || 4.0e11 Cu 20 ppm 4.04e08 kg Copper 4.04e08 kg 2.00/kg || 8.1e08 U 22 ppb 4.44e05 kg Uranium 4.44e05 kg 140/kg || 6.2e07 In 13 ppb 2.63e05 kg Indium 2.63e05 kg 700/kg || 1.8e08 Core 32.2%, my share is 9.8e12 kg, 2900 to 6370 km Fe 85% 8.33e12 kg Cast Iron 8.33e12 kg 0.20/kg || 1.7e12 Ni 5% 4.90e11 kg Nickel 4.90e11 kg 3.90/kg || 1.9e12 Estimated market value of my minerals, 34 trillion dollars$3.4e13

Lift Energy compared to Asteroid Re-orbit Energy

The material is in a tall, skinny, inverted pyramid. We will optimistically assume uniform density to make the calculation easier, though this is hugely less optimistic than the absurd idea that we can actually do anything like this.

Assume a perfect sphere with a surface gravity of g, radius R_e , and density \rho. Assume the surface area is A . How much energy d E does it take to lift a slab from depth R with thickness d R to the surface?

The area of a slab at depth R is A_r = A ~ ( R / R_e )^2 . The mass of this slab is d m = \rho ~ A_r ~ dR = \rho ~ A ~ ( R / R_e )^2 ~ d R

Gravity inside a uniform sphere is linearly proportional to radius, so the gravity at radius R_y is g_y = g ~ R_y / R_e . The amount of energy to lift the slab form depth R a distance of d R_y is

d E_R ~ = ~ g_y ~ d m ~ d R = ( g ~ R_y / R_e ) ( \rho ~ A ~ ( R / R_e )^2 ) ~ d_R ~ d R_y ~ = ~ ( \rho ~ g ~ A ~ R^2 / {R_e}^3 ) ~ R_y d_R ~ d R_y

Let's integrate that over R_y from R to the surface R e :

E_R ~ = ~ ( \rho ~ g ~ A ~ R^2 / {R_e}^3 ) ( {R_e}^2 - R^2 / 2 ) = ( \rho ~ g ~ A / 2 ) ( ( R^2 / R_e ) - ( R^4 / {R_e}^3 ) )

The total energy for the entire column of material integrates over R from 0 to R_e :

E ~ = ~ \rho ~ g ~ A ( R_e )^2 / 15

The volume of the inverted pyramid is A R_e / 3 , so the assumed-uniform density \rho is the total mass M divided by volume:

\rho ~ = ~ 3 ~ M / ( A ~ R_e )

Producing a very simple result for the whole column:

E ~ = ~ g ~ M ~ R_e / 5

If g = 9.8 m/s², M = 3e13 kg, and R_e = 6371 km, the total lift energy E is 3.7e20 J, about 50 minutes of earth solar illumination, or 9 hours of 10% efficient global PV. The energy per kilogram is 12.5 MJ, or about 3.47 kilowatt hours, or a \delta V equivalent of 5000 m/s. Assuming that the heat capacity of earth mantle material is 500 J/kg-K (POMA), a kilogram brought up from the 4000 K lower mantle boundary would add about 2 MJ/kg, a total of 6e19 J, about 8 minutes of total earth surface solar energy.

Asteroid Delivery Energy

If the volume of material was shaped into a spherical asteroid, it would have a radius \$ R_A ~ = ~ ( A ~ R_e / 4 \pi )^{1/3} ~ = ~ 1.11 km. But since we need reaction mass, let's assume we are tossing the material off a much larger asteroid with a superhigh acceleration launcher.

Moving a chunk of asteroid from the asteroid belt into low earth orbit is difficult - let's assume we will merely change its orbit from a circular 2 AU to an elliptical orbit between 2 AU and 1 AU, and we will use the atmosphere to bring the material down, and that the material somehow remains intact and impacts somewhere safe. We will assume it has the same composition as our hypothetical wedge of earth and is equally valuable.

The 1 AU earth orbits at 30 km/s, and the 2 AU asteroid belt orbits \sqrt{2} slower, 21.2 km/s. An elliptical orbit between 1 AU and 2 AU has an eccentricity e = 1/3, a semimajor axis a = 1.5 AU, a characteristic velocity v_0 = 26 km/s, an aphelion velocity v_a = 17.3 km/s, and a perihelion velocity v_p = 34.6 km/s . So, assuming a relatively small escape velocity from the source asteroid, the asteroid launch velocity is 21.2 - 17.3 = 3.9 km/s, only 61% of the energy of extraction of the wedge from the earth. Note that if both operations are solar powered, the Earth's surface gets about 2.8x more solar illumination.

The asteroidal material arrives at the earth's orbit with a relative velocity of 4.6 km/s, an energy of 10.6 MJ/kg, which is added to the earth escape velocity ( 10.7 km/s ) energy of 57.3 MJ/kg, totalling 68 MJ/kg. That energy would be divided between impact energy and atmospheric heating, but would all become atmospheric heat after the impact energy dissipates. Total energy for 3e13 kg is 2e21 joules, equivalent to 4.5 hours of earth surface solar energy.

Conclusion

Overall, the two processes are within an order of magnitude of each other, both delivering gravitationally sorted but otherwise non-beneficiated rock of approximately equal (low) value to the earth's surface. The "core the earth" approach is obviously silly - besides access to a nickel-rich core, there many disadvantages compared to a mine 20 km deep and 67 acres in area, or a mine 200 meters deep and 6700 acres in area, which would require far less energy to remove. The product is the same - uninteresting rock, unless this was done around a concentrated ore body.

Asteroid mining to provide raw materials to Earth is ridiculous.

BackyardMinerals (last edited 2021-04-18 00:58:37 by KeithLofstrom)


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## Dipole-dipole interactions in NMR: explained

Interactions between spins are fundamental for understanding magnetic resonance. One of the most important ones is the magnetic dipole-dipole interaction. Spins act as tiny magnets, thus, they can interact with each other directly through space, pretty much the same way as classical magnetic dipoles (Figure 1). In NMR, dipole-dipole interactions very often determine lineshapes of solid-state samples and relaxation rates of nuclei in the liquid state.

In many NMR textbooks, you can find the following expression for dipole-dipole (DD) Hamiltonian between two spins 1 and 2:

$\hat{{H}}_{\rm DD} = d_{12} \left( 3 \hat{I}_{1z} \hat{I}_{2z} - \hat{\mathbf{I}}_1 \cdot \hat{\mathbf{I}}_2 \right)$

where $d_{12} = - \displaystyle\frac{\mu_0}{4 \pi} \displaystyle\frac{\gamma_1 \gamma_2 \hbar}{r^3} \frac{\left( 3 \cos^2{\theta} - 1 \right)}{2}$, $\gamma_1$ and $\gamma_2$ are gyromagnetic ratios of the spins, $r$ is the distance between them.  Meaning of the angle $\theta$ can be seen from the Figure 2. But how was this expression derived?

In physics, I rarely struggled with imagining abstract things and concepts, but rather, I was often lazy to do math thoroughly and derive equations from the beginning to the end. That’s why I decided to write here a full derivation, from the beginning to the end, for the Hamiltonian of interacting nuclear spins. We will start from a classical expression of the magnetic field produced by the dipole and finish by the analysis of a truncated dipole-dipole Hamiltonian. Usually, you don’t see such a full derivation in textbooks. Maybe, this is because textbooks have limited space… Luckily, here we do not have such limitations, thus, we can have some fun here! Let’s go then! 🙂

Classical equation describing the magnetic field $\vec{\boldsymbol{B}}_{\rm dip} (\vec{\boldsymbol{r}})$ produced by a magnetic dipole moment $\vec{\boldsymbol{\mu}}$ is

$\vec{\boldsymbol{B}}_{\rm dip} (\vec{\boldsymbol{r}}) = \displaystyle\frac{\mu_0}{4 \pi r^3} \left( 3 \left( \vec{\boldsymbol{\mu}} \cdot \hat{\boldsymbol{r}} \right) \hat{\boldsymbol{r}} - \vec{\boldsymbol{\mu}} \right)$

where $\hat{\boldsymbol{r}} = \displaystyle\frac{\vec{\boldsymbol{r}}}{|\vec{\boldsymbol{r}}|}$ is a unit vector. It is easy to show that $\left( \vec{\boldsymbol{\mu}} \cdot \vec{\boldsymbol{r}} \right) \vec{\boldsymbol{r}} = \left( \vec{\boldsymbol{r}} \cdot \vec{\boldsymbol{r}}^{\: \intercal} \right) \vec{\boldsymbol{\mu}}$.

