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# Independent phase modulation for quadruplex polarization channels enabled by chirality-assisted geometric-phase metasurfaces

## Abstract

Geometric-phase metasurfaces, recently utilized for controlling wavefronts of circular polarized (CP) electromagnetic waves, are drastically limited to the cross-polarization modality. Combining geometric with propagation phase allows to further control the co-polarized output channel, nevertheless addressing only similar functionality on both co-polarized outputs for the two different CP incident beams. Here we introduce the concept of chirality-assisted phase as a degree of freedom, which could decouple the two co-polarized outputs, and thus be an alternative solution for designing arbitrary modulated-phase metasurfaces with distinct wavefront manipulation in all four CP output channels. Two metasurfaces are demonstrated with four arbitrary refraction wavefronts, and orbital angular momentum modes with four independent topological charge, showcasing complete and independent manipulation of all possible CP channels in transmission. This additional phase addressing mechanism will lead to new components, ranging from broadband achromatic devices to the multiplexing of wavefronts for application in reconfigurable-beam antenna and wireless communication systems.

## Introduction

Artificially engineered metasurfaces have attracted great interests during the last two decades1,2,3,4,5. Due to the unique abilities to provide a large degree of control over the amplitude, phase, and polarization of local fields, metasurfaces find various applications in frequency regimes extending from microwaves to visible domains6,7. In particular, phase-gradient metasurfaces have emerged as a versatile platform for wave front engineering8. Compared to traditional bulky optical lenses relying on phase accumulation along propagation, metasurfaces can attain abrupt phase change within subwavelength thickness by suitably engineered inclusions that can enhance local interaction with waves, and thus have been explored to construct low-profile lenses9,10,11,12, holograms13,14,15,16,17,18,19, beam deflectors20,21,22, spin-orbital angular momentum manipulation23,24,25,26,27,28, information processing, and analog computations29,30, among many others.

Since proposed by Pancharatnam and Berry31,32, geometric phase, namely Pancharatnam–Berry (PB) phase, has been adopted in the manipulation of circularly polarized (CP) waves33,34,35,36,37,38,39. Especially on the platform of metasurfaces, PB phase can be conveniently introduced into transmitted (or reflected) waves by gradually varying local orientation from unit to unit. Ideally, there should be four channels by switching the handedness of CP input and output beams for PB phase based metasurfaces operating in transmission manner, including two cross-polarized channels (L–R, right-handed circular polarization (RHCP) output under left-handed circular polarization (LHCP) input, and R–L, LHCP output under RHCP input) and two co-polarized channels (L–L, LHCP output under LHCP input, and R–R, RHCP output under RHCP input). The inherent symmetrical response of PB phase results in that the functionality exhibited by metasurfaces in L–R channel and R–L channel would be equal and opposite. For instance, a geometric phase converging metalens for LHCP incidence will act as a diverging metalens for RHCP incidence. More recently, the combination of propagation and geometric phases is proposed to independently modulate the two cross-polarized output channels (L–R and R–L) under the orthogonal CP incidence states40,41,42,43,44, but it has to be noted that the co-polarized output field are still left behind. The propagation phase along the fast and slow axis could affect the phase of co-polarized output fields, but it owns no selectivity for the handedness, which would result in the similar response for L–L and R–R output channels45. Obviously, a third phase resource is needed for the full phase modulation of CP beam. A direct property that can be utilized is the chirality, which can effectively generate circular dichroism with the illumination of orthogonal CP incidences46,47. It is demonstrated that chirality can be used to decouple the coherence between linearly polarized phase and amplitude response of metasurface to generate finite-energy airy beam48. Nevertheless, there still lacks a methodology that could integrate chirality into the phase modulation scheme as an additional degree of freedom to discriminate the wavefront of co-polarized L–L and R–R channels.

In this work, a general phase modulation scheme for the construction of metasurfaces enabling distinct wavefronts in all four CP channels is established. Chirality-assisted phase response, instead of amplitude response, namely circular dichroism, is proposed and then synthesized with propagation phase and PB phase to activate and distinguish wavefront engineering in all four co-polarized and cross-polarized channels under LHCP and RHCP wave illuminations. Accordingly, two metasurfaces are proposed and experimentally verified with different functionalities that in one case, beam deflection with arbitrary designer four refraction angles are obtained. With incident polarization state following the trace from RHCP to LHCP on the Poincaré sphere, evolution of diffraction order further confirms that all four CP channels are fully used. In another case, orbital angular momentum (OAM) modes with four independent topological charges are proposed. This work provides an additional step on the prospective way towards multifunctional metasurface devices and can be possibly extended to other frequency regimes including visible wavelength.

## Results

### Principle of phase manipulation of full CP channels

Here, we propose a general formalism for complete and separate phase manipulation of all four CP channels, where arbitrary and independent wavefronts can be achieved by altering the CP states of input and output ends, as schematically illustrated in Fig. 1. Taking generation of OAM modes as a demonstration example, four OAM mode with topological charge l = 0, 1, 2, and 3 are realized through different CP channels. The corresponding theoretical spatial phase distributions of the different output OAM modes and the schematic CP conversion process from input to output waves are exhibited in the insets of Fig. 1.

Different from previous works on birefringent metasurfaces41,45, the aim of this work is to perform distinct phase-modulation in four CP transmission channels, which can be described by the four element transmission coefficients in the Jones matrix $$T_{{\mathrm{cir}}} = \left[ {\begin{array}{*{20}{c}} {t_{{\mathrm{LL}}}^{}} & {t_{{\mathrm{LR}}}^{}} \\ {t_{{\mathrm{RL}}}^{}} & {t_{{\mathrm{RR}}}^{}} \end{array}} \right]$$. The equivalent metasurface system is supposed to be passive, lossless, matched, and reciprocal. Therefore, the four CP transmission coefficients (the first/second subscript represents the input/output CP state, L/R denotes the LHCP/RHCP state) with linear base can be described as follows:

$$t_{{\mathrm{LL}}} = \frac{1}{2}\left[ {( {t_{{\mathrm{xx}}} + t_{{\mathrm{yy}}}} ) + {\mathrm{i}} \cdot ( {t_{{\mathrm{xy}}} - t_{{\mathrm{yx}}}} )} \right],$$
(1a)
$$t_{{\mathrm{LR}}} = \frac{1}{2}\left[ {( {t_{{\mathrm{xx}}} - t_{{\mathrm{yy}}}} ) - {\mathrm{i}} \cdot ( {t_{{\mathrm{xy}}} + t_{{\mathrm{yx}}}} )} \right] \cdot {\mathrm{e}}^{{\mathrm{i}} \cdot 2\theta },$$
(1b)
$$t_{{\mathrm{RL}}} = \frac{1}{2}\left[ {( {t_{{\mathrm{xx}}} - t_{{\mathrm{yy}}}} ) + {\mathrm{i}} \cdot ( {t_{{\mathrm{xy}}} + t_{{\mathrm{yx}}}} )} \right] \cdot {\mathrm{e}}^{ - {\mathrm{i}} \cdot 2\theta },$$
(1c)
$$t_{{\mathrm{RR}}} = \frac{1}{2}\left[ {( {t_{{\mathrm{xx}}} + t_{{\mathrm{yy}}}} ) - {\mathrm{i}} \cdot ( {t_{{\mathrm{xy}}} - t_{{\mathrm{yx}}}} )} \right],$$
(1d)

where $$t_{{\mathrm{xx}}} = \left| {t_{{\mathrm{xx}}}} \right| \cdot {\mathrm{e}}^{{\mathrm{i}} \cdot \varphi _{{\mathrm{xx}}}}$$ and $$t_{{\mathrm{yy}}} = | {t_{{\mathrm{yy}}}} | \cdot {\mathrm{e}}^{{\mathrm{i}} \cdot \varphi _{{\mathrm{yy}}}}$$ are the diagonal linear transmission coefficients, and $$t_{{\mathrm{xy}}} = | {t_{{\mathrm{xy}}}} | \cdot {\mathrm{e}}^{{\mathrm{i}} \cdot \varphi _{{\mathrm{xy}}}}$$ and $$t_{{\mathrm{yx}}} = | {t_{{\mathrm{yx}}}} | \cdot {\mathrm{e}}^{{\mathrm{i}} \cdot \varphi _{{\mathrm{yx}}}}$$ are the off-diagonal linear transmission coefficients. θ is the exterior rotation angle introduced by rotation matrix $$M(\theta ) = \left[ {\begin{array}{*{20}{c}} {\cos \theta } & {\sin \theta } \\ { - \sin \theta } & {\cos \theta } \end{array}} \right]$$. Here, tLL and tRR are defined as co-polarized transmission channels, which maintain the polarization state of input waves. tLR and tRL represent cross-polarized channels, which flip the output fields into opposite CP state. The first components with totally same expression in tLL and tRR (or tLR and tRL) can be labeled as $$t_{{\mathrm{propa}}}^{{\mathrm{co}}} = | {t_{{\mathrm{propa}}}^{{\mathrm{co}}}} |{\mathrm{e}}^{{\mathrm{i}} \cdot \varphi _{{\mathrm{propa}}}^{{\mathrm{co}}}} = \frac{1}{2}(t_{{\mathrm{xx}}} + t_{{\mathrm{yy}}})$$$$( {t_{{\mathrm{propa}}}^{{\mathrm{cross}}} = | {t_{{\mathrm{propa}}}^{{\mathrm{cross}}}} |{\mathrm{e}}^{{\mathrm{i}} \cdot \varphi _{{\mathrm{propa}}}^{{\mathrm{cross}}}} = \frac{1}{2}( {t_{{\mathrm{xx}}} - t_{{\mathrm{yy}}}} )} )$$. The phase pattern of the two transmission components $$\varphi _{{\mathrm{propa}}}^{{\mathrm{co}}}$$ and $$\varphi _{{\mathrm{propa}}}^{{\mathrm{cross}}}$$ are uniquely dependent to propagation phase modulation, which would produce initial influence in both co-polarized and cross-polarized fields regardless of the incident CP state. Meanwhile, the second components carrying opposite symbols in tLL and tRR (or tLR and tRL) can be extracted as $$t_{{\mathrm{chiral}}}^{{\mathrm{co}}} = | {t_{{\mathrm{chiral}}}^{{\mathrm{co}}}} |{\mathrm{e}}^{{\mathrm{i}} \cdot \varphi _{{\mathrm{chiral}}}^{{\mathrm{co}}}} = \frac{1}{2} \cdot {\mathrm{i}} \cdot (t_{{\mathrm{xy}}} - t_{{\mathrm{yx}}})$$$$( {t_{{\mathrm{chiral}}}^{{\mathrm{cross}}} = | {t_{{\mathrm{chiral}}}^{{\mathrm{cross}}}} |{\mathrm{e}}^{{\mathrm{i}} \cdot \varphi _{{\mathrm{chiral}}}^{{\mathrm{cross}}}} = \frac{1}{2} \cdot {\mathrm{i}} \cdot ( {t_{{\mathrm{xy}}} + t_{{\mathrm{yx}}}} )} )$$. The phase pattern of these two components $$\varphi _{{\mathrm{chiral}}}^{{\mathrm{co}}}$$ and $$\varphi _{{\mathrm{chiral}}}^{{\mathrm{cross}}}$$ are determined by chirality-assisted phase, indicating that the two components would be an additional degree of freedom to decouple inherent consistency between co-polarized channels. Moreover, the PB phase pattern generated by rotating matrix φgeo = 2θ, which is sensitive to the incidence polarization state, influence only the two cross-polarized fields. Here, the plus or minus (±) sign represents the conjugate responses to LHCP or RHCP input states. According to Eq. (1a–d), it can be concluded that the full phase-modulation scheme is based on three phases that constitute three degrees of freedom to decouple the inherent consistencies between CP transmission coefficients: (i) propagation phase modulation is applied to define the initial phase profiles of two diagonal and two off-diagonal transmission elements tLL (=tRR) and tLR (=tRL) when there is no chirality-assisted and geometric phase responses, (ii) the chirality-assisted phase modulation is introduced to decouple the consistency between tLL and tRR while keeping tLR and tRL unchanged, and (iii) the geometric phase modulation can further realize distinct profiles of tLR and tRL, meanwhile producing no effect on tLL and tRR.