Indeed,

$\left( \vec{\boldsymbol{\mu}} \cdot \vec{\boldsymbol{r}} \right) \vec{\boldsymbol{r}} = \left( \mu_x r_x + \mu_y r_y + \mu_z r_z \right) \cdot \begin{pmatrix} r_x \\ r_y \\ r_z \end{pmatrix} = \begin{pmatrix} \mu_{x} r_{x} r_{x} + \mu_{y} r_{y} r_{x} + \mu_{z} r_{z} r_{x} \\ \mu_{x} r_{x} r_{y} + \mu_{y} r_{y} r_{y} + \mu_{z} r_{z} r_{y} \\ \mu_{x} r_{x} r_{z} + \mu_{y} r_{y} r_{z} + \mu_{z} r_{z} r_{z} \end{pmatrix}$

which is the same as

$\left( \vec{\boldsymbol{r}} \cdot \vec{\boldsymbol{r}}^{\: \intercal} \right) \vec{\boldsymbol{\mu}} = \begin{pmatrix} r_{x} r_{x} & r_{x} r_{y} & r_{x} r_{z} \\ r_{y} r_{x} & r_{y} r_{y} & r_{y} r_{z} \\ r_{z} r_{x} & r_{z} r_{y} & r_{z} r_{z} \end{pmatrix} \cdot \begin{pmatrix} \mu_x \\ \mu_y \\ \mu_z \end{pmatrix} = \begin{pmatrix} r_{x} r_{x} \mu_{x} + r_{x} r_{y} \mu_{y} + r_{x} r_{z} \mu_{z} \\ r_{y} r_{x} \mu_{x} + r_{y} r_{y} \mu_{y} + r_{y} r_{z} \mu_{z} \\ r_{z} r_{x} \mu_{x} + r_{z} r_{y} \mu_{y} + r_{z} r_{z} \mu_{z} \end{pmatrix}$

So, we write the magnetic field produced by a magnetic dipole $\vec{\boldsymbol{\mu}}_2$ as

$\vec{\boldsymbol{B}}_{\mu_2} = \displaystyle\frac{\mu_0}{4 \pi r^3} \left( 3 \left( \hat{\boldsymbol{r}} \cdot \hat{\boldsymbol{r}}^{\: \intercal} \right) \vec{\boldsymbol{\mu}}_2 - \vec{\boldsymbol{\mu}}_2 \right)$

The energy of a magnetic dipole $\vec{\boldsymbol{\mu}}_1$ interacting with the magnetic field $\vec{\boldsymbol{B}}_{\mu_2}$ produced by a magnetic dipole $\vec{\boldsymbol{\mu}}_2$ (dipole-dipole interaction) is therefore

$E_{\rm DD} = - \left( \vec{\boldsymbol{\mu}}_1 \cdot \vec{\boldsymbol{B}}_{\mu_2} \right) = -\displaystyle\frac{\mu_0}{4 \pi r^3} \left( 3 \cdot \vec{\boldsymbol{\mu}}_1 \left( \hat{\boldsymbol{r}} \cdot \hat{\boldsymbol{r}}^{\: \intercal} \right) \vec{\boldsymbol{\mu}}_2 - \left( \vec{\boldsymbol{\mu}}_1 \cdot \vec{\boldsymbol{\mu}}_2 \right) \right)$

The transition from classical to quantum mechanics is realized by substituting the measurable quantities by corresponding quantum mechanical operators:

$E_{\rm DD} \rightarrow \hat{H}_{\rm DD} \quad \vec{\boldsymbol{\mu}}_1 \rightarrow \gamma_1 \hbar \hat{\mathbf{I}}_1\quad \vec{\boldsymbol{\mu}}_2 \rightarrow \gamma_2 \hbar \hat{\mathbf{I}}_2$

$\label{Eq_Hdd} \hat{H}_{\rm DD} = - \displaystyle\frac{\mu_0}{4 \pi } \displaystyle\frac{\gamma_1 \gamma_1 \hbar}{r^3} \left( 3 \cdot \hat{\mathbf{I}}_1 \left( \hat{\boldsymbol{r}} \cdot \hat{\boldsymbol{r}}^{\: \intercal} \right) \hat{\mathbf{I}}_2 - \left( \hat{\mathbf{I}}_1 \cdot \hat{\mathbf{I}}_2 \right) \right) = b_{12} \hat{\mathbf{I}}_1 \hat{\mathbf{D}} \hat{\mathbf{I}}_2$

here $b_{12}$ is a factor which depends only on the types of the nuclear spins and the distance between them, and a tensor of dipole-dipole interactions contains information about the mutual orientation of two spins:

$\hat{\mathbf{D}} = 3 \cdot \left( \hat{\boldsymbol{r}} \cdot \hat{\boldsymbol{r}}^{\: \intercal} \right) - \hat{1}$

here $\hat{1}$ is a unit matrix. Note that we write Hamiltonian $\hat{H}_{\rm DD}$ in units of [rad/s], that is why one $\hbar$ is missing. In spherical coordinates:

$\hat{\boldsymbol{r}} = \begin{pmatrix} \sin{\theta} \cos{\phi} \\ \sin{\theta} \sin{\phi} \\ \cos{\theta} \end{pmatrix}$

Therefore,

$\hat{\mathbf{D}} = \begin{pmatrix} 3 \sin^2{\theta} \cos^2{\phi} - 1 & 3 \sin^2{\theta} \cos{\phi} \sin{\phi} & 3 \sin{\theta} \cos{\theta} \cos{\phi} \\ 3 \sin^2{\theta} \cos{\phi} \sin{\phi} & 3 \sin^2{\theta} \sin^2{\phi} - 1 & 3 \sin{\theta} \cos{\theta} \sin{\phi} \\ 3 \sin{\theta} \cos{\theta} \cos{\phi} & 3 \sin{\theta} \cos{\theta} \sin{\phi} & 3 \cos^2{\theta} - 1 \end{pmatrix}$

Looks good, doesn’t it? Now, let’s evaluate the product $\hat{\mathbf{I}}_1 \hat{\mathbf{D}} \hat{\mathbf{I}}_2$:

$\hat{\mathbf{I}}_1 \hat{\mathbf{D}} \hat{\mathbf{I}}_2 = \\ \begin{pmatrix} \hat{I}_{1x} & \hat{I}_{1y} & \hat{I}_{1z} \end{pmatrix} \begin{pmatrix} 3 \sin^2{\theta} \cos^2{\phi} - 1 & 3 \sin^2{\theta} \cos{\phi} \sin{\phi} & 3 \sin{\theta} \cos{\theta} \cos{\phi} \\ 3 \sin^2{\theta} \cos{\phi} \sin{\phi} & 3 \sin^2{\theta} \sin^2{\phi} - 1 & 3 \sin{\theta} \cos{\theta} \sin{\phi} \\ 3 \sin{\theta} \cos{\theta} \cos{\phi} & 3 \sin{\theta} \cos{\theta} \sin{\phi} & 3 \cos^2{\theta} - 1 \end{pmatrix} \begin{pmatrix} \hat{I}_{2x} \\ \hat{I}_{2y} \\ \hat{I}_{2z} \end{pmatrix} =$

$\begin{pmatrix} \hat{I}_{1x} & \hat{I}_{1y} & \hat{I}_{1z} \end{pmatrix} \begin{pmatrix} \hat{I}_{2x} \left( 3 \sin^2{\theta} \cos^2{\phi} - 1 \right) + \hat{I}_{2y} \left( 3 \sin^2{\theta} \cos{\phi} \sin{\phi} \right) + \hat{I}_{2z} \left( 3 \sin{\theta} \cos{\theta} \cos{\phi} \right) \\ \hat{I}_{2x} \left( 3 \sin^2{\theta} \cos{\phi} \sin{\phi} \right) + \hat{I}_{2y} \left( 3 \sin^2{\theta} \sin^2{\phi} - 1 \right) + \hat{I}_{2z} \left( 3 \sin{\theta} \cos{\theta} \sin{\phi} \right) \\ \hat{I}_{2x} \left( 3 \sin{\theta} \cos{\theta} \cos{\phi} \right) + \hat{I}_{2y} \left( 3 \sin{\theta} \cos{\theta} \sin{\phi} \right) + \hat{I}_{2z} \left( 3 \cos^2{\theta} - 1 \right) \end{pmatrix} =$

$= \hat{I}_{1x} \hat{I}_{2x} \left( 3 \sin^2{\theta} \cos^2{\phi} - 1 \right) + \hat{I}_{1x} \hat{I}_{2y} \left( 3 \sin^2{\theta} \cos{\phi} \sin{\phi} \right) + \hat{I}_{1x} \hat{I}_{2z} \left( 3 \sin{\theta} \cos{\theta} \cos{\phi} \right) + + \hat{I}_{1y} \hat{I}_{2x} \left( 3 \sin^2{\theta} \cos{\phi} \sin{\phi} \right) + \hat{I}_{1y} \hat{I}_{2y} \left( 3 \sin^2{\theta} \sin^2{\phi} - 1 \right) + \hat{I}_{1y} \hat{I}_{2z} \left( 3 \sin{\theta} \cos{\theta} \sin{\phi} \right) + + \hat{I}_{1z} \hat{I}_{2x} \left( 3 \sin{\theta} \cos{\theta} \cos{\phi} \right) + \hat{I}_{1z} \hat{I}_{2y} \left( 3 \sin{\theta} \cos{\theta} \sin{\phi} \right) + \hat{I}_{1z} \hat{I}_{2z} \left( 3 \cos^2{\theta} - 1 \right)$

Let’s color terms to make it easier grouping them:

$\hat{I}_{1x} \hat{I}_{2x} \left( 3 \sin^2{\theta} \cos^2{\phi} - 1 \right) +$ $\hat{I}_{1x} \hat{I}_{2y} \left( 3 \sin^2{\theta} \cos{\phi} \sin{\phi} \right) +$ $\hat{I}_{1x} \hat{I}_{2z} \left( 3 \sin{\theta} \cos{\theta} \cos{\phi} \right) +$ $\hat{I}_{1y} \hat{I}_{2x} \left( 3 \sin^2{\theta} \cos{\phi} \sin{\phi} \right) +$ $\hat{I}_{1y} \hat{I}_{2y} \left( 3 \sin^2{\theta} \sin^2{\phi} - 1 \right) +$ $\hat{I}_{1y} \hat{I}_{2z} \left( 3 \sin{\theta} \cos{\theta} \sin{\phi} \right) +$ $\hat{I}_{1z} \hat{I}_{2x} \left( 3 \sin{\theta} \cos{\theta} \cos{\phi} \right) + \hat{I}_{1z} \hat{I}_{2y} \left( 3 \sin{\theta} \cos{\theta} \sin{\phi} \right) +$ $\hat{I}_{1z} \hat{I}_{2z} \left( 3 \cos^2{\theta} - 1 \right)$

Groupling the red terms gives

$\left( \hat{I}_{1x} \hat{I}_{2x} \cos^2{\phi} + \hat{I}_{1y} \hat{I}_{2y} \sin^2{\phi} \right) 3 \sin^2{\theta} - \left( \hat{I}_{1x} \hat{I}_{2x} + \hat{I}_{1y} \hat{I}_{2y} \right)$

Let’s not forget about intrinsic connections of spin angular momentum with raising and lowering operators:

$\hat{I}_{1x} \hat{I}_{2x} = \displaystyle\frac{\left( \hat{I}_{1+} + \hat{I}_{1-} \right)}{2}\displaystyle\frac{\left( \hat{I}_{2+} + \hat{I}_{2-} \right)}{2} = \displaystyle\frac{1}{4} \left( \hat{I}_{1+} \hat{I}_{2+} + \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} + \hat{I}_{1-} \hat{I}_{2-} \right)$

$\hat{I}_{1y} \hat{I}_{2y} = \displaystyle\frac{\left( \hat{I}_{1+} - \hat{I}_{1-} \right)}{2 i}\displaystyle\frac{\left( \hat{I}_{2+} - \hat{I}_{2-} \right)}{2 i} = -\displaystyle\frac{1}{4} \left( \hat{I}_{1+} \hat{I}_{2+} - \hat{I}_{1+} \hat{I}_{2-} - \hat{I}_{1-} \hat{I}_{2+} + \hat{I}_{1-} \hat{I}_{2-} \right)$

Therefore, grouping the red terms gives

$\left( \hat{I}_{1x} \hat{I}_{2x} \cos^2{\phi} + \hat{I}_{1y} \hat{I}_{2y} \sin^2{\phi} \right) 3 \sin^2{\theta} - \left( \hat{I}_{1x} \hat{I}_{2x} + \hat{I}_{1y} \hat{I}_{2y} \right) = \left( \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} \right) \frac{3}{4} \sin^2{\theta} + \left( \hat{I}_{1+} \hat{I}_{2+} + \hat{I}_{1-} \hat{I}_{2-} \right) \frac{3}{4} \sin^2{\theta} \cdot \left( \cos{ 2 \phi} \right) - \frac{1}{2} \left( \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} \right) = \left( \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} \right) \frac{1}{4} \left(1 - 3 \cos^2{\theta} \right) + \left( \hat{I}_{1+} \hat{I}_{2+} + \hat{I}_{1-} \hat{I}_{2-} \right) \frac{3}{4} \sin^2{\theta} \cdot \left( \cos{ 2 \phi} \right)$

Groupling the blue terms gives

$\left( \hat{I}_{1x} \hat{I}_{2y} + \hat{I}_{1y} \hat{I}_{2x} \right) 3 \sin^2{\theta} \cos{\phi} \sin{\phi} = \left( \hat{I}_{1+} \hat{I}_{2+} - \hat{I}_{1-} \hat{I}_{2-} \right)\displaystyle\frac{3}{4} \sin^2{\theta} \cdot \left( - i \sin{2 \phi} \right)$

where we took into consideration that

$\hat{I}_{1x} \hat{I}_{2y} =\displaystyle\frac{\left( \hat{I}_{1+} + \hat{I}_{1-} \right)}{2} \frac{\left( \hat{I}_{2+} - \hat{I}_{2-} \right)}{2 i} = \frac{1}{4 i} \left( \hat{I}_{1+} \hat{I}_{2+} - \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} - \hat{I}_{1-} \hat{I}_{2-} \right)$
$\hat{I}_{1y} \hat{I}_{2x} = \displaystyle\frac{\left( \hat{I}_{1+} - \hat{I}_{1-} \right)}{2 i} \frac{\left( \hat{I}_{2+} + \hat{I}_{2-} \right)}{2} = \frac{1}{4 i} \left( \hat{I}_{1+} \hat{I}_{2+} + \hat{I}_{1+} \hat{I}_{2-} - \hat{I}_{1-} \hat{I}_{2+} - \hat{I}_{1-} \hat{I}_{2-} \right)$

Red and blue terms can be combined nicely to form

$\left( \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} \right) \displaystyle\frac{1}{4} \left(1 - 3 \cos^2{\theta} \right) + \hat{I}_{1+} \hat{I}_{2+} \left( \frac{3}{4} \sin^2{\theta} \cdot e^{- 2 i \phi} \right) + \hat{I}_{1-} \hat{I}_{2-} \left( \frac{3}{4} \sin^2{\theta} \cdot e^{+ 2 i \phi} \right)$

Now let’s focus on purple terms:

$\left( \hat{I}_{1x} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2x} \right) \left( \hat{I}_{1x} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2x} \right) \left( 3 \sin{\theta} \cos{\theta} \cos{\phi} \right) + \left( \hat{I}_{1y} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2y} \right) \left( 3 \sin{\theta} \cos{\theta} \sin{\phi} \right) = \left( \left( \hat{I}_{1+} \hat{I}_{2z} + \hat{I}_{1-} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2+} + \hat{I}_{1z}\hat{I}_{2-} \right) \cos{\phi} \right) \displaystyle\frac{3}{4} \sin{2 \theta} + \left( \left( -i \hat{I}_{1+} \hat{I}_{2z} + i \hat{I}_{1-} \hat{I}_{2z} -i \hat{I}_{1z} \hat{I}_{2+} + i \hat{I}_{1z} \hat{I}_{2-} \right) \sin{\phi} \right) \frac{3}{4} \sin{2 \theta} = \left( \hat{I}_{1+} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2+} \right) \left( \frac{3}{4} \sin{2 \theta} \right) e^{-i \phi} + \left( \hat{I}_{1-} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2-} \right) \left( \frac{3}{4} \sin{2 \theta} \right) e^{+i \phi}$

Overall, we have split our dipolar Hamiltonian into 6 term, so-called “Dipolar Alphabet”:

$\hat{\mathbf{I}}_1 \hat{\mathbf{D}} \hat{\mathbf{I}}_2 = \hat{A} + \hat{B} + \hat{C} + \hat{D} + \hat{E} + \hat{F}$

where

$\hat{A} \quad = \quad \hat{I}_{1z} \hat{I}_{2z} \left( 3 \cos^2{\theta} - 1 \right)$

$\quad\quad \hat{B} \quad = \quad \left( \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} \right) \cdot \displaystyle\frac{\left(1 - 3 \cos^2{\theta} \right)}{4}$

$\quad\quad \hat{C} \quad = \quad \left( \hat{I}_{1+} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2+} \right) \left( \displaystyle\frac{3}{4} \sin{2 \theta} \right) e^{-i \phi}$

$\quad\quad \hat{D} \quad = \quad \left( \hat{I}_{1-} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2-} \right) \left( \displaystyle\frac{3}{4} \sin{2 \theta} \right) e^{+i \phi}$

$\hat{E} \quad = \quad \hat{I}_{1+} \hat{I}_{2+} \left( \displaystyle\frac{3}{4} \sin^2{\theta} \right) e^{- 2 i \phi}$

$\quad\quad \hat{F} \quad = \quad \hat{I}_{1-} \hat{I}_{2-} \left( \displaystyle\frac{3}{4} \sin^2{\theta} \right) e^{+ 2 i \phi}$

To summarize, the Hamiltonian of two interacting spins is a $4 \times 4$ matrix composed of 6 operators. Each of the letters of the dipolar alphabet corresponds to certain matrix elements in the final Hamiltonian (Figure 3).

Without an externally imposed direction in space (for example, in the case of two equivalent spins in zero magnetic field), all of the terms of the dipole-dipole Hamiltonian need to be used for calculating an NMR spectrum. This is because all orientations in space are equivalent. However, in the presence of the external high magnetic field, the Hamiltonian can be simplified via the use of so-called “secular approximation”.

The secular approximation concerns the case where the Hamiltonian is the sum of two terms:

$\hat{H} = \hat{A} + \hat{B}$

where $\displaystyle\hat{A}$ is a “large” operator and $\hat{B}$ is a “small” operator. In our case, $\hat{A}$ can be an operator describing the interaction with the magnetic field (Zeeman Hamiltonian) and $\hat{B}$ is DD Hamiltonian. Eigenstates of the Zeeman Hamiltonian are familiar αααβ, βα, ββGenerally, $\hat{B}$ does not commute with $\hat{A}$, therefore, if written in the eigenbasis of $\hat{A}$, it has finite elements everywhere.

The secular approximation for $\hat{B}$ means that we leave only the blocks that correspond to the eigenvalue structure of the operator $\hat{A}$ (Figure 4) and disregard all other elements.