When the proposed metasurface is illuminated by LHCP incident wave $$| {\overrightarrow {\mathrm{L}} } \rangle = \left[ {\begin{array}{*{20}{c}} {\mathrm{1}} \\ {\mathrm{i}} \end{array}} \right]$$ and RHCP incident wave $$| {\overrightarrow {\mathrm{R}} } \rangle {\mathrm{ = }}\left[ {\begin{array}{*{20}{c}} {\mathrm{1}} \\ {{\mathrm{ - i}}} \end{array}} \right]$$ respectively, the output electric fields can be expressed as:

$$\overrightarrow E _{{\mathrm{out}}}^{{\mathrm{L,in}}} = T \cdot \left| {\mathrm{L}} \right\rangle \\ = \left( {t_{{\mathrm{propa}}}^{{\mathrm{co}}} + t_{{\mathrm{chiral}}}^{{\mathrm{co}}}} \right) \cdot \left| {\mathrm{L}} \right\rangle + \left( {t_{{\mathrm{propa}}}^{{\mathrm{cross}}} - t_{{\mathrm{chiral}}}^{{\mathrm{cross}}}} \right) \cdot {\mathrm{e}}^{{\mathrm{i}} \cdot \varphi _{{\mathrm{geo}}}} \cdot \left| {\mathrm{R}} \right\rangle \\ = \overrightarrow E _{{\mathrm{LL}}} + \overrightarrow E _{{\mathrm{LR}}}$$
(2a)
$$\overrightarrow E _{{\mathrm{out}}}^{{\mathrm{R,in}}} = T \cdot \left| {\mathrm{R}} \right\rangle \\ = \left( {t_{{\mathrm{propa}}}^{{\mathrm{co}}} - t_{{\mathrm{chiral}}}^{{\mathrm{co}}}} \right) \cdot \left| {\mathrm{R}} \right\rangle + \left( {t_{{\mathrm{propa}}}^{{\mathrm{cross}}} + t_{{\mathrm{chiral}}}^{{\mathrm{cross}}}} \right) \cdot {\mathrm{e}}^{ - {\mathrm{i}} \cdot \varphi _{{\mathrm{geo}}}} \cdot \left| {\mathrm{L}} \right\rangle \\ = \overrightarrow E _{{\mathrm{RR}}} + \overrightarrow E _{{\mathrm{RL}}}$$
(2b)

It can be observed from Eq. (2a, b) that under the orthogonal CP illuminations, the metasurface can produce four different output fields, including two co-polarized output components $$\overrightarrow E _{{\mathrm{LL}}}$$ and $$\overrightarrow E _{{\mathrm{RR}}}$$, and two cross-polarized components $$\overrightarrow E _{{\mathrm{LR}}}$$ and $$\overrightarrow E _{{\mathrm{RL}}}$$. Based on the analyses above, four CP transmission coefficients can be independently phase-modulated, where the inherent consistencies between four output phase patterns φLL, φLR, φRL, and φRR are independently decoupled by propagation phase, chirality-assisted phase and geometric phase, respectively.

### Meta-atom design and verification

Here a passive, reciprocal transmission-type meta-atom is applied to implement all these three phase modulation schemes, namely chirality-assisted phase, propagation phase, and PB phase. The proposed meta-atom is composed of five metallic layers and four dielectric substrates for realizing a 4th-order LC band-pass filter, whose geometric structure of meta-atom is introduced and schematically shown in Fig. 2a and the detailed operating mechanism is explained in Supplementary Note 1, Supplementary Figs. 1 and  2. In our designed meta-atom, propagation phase is provided by tailoring the width px and length py of the rectangular patch layers as shown in Fig. 2b. Figure 2c exhibits the phase profiles of all four outputs against px, where the phase response of the two co-polarized output channels show the same tendency φLL = φRR, as well as the two cross-polarized output phase delays φLR = φRL, while the co-polarized and cross-polarized phase responses exhibit differently φLL ≠ φLR (φRR ≠ φRL) (the co-polarized and cross-polarized phase profile tendencies against both px and py are shown in Supplementary Fig. 3). That is to say, propagation phase can achieve independent manipulation of co-polarized and cross-polarized phase delays but does not show any selectivity on the spin state of incidence. In order to conquer this limit, chirality-assisted phase is introduced by the relative rotation of three similar patch layers of meta-atom with different angles, regarded as an interior rotation. Here orientation of middle layer is fixed with no further rotation, while the upper and lower layers are rotated with opposite angles ϑ respectively, as shown in Fig. 2d. A representative meta-atom selected with specific dimensions and different interior angle ϑ is simulated and presented in Fig. 2e. It can be seen that the two co-polarized phase profiles exhibit distinct tendency φLL ≠ φRR, while the two cross-polarized phase responses keep same tendency with the interior rotation φLR = φRL, which is due to the symmetrical rotation angle in upper and lower patch layers (derived and explained in Supplementary Note 2 and Supplementary Note 3). It is verified that the chirality-assisted phase can decouple the intrinsic coherence between co-polarized output under orthogonal CP states, while makes no disparate effect in two cross-polarized channels. As for the geometric phase, it is performed through rotation of the whole meta-atom structure with angle θ named exterior rotation as shown in Fig. 2f. Figure 2g presents the four output phase profiles of the representative meta-atom with specific dimensions and interior angle but various exterior angles, where two cross-polarized phases exhibit different tendency φLR ≠ φRL and two co-polarized phases are the same φLL = φRR.

Based on all above theoretical and simulated results, it can be confirmed that all four phase responses of the CP channels can be separately tuned with independent geometrical parameters of the proposed meta-atom. For meta-devices pursuing complete manipulation of the four CP conversion channels to achieve multiple functionalities simultaneously, the meta-atom library to endow the required phase gradients of all CP conversion channels, is established with sweeping of length px, width py and interior angle ϑ presented in Supplementary Note 4. Once the required phase profiles in different channels are confirmed, meta-atoms can be selected from the library and further optimized to construct the desired multifunctional metasurfaces. It is noted that, in this scheme we applied four parameter degrees of freedom (px, py, ϑ, and θ) to independently manipulate four phase patterns of transmission channels (φLL, φLR, φRL, and φRR). However, within this phase-modulation process, the amplitude responses are not considered and the efficiency is partly sacrificed due to the particularity of the proposed meta-atom structure. This limitation is detailed in Supplementary Note 5 and Supplementary Fig. 4. The exploration of an extra degree of freedom (e.g., material losses, active components, and even time) for independent amplitude modulation is therefore desired in future works.

### Simulation and measurement results on two metasurfaces

Based on the above specific explanation of general principle and construction of the meta-atom library, here we briefly clarify the construction of a single metasurface to achieve a distinct wavefront from each of the four CP channels. It is supposed that the desired spatial phase distributions can be described by $$F_{{\mathrm{LL}}}(x,y)$$, $$F_{{\mathrm{LR}}}(x,y)$$, $$F_{{\mathrm{RL}}}(x,y)$$, and $$F_{{\mathrm{RR}}}(x,y)$$ respectively. To implement the synthesized three phase schemes, chirality-assisted phase, propagation phase and PB phase components should be established by desired spatial phase distributions simultaneously (detailed in Supplementary Note 3). Following the phase modulation scheme, here we construct two metasurfaces with distinct functionalities to demonstrate the proposed phase responses for independent manipulation of full CP conversion channels.