In general, we can omit a matrix element $b_{nm}$ that is much smaller than

$|b_{mn}| \ll |E_m - E_n|$

For homonuclear case (e.g., two interacting protons), this means that only the first two terms of the dipolar Alphabet will survive:

$\hat{\mathbf{I}}_1 \hat{\mathbf{D}} \hat{\mathbf{I}}_2 = \hat{A} + \hat{B} =$

$= \hat{I}_{1z} \hat{I}_{2z} \left( 3 \cos^2{\theta} - 1 \right) + \left( \hat{I}_{1+} \hat{I}_{2-} + \hat{I}_{1-} \hat{I}_{2+} \right) \cdot \displaystyle\frac{\left(1 - 3 \cos^2{\theta} \right)}{4} =$

$= \hat{I}_{1z} \hat{I}_{2z} \left( 3 \cos^2{\theta} - 1 \right) - \left( \hat{I}_{1x} \hat{I}_{2x} + \hat{I}_{2y} \hat{I}_{2y} \right) \cdot \displaystyle\frac{\left(3 \cos^2{\theta} - 1\right)}{2} =$

$= \displaystyle\frac{\left( 3 \cos^2{\theta} - 1\right)}{2} \cdot \left( 2 \hat{I}_{1z} \hat{I}_{2z} + \hat{I}_{1z} \hat{I}_{2z} - \hat{I}_{1z} \hat{I}_{2z} - \left( \hat{I}_{1x} \hat{I}_{2x} + \hat{I}_{2y} \hat{I}_{2y} \right) \right) =$

$= \displaystyle\frac{\left(3 \cos^2{\theta} - 1\right)}{2} \cdot \left( 3 \hat{I}_{1z} \hat{I}_{2z} - \hat{\mathbf{I}}_1 \cdot \hat{\mathbf{I}}_2 \right)$

Overall, this is how you go from the classical description of the magnetic field of the dipole to the truncated form of the Hamiltonian in the high nagnetic field. In the next post I will show how this Hamiltonian leads to the characteristic lineshape of the NMR line for solids.

## Parahydrogen

To begin my blog, let’s introduce parahydrogen. Lately, this little molecule has been attracting a lot of attention in the magnetic resonance community due to tremendous opportunities it brings for NMR/MRI signal enhancement. I will explain a bit later how this parahydrogen-based NMR signal enhancement works. But first, let’s talk about physical origins of parahydrogen!

Parahydrogen (para-H2) is a nuclear spin isomer of a hydrogen molecule. Nuclear spin isomerism is a very special form of isomerism. Unlike “traditional” molecular isomers (molecules having the same atomic composition but different chemical structure) and isotopologues (isomers that differ only in their isotopic composition), nuclear spin isomers are chemically identical: they have exactly the same atomic (and even isotopic) structure. However, nuclear spin isomers differ in the nuclear spin state of their atoms. It turns out that this tiny change (energy difference associated with nuclear spin transitions is only ~0.1 J/mol) may lead to different thermodynamic and spectroscopic properties of molecules. So, how does this work?

Unfortunately (or fortunately), we will have to use rules of quantum mechanics and some math. In quantum mechanics, in order to describe properties of quantum systems (atoms, molecules, etc.), physicists use wavefunctions. By knowing a wavefunction one will be able to calculate probabilities to find a quantum system in different states (namely, a squared modulus of the wavefunction determines the probability to find a system in a given state). Let’s look how it works taking as an example hydrogen molecule.

Hydrogen molecule consists of two hydrogen atoms (H) and is denoted as Н2. Each atom has a nucleus – a proton which is a spin-1/2 particle. Physicists say that hydrogen molecule has several degrees of freedom: translational, rotational, vibrational, etc., and these degrees of freedom can be considered independent. In other words, rotation of the hydrogen molecule does not depend on how and where the molecule is moving and how it is vibrating. Each degree of freedom has a wavefunction associated with it. I will use different colors to describe electron and nuclear wavefunctions. A position of the molecule in space, as well as its rotation and vibrations, are determined by the position and movements of nuclei, therefore, these degrees of freedom are described by translational (ψtr), rotational (ψrot), vibrational (ψvib), and nuclear spin (ψspin) wavefunctions. Atomic nuclei are surrounded by electrons which provides the bonding between the nuclei. The wavefunction describing movements of electrons is called orbital wavefunction ψorb, and state of the electron spins is described by the electron spin wavefunction ψspin.

Since probabilities of independent events are multiplied, the total wavefunction is a product of the above-mentioned wavefunctions:

ψtot = ψtr·ψrot·ψvib·ψspin·ψorb·ψspin

However, rules of quantum mechanics are trickier than they may sound. According to Pauli’s principle, the total wavefunction of the system of spins-1/2 particles has to be antisymmetric with respect to the exchange (also called permutation) of two identical particles. What does this mean?

Let’s take for example ψspin. A system consisting of two spins-1/2 can be described as α1α2, β1β2 or combinations α1β2+β1α2, α1β2β1α2. Here α and β denote the projection of nuclear spin angular momentum along the quantization axis (more on this stuff later, for now, one can imagine the state α as a magnetic moment – spin – pointing up along the external magnetic field and the state β as a magnetic moment pointing down, opposite to the field). Indexes 1 and 2 say to which nucleus the spin belongs. For example, the state α1α2 means that both nuclear spins point along the field while the state β1β2 means that both spins point opposite to the field. The combination states α1β2+β1α2 and α1β2β1α2 are more interesting. Neither of spins points along or opposite to the field but if we take one spin and determine its orientation, the second spin will take the opposite orientation. We can see now that two spins are correlated: the state of the second spin depends on the state of the first one.

Now let’s look what happens if we exchange (permute) particles. Mathematically, permutation simply means interchange of indexes (1→2, 2→1). One can see that upon permutation of indexes the first three states do not change: α2α1= α1α2β2β1 = β1β2, (α2β1+β2α1) = (α1β2+β1α2), but the last state changes the sign: (α2β1β2α1) = –(α1β2β1α2). Therefore, the first three states are called symmetric wavefunctions and the last one – antisymmetric with respect to permutation of particles.

So, our hydrogen molecule contains four spin-1/2 particles: two electrons and two nuclei). Permutation of electrons can only affect ψorb and ψspin. The first wave function, corresponding to the electronic ground state, is symmetric with respect to the electrons, the second, the electron spin wavefunction, is antisymmetric, and the rest are independent of the electrons’ variables and, thus, symmetric. Therefore, Pauli’s principle is fulfilled for electrons: the total wavefunction is antisymmetric with respect to permutation of electrons, thanks to antisymmetric ψspin. Permutation of nuclei can affect two wavefunctions: ψspin (as we just saw above) and ψrot. A mathematical expression for ψrot is rather complicated but it is not necessary to know its full form to understand the symmetry properties.

Figure 1. Schematic energy diagram of rotational levels of the hydrogen molecule.

This is because rotating diatomic molecules possess a set of stable rotational states, which can be described by only one parameter – the rotational quantum number J. This number can take integer values 0, 1, 2, 3, … This means that molecule can be in a stable state with J = 0, J = 1, J = 2, etc. (Figure 1). It turns out that the symmetry (with respect to permutation of nuclei) of the rotational wavefunction can be described as

P12·ψrot = (-1)J·ψrot

where P12 represents the permutation operator that interchanges the nuclei’s positions (indexes). This means that the rotational wavefunction is symmetric for even rotational states (J = 0, 2, 4, …) and antisymmetric for odd rotational states (J = 1, 3, 5, ).

Coming back to Pauli’s principle, permutation of nuclei should lead to the change of sign of the total wavefunction. Since only ψspin and ψrot can change sign upon such permutation, these two wavefunctions become connected: even (symmetric) rotational wavefunctions must be combined with the antisymmetric nuclear wavefunction (α1β2β1α2), whereas each antisymmetric rotational wavefunction has to be associated with one of the three symmetric spin functions. All this is required to yield a total wavefunction being antisymmetric with respect to the exchange of the nuclei. This is where two hydrogen spin isomers come from. One is called parahydrogen (para-H2), having an antisymmetric nuclear spin wavefunction 1β2β1α2) and existing only in even rotational states, and the other called orthohydrogen (ortho-H2), having a symmetric nuclear spin wavefunction and existing only in the odd rotational states.

It follows from the Pauli’s principle that nuclear spin state and rotational state of the hydrogen molecule are strictly correlated. This is remarkable, because the notion of independence (which allowed us to write a wavefunction as a product of individual wavefunctions) has led to complete dependence of these degrees of freedom from each other!

Remarkably, parahydrogen and orthohydrogen can be seen as two individual gases because their thermodynamic properties (boiling point, heat capacity, etc.) are slightly different. This is not surprising taking into account the fact that molecules constituting these two gases are rotating differently!

Importantly, conversion between the two states occurs extremely slowly because the transition between symmetric and antisymmetric nuclear spin states are forbidden by the selection rules of quantum mechanics. Therefore, after its production parahydrogen may be stored for long periods before use in a tank as an individual gas, as the relaxation rate of the parahydrogen back to room-temperature equilibrium can be on the order of months.

However, the use of paramagnetic catalysts (i.e., activated charcoal, nickel, hydrated iron(III) oxide) promotes the establishment of Boltzmann thermodynamic equilibrium between ortho-H2/para-H2 states for a given temperature at greatly accelerated rates. This happens because paramagnetic materials can create a strong inhomogeneous magnetic field on the atomic scale. In such fields the two hydrogen atoms are no longer equivalent, thus, spin-flip transitions between ortho-H2 and para-H2 are no longer forbidden. In practice, normal hydrogen gas (i.e., equilibrium ratio of spin isomers at room temperature) consisting of 75% ortho– and 25% para-hydrogen is passed through a chamber filled with paramagnetic catalyst and maintained at cryogenic temperatures, where the equilibration to the isomer ratio governed by the Boltzmann distribution occurs. For example, a parahydrogen generator operating at 77 K (obtained conveniently by a liquid-N2 bath) yields a mixture with ~50% parahydrogen, whereas the designs based on cryo-chillers (e.g. T~20 K) yield >99% parahydrogen (Figure 2). I should note that the enrichment of hydrogen with para-isomer happens so easily only because of the big energy gap between rotational spin states. This, in turn, is due to the small mass of molecular hydrogen (in general, the energy difference between rotational spin states is inversely proportional to the moment of inertia of a rotating molecule).

The existence of nuclear spin isomers of molecular hydrogen (which was experimentally confirmed by the early 1930s) was one of the first triumphs of quantum mechanics. Indeed, the citation of the Nobel Prize awarded to Werner Heisenberg in 1932 stated that he had “created quantum mechanics, the application of which led to the discovery of the two allotropic forms of hydrogen”!

Knowledge about ortho– to para-H2 conversion is important for the storage of liquid hydrogen (especially as a rocket fuel). The difference in energy associated with the different rotational levels means that energy is released when ortho-H2 converts to para-H2, and energy is absorbed in the reverse process. This phenomenon can be thought of as a latent heat of conversion. If one quickly liquefies normal hydrogen, it will still have 3:1 ortho:para composition which will eventually lead to the heat release. This can vaporize a significant portion of hydrogen and break the impermeability of the storage container. At the dawn of industrial liquid hydrogen production, this presented a major problem. Modern hydrogen liquefying processes now ensure that the liquid hydrogen has reached equilibrium concentration at 99.8% para-H2 before being transported and stored for use.

One may ask how can para-H2 be important for NMR? Indeed, this spin isomer has a zero total nuclear spin and, thus, it does not possess 1H NMR spectrum. However, para-H2 is a pure quantum mechanical state and a highly organized spin order which is readily achievable simply by cooling. Pure state means that all para-H2 molecules are described by the same wavefunction – 1β2β1α2). For comparison, ortho-H2 is a mixture of three wavefunctions, α1α2α1β2+β1α2 and β1β2 and, thus, it is not a pure state. It turns out that once you have a quantum mechanically pure state, you can manipulate it and transfer the spin order from one form to another. This is how parahydrogen-induced polarization (PHIP) and signal amplification by reversible exchange (SABRE) work: they transfer NMR-silent singlet spin order of para-H2 into observable nuclear magnetization.

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#### DOCUMENTA MATHEMATICA, Vol. 12 (2007), 441-471

Alexander Schmidt

Rings of Integers of Type \$K(,1)\$

We investigate the Galois group \$G_S(p)\$ of the maximal \$p\$-extension unramified outside a finite set \$S\$ of primes of a number field in the (tame) case, when no prime dividing \$p\$ is in \$S\$. We show that the cohomology of \$G_S(p)\$ is `often' isomorphic to the étale cohomology of the scheme \$\Spec(Ø_k \sm S)\$, in particular, \$G_S(p)\$ is of cohomological dimension \$2\$ then.

2000 Mathematics Subject Classification: 11R34, 12G10

Keywords and Phrases: Galois cohomology, restricted ramification, cohomological dimension

Full text: dvi.gz 55 k, dvi 140 k, ps.gz 917 k, pdf 302 k.


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## Browse

Boundary behavior of harmonic functions for subordinate Brownian motion

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Authors
이윤주
김판기
Major
자연과학대학 수리과학부
Issue Date
2013-02
Publisher
서울대학교 대학원
Keywords
subordinate Brownian motionrelative Fatou type theoremMartin kernelMartin boundaryharmonic functionMartin representation
Description
학위논문 (박사)-- 서울대학교 대학원 : 수리과학부, 2013. 2. 김판기.
Abstract
In this thesis, we establish an oscillation estimate of nonnegative harmonic functions for a pure-jump subordinate Brownian motion. The infinitesimal generator of such subordinate Brownian motion $X$ is an integro-differential operator. As an application, we give a probabilistic proof of the following form of relative Fatou theorem for such subordinate Brownian motion $X$ in a bounded $\kappa$-fat open set
if $u$ is a positive harmonic function with respect to $X$ in a bounded $\kappa$-fat open set $D$ and $h$ is a positive harmonic function in $D$ vanishing on $D^c$, then the non-tangential limit of $u/h$ exists almost everywhere with respect to the Martin-representing measure of $h$. Under the gaugeability assumption, relative Fatou theorem is true for operators obtained from the generator of pure-jump subordinate Brownian motion in bounded $\kappa$-fat open set $D$ through non-local Feynman-Kac transforms.
Language
English
URI
https://hdl.handle.net/10371/121260
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College of Natural Sciences (자연과학대학)Dept. of Mathematical Sciences (수리과학부)Theses (Ph.D. / Sc.D._수리과학부)


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Definitions

# Diesel cycle

The Diesel cycle is the thermodynamic cycle which approximates the pressure and volume of the combustion chamber of the Diesel engine, invented by Rudolph Diesel in 1897. It is characterised by having constant pressure during the "combustion" phase. This is in contrast to the Otto cycle which is an idealization of the process in a petrol engine, which approximates the real behavior with constant volume during that phase.

## The Ideal Diesel Cycle

The image on the left shows a P-v diagram for the ideal diesel cycle; where P is pressure and v is specific volume. The ideal diesel cycle follows the following four distinct processes (The color references refer to the color of the line on the diagram.):

• Process 1 to 2 is isentropic compression (blue)
• Process 2 to 3 is reversible constant pressure heating (red)
• Process 3 to 4 is isentropic expansion (yellow)
• Process 4 to 1 is reversible constant volume cooling (green)

The diesel is a heat engine: it converts heat into work.

• Work in (Win) is done by the piston compressing the working fluid
• Heat in (Qin) is done by the combustion of the fuel
• Work out (Wout) is done by the working fluid expanding on to the piston, this produces usable torque
• Heat out (Qout) is done by venting the air

### Maximum thermal efficiency

The maximum thermal efficiency of a diesel cycle is dependent on the compression ratio and the cut-off ratio. It has the following formula:

$eta_\left\{th\right\}=1-frac\left\{1\right\}\left\{r^\left\{gamma-1\right\}\right\}left \left(frac\left\{alpha^\left\{gamma\right\}-1\right\}\left\{gamma\left(alpha-1\right)\right\} right \right)$

Where

$eta_\left\{th\right\}$ is thermal efficiency
$alpha$ is the cut-off ratio $frac\left\{V_3\right\}\left\{V_2\right\}$ (ratio between the end and start volume for the combustion phase)
r is the compression ratio $frac\left\{V_1\right\}\left\{V_2\right\}$
$gamma$ is ratio of specific heats (Cp/Cv)

The cut-off ration can be expressed in terms of temperature as shown below:

$frac\left\{T_2\right\}\left\{T_1\right\} =\left\{\left(frac\left\{V_1\right\}\left\{V_2\right\}\right)^\left\{gamma-1\right\}\right\} = r^\left\{gamma-1\right\}$
$\left\{T_2\right\} =\left\{T_1\right\} r^\left\{gamma-1\right\}$
$frac\left\{V_3\right\}\left\{V_2\right\} = frac\left\{T_3\right\}\left\{T_2\right\}$
$alpha = \left(frac\left\{T_3\right\}\left\{T_1\right\}\right)\left(frac\left\{1\right\}\left\{r^\left\{gamma-1\right\}\right\}\right)$

$\left\{T_3\right\}$ can be approximated to the flame temperature of the fuel used. The flame temperature can be approximated to the adiabatic flame temperature of the fuel with corresponding air-to-fuel ratio and compression pressure, $\left\{P_3\right\}$. $\left\{T_1\right\}$ can be approximated to the inlet air temperature.

This formula only gives the ideal thermal efficiency. The actual thermal efficiency will be significantly lower due to heat and friction losses. The formula is more complex than the Otto cycle (petrol/gasoline engine) relation that has the following formula;

$eta_\left\{otto,th\right\}=1-frac\left\{1\right\}\left\{r^\left\{gamma-1\right\}\right\}$

The additional complexity for the diesel formula comes around since the heat addition is at constant pressure and the heat rejection is at constant volume. The Otto cycle by comparison has both the heat addition and rejection at constant volume.

Comparing the two formulae it can be seen that for a given compression ratio (r), the ideal Otto cycle will be more efficient. However, a diesel engine will be more efficient overall since it will have the ability to operate at higher compression ratios. If a petrol engine was to have the same compression ratio, then knocking (self-ignition) would occur and this would severely reduce the efficiency, whereas in a diesel engine, the self ignition is the desired behavior. Additionally, both of these cycles are only idealizations, and the actual behavior does not divide as clearly or sharply. And the ideal Otto cycle formula stated above does not include throttling losses, which do not apply to diesel engines.

The Diesel cycle is a combustion process of a reciprocating internal combustion engine. In it, fuel is ignited by heat generated by compressing air in the combustion chamber, into which fuel is injected. This is in contrast to igniting it with a spark plug as in the Otto cycle (four-stroke/petrol) engine. Diesel engines (heat engines using the Diesel cycle) are used in automobiles, power generation, diesel-electric locomotives, and submarines.

Invented by Rudolph Diesel in 1897, it was originally intended to run on coal dust. This engine generates power (while continuously injecting fuel) to maintain the cylinder at constant pressure during its power stroke.

The Diesel cycle refers to a compression ignition engine, called the Diesel engine that can have a two or four stroke power cycle, drawing in air by its piston, or by a mechanically or exhaust driven supercharger. As air is compressed, its temperature rises from adiabatic compression until the piston reaches the top of its compression stroke. At that point fuel is injected directly into the cylinder with an atomizing fuel injector and ignites immediately; however, because diesel fuel has a higher molecular weight than gasoline, it vaporizes and burns more slowly. The piston is already moving down by the time combustion begins. Fuel injection can be limited to a short part, or continue to near the bottom of the power stroke. If combustion is incomplete when the piston reaches the bottom of its stroke, smoke is generated and fuel is wasted.