As a first device concept, a meta-deflector able to separate all the co-polarized and cross-polarized output beams to four independent directions, is designed and the functional schematic is shown in Fig. 3a. The phase distribution of the proposed meta-deflector is based on the generalized Snell’s Law and is described as:

$${\mathrm{{\Phi}}}_i^{{\mathrm{refr}}}(x,y) = {\Delta}{\Phi}_i \cdot x = \frac{{2{\uppi}}}{{\lambda _0}}(\sin \varsigma _i^{{\mathrm{out}}} - \sin \varsigma ^{{\mathrm{in}}}) \cdot x,\,\\ {\mathrm{with}}\,i = 1,\,2,\,3,\,{\mathrm{and}}\,4,$$
(3)

where the normal direction is set along +z-axis and $${\Delta}{\Phi}_i$$ represents the phase gradient along x-axis. ςout and ςin are the angles of output and incident waves relative to +z-axis, and we consider the normal incident case where $$\varsigma ^{{\mathrm{in}}} = 0^\circ$$. In this section, the output tilting angles are preset as $$\varsigma _{\mathrm{1}}^{{\mathrm{out}}} = - 35^\circ$$ in L–L channel, $$\varsigma _{\mathrm{2}}^{{\mathrm{out}}} = 0^\circ$$ in L–R channel, $$\varsigma _{\mathrm{3}}^{{\mathrm{out}}} = 58^\circ$$ in R–L channel and $$\varsigma _{\mathrm{4}}^{{\mathrm{out}}} = - 16^\circ$$ in R–R channel. The corresponding required phase gradients are $${\Delta}{\Phi}_1 = \frac{{\uppi }}{{\mathrm{3}}}$$ for FLL(x,y), $${\Delta}{\Phi}_2 = 0$$ for FLR(x,y), $${\Delta}{\Phi}_3 = -\frac{{ {\uppi}}}{{\mathrm{2}}}$$ for FRL(x,y), and $${\Delta}{\Phi}_4 = \frac{{\uppi }}{{\mathrm{6}}}$$ for FRR(x,y). As for the construction of metasurface, the desired phase gradients for different channels along x-axis are discretized and implemented by 25 meta-atoms. The theoretical phase profiles and simulated phase value of all these meta-atoms are exhibited in Fig. 3b, indicating the feasibility of metasurface construction. Figure 3c displays the simulation and measurement results of the meta-deflector for four independent refracted wavefronts, whose far-field patterns at the center frequency of 10 GHz and the results clearly describe the peak direction of the output field intensities. It can be seen that under LHCP wave illumination, the co-polarized output wavefront tilts to −35° (in L–L channel), while the cross-polarized component indeed does not show any refraction effect (in L–R channel). Upon flipping the incidence into RHCP, the LHCP and RHCP output components are reconstructed with deflected wavefronts of 58° and −15° as verified by simulation and measurement (in R–L and R–R channel), respectively. It is worth-noting that in this part, the cross-polarized output wavefront under LHCP incident wave is preset with refraction angle of 0°, indicating that the energy in this channel is manipulated with converted polarization and kept unchanged wavefront state with the input. Another meta-deflector design is presented in the Supplementary Note 8 and Supplementary Fig. 6 to verify that all four output wavefronts can be imposed with different and arbitrary refraction angles. For further demonstration, the meta-deflector is fabricated and experimentally measured in microwave region as detailed in “Methods” section, A photograph of the sample is shown in Fig. 3d and measured results are displayed in Fig. 3e. The far-field pattern measurements have further been performed in frequency band from 9 to 11 GHz, which are function against frequency and detection angles. It can be seen that the deflecting angle slightly changes within the whole bandwidth, which can be considered as the frequency dispersion in Eq. (3). However, at the center frequency of 10 GHz, all the simulated and measured wavefronts generated in channels L–L, L–R, R–L, and R–R present the desired refraction directions, verifying the availability of proposed mechanism for simultaneous and independent manipulation of full spin conversion channels.

Additionally, we further examine the evolution of diffraction order against the variation of incident polarization state. Any arbitrary polarized wave κ can be considered as the superposition of two orthogonal CP waves with different proportionality coefficient, as described as:

$$\kappa = \eta _{\mathrm{L}} \cdot \left| {\mathrm{L}} \right\rangle + \eta _{\mathrm{R}} \cdot \left| {\mathrm{R}} \right\rangle,$$
(4)

where ηL and ηR denote the coefficient of LHCP and RHCP components. Here, we select five different polarization states as the incident wave, and the polarization variation path can be described as the black line on the Poincaré sphere in Fig. 4. From the north pole (point A) representing RHCP to the south pole (point E) for LHCP, continuous evolution is experienced through right-handed elliptical polarization (RHEP, point B), linear polarization (x-LP, point C), and left-handed elliptical polarization (LHEP, point D), where the ratios between ηL and ηR of these five polarized incidence A, B, C, D, and E are 0, 0.3, 1, 3, and ∞.

The simulated and measured output far-field intensities for five input polarization states are exhibited in Fig. 4. When the input polarization is RHCP, the output far-field intensity contains two main lobes, co-polarized component R–R with −16° refraction and cross-polarized component R–L with 58° refraction, as shown in Fig. 4a. With the polarization of incidence changing to state B, it can be seen from Fig. 4b that the output wavefront includes four lobes located at approximately, −35°, 0°, 58°, and −16° directions, which are the individual contributions of the four output channels L–L, L–R, R–L, and R–R. The proportion between the sum of energy distributed in both L–L and L–R channels to that in R–L and R–R channels is approximately 1:3, corresponding to the ratio between LHCP and RHCP component at the input end with RHEP state. Figure 4c displays the far-field intensity under LP wave illumination, where the far-field lobes in L–L and L–R channels exhibit nearly equal amplitude to that in R–L and R–R channels. When the input state is switched to LHEP, the proportion of output sum energy in both L–L and L–R channels to that in R–L, R–R channels is changed to 3:1 and displayed in Fig. 4d, as reversely symmetric situation to that in RHEP results. Finally, under LHCP wave illumination, only two main lobes with −35° and 0° refraction are left as in Fig. 4e. According to the above evolution process, it can be confirmed that all four CP channels can be fully utilized, and the ratio between different diffraction orders can be considered as function of the incident polarization.

For further demonstration of complete manipulation of the four CP conversion channels, a spin-to-orbital angular momentum meta-convertor is proposed and designed, where the orbital angular momentum (OAM) with different topological charge is achieved in each CP channel. The spatial helical phase distribution of optical spiral phase plate for OAM generation is described as:

$${\Phi}_i^{{\mathrm{OAM}}}(x,y) = l_{\mathrm{i}} \cdot \arctan (y/x)\,{\mathrm{with}}\,i = 1,\,2,\,3,\,{\mathrm{and}}\,4,$$
(5)

where (x, y) is the required coordinate of meta-atom and li = i − 1 is the topological charge of OAM mode. $${\Phi}_i^{{\mathrm{OAM}}}(x,y)$$ with i = 1, 2, 3, and 4 are the corresponding phase distributions for $$F_{{\mathrm{LL}}}(x,y)$$, $$F_{{\mathrm{LR}}}(x,y)$$, $$F_{{\mathrm{RL}}}(x,y)$$, and $$F_{{\mathrm{RR}}}(x,y)$$, respectively.

Figure 5a displays the simulated energy distributions in xoy plane at z = 150 mm (corresponding to 5λ0 for the center frequency of 10 GHz) of transmitted output vortex beams carrying OAM modes with l = 0, 1, 2, and 3 through the CP channels L–L, L–R, R–L, and R–R orderly, where all the typical doughnut-shape energy rings are exhibited and compared. It can be obviously seen that in L–L channel, the OAM mode is 0, whose energy intensity is as a plane wave with no hollow distribution in the center. Meanwhile with the topological charge increasing from l = 1–3 in L–R, R–L, and R–R channels, respectively, the radius of energy rings is gradually enlarged, which experimentally verifies the divergent properties of typical vortex beam. Figure 5b presents the corresponding measured energy distributions of vortex beams generated from the CP conversion channels, exhibiting four vortex beams with increasing energy ring radius, which is in good agreement with the simulations. Meanwhile, it can be seen that there exist some nonuniform and discrepancies in simulated and measured energy distributions, which are detailed discussed in Supplementary Note 6. The simulated and measured phase distributions of vortex beams carrying the corresponding four OAM modes are shown in Fig. 5c, d. It can be obviously observed that the helical phase pattern of OAM modes are totally distinct, verifying that the functionalities in all four CP channels can be independently modulated and the helical patterns are in accordance with the topological charge l · 2π. The measured phase patterns agree well with the simulated results, which further indicates the feasibility of independent multiple OAM modes generation based on full CP conversion channels.

Here, it should be noted that the phase manipulation to achieve distinct responses in all four transmission CP channels is a general method, which is not limited by operating wavelength. To extend the method to shorter wavelengths, efforts in high precision lithography would be required. Moreover, metallic lossy plasmonic resonators to be used in the replacement of current metallic patches, would have to be replaced by chiral all-dielectric nanoantennas.

## Discussion

We propose a general criterion to construct metasurfaces, which activate all CP channels and make full utilization of transmitted energy simultaneously. Through the synthesis of chirality-assisted phase, propagation phase and geometric phase, all the components in the Jones matrix can be decoupled and independently tuned. Two meta-devices are proposed based on synthesized three phase modulations to achieve the multifunctional wavefronts through four CP channels. Additionally, the evolution of diffraction orders of output far-field patterns against the variation of input polarization state is verified, further indicating the feasibility of proposed schemes for full channels utilization. This paradigm can be further extended and applied to other frequency spectrum with appropriate chiral meta-structures.

## Methods

### Microwave sample fabrication

The prototypes proposed in this paper are fabricated by using classical printed circuit board (PCB) technique, and which is briefly introduced as follows. In fabrication process, meta-structures are integrated with four pieces of polyfluortetraethylene dielectric-slabs and three adhesive layers. The 1 mm-thick dielectric substrates used have a relative permittivity εr = 3.5 and double copper-cladding layer of 0.035 mm-thick, while the adhesive layer has a relative permittivity εr = 2.74 and thickness of 0.1 mm. The total thickness of fabricated samples is 4.475 mm (about 0.15 λ0 at 10 GHz). The proposed metasurfaces all consist of 25 × 25 meta-atoms, which exhibit total size of 220 mm × 220 mm, and the fabrication tolerance is in accordance with initial design requirements.

### Experimental measurements

Measurements are conducted by a setup surrounded by microwave absorbers in order to minimize parasitic reflections. A 2–18 GHz dual-polarized wideband horn antenna is used as the feeding source to launch the circularly polarized quasi-plane waves (left-handed circular polarization, and right-handed circular polarization). For the near-filed mapping measurement, a fiber optic active antenna is used as field probe to measure both the amplitude and phase of the electric field. The probe receiver is fixed on two translation stages controlled by a motion controller and its position is incremented by a step of 2 mm. For far-field pattern measurement, the receiver is set as a horn antenna that is similar to the transmitting antenna. The fabricated sample and transmitting antenna are placed on a turntable to provide an angular scanning, the receiving antenna is fixed and placed at a distance d > 20λ0 away from the meta-device. In both measurement systems, the transmitting and receiving antennas are connected to an Agilent 8722ES vector network analyzer, which is adopted to measure the complex S11 and S21 parameters including the amplitude and phase. The schematic of microwave experimental setups and measurement calculation methods are shown in Supplementary Note 7 and Supplementary Fig. 5.