Diesel engines are nevertheless more efficient than Otto cycle engines overall, but only during partial load with fuel cut-off at part of the power stroke. Most land vehicles rarely run at the maximum rated power of an engine. Unless the vehicle is at full stroke injection, when the pedal is 'floored', it is at partial rated power. Since diesel engines use the heating effect of compressing air to ignite fuel, it can inject as little or as much fuel as the situation demands. It is important to note that Otto cycle engines can be more efficient than Diesel cycle engines, but only when the engine is running at or near maximum power.

## General information

The diesel engine has the lowest specific fuel consumption of any large internal combustion engine, 0.26 lb/hp.h (0.16 kg/kWh) for very large marine engines. In fact, two-stroke diesels with high pressure forced induction, particularly turbocharging, make up a large percentage of the very largest diesel engines.

In North America, diesel engines are primarily used in large trucks, where the low-stress, high-efficiency cycle leads to much longer engine life and lower operational costs. These advantages also make the diesel engine ideal for use in the heavy-haul railroad environment.

## Other internal combustion engines without spark plugs

Many model airplanes use very simple "glow" and "diesel" engines. Glow engines use glow plugs. "Diesel" model airplane engines have variable compression ratios. Both types depend on special fuels (easily obtainable in such limited quantities) for their ignition timing.

Some 19th century or earlier experimental engines used external flames, exposed by valves, for ignition, but this becomes less attractive with increasing compression. (It was not until Nicolas Léonard Sadi Carnot that the thermodynamic value of compression was known.) An historical implication of this is that the diesel engine would eventually have been invented without the aid of electricity.


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# Confusion about what the Euler-Lagrange equation says

I roughly understand the concept of the Lagrangian $L = T - V$ as well as the idea of stationary action $\delta \mathcal{S} =0$. However, I am confused what the Euler-Lagrange equation actually says.

Consider the Euler-Lagrange equation: $$\frac{\partial L}{\partial q} - \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}}\right) = 0$$

Here's my confusion:

To me, this looks like an empty mathematical exercise. If I know the partial derivatives $\frac{\partial L}{\partial q}$ and $\frac{\partial L}{\partial \dot{q}}$ and can take the derivative with respect to $t$ of the latter, what is the use of plugging all that into this setup? That's like saying after I show $2 + 3 = 5$, then show $x + y = z$, where $x=2,y=3,z=5$. In short, an empty exercise since it would be the same proof showing $x + y = z$ as showing $2 +3 = 5$ in this case.

Can someone explain what I am misunderstanding?

• I'm not sure I understand what you're asking. Given a Lagrangian, the E-L equations tell you what the equations of motion are.
– Jold
Jan 22 '15 at 5:11
• Yeah, I think that's what I'm not grasping. For instance, consider dropping an object of mass $m$ from a height $h$ off the surface of the earth. $mgh$ is PE $\frac{1}{2}mv^2$ is KE. What does E-L tell me about the path that I don't already know given a vector field model of gravity. I know the path is towards the earths surface already. What can E-L tell me? Jan 22 '15 at 5:14
• Here is a good introduction. damtp.cam.ac.uk/user/tong/dynamics.html Jan 22 '15 at 5:28

The reason algebra ($x + y = z$) is more useful than arithmetic ($2 + 3 = 5$) is that it allows you to treat large classes of problems in the same way, and to obtain very general results about them without having to do a separate calculation for every individual case. Same goes for the stationary action principle (a.k.a. action minimization). It allows you to treat a large class of problems - in this case, all of physics - in the same way, by solving $\delta S = 0$. Different choices of $S$ correspond to different areas of physics, but once you have $S$, the procedure for solving is always the same, namely the Euler-Lagrange equations.

Perhaps you've only been exposed to one formula for the action, $S = \int (T - V)\,\mathrm{d}t$ (or equivalently $L = T - V$), which is the action for a Newtonian system, and in that case it's understandable that the stationary action principle seems kind of useless. Similarly, if you had only been exposed to the one example of $x + y = z$ where $x = 2$, $y = 3$, and $z = 5$, it would seem pretty pointless to use letters. But it makes more sense once you see how general the formalism is.

• Oh, that is probably exactly the problem. I learned about applications in my geodesic exercises in general relativity today and for the first time I started to see how it could be more powerful than just the Newtonian case, as you say. Jan 22 '15 at 5:57
• What are some fundamental example uses for the Euler-Lagrange equation that go beyond the Newtonian cases so I can start to expand my understanding? I assume any elementary text will cover them, but will help if I know which examples to pay particular attention to. Jan 22 '15 at 5:59
• Any area of physics you want to explore, there's probably an action for it. For example general relativity uses the Einstein-Hilbert action. In this case the Euler-Lagrange equations give you the Einstein field equations. There are also actions for electromagnetism, relativistic particles, quantum field theory, etc. Jan 22 '15 at 7:36

Euler-Lagrange equations are a necessary condition if the action integral reaches an extremal through the trajectory described by $q(t)$ and you should obtain an expression for $q(t)$ for a given Lagrangian $L$ by solving Euler-Lagrange equations. Take for example the simple Lagrangian for a free particle with constant mass: $L=m\dot{q}^2/2$ then E-L equations are

$$\frac{d}{dt}\left(m\dot{q}\right)=m\ddot{q}=0$$

which has for solution

$$q(t)=at+b$$

a straight line, something you already knew from Newton first law of motion.

Try solving E-L equations with the Lagrangian $L=m\dot{z}^2/2-mgz$ for the particle in free fall (taking partial derivatives with respect to $z$ and $\dot{z}$ instead of $q$ and $\dot{q}$) to obtain the well known trajectory $z(t)=gt^2/2+\dot{z}_0t+z_0$, where $z_0$ and $\dot{z}_0$ are the initial position and velocity.

• Ohh, wait. Okay, maybe I am starting to understand. So are they merely a necessary condition or are they a sufficient condition to tell us if the action integral reaches an extremal through the trajectory described by $q(t)$? Jan 22 '15 at 5:37
• They are not a sufficient condition, but i can't remember of a counterexample right now. To derive the E-L equations you assume that there exist a trajectory which minimizes the action and then you find out that this trajectory must satisfy the E-L equations. In fact there are several ways to derive these equations, look at the book "Emmy Noether's Wonderful Theorem" the author is Dwight E. Neuenschwander, there you'll find a good discussion about the equivalence of the principle of minimum action and Newton's second law. Jan 22 '15 at 7:40


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# What is the closed form for this?

1. Oct 27, 2007

### Sam_

Infinite Sum [n=1] 1/Fibonacci[n]

Or this,

Infinite Sum[n=1] 1/(n^n)

Using some known mathematical constants.

2. Oct 27, 2007

### Gib Z

Welcome to Physics forums sam! Perhaps you don't exactly know, or you just want to give us problems we can't do to torture us, but:

The first sum is equal to a known mathematical constant, and its called the Reciprocal Fibonacci Constant:
$$\psi = \sum_{k=1}^{\infty} \frac{1}{F_k} = 3.359885666243 \dots$$, which has been proven to be irrational, but it is currently not known if that constant is expressible in terms of more elementary or common constants.

The second One I know converges, though I have never seen a closed form solution for.

3. Oct 27, 2007

### Sam_

Hi Gib_Z,

Sorry, I didn't mean to torture you

4. Oct 27, 2007

### Gib Z

This place is about the best place to ask in my opinion as well =] You are really quite lucky, if I hadn't read through the wikipedia article on the Fibonacci numbers just last week, I would not have even remembered about that constant =]

5. Oct 27, 2007

### Gib Z

Doing some crawling on the internet I have found out for sure that there is no closed form for the second series, though there is a nice identity that I am attempting at this moment to prove, the derivation isn't given...I should have it by tomorrow, but heres the identity now:

$$\sum_{n=1}^{\infty} \frac{1}{n^n} = \int^1_0 x^{-x} dx$$


text:
Last edited by Kidal
Wednesday, May 13, 2020 | History

2 edition of Fan/rotor noise found in the catalog.

Fan/rotor noise

G. W. Johnston

# Fan/rotor noise

## by G. W. Johnston

Written in English

Subjects:
• Aerodynamic noise.

• Edition Notes

The Physical Object ID Numbers Statement G.W. Johnston. Pagination 26, [23] leaves : Number of Pages 26 Open Library OL21236565M

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# B mode detected by South Pole Telescope

Tags:
1. Jul 23, 2013

### skydivephil

http://arxiv.org/abs/1307.5830

but its from gravitational lensing not primordial gravity waves so guess we are still waiting for those, but a good sign methinks.

Last edited: Jul 23, 2013
2. Jul 23, 2013

### Mordred

thanks I found it an interesting read

3. Jul 24, 2013

### Chalnoth

That's really cool :)

Unfortunately, getting at the gravitational lensing portion of the B-mode requires going to very large scales (low $\ell$), and the South Pole Telescope isn't equipped for measuring very large scales (it can effectively see roughly 1/10th of the sky, and the survey area is just over half that).

I'm hoping that the EBEX results will be available before too long. It flew late last year/early this year, and was designed explicitly for the detection of large-scale polarization signals. I'm rather excited to see what they find.


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# $Male anopheles acts as the carrier host for malarial parasite ! Sleeping Sickness is caused by Leishmonia. 0 votes 19 views in Biology$ Male anopheles acts as the carrier host for malarial parasite
! Sleeping Sickness is caused by Leishmonia.
A. If both As and R are correct and R is correct explanation of As
B. If both As and R are correct but R is not correct explanation of As
C. If As is correct, R is wrong
D. If both As and R are wrong.

by (24.9k points)


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# Implementing FIR Hilbert Transformer in Python

I want to implement FIR Hilbert Transformer in Python using window method. I'm new to digital signal processing so please point out to me all mistakes you notice.

Ideal Hilbert transformer is described as: $$H(e^{j\omega}) = \begin{cases} -j & \text{, 0 < \omega < \pi} \\ j & \text{, -\pi < \omega < 0} \\ \end{cases}$$

from which we conclude that ideal Hilbert Transformer is all pass filter $$|H(e^{j\omega})| = 1$$.

I know that FIR filter can be implemented using scipy.signal.firwin function.

What I don't know is which parameters to pass to function in order to get Hilbert filter aproximation.

• Being new to signal processing, I recommend starting with easier problems. Also, SciPy already includes a Hilbert transformer; why do you want to implement your own?
– MBaz
May 18 at 17:39
• That's one of tasks I need to do, I understand theory, but I'm not sure how to implement. I would appreciate help. Thanks May 18 at 17:41
• Hilbert transformers are not, strictly speaking, all-pass filters (APF) because they don't pass DC. May 19 at 3:02

You can't implement an ideal Hilbert transformer: its impulse response is non causal and infinite in time.

So you can only implement an "approximate" one and the best way to do this depends on the specific requirement of your application: what's your frequency range of interest, how much magnitude and/or phase deviation can you tolerate, are you constrained in latency or computational resources, do you need to maintain causality and transients, etc.

Python has a Hilbert transform function https://docs.scipy.org/doc/scipy/reference/generated/scipy.signal.hilbert.html but without understanding your requirements first, it will be difficult to control the parameters properly.

Very good article on Hilbert transform and analytical signals in the discrete domain: http://andrewduncan.net/air/

• These are the requirements that I have: Oscillations in bandwidth must be less than 0.05dB, Attenuation less than 3db in frequency range [0.01pi, 0.99pi] May 18 at 19:55
• This is what i wrote: N = 50 hh = signal.firwin(2*N+1, [0.01,0.99], window='hamming', pass_zero = False) May 18 at 19:56
• That's a very nice link (the 2nd). +1. @AleksandarSimonović That link should make you understand everything. May 18 at 22:03
• I was happy to see the reference to Andrew Duncan's analytic impulse stuff. I was both impressed and jealous when that came out. May 20 at 21:25

There are routines that will provide the Hilbert coefficients directly, but an approach I like to use given its simplicity and clarity in functionality is to transform a Half Band filter to a Hilbert as follows:

Step 1: Estimate the number of taps needed from the specifications using these commonly used estimators. The one Marcus Mueller provided would be well suited for the OP's case given there is a passband ripple and stopband requirement. Using that along with 45 dB rejection, the estimated number of coefficients (taps) needed came to 269 (also for the estimator, the passband edge of 0.99 is changed to 0.98 due to the Half Band to Hilbert transformation). Nearly half of these 269 coefficients will be zero, and the coefficients are symmetric such that it can be implemented with approximately 69 multiplications.)

Step 2: If an absolute min/max requirement is needed, use the Remez algorithm to design an optimum half band filter in minimax sense. (Typically I would prefer to use the least squares algorithm for an optimum solution in the least squared sense which would result in fewer taps but have higher peak error. A windowing design approach can be done as well in which case I would choose the Kaiser window and window the Sinc function that defines this HB filter; the results will be quite similar to least squares). In Python the Remez solution is done as follows:

coeff = sig.remez(N, [0, pb, 0.5- pb, .5], [1, 1, 0, 0])

Alternatively for least squares use:

coeff = sig.firls(N, [0, 2 * pb, 2* (0.5- pb), 1], [1, 1, 0, 0])

Where pb is the pass band edge normalized to $$F_s = 1$$, where $$F_s$$ is the sampling rate and $$N$$ is the total number of coefficients which must be an odd number.

Step 3: Transform the $$N$$ coefficients numbered $$1:N$$ for a Half Band filter (which should be odd) to the coefficients for a Hilbert Transform by setting the center coefficient $$\lceil N/2 \rceil$$ of the Half Band to 0, all coefficients that appear earlier than the center coefficient (coefficients $$1: \lfloor N/2 \rfloor$$) to be negative the absolute value of the half band, and all the coefficients that appear later than the center coefficient (coefficients $$\lceil N/2 \rceil+1: N$$) to be positive. Multiply all coefficients by two to normalize the scale.

The reason this works is the Half Band filter is (a windowed version of) a Sinc function where the coefficients are all the peaks of the Sinc. The peaks of the Sinc go down at rate $$1/n$$. The impulse response of a Hibert is a non-causal $$1/n$$ (with $$\pi$$ scaling but for clarity of the main point I leave that out). For for n positive it would all be positive and for n negative it would all be negative. So we get that directly through the modification of the Half Band filter coefficients as optimally derived (and as RBJ further details mathematically in the comments). We make it causal by truncating the Sinc and shifting in out to $$n= \lceil N/2 \rceil$$. For this reason we need to feed the input into a tracking filter which is just a simple delay by $$n= \lceil N/2 \rceil$$ samples, which will be in quadrature (as desired) from the output of the filter above.

Doing the above came up with the following result for amplitude balance with a perfect quadrature balance across the band:

• Bandlimited wire: $$G(f) = \operatorname{rect}(f)$$ then $g(t) = \operatorname{sinc}(t)$ Then the bandlimited Hilbert: $$H(f) = -j\operatorname{sgn}(f)\operatorname{rect}(f)$$ and then $h(t)=\frac{1-\cos(\pi t)}{\pi t}$ May 20 at 17:02
• Look, Dan: If you're going to be talking about the sinc function (in the time domain), it's about a corresponding bandlimited counterpart in the frequency domain. But the Hilbert is not one single rect() function but two of them with opposite sign. Now we know that $$\frac{1-\cos(\pi t)}{\pi t} = \frac{2\sin^2(\frac{\pi}{2}t)}{\pi t} = \sin(\tfrac{\pi}{2}t)\operatorname{sinc}(\tfrac{t}{2})$$ but that is a consequence of the original analysis (and it's at half the frequency). May 23 at 2:18
• then notice that the $\sin(\tfrac{\pi}{2}t)$ puts in the zeros for every even integer and flips the sign for every odd integer, which is what accomplishes your abs() operation. but the abs() is not the correct math. it just happens to be correct for integer n, but for integer n, the sinc() function for the bandlimited wire is simply the Kronecker delta. May 23 at 2:21
• So here's the cool thing: when you window either (the wire and the Hilbert transformer) to get an FIR, that window spectrum is convolved with the edges of the rect functions, and at high frequencies that convolving does the same thing to both real (wire) and imaginary (Hilbert) parts, so the angle is unaffected. that means that the phase increment (which is instantaneous frequency) is also not affected near the bandedge. May 23 at 2:34
• It also means that multiplying by $e^{j \omega_0 n}$ will correctly offset each frequency component cleanly, even those close to Nyquist that have amplitude affected by that convolving by the window spectrum. So you get a nice frequency shifter that doesn't have aliasing. But down by DC the wire and the Hilbert work differently, so to prevent crap down there, you precede both the wire and the Hilbert transformer with a simple DC-blocking HPF so that there is no content down there by DC. May 23 at 2:38


text:
# Statistical time-domain characterization of non-periodic optical clocks

Dario Cilluffo

Institute of Theoretical Physics & IQST, Ulm University, Albert-Einstein-Allee 11 89081, Ulm, Germany
Universit$\grave{a}$ degli Studi di Palermo, Dipartimento di Fisica e Chimica - Emilio Segrè, via Archirafi 36, I-90123 Palermo, Italy

### Abstract

Measuring time means counting the occurrence of periodic phenomena. Over the past centuries a major effort was put to make stable and precise oscillators to be used as clock regulators. Here we consider a different class of clocks based on stochastic clicking processes. We provide a rigorous statistical framework to study the performances of such devices and apply our results to a single coherently driven two-level atom under photodetection as an extreme example of non-periodic clock. Quantum Jump MonteCarlo simulations and photon counting waiting time distribution will provide independent checks on the main results.

Using a simplified optical model, we show that the large deviation formalism of quantum trajectories can be easily exploited to study the performances of a particular class of clocks relying on stochastic clicking processes. The proof of principle presented here provides a clear application of thermodynamics of quantum trajectories to practical problems and, at the same time, it suggests further connections with metrology.

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### Cited by

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text:
# 21. Complex Numbers Problems Part 3

1. Two different non-parallel lines cut the circle $$|z|= r$$ at points $$a, b, c, d$$ respectively. Prove that these two lines meet at point given by $$\frac{a^{-1} + b^{-1} + c^{-1} + d^{-1}}{a^{-1}b^{-1}c^{-1}d^{-1}}$$.

2. If $$z = 2 + t + i\sqrt{3 - t^2},$$ where $$t$$ is real and $$t^2 < 3,$$ show that the modulus of $$(z+1)(z-1)$$ is independent of $$t.$$ Also, show that the locus of the point $$z$$ for different value of $$t$$ is a circle and find its center and radius.