## Data availability

The data that support the plots within this paper and other findings of this study are available from the corresponding author upon reasonable request.

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15. Arbabi, A., Horie, Y., Bagheri, M. & Faraon, A. Dielectric metasurfaces for complete control of phase and polarization with subwavelength spatial resolution and high transmission. Nat. Nanotechnol. 10, 937–U190 (2015).

16. Wang, L. et al. Grayscale transparent metasurface holograms. Optica 3, 1504–1505 (2016).

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18. Yifat, Y. et al. Highly efficient and broadband wide-angle holography using patch-dipole nanoantenna reflectarrays. Nano Lett. 14, 2485–2490 (2014).

19. Li, X. et al. Multicolor 3D meta-holography by broadband plasmonic modulation. Sci. Adv. 2, e1601102 (2016).

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22. Yuan, Y. Y. et al. Complementary transmissive ultra-thin meta-deflectors for broadband polarization-independent refractions in the microwave region. Photonics Res. 7, 80–88 (2019).

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## Acknowledgements

K.Z. acknowledges funding from National Natural Science Foundation of China under Grant No. 61771172. Q.S. and P.G. acknowledge funding from the European Research Council (ERC) under the European Union’s Horizon 2020 research and innovation programme (Grant agreements no. 639109).

## Author information

Authors

### Contributions

K.Z., S.N.B., and P.G. conceived the idea. Y.Y. and B.R. conducted the numerical simulations and theoretical analysis. K.Z., Y.Y., Q.S., S.N.B., X.D., Q.W., and P.G. wrote the manuscript. All authors participated in the experiments and data analysis and read the manuscript.

### Corresponding authors

Correspondence to Kuang Zhang, Xumin Ding, Shah Nawaz Burokur or Patrice Genevet.

## Ethics declarations

### Competing interests

The authors declare no competing interests.

Peer review information Nature Communications thanks Adam Overvig and Changxi Zheng for their contribution to the peer review of this work. Peer reviewer reports are available.

Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

## Rights and permissions

Reprints and Permissions

Yuan, Y., Zhang, K., Ratni, B. et al. Independent phase modulation for quadruplex polarization channels enabled by chirality-assisted geometric-phase metasurfaces. Nat Commun 11, 4186 (2020). https://doi.org/10.1038/s41467-020-17773-6

• Accepted:

• Published:

• DOI: https://doi.org/10.1038/s41467-020-17773-6

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# Proof related to breadth first search

Suppose a connected graph G has a cycle C of length n. Prove that in any breadth-first search tree of G, any two vertices in C are at most $\lfloor\frac{n}{2}\rfloor$ levels apart.

No idea how to work this one. some hints on where to start will be appreciated!

-

The level of a vertex in such a tree is the number edges in a shortest path to the root. Any two vertices in $C$ are at most $\lfloor \tfrac{n}{2} \rfloor$ edges apart. So traveling from one vertex in $C$ via another one in $C$ to the root...

format: plain


text:
kidzsearch.com > wiki   Explore:web images videos games

# British Columbia

 Capital Largest city British ColumbiaColombie-Britannique  (French) Motto: Latin: Splendor sine occasu(English: Splendour without diminishment) BC AB SK MB ON QC NB PE NS NL YT NT NU [[File:||Canadian Provinces and Territories]] Victoria Vancouver Metro Vancouver None British Columbian[1] Type Constitutional monarchy Lieutenant Governor Janet Austin Premier John Horgan (NDP) Legislative Assembly of British Columbia (in Canadian Parliament) House seats 42 of 308 (13.6%) Senate seats 6 of 105 (5.7%) July 20, 1871 (7th) Ranked 5th Total 944,735 km2 (364,764 sq mi) Land 925,186 km2 (357,216 sq mi) Water (%) 19,548.9 km2 (7,547.9 sq mi) (2.1%) Proportion of Canada 9.5% of 9,984,670 km2 Ranked 3rd Total (2016) 4,648,055 [2] Density (2016) 5.02/km2 (13.0/sq mi) Ranked 4th Total (2015) C$249.981 billion[3] Per capita C$53,267 (8th) Postal BC ISO 3166-2 CA-BC Pacific (most of province) UTC −8/−7Mountain (far eastern) UTC −7/−6 V Pacific dogwood Western red cedar Steller's jay www2.gov.bc.ca Rankings include all provinces and territories

British Columbia (fr: Colombie-Britannique) is the westernmost province in Canada. It was originally two separate colonies of Great Britain, the colony of Vancouver Island, and the colony of British Columbia. British Columbia joined the Canadian Confederation In 1871. The capital of British Columbia is Victoria, but the city with the most people is Vancouver.

## History

### Early History

Before European contact, British Columbia was home to large numbers of indigenous people. These people, for the large part, still live in the province. Because BC has a large amount of resources, the native people who dwelled here lived comfortably enough. The large number of resources, in particular, led to the creation of very complex hierarchical societies in British Columbia, long before the government of the Europeans.

### Oregon Dispute

A map of the disputed region. The 54*40' parallel is visible.

The Oregon dispute was an important factor in determining the physical shape of British Columbia, and as the political impetus for British investment in protecting her lands in British Columbia. During the middle of the 19th century, the Democratic Party of the United States, appealing to the public attraction to the Manifest Destiny (a philosophy preaching American Supremacy and expansionism), declared that the United States had a claim over the entirety of Columbia District/Oregon Territory. This attitude was reflected even further, as 54*-40' or fight, referring to the end of Columbia District, and the start of Russian Territory of Alaska. This alarming attitude, caused many in London to fear that yet another war, in less than a century, was brewing. Ultimately, however, it was resolved, at the 49th parallel. British Columbia lost Fort Vancouver (now Vancouver, Washington), and lots of the gulf islands. To this day, the Oregon Treaty is criticized by historians, as being too lenient for the Americans.[4]

### Hudson's Bay Corporation

While British Columbia was technically a British colony, it was the private corporation, the Hudson's Bay Company that really ran the place. The first Governor of Vancouver Island (then a separate colony), actually resigned over the fact that the Queen, had no say, above the interests of the Colony. After him, the next governor of the Island, was also, coincidentally, the head of the HBC's operations in British Columbia and Vancouver Island.

### Gold Rush

The first major gold rush in British Columbia started in 1858, after gold was discovered, on the upper reaches of the Fraser River. The sudden onslaught of uncouth Americans led to the declaration that British Columbia shall be a crown colony, instead of the unincorporated region it was, after the Oregon spat. James Douglas was appointed the governor of the newly formed colony, as well as the Colony of Vancouver Island, and for military convenience, the Capital of British Columbia was declared Queensboro, then renamed New Westminster. As massive numbers of Americans came, fears that the Americans would try to bluster themselves another chunk of territory became more and more concrete. Thus, a contingent of Royal Engineers and Marines were deployed to New Westminster. To this day, the neighbourhood of Sapperton, is known, for its history as a site of the barracks for the Engineers, nicknamed the Sappers. The infrastructure ordered by James Douglas, to support the vast amount of capital flowing from the gold industry, caused high debt for British Columbia. By 1864, British Columbia was £200,000 in debt, plus an additional £18,000, from various spats with natives, and constant policing required to maintain law and order. Its vast amount of gold was useless, while BC was in debt. British Columbia amalgamated with Vancouver Island, forming one super-colony. Vancouver Island would undertake the debt, and British Columbia's capital would be in Victoria.

## Industry

British Columbia's industry is primarily, and has always been, resource centered. For large parts of its history, including the entire time it was a colony, BC was dominated by the big 4: Forestry, Mining, Logging, and Agriculture, which have shaped British Columbia's culture, and society.

### Forestry in British Columbia

British Columbia's forestry industry has always been booming. BC had a bonanza of massive trees, thanks to its location in the Pacific Rainforest region. This provided ample opportunity for entrepreneurs to build businesses. Massive sawmills sprung up, all over British Columbia, especially in what is now Vancouver and surrounding regions. However, this required massive amounts of human capital - which was surprisingly in short supply. To overcome this, massive numbers of immigrants were invited to Vancouver, and places like so. This changed BC's demographic makeup so much, that to try to stop it, the government passed several acts, to try and limit the number of immigrants. However, the change was already underway - BC subsequently became one of the most diverse places in North America, with people calling it home, from places like Asia, Europe, and even the Pacific Islands.

### Mining in British Columbia

British Columbia's large mineral resources also led to a large mining industry. For the most of the 19th century, and well into the 20th century, mining was highly profitable. For the immigrants, once again, this was another opportunity to exploit. Large numbers of immigrants, skilled from their experience mining at home, filled up the labour demands, and quickly, new towns were formed. The city of Nanaimo, for instance, was founded thanks to a large amount of coal found in its harbour. Coal especially, was the fuel of the industry, as coal was required for everything it the time, from ships, to trains, to the heater in everyone's homes.

### Fishing in British Columbia

Thanks to BC's location on the coast, fishing has been very prevalent in British Columbia. Before agriculture had developed very far in BC, and well before farming fish was even a viable option, fishing was the only reliable way to get food. The native people of BC already had a very active fishing industry, with large hauls of salmon, and other creatures sustaining them. When civilization came, the fishing industry took off, as everybody needed to eat, and there was not enough agricultural goods to go around. In fishing, once again, it was the immigrants that pulled it off. Mostly Japanese workers, and their families, came. Using their rich heritage of industrialized fishing, they were able to out compete every other fisherman, that by the turn of the century, the dockyards, canneries, and fishing fleets were filled to the brim with the Japanese. Only the forced Japanese internment in WW2 put a halt to the monopoly the Japanese had on fishing in British Columbia. Today, fishing is still a powerful force, although fettered by ordinances to prevent overfishing.

### Agriculture in British Columbia

The agriculture industry in BC is, and always has been humble. However, as they say, good things come in small packages, and the agricultural goods of BC, are about as good as they come. Due to the fact that BC has only about 2% arable land, the remainder being too dry, or too cold, or too remote, or a plethora of other reasons, BC has never been able to sustain a primary food agricultural industry. Even today, BC cannot grow enough food to feed its entire population, and imports foods instead, from the eastern provinces, such as Alberta, and Saskatchewan. However, where BC does grow food, the food is scrumptious, and award-winning. The Okanagan valley, is known for its award-winning fruits, and wine-related products. However, that is not where all the food is produced. In the Fraser valley delta, large amounts of other food crops are grown, such as wheat, and berries.

British Columbia Parliament Buildings

### Housing in British Columbia

The value of real estate in British Columbia has increased a lot in recent years. Property assessed values have increased by $1-trillion since 2007. In 2018 the value was$1.861 trillion. In 2007 it was only \$808 billion.[5]

## Health care

The provincial health insurance plan in British Columbia is called Medical Services Plan (MSP). In order to join MSP a person:

• must make his or her home in B.C.; and
• Eligible B.C. residents (citizens of Canada or persons who are lawfully admitted to Canada for permanent residence) who are outside B.C. for vacation purposes only, are allowed a total absence of up to seven months in a calendar year.[6]

## Politics

British Columbia's premier is John Horgan, leader of the provincial New Democratic Party, a social democratic party. The NDP governs British Columbia through a deal with the provincial Green Party, led by Andrew Weaver.[7] The BC Liberal Party is the main opponent to the NDP, and is led by Andrew Wilkinson. Other parties are active in British Columbia, such as the Conservative Party, led by Scott Anderson, although only the NDP, BC Liberal Party, and Green Party currently have seats.[8]

## Flags

The flag of British Columbia was adopted in 1960. It is based on the provincial arms and the union jack.
British blue ensign used between 1870 & 1906

## References

1. According to the Oxford Guide to Canadian English Usage (

Template-specific style sheet:

ISBN 0-19-541619-8; p. 335), BCer(s) is an informal demonym that is sometimes used for residents of BC
2. "Population and dwelling counts, for Canada, provinces and territories, 2016 and 2011censuses". Statistics Canada. February 8, 2017. Retrieved February 8, 2012.
3. Statistics Canada. Gross domestic product, expenditure-based, by province and territory (2015); November 9, 2016 [cited January 26, 2017].
4. http://rside.f11.us/honeypot/index.php/History_of_BC
5. https://vancouversun.com/news/local-news/billionaire-bash-calls-on-government-to-tax-the-rich-to-house-the-poor
6. https://www2.gov.bc.ca/gov/content/health/health-drug-coverage/msp/bc-residents


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# Revision history [back]

Here's a simple example over $\mathbb{Q}$ with $f(x) = x^3+1$ and $g(x) = x^2-1$. First, define a polynomial ring over $\mathbb{Q}$:

sage: R.<x> = PolynomialRing(QQ,'x')
sage: f = x^3+1; g = x^2-1
sage: f.gcd(g)
x+1

One can also do this with extension fields. (By the way, there's probably a faster way to do the following.) First, let's construct $\mathbb{Q}(\sqrt{2},\sqrt{3})$ using the ring R constructed above

sage: p = x^2-2        # using the x defined above
sage: q = x^2-3
sage: K.<a,b> = QQ.extension([p,q])
sage: print a^2, b^2   # testing variable assignment
2  3

Now create a polynomial ring over K, define some elements, and compute their GCD:

sage: S.<x> = PolynomialRing(K,'x')
sage: f = a*x^2 * (x-1); g = a*x^2 * (x-b)
sage: f.gcd(g)
x^2

Hrm, it seems like there's either a bit of a bug, here, or I constructed something incorrectly. (The result should be $\sqrt{2}x^2$.) I'll submit this as a bug report. However, it seems like you'll at least get a common factor "modulo" the elements used to create the extension field.

sage: f/x^2
a*x - a
sage: g/x^2
a*x - b*a

So perhaps by inspection you can compute the GCD. Hopefully this will be resolved soon.


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# Statistics $X_{(1)}$ complete for a Uniform Distribution?

Someone had asked this earlier, but since it was good practice for my qualifying exam coming up, I figured I would ask and share my work on the problem. The problem is:

Suppose $X$ is Unif$(0,\theta)$. Let $T = X_{(1)}$ (minimum order statistic). Is $T$ complete?

My initial thought was no since usually lectures show $X_{(n)}$ is, but here was my attempt:

I assume that $E[g(T)]=0$ for some function $g$ and want to find if $P(g =0)=1$.

$$P(T \leq t) = 1 - P(T > t) = 1-\left(1-\frac{t}{n} \right)^n$$

So taking derivatives to get the pdf:

$$f_T(t) = \left(1-\frac{t}{n}\right)^{n-1}$$

So now working with expectations:

$$0=E[g(T)] = \int_0^\theta g(t)\left(1-\frac{t}{n}\right)^{n-1} dt$$

This integral is differentiable, and since the integral is $0$, then the derivative is $0$, so taking the derivative with respect to $\theta$:

$$0= g(\theta)\left(1-\frac{\theta}{n}\right)^{n-1}$$

Hence $g(\theta) =0$, so $T$ is a complete statistics.

Does this argument work? I basically modeled it after the proof of showing the max order statistic is complete. If $X_{(1)}$ is not complete, where does the proof go wrong?

• First mistake: one wants to prove that $P_\theta(g(T) =0)=1$ for every $\theta$ ($g(T)$, not $g$). Second mistake: $P_\theta(T \leqslant t) = 1-\left(1-\frac{t}{\theta} \right)^n$ for every $t$ in $(0,\theta)$ ($\frac{t}{\theta}$, not $\frac{t}n$). Can you correct these and pursue? – Did May 21 '16 at 19:19
• @Did Thanks for the corrections! So using the fixed pdf and getting to the integral, I get $0=E[g(T)] = \int_0^{\theta}g(t)\frac{n}{\theta}\left(1-\frac{t}{\theta} \right) dt$. So now when I take a derivative with respect to $\theta$, the $\left(1-\frac{t}{\theta} \right)$ term becomes $0$, so $g(T)$ does not have to be $0$ with probability $1$, and thus it's not a complete statistic? – Brenton May 21 '16 at 19:28
• Huh? First, this is $\left(1-\frac{t}\theta\right)^{n-1}$, not $\left(1-\frac{t}\theta\right)$. Second, the derivative of the integral is not what you say. – Did May 21 '16 at 19:40
• @Did Whoops, left out the $n-1$. So I have $\int_0^{\theta} g(t)\frac{n}{\theta}\left(1-\frac{t}{\theta}\right)^{n-1} dt$. So taking the derivative w.r.t $\theta$... don't I get $g(\theta)\frac{n}{\theta}\left(1-\frac{\theta}{\theta} \right)^{n-1}$? – Brenton May 21 '16 at 19:45
• No. What is the formula for $$\frac{d}{d\theta}\left(\int_{a(\theta)}^{b(\theta)}h(\theta,t)dt\right),$$ already? – Did May 21 '16 at 19:47


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On functional determinants of matrix differential operators with multiple zero modes

G.M. Falco, Andrey A Fedorenko, Ilya A Gruzberg

8 Citations (Scopus)
We generalize the method of computing functional determinants with a single excluded zero eigenvalue developed by McKane and Tarlie to differential operators with multiple zero eigenvalues. We derive general formulas for such functional determinants of $r\times r$ matrix second order differential operators O with $0 < n \leqslant 2r$ linearly independent zero modes. We separately discuss the cases of the homogeneous Dirichlet boundary conditions, when the number of zero modes cannot exceed r, and the case of twisted boundary conditions, including the periodic and anti-periodic ones, when the number of zero modes is bounded above by 2r. In all cases the determinants with excluded zero eigenvalues can be expressed only in terms of the n zero modes and other $r-n$ or $2r-n$ (depending on the boundary conditions) solutions of the homogeneous equation $O h=0$ , in the spirit of Gel'fand–Yaglom approach. In instanton calculations, the contribution of the zero modes is taken into account by introducing the so-called collective coordinates. We show that there is a remarkable cancellation of a factor (involving scalar products of zero modes) between the Jacobian of the transformation to the collective coordinates and the functional fluctuation determinant with excluded zero eigenvalues. This cancellation drastically simplifies instanton calculations when one uses our formulas.


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X

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Copied to Clipboard
Circle of strings
Medium Accuracy: 44.12% Submissions: 10667 Points: 4

Given an array of lowercase strings A[] of size N, determine if the strings can be chained together to form a circle.
A string X can be chained together with another string Y if the last character of X is same as first
character of Y. If every string of the array can be chained, it will form a circle.

For example, for the array arr[] = {"for", "geek", "rig", "kaf"} the answer will be Yes as the given strings can be chained as "for", "rig", "geek" and "kaf"

Example 1:

Input:
N = 3
A[] = { "abc", "bcd", "cdf" }
Output:
0
Explaination:
These strings can't form a circle
because no string has 'd'at the starting index.

Example 2:

Input:
N = 4
A[] = { "ab" , "bc", "cd", "da" }
Output:
1
Explaination:
These strings can form a circle
of strings.

You don't need to read input or print output. Your task is to complete the function isCircle() which takes the length of the array N and the array A as input parameters and returns 1 if we can form a circle or 0 if we cannot.

Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)

Constraints:
1 ≤ N ≤ 104
1 ≤ Length of strings ≤ 20

We are replacing the old Disqus forum with the new Discussions section given below.


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# Shouldn't the upwash on an airfoil increase the effective angle of attack?

From what I understand, induced drag comes from a reduction of the effective angle seen by the airfoil.

However, from wind tunnel smoke visualization, we can see that the streamlines at the leading edges and stagnation point tend to "go upward'. Which is the opposite of what the effective relative wind is supposed to be and do.

Is this "upward tilt angle" of the streamlines, the result of suction from the upper surface minus the induced angle?

I've included two pictures:


text:
How did inflation affect time-like space-time?

Alan Guth gave an overview of inflation in a series of videos for World Science U. The emphasis is on how inflation expanded a patch of space to the size of a marble, with no mention of how time-like space-time was affected.

• What is time-like space-time!? – MBN Nov 12 '18 at 12:01
• @MBN I sloppily conflated space-time with space-time interval. I'll leave it in since an answer points out this flaw. – Larry Harson Nov 13 '18 at 0:24

During the brief period of cosmic inflation, spacetime had, to a reasonable approximation, the geometry of deSitter space. See

https://en.m.wikipedia.org/wiki/De_Sitter_universe#Modelling_cosmic_inflation

In the distant future, the geometry will again approach that of de Sitter space as dark energy dominates more and more over matter.