3. Let $$z_1, z_2, z_3$$ be the three non-zero complex numbers such that $$z_2 \ne 1, a = |z_1|, b = |z_2|$$ and $$c = |z_3|.$$

$\text{Let } \begin{vmatrix} a & b & c\\ b & c & a\\ c & a & b \end{vmatrix} = 0,$

then show that $$arg\left( \frac{z_3}{z_2}\right) = arg\left( \frac{z_3 - z_1}{z_2 - z_1}\right)^2.$$

4. $$P$$ is such a point that on a circle with $$OP$$ as diameter two points $$Q$$ and $$R$$ are taken such that $$\angle POQ = \angle QOR = \theta.$$ If $$O$$ is the origin and $$P, Q \text{ and }R$$ are represented by the complex numbers $$z_1, z_2 \text{ and } z_3$$ respectively, show that $$z_2^2 cos2\theta = z_1z_3 cos^2\theta.$$

5. Find the equation in complex variables of all circles which are orthogonal to $$|z| = 1$$ and $$|z - 1| = 4.$$

6. Find the real values real value of the parameter $$t$$ for which there is at least one complex number $$z = x + iy$$ satisfying the condition $$|z+3| = t^2 - 2i + 6$$ and the inequality $$z - 3\sqrt{3}i < t^2.$$

7. If $$a, b, c$$ and $$d$$ are real and $$ad > bc,$$ show that the imaginary parts of the complex number $$z$$ and $$\frac{az + b}{cz + d}$$ have the same sign.

8. If $$z_1 = x_1 + iy_1, z_2 = x_2 + iy_2$$ and $$z_1 = \frac{i(z_2 + 1)}{z_2 - 1}$$ prove that

$x_1^2 + y_1^2 - x_1 = \frac{x_2^2 + y_2^2 + 2x_2 - 2y_2 +1}{(x_2 - 1)^2 + y_2^2}$
9. Simplify the following:

$\frac{(cos3\theta - isin3\theta)^6(sin\theta - icos\theta)^3}{(cos2\theta + isin2\theta)^5}$
10. Find all complex numbers such that $$z^2 + |z| = 0.$$

11. Solve the equation $$z^2 + z|z| + |z^2| = 0.$$

12. Solve the equation $$2z = |z| + 2i$$ in complex numbers.

13. If $$a > 0$$ and $$z|z| + az + 1 = 0$$, show that $$z$$ is a negative real number.

14. For every real number $$a > 0$$ find all complex numbers $$z$$ satisfying the equation $$z|z| + az + i = 0$$

15. For every real number $$a > 0$$ determine the complex numbers $$z$$ which will satisfy the equation $$|z|^2 -2iz + 2a(1 + i) = 0$$

16. For any two complex numbers $$z_1$$ and $$z_2$$ and any real numbers $$a$$ and $$b$$, show that $$|az_1 -bz_2|^2 + |bz_1 - az_2|^2 = (a^2 + b^2)(|z_1|^2 + |z_2|^2)$$

17. If $$\alpha$$ and $$\beta$$ are any two complex numbers, show that $$|\alpha + \beta|^2 = |\alpha|^2 + |\beta|^2 + Re(\alpha\overline{\beta}) + Re(\overline{\alpha}\beta)$$

18. Find the integral solution of the following equations (i) $$(3 + 4i)^x = 5^\frac{x}{2}$$ (ii) $$(1 - i)^x = 2^x$$ (iii) $$(1 - i)^x = (1 + i)^x$$

19. Prove that $$|1 - \overline{z_1}z_2|^2 - |z_1 - z_2|^2 = (1 - |z_1|^2)(1 - |z_2|^2)$$

20. If $$a_i, b_i \in R, i = 1, 2, 3, ..., n$$ show that

$\left(\sum_{n=1}^na_i\right)^2 + \left(\sum_{n=1}^nb_i\right)^2 \le \left(\sum_{n=1}^n\sqrt{a_i^2 + b_i^2}\right)^2$
21. Let $$\left|\frac{\overline{z_1} - 2\overline{z_2}}{2 - z_1\overline{z_2}}\right| = 1$$ and $$|z_2| \ne 1,$$ where $$z_1$$ and $$z_2$$ are complex numbers, show that $$|z_1| = 2.$$

22. If $$|z_1| < 1$$ and $$\left|\frac{z_1 - z_2}{1 - \overline{z_1}z_2}\right| < 1,$$ then show that $$|z_2| < 1$$

23. If $$z_1$$ and $$z_2$$ are complex numbers and $$u = \sqrt{z_1z_2},$$ prove that

$|z_1| + |z_2| = \left|\frac{z_1 + z_2}{2} + u\right| + \left|\frac{z_1 + z_2}{2} - u\right|$
24. If $$z_1$$ and $$z_2$$ are the roots of the equation $$\alpha z^2 + 2\beta z + \gamma = 0,$$ then prove that $$|\alpha||(|z_1| + |z_2|) = |\beta + \sqrt{\alpha \gamma}| + |\beta - \sqrt{\alpha \gamma}|$$

25. If $$a, b, c$$ are complex numbers such that $$a + b + c = 0$$ and $$|a| = |b| = |c| = 1,$$ find the value of $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$

26. If $$|z + 4| \le 3,$$ find the least and greatest value of $$|z + 1|.$$

27. Show that for any two non-zero complex numbers $$z_1, z_2$$

$(|z_1| + |z_2|)\left(\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}\right) \le 2|z_1 + z_2|$
28. Show that the necessary and sufficient condition for both the roots of the equation $$z^2 + az + b = 0$$ to be of unit modulus are $$|a| \le 2, |b| = 1, argb = 2arga$$

29. If $$z$$ is a complex number, show that $$|z| \le |R(z)| + |I(z)| \le \sqrt{2}|z|$$.

30. If $$\left|z - \frac{4}{z}\right| = 2$$ show that the greatest value of $$|z|$$ is $$\sqrt{5} + 1.$$

31. If $$\alpha, \beta, \gamma, \delta$$ be the real roots of the equation $$ax^4 + bx^3 + cx^2 + dx + e = 0,$$ show that $$a^2(1 + \alpha^2)(1 + \beta)^2(1 + \gamma)^2(1 + \delta)^2 = (a - c + e)^2 + (b - d)^2.$$

32. If $$a_i \in R, i = 1, 2, ..., n$$ and $$\alpha_1, \alpha_2, ..., \alpha_n$$ are the roots of the equation

$x^n + a_1x^{n - 1} + a_2x^{n - 2} + ... + a_{n - 1}x + a)n = 0,$

show that

$\prod_{i = 1}^n(1 + \alpha_i^2) = (1 - a_2 + a_4 - ...)^2 + (a_1 - a_3 + ...)^2$
33. If the complex numbers $$z_1, z_2, z_2$$ are the vertices of an equilateral triangle such that $$|z_1| = |z_2| = |z_3|,$$ prove that $$z_1 + z_2 + z_3 = 0.$$

34. If $$z_1 + z_2 + z_3 = 0$$ and $$|z_1| = |z_2| = |z_3| = 1,$$ then prove that the points $$z_1, z_2, z_3$$ are the vertices of an equilateral triangle inscribed in an unit circle.

35. If $$z_1, z_2, z_3$$ be the vertices of of an equilateral triangle in the Argand plane whose circumcenter is $$z_0$$ then prove that $$z_1^2 + z_2^2 + z_3^2 = 3z_0^2.$$

36. Prove that the complex numbers $$z_1$$ and $$z_2$$ and the origin form an equilateral triangle if $$z_1^2 + z_2^2 - z_1z_2 = 0.$$

37. If $$z_1$$ and $$z_2$$ be the roots of the equation $$z^2 + az + b = 0,$$ then prove that the origin, $$z_1$$ and $$z_2$$ form an equilateral triangle if $$a^2 = 3b.$$

38. Let $$z_1, z_2$$ and $$z_3$$ be the roots of the equation $$z^3 + 3\alpha z^2 + 3\beta z + \gamma = 0,$$ where $$\alpha, \beta$$ and $$\gamma$$ are complex numbers and that these represent the vertices of $$A, B$$ and $$C$$ of a triangle. Find the centroid of $$\triangle ABC.$$ Show that the triangle will be equilateral, if $$\alpha^2 = \beta.$$

39. If $$z_1, z_2$$ and $$z_3$$ are in A.P., then prove that they are collinear.

40. If $$z_1, z_2$$ and $$z_3$$ are collinear points in Argand plane then show that one of the following holds

$-z_1|z_2 - z_3| + z_2|z_3 - z_1| + z_3|z_1 - z_2| = 0 z_1|z_2 - z_3| - z_2|z_3 - z_1| + z_3|z_1 - z_2| = 0 z_1|z_2 - z_3| + z_2|z_3 - z_1| - z_3|z_1 - z_2| = 0$
41. Find the locus of point $$z$$ if $$\frac{z - i}{z + i}$$ is purely imaginary.

42. What region in the Argand plane is represented by the inequality $$1 < |z - 3 -4i| < 2.$$

43. Find the locus of point $$z$$ if $$|z - 1| + |z + 1| \le 4.$$

44. If $$z = t + 5 + i\sqrt{4 -t^2}$$ and $$t$$ is real, find the locus of $$z.$$

45. If $$\frac{z^2}{z - 1}$$ is real show that locus of $$z$$ is a circle with center $$(1, 0)$$ and radius unity and the $$x$$-axis.

46. If $$|z^2 - 1| = |z|^2 + 1,$$ show that locus of $$z$$ is a straight line.

47. Find the locus of point $$z$$ if $$\frac{\pi}{3} \le arg~z \le \frac{3\pi}{2}.$$

48. Find the locus of the point $$z$$ if $$arg\left(\frac{z - 2}{z + 2}\right) = \frac{\pi}{3}.$$

49. Show that the locus of the point $$z$$ satisfying the condition $$arg\left(\frac{z - 1}{z + 1}\right) = \frac{\pi}{2}$$ is the semicircle above $$x$$-axis whose diameter is the joins of the points $$(-1, 0)$$ and $$(1, 0)$$ excluding those points.

50. Find the locus of the point $$z$$ if $$\log_{\sqrt{3}}\frac{|z|^2 - |z| + 1}{2 + |z|} < 2.$$