It is not meaningful to talk about “time-like spacetime”. It is intervals and paths in spacetime that can be time-like, space-like, or light-like.

The reason that Guth focused on what was happening to space and not time is that the metric of a homogeneous and isotropic universe can be written in the simple form

$$ds^2=a(t)^2ds_3^2-c^2dt^2$$

where $$ds_3^2$$ is the 3-d metric for space... either flat Euclidean space or a space that has uniform positive or negative curvature. (The positive case, for example, is a 3-sphere.)

As you can see from this metric, nothing is happening to the size of spacetime in the time direction. All that happens is that the size of spatial directions grows or shrinks with time. In the case of inflation and deSitter space, the scale factor $$a(t)$$ actually grows exponentially with time!


text:
###### expression 1

$$1\frac{3}{4}-1\frac{11}{17}\times\frac{51}{56}$$

Compute

$$1\frac{3}{4}-1\frac{11}{17}\times\frac{51}{56}=\frac{7}{4}-\frac{28}{17}\times\frac{51}{56}=\frac{7}{4}-\frac{6}{4}=\frac{1}{4}=0,25$$

200 g of an alloy contained 80 g of copper, then how much copper did 300 g of alloy contain?

Compute

If 200 g of an alloy contained 80 g of cooper, then 300 g of an alloy contained X g of cooper. Then

$$\frac{200}{300}=\frac{80}{x}$$

$$x=\frac{300\times80}{200}=120$$ g

###### expression 3

$$(1,5+\frac{1}{4}):18\frac{1}{3}$$

Compute

$$(1,5+\frac{1}{4}):18\frac{1}{3}=(\frac{3}{4}+\frac{1}{4}):\frac{55}{3}=\frac{7}{4}\times\frac{3}{55}=\frac{21}{220}$$

###### expression 4

$$(\frac{5}{6}-\frac{1}{3})+(\frac{1}{2}+\frac{9}{10})$$

Compute

$$(\frac{5}{6}-\frac{1}{3})+(\frac{1}{2}+\frac{9}{10})=\frac{1}{2}+\frac{14}{10}=\frac{5+14}{10}=1,9$$

###### expression 5

$$2\frac{2}{3}:3\frac{17}{21}$$

Compute

$$2\frac{2}{3}:3\frac{17}{21}=\frac{8}{3}:\frac{80}{21}=\frac{8\times{21}}{3\times80}=\frac{7}{10}=0,7$$

The plan is 35 parts per day. The worker made 42 parts. For what percentage did he exceed the plan?

Compute

He made 42 parts. The plan was 35 parts. Then he exeeded the plan on $$\frac{42}{35}=1,2=120%$$%

Into the store they brought 14 tons of cabbage. 30% of the cabbage was sold. How many cabbage is left?

Compute

They sold 30% of the cabbage. The 70% is left.

$$14\times0,7=9,8$$ tons

For 5 tables you have to pay $120. How much do 7 tables cost? Compute 5 tables -$120

7 tables - \$X

$$\frac{5}{7}=\frac{120}{x}$$

$$x=\frac{7\times120}{5}=168$$

For sowing 8 hectares the farmer spent 560 kg of peas. How many kilograms of peas does it require to seed 11 hectares?

Compute

8 ha - 560 kg of peas

11 ha - x kg of peas.

Then $$\frac{8}{11}=\frac{560}{x}; x=\frac{11\times560}{8}=770$$ kg

The farmer planted 150 hectares with wheat and oats in a ratio of 4:2. For how many hectares more did he seed with wheat than oats?

Compute

If x - is wheat and y - oat, then

$$\frac{x}{y}=\frac{4}{2}$$ and $$x+y=150$$

Then $$x=2y$$ (from the first equation)

$$2y+y=150; 3y=150; y=50$$ ha

$$x=100$$ ha

Then he seed wheat on 50 ha more.

###### Ratio

Find the ratio of 4 km to 80 meters

Find the ratio of 4 km to 80 meters

4 km is 4000 m.

Then $$\frac{4000}{80}=50$$

###### Number 1

Find the number if 13% of it is 1,69?

Find the number if 13% of it is 1,69?

$$1,69:0,13=13$$

###### Number 2

Anna, increasing the number 145 by 60%, calculated 25% from the last number. What is the number calculated by Anna?

$$(145+145\times0,6)\times0,25=1,6\times145\times0,25=58$$

###### expresion 21

Compute $$x^3+x^2+x+1$$ when x=4

Compute

$$x^3+x^2+x+1=4^3+4^2+4+1=85$$

###### fraction

Get rid of irrationality in the denominator of the fraction $$\frac{10}{3\sqrt5}$$

$$\frac{10}{3\sqrt5}=\frac{10\sqrt5}{3\times5}=\frac{2\sqrt5}{3}$$

###### expression 29

$$\sqrt{9+4\sqrt5}$$

Compute

$$\sqrt{9+4\sqrt5}=\sqrt{4+4\sqrt5+5}=\sqrt{(2+\sqrt5)^2}=2+\sqrt5$$

###### Number 3

Find the minimal common multiple of 525 and 588

$$525=5^2\times3\times7$$

$$588=2^2\times7^2\times3$$

the minimal common multiple $$525\times7\times2^2=14700$$

###### test

At a test operation 18 people were rated good or excellent. Number of ratings excellent and good is proportional to the numbers 2 and 1. How much less good ratings than excellent?

x - excellent

y - good

x/y=2/1; x=2y

x+y=18; 3y=18; y=6

x=12

12-6=6

###### Number 4

Which figure can be put in place of * in 46 * three-digit number to the resulting number is divisible by 2, 3, 6, 9 at the same time?

The resulting number 46* must be divisible by only 2 and 9 (and then it will be divisible by 2,3,6,9).

Then 46* must be even (only 6 and 8 are fit, because 10 is not a figure).

To be divisible by 9, the sum 4+6+* must be divisible by 9.

4+6+*=10+*

* could be only 8, because 18 is divisible by 9.

###### proportion 2

$$15:2\frac{1}{2}=x:8\frac{1}{3}$$

Find x

$$\frac{15}{\frac{5}{2}}=\frac{x}{\frac{25}{3}}$$

$$\frac{30}{5}=\frac{3x}{25}$$

$$30\times25=3x\times5$$

x=50

###### expresion 22

5,6a+8,4a+186,4

Compute when a=3,5

$$5,6a+8,4a+186,4=14a+186,4=14\times3,5+186,4=235,4$$

###### expresion 25

(0,7*0,4+0,32)*5

Compute

(0,7*0,4+0,32)*5=(0,28+0,32)*5=0,6*5=0,3

###### Z

Find Z if 60% of Z is equal to 108.

108:0,6=180

###### expresion 37

$$(\sqrt2-\sqrt5)(\sqrt2+\sqrt5)$$

Compute

$$(\sqrt2-\sqrt5)(\sqrt2+\sqrt5)=(\sqrt2)^2-(\sqrt5)^2=2-5=-3$$

###### find x

Find x if 40% of x is equal to 52.

Find x if 40% of x is equal to 52.

0,4x=52

x=52:0,4

x=130

###### expression 38

$$\sqrt{3\frac17}\times\sqrt{\frac{7}{88}}$$

Compute

$$\sqrt{3\frac17}\times\sqrt{\frac{7}{88}}=\sqrt{\frac{22}{7}\times\frac{7}{88}}=\sqrt{\frac14}=\frac12=0,5$$

###### expression 38

Compute

$$(81108:27-125\times12):4$$

$$(81108:27-125\times12):4=(3004-1500):4=1504:4=376$$

###### Root

$$\sqrt[4]{\frac{0,0016a^4}{81b^8c^12}}$$

Compute when a=3; b=2; c=1

$$\sqrt[4]{\frac{0,0016a^4}{81b^8c^12}}=\frac{0,2a}{3b^2c^3}=\frac{0,2\times3}{3\times4\times1}=\frac{1}{20}$$

###### expression 44

$$(a^4)^{-\frac34}\times(b^{-\frac23})^{-6}$$

Simplify

$$(a^4)^{-\frac34}\times(b^{-\frac23})^{-6}=a^{4\times(-\frac34)}\times{b^{(-\frac23)\times(-6)}}=a^{-3}b^4=\frac{b^4}{a^3}$$

###### Number 5

Find 12% from 900

Find 12% from 900

900*0,12=108

###### Expression 47

$$1-0,15:(\frac{11}{12}-0,75)$$

Compute

$$1-0,15:(\frac{11}{12}-0,75)=1-\frac{15}{100}:(\frac{11}{12}-\frac{9}{12})=1-\frac{15}{100}\times6=1-0,9=0,1$$


text:
# Chapter 31. Features, Limits, and Formats

Vulkan is designed to support a wide range of hardware and as such there are a number of features, limits, and formats which are not supported on all hardware. Features describe functionality that is not required and which must be explicitly enabled. Limits describe implementation-dependent minimums, maximums, and other device characteristics that an application may need to be aware of. Supported buffer and image formats may vary across implementations. A minimum set of format features are guaranteed, but others must be explicitly queried before use to ensure they are supported by the implementation.

Note on extensibility The features and limits are reported via basic structures (that is VkPhysicalDeviceFeatures and VkPhysicalDeviceLimits). It is expected that when new features or limits are added in a future Vulkan version, new structure(s) and entry point(s) will be added as necessary to query these. New functionality added by extensions is not expected to modify the core feature and limit structures.


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# In the figure above, the number of shaded squares is what percent grea

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In the figure above, the number of shaded squares is what percent grea [#permalink]

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11 Nov 2017, 06:47
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In the figure above, the number of shaded squares is what percent greater than the number of unshaded squares?

(A) 25%
(B) 40%
(C) 50%
(D) 60%
(E) 75%

[Reveal] Spoiler:
Attachment:

2017-11-11_1743.png [ 2.01 KiB | Viewed 450 times ]
[Reveal] Spoiler: OA

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Re: In the figure above, the number of shaded squares is what percent grea [#permalink]

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11 Nov 2017, 08:34
Bunuel wrote:

In the figure above, the number of shaded squares is what percent greater than the number of unshaded squares?

(A) 25%
(B) 40%
(C) 50%
(D) 60%
(E) 75%

[Reveal] Spoiler:
Attachment:
2017-11-11_1743.png

Total No of squares = 25
Unshaped = 10

Quote:
the number of shaded squares is what percent greater than the number of unshaded squares

$$\frac{5}{10}*100$$

= $$50$$ %

Thus, answer will be (C) 50%
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In the figure above, the number of shaded squares is what percent grea [#permalink]

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11 Nov 2017, 09:03
Bunuel wrote:

In the figure above, the number of shaded squares is what percent greater than the number of unshaded squares?

(A) 25%
(B) 40%
(C) 50%
(D) 60%
(E) 75%

[Reveal] Spoiler:
Attachment:
2017-11-11_1743.png

1) Half of 10 is 5
50 percent of 10 is 5
15 - 10 = 5
15 is 50 percent greater than 10

OR

2) Percent greater than:

$$\frac{New - Old}{Old} * 100$$

$$(\frac{15 -10}{10}) = (\frac{5}{10}) = (.5 * 100)$$ = 50 percent

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In the figure above, the number of shaded squares is what percent grea   [#permalink] 11 Nov 2017, 09:03
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# In the figure above, the number of shaded squares is what percent grea

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Optimization of media of rhdotorula glutinis, a oleaginous yeast, in continuous culture = 연속식 배양에서유지 생산 효모의 균체량 증대를 위한 배지 최적화에 대한 연구

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Rhodotorula glutinis NRRL-Y-1091, a red oleaginous yeast, was grown in batch and continuous culture system. At batch culture, the accumulaion of lipid is favoured not only by nitrogen limitation but also by deficiency in magnesium and phosphate. It would appear from the above observations that any constraint which limits growth but not assimilation of the carbon source will tend to cause accumulation of lipid. At continuous culture, mass and energy balance method was used to analyze that the cell concentration decreased rapidly with increasing dilution rate. Energetic yield, $\eta$, decreased with increasing dilution rate, but metabolic heat fraction, $\epsilon$, increased with increasing dilution rate. Since $\eta + \epsilon$, is almost constant at all dilution rate experimented, it may be said that product formation may not be correlated with that result. Since in experiment, the dissolved oxygen concentration was maintained at 50\% at higher dilution rate, incomplete aeration and agitation were not correlated with that result. The medium may be inadequate in composition and preparation procedure. Therefore, medium optimization in continuous culture was carried out using a pulse method.
Kim, Jung-HoeresearcherRhee, Joon-Shickresearcher김정회researcher이준식researcher
Description
한국과학기술원 : 생물공학과,
Publisher
한국과학기술원
Issue Date
1987
Identifier
65590/325007 / 000851245
Language
eng
Description

학위논문(석사) - 한국과학기술원 : 생물공학과, 1987.2, [ v, 48 p. ]

URI
http://hdl.handle.net/10203/28265


text:
# Linear Algebra Dual Space

## Homework Statement

Let Hom(V,W) be the set of linear transformations from V to W. Define addition on Hom(V,W) by (f + g)(v) = f(v) + g(v) and scalar multiplication by (af)(v) = af(v.

If V is a vector space over a field K, define V* = Hom(V,K). This is called the dual space of V. If <v1,.....,vn> is a basis of V, show that for each I there is an vi* that is an element of V* satisfying vi*(vj) = d_ij (Kronecker delta function).

*Hopefully I typed this question out clear enough, I'm hoping that a real maverick of linear algebra happens upon this thread and understands the question*

## The Attempt at a Solution

I've put a lot of thought into this problem and wanted to discuss my thoughts with someone who knows what's going on. So given vi* is an element of V*, we need find what mapping vi* must be such that vi*(vj) = d_ij (Kronecker delta function).

what if vi*(vj) = ( vj / |vj| )e_i where e_i is a unit vector in the I direction. The (vj / |vj|) will equal one, since this is the equation for normalizing a vector, and then multiplying this by e_i will be zero if j does not equal I.

But then I realized that multiplying by e_i will be zero only if I apply grand Schmitt's orthonormalization process first, right?

Anyone, I'm knew to the subject and hope somebody can somewhat understand my thoughts and provide some insight! Thanks PF

LCKurtz
Homework Helper
Gold Member

## Homework Statement

Let Hom(V,W) be the set of linear transformations from V to W. Define addition on Hom(V,W) by (f + g)(v) = f(v) + g(v) and scalar multiplication by (af)(v) = af(v.

If V is a vector space over a field K, define V* = Hom(V,K). This is called the dual space of V. If <v1,.....,vn> is a basis of V, show that for each I there is an vi* that is an element of V* satisfying vi*(vj) = d_ij (Kronecker delta function).

*Hopefully I typed this question out clear enough, I'm hoping that a real maverick of linear algebra happens upon this thread and understands the question*

## The Attempt at a Solution

I've put a lot of thought into this problem and wanted to discuss my thoughts with someone who knows what's going on. So given vi* is an element of V*, we need find what mapping vi* must be such that vi*(vj) = d_ij (Kronecker delta function).

what if vi*(vj) = ( vj / |vj| )e_i where e_i is a unit vector in the I direction. The (vj / |vj|) will equal one, since this is the equation for normalizing a vector, and then multiplying this by e_i will be zero if j does not equal I.

But then I realized that multiplying by e_i will be zero only if I apply grand Schmitt's orthonormalization process first, right?

Anyone, I'm knew to the subject and hope somebody can somewhat understand my thoughts and provide some insight! Thanks PF

If ##v = \sum_{i=1}^n c_i v_i##, think about defining ##v_j^*## as picking out the ##j##th coefficient:$$v_j^*(v) = c_j$$See if you can do something with that.

PsychonautQQ
Stephen Tashi
Your thinking is reasonable. It depends on having a "standard basis" {e1, e2,...eN} available. Some proofs can be effectively be done by thinking of the abstract vector space as being Euclidean space in a diagram where the illustrator has set up {e1,e2,..eN} for us. However, statistically speaking, abstractly phrased problems more often expect students to abstract methods if they do the exercise "the easy way"

My thought on this problem is that you should define a transformation P that has the desired properties, without initially claiming it is a linear transformation. Then prove P is a linear transformation. According to the traditions of how writing mathematics is interpreted, the definition:

Let Hom(V,W) be the set of linear transformations from V to W.

Implies that any linear transformation you create from V into W is an element of Hom(V,W). So if you can make up a linear transformation P with the desired properties, it will be an element in Hom(V,W).

Give the vector vj in V, you could say "Let Pj be the transformation Pj(vi) = d_ij". However, this does not "well define" Pj since it doesn't tell what Pj does to an arbitrary vector R. To "well define" Pj(R) you could say that Pj is defined as "Express R as its unique linear combination of the vectors in the basis {v1, v2,...vN} and let Pj(R) be the coefficient of vj in that linear combination". (I'm sure you can find a more dignifed way of putting it.) You have to prove Pj is a linear transformation. (Pj is "projection on the j-direction" if you are thinking of Euclidean space, but I wouldn't use the word "direction" when you write the proof.)

It interesting that your course materials, within a single problem, define Hom(V,W) as a set of linear transformations between vector spaces and then specialize to the case Hom(V,K) where K is the scalar field. The special case of Hom(V,K) is very important. The linear transformations of Hom(V,K) can be called "linear functionals". You can think of certain sets of functions as vector spaces. In advanced calculus, a linear functional is defined as linear function that maps functions to numbers. (For example $L(f(x)) = \int_0^\infty x f(x) dx$ ).

There are also examples of Hom(V,W) in advanced calculus - many of those mysterious "transforms" that take a function f(x) to a function F(s) in a different variable are linear transformations from one vector space of functions to another vector space of functions.

Last edited:
PsychonautQQ


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text:
Sciencemadness Discussion Board » Fundamentals » Organic Chemistry » Diethylamine Synthesis Select A Forum Fundamentals   » Chemistry in General   » Organic Chemistry   » Reagents and Apparatus Acquisition   » Beginnings   » Responsible Practices   » Miscellaneous   » The Wiki Special topics   » Technochemistry   » Energetic Materials   » Biochemistry   » Radiochemistry   » Computational Models and Techniques   » Prepublication Non-chemistry   » Forum Matters   » Legal and Societal Issues

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Author: Subject: Diethylamine Synthesis
ScienceSquirrel
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 Quote: Originally posted by mbrown3391 Sorry, i was thinking of the amine. so does the ethylene glycol raise the boiling point of the diethylamine?

No.

It is just a higher boiling solvent to carry out the hydrolysis in.
The higher the temperature the quicker the reaction.
Watch out for the hot potassium hydroxide in ethylene glycol and water though. It will strip skin like hot water will remove wallpaper.
Ritter
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 Quote: Originally posted by ScienceSquirrel If you want to make anhydrous diethylamine ( eg free from water or alcohol ) then you are getting into a whole new ballgame.

Yes. The alkaline hydrolysis would allow you to distill over the amine but it would be wet with water. $17 would get you 5 lbs of anhydrous sodium sulfate on eBay item # 270268727018 . Ritter ============================= \"The production of too many useful things results in too many useless people.\" Karl Marx Ritter National Hazard Posts: 370 Registered: 20-6-2008 Location: Earth Member Is Offline Mood: Curious  Quote: Originally posted by mbrown3391 Sorry, i was thinking of the amine. so does the ethylene glycol raise the boiling point of the diethylamine? I think you should consider taking a couple of basic chemistry courses including organic before you injure yourself or others. Basic rule of thumb for insuring personal safety: "If you don't know what you're doing, don't do it." [Edited on 28-8-2008 by Ritter] Ritter ============================= \"The production of too many useful things results in too many useless people.\" Karl Marx ScienceSquirrel International Hazard Posts: 1863 Registered: 18-6-2008 Location: Brittany Member Is Offline Mood: Dogs are pets but cats are little furry humans with four feet and self determination! Quote: Originally posted by Ritter  Quote: Originally posted by ScienceSquirrel If you want to make anhydrous diethylamine ( eg free from water or alcohol ) then you are getting into a whole new ballgame. Yes. The alkaline hydrolysis would allow you to distill over the amine but it would be wet with water.$17 would get you 5 lbs of anhydrous sodium sulfate on eBay item # 270268727018 .

Sodium sulphate will not dry the amine to any appreciable degree.
The only practical home method is to prepare the hydrochloride and then dry distill from solid potassium hydroxide.
It f**ks up the flask but it works!
Alternatively you can reflux it with loads of calcium hydride but that is like rocking horse sh*t to the average home chemist.

And I would echo Ritter's remark above.
You need a lot of knowledge and experience before you attempt this!

[Edited on 28-8-2008 by ScienceSquirrel]
mbrown3391
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i read that the amine was water soluble and remembered it as the amide. That doesn't mean i cant carry out a simple distillation at 55 degrees c. The dangers of this distillation aren't more than many other distillations.
ScienceSquirrel
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 Quote: Originally posted by mbrown3391 i read that the amine was water soluble and remembered it as the amide. That doesn't mean i cant carry out a simple distillation at 55 degrees c. The dangers of this distillation aren't more than many other distillations.

If you ignore the fact that you are distilling from a hot solution of potassium hydroxide in ethylene glycol.
It will remove your skin and give you a deep flesh burn in seconds if you spill it on yourself.
mbrown3391
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First of all, there is no reason to believe that experience in chemistry is proportional to ability not to spill something on yourself.

Second of all, if used in stoichiometrically correct proportions, will all the KOH not be neutralized by the m-Toluic acid?

[Edited on 28-8-2008 by mbrown3391]
ScienceSquirrel
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 Quote: Originally posted by mbrown3391 First of all, there is no reason to believe that experience in chemistry is proportional to ability not to spill something on yourself. Second of all, if used in stoichiometrically correct proportions, will all the KOH not be neutralized by the m-Toluic acid? [Edited on 28-8-2008 by mbrown3391]

You are going to have to use say two moles of potassium hydroxide for each mole of DEET.
This will ensure complete hydrolysis of the DEET which is your most expensive reagent and it will suppress protonation of the amine making it easier to distill it from the reaction mixture.
The waste caustic component left in the flask should be allowed to cool, treated with acid until neutral and disposed of appropriately.

[Edited on 28-8-2008 by ScienceSquirrel]
mbrown3391
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Good point. So are there any known, interesting uses of Sodium Phenylacetate, which will be the waste salt i believe.
ScienceSquirrel
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Mood: Dogs are pets but cats are little furry humans with four feet and self determination!

 Quote: Originally posted by mbrown3391 Good point. So are there any known, interesting uses of Sodium Phenylacetate, which will be the waste salt i believe.

Your waste salt will be potassium 3-methylbenzoate.
If you could get sodium phenylacetate out of potassium hydroxide and DEET you would be a miracle worker and a friend to every drug cook in the world.

This is the structure of DEET

http://en.wikipedia.org/wiki/DEET

This is 3-methylbenzoic acid

http://en.wikipedia.org/wiki/M-Toluic_acid

This is phenylacetic acid

http://en.wikipedia.org/wiki/Phenylacetic_acid

[Edited on 28-8-2008 by ScienceSquirrel]
mbrown3391
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1. Prepare a solution of 80g NaOH in 73ml H2O.
2. To this solution, add 192ml of DEET. Two layers will form.
3. While stirring, slowly add propylene glycol to the mixture until homogeneous.
4. Heat to 55 degrees C in a distillation apparatus and maintain this temperature until boiling ceases.
5. Theoretical Yield: 103.39ml Diethylamine

Any problems with this procedure?
stoichiometric_steve
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 Quote: Originally posted by mbrown3391 First of all, there is no reason to believe that experience in chemistry is proportional to ability not to spill something on yourself.

You bet there is. Even more so for people who cannot discern phenylacetates from benzoates...

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 Quote: Originally posted by mbrown3391 i read that the amine was water soluble and remembered it as the amide.

An example of not knowing what you are doing. If you can't visualize the difference between an amide and an amine or between your target product and your starting material you should not be attempting this.

 Quote: That doesn't mean i cant carry out a simple distillation at 55 degrees c. The dangers of this distillation aren't more than many other distillations.

This, of course, depends on your apparatus setup, what kind of glassware you'll be using, what kind of heating you will be using, and most especially your technique. Assuming for the moment that you will not be using a magnetic stirrer, how will you be preventing 'bumping' that could propel hot caustic glycol all over you and your work area?

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 Quote: Originally posted by mbrown3391 1. Prepare a solution of 80g NaOH in 73ml H2O. 2. To this solution, add 192ml of DEET. Two layers will form. 3. While stirring, slowly add propylene glycol to the mixture until homogeneous. 4. Heat to 55 degrees C in a distillation apparatus and maintain this temperature until boiling ceases. 5. Theoretical Yield: 103.39ml Diethylamine Any problems with this procedure?

Where did you get it from?
mbrown3391
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C12H17NO + H2O --> C4H11N + C7H7COOH
C7H7COOH + NaOH --> NaC8H7O2 + H2O

NaC8H7O2 is Sodium Phenylacetate according to http://www.encyclo.co.uk/define/NaC8H7O2
Ritter
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Quote:
Originally posted by ScienceSquirrel
 Quote: Originally posted by mbrown3391 1. Prepare a solution of 80g NaOH in 73ml H2O. 2. To this solution, add 192ml of DEET. Two layers will form. 3. While stirring, slowly add propylene glycol to the mixture until homogeneous. 4. Heat to 55 degrees C in a distillation apparatus and maintain this temperature until boiling ceases. 5. Theoretical Yield: 103.39ml Diethylamine Any problems with this procedure?

Sounds like an Internet recipe.

[Edited on 28-8-2008 by Ritter]

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mbrown3391
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i just wrote it based on the discussion in this thread.
Im assuming that NaOH can be used in place of KOH since i already have plenty of NaOH

[Edited on 28-8-2008 by mbrown3391]
Ritter
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 Quote: Originally posted by mbrown3391 C12H17NO + H2O --> C4H11N + C7H7COOH C7H7COOH + NaOH --> NaC8H7O2 + H2O NaC8H7O2 is Sodium Phenylacetate according to http://www.encyclo.co.uk/define/NaC8H7O2

What does this have to do with making diethylamine? I think you're lost.

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ScienceSquirrel
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 Quote: Originally posted by mbrown3391 C12H17NO + H2O --> C4H11N + C7H7COOH C7H7COOH + NaOH --> NaC8H7O2 + H2O NaC8H7O2 is Sodium Phenylacetate according to http://www.encyclo.co.uk/define/NaC8H7O2

Have a look at my post above!

3-Methylbenzoic acid and phenylacetic acid have the same molecular formulas but different structural formulas.
mbrown3391
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to be fair, you edited your post after i posted that reply
mbrown3391
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but thats irrelevant. Can anyone point out any mistakes with the procedure i posted?
ScienceSquirrel
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 Quote: Originally posted by mbrown3391 to be fair, you edited your post after i posted that reply

I edited the post to include the structures so that it was clearer for you.
mbrown3391
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Quote:
Originally posted by stoichiometric_steve
 Quote: Originally posted by mbrown3391 First of all, there is no reason to believe that experience in chemistry is proportional to ability not to spill something on yourself.

You bet there is. Even more so for people who cannot discern phenylacetates from benzoates...

Good one!
ScienceSquirrel
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 Quote: Originally posted by mbrown3391 but thats irrelevant. Can anyone point out any mistakes with the procedure i posted?

Where do I start;

1) Too big a scale, a reaction like this should be executed on a scale of say 10 - 50 millimoles to start with. You do not know how hot it is going to get or how quickly it is going to react.

3) A mixture like that will have to heated above 55 C to get a hydrophilic substance like diethylamine to distill out. I would guess that 80 - 100 C would be reasonable.

2) You have substituted sodium hydroxide for potassium hydroxide. It is a lot less soluble in alcohols, it may not matter but it is still a stupid change.

3) You have switched ethylene glycol for propylene gycol

4) Your product will be a low boiling, corrosive and irritant liquid.The apparatus should have a cooled receiver and a gas take off connected to a pipe that goes outside your workplace.

[Edited on 28-8-2008 by ScienceSquirrel]
mbrown3391
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1. Prepare a solution of 80g NaOH in 73ml H2O.
2. To this solution, add 192ml of DEET. Two layers will form.
3. While stirring, slowly add ethylene glycol to the mixture until homogeneous.
4. Add a small amount to a distillation apparatus and heat to 55 degrees C. If all goes well, add the rest and maintain 55 degrees C until boiling ceases
5. Theoretical Yield: 103.39ml Diethylamine

[Edited on 28-8-2008 by mbrown3391]
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Maximum-weight set of cliques of size 3 with no common vertices in undirected graph

I'm looking for an algorithm/insight into a problem that's an extension of the Maximum Weight Matching problem. The maximum weight matching problem looks for the max-weight set of edges that contain 0 common vertices. The problem can be thought of as grouping the vertices into sets of 2 such that no vertex belongs to more than 1 group.

I'm looking to extend the problem such that the vertices can be split into groups of 3. For an undirected graph, this translates into looking for cliques/cycles of size 3 that do not contain common vertices. Any guidance on how to approach this problem either deterministically or with an approximation/heuristic algorithm?

Note: the graphs I'm working with are typically dense

So, no efficient algorithm can be expected. For an exponential-time exact algorithm, I would probably try branch and bound intertwined with maximum bipartite (2-)matching, as follows: At each stage, we have a vertex $$v_i$$, which we will take to be the lowest-numbered vertex of a triangle, and the decision is which of its neighbour vertices $$v_j$$ should become the highest-numbered vertex in the triangle. For each possible $$v_j$$, to test feasibility of the choice, create a new, bipartite graph with a vertex for each low+high vertex pair decided so far, as well as a vertex for each as-yet unprocessed vertex, and an edge between two vertices whenever the low+high pair represented by one would form a triangle in the original graph with the other and the latter vertex is numbered between the low and high vertices represented by the former. Run maximum bipartite matching on this new graph, and check that every low+high pair decided so far is saturated (adjacent to a matched edge) -- if this fails, then the current assignment of low+high vertices cannot be extended to a full triangle partition and the branch can be pruned. I think this algorithm, which chooses just 2 vertices within each triangle, will result in B&B tree depths of around 2/3rds the fully naive approach in which all 3 vertices of each triangle are chosen.