Category Archives: Spring 2012

Derivation of Transmitted Wave Equation

(1)   \begin{equation*} \bigtriangledown \times \bigtriangledown \times \widetilde{E}_{n} + \frac{\epsilon^{(1)}(\omega_{n})}{c^{2}} \bullet \frac{\partial^2 \widetilde{E}_{n}}{\partial t^2} = \frac{-4\pi}{c^2} \frac{\partial^2 \widetilde{P}^{NL}_{n}}{\partial t^2} \end{equation*}

$\widetilde{P}^{NL}_{n} =$ Non-Linear part of Polarization Vector

$ \widetilde{E}_{n} =$ Electric Field vector

$\epsilon^{(1)}(\omega_{n}) =$ Frequency dependent dielectric Tensor

Equation (1) is derived from Maxwell’s equation and is the equation for waves in medium. It is valid for each frequency component of the field.

$\widetilde{E}_{2}(z,t) =A_{2}e^{i(k_{2}z-wt)},   \widetilde{P}_{j}(z,t) =P_{j}e^{-i\omega_{j}t},$    $P_{1} =4dA_{2}A^{*}_{1}e^{i(k_{2}-k_{1})z}, P_{2} =2dA^{2}_{1}e^{i2k_{1}z}$

$\widetilde{E}_{2}(z,t)$ will be my equation for the transmitted wave at frequency  propagating in the z direction,$ \widetilde{P}_{2}(z,t)$ the nonlinear source term and $P_{2}, P_{1}$  the amplitude of the nonlinear polarization and amplitude of incident beam respectively.

(2)   \begin{equation*} -\bigtriangledown^{2} \widetilde{E}_{n}-\frac{\omega^{2}_{n}}{c^{2}}\epsilon^{(1)}(\omega_{n}) \bullet \widetilde{E}_{n}= \frac{4\pi \omega^{2}_{n}}{c^2} \widetilde{P}^{NL}_{n}} \end{equation*}

$\bigtriangledown^{2}$ can be replaced with $ \frac{\partial^2}{\partial z^{2} }$

(3)   \begin{equation*} \frac{\partial^2}{\partial z^{2}}\widetilde{E}_{2}= \frac{\partial}{\partial z} [ \frac{\partial A_{2}}{\partial z}e^{i(k_{2}z-wt)} +ik_{2}A_{2}e^{i(k_{2}z-wt)}] \end{equation*}

(4)   \begin{equation*} \frac{\partial^2}{\partial z^{2}}\widetilde{E}_{2}= [ \frac{\partial^{2} A_{2}}{\partial z^{2}} +ik_{2}\frac{\partial A_{2}}{\partial z}+ ik_{2} \frac{\partial A_{2}}{\partial z} -k^{2}_{2}A_{2}]e^{i(k_{2}z-wt)} \end{equation*}

So the wave equation becomes

(5)   \begin{equation*} -[ \frac{\partial^{2} A_{2}}{\partial z^{2}} + 2ik_{2} \frac{\partial A_{2}}{\partial z}-k^{2}_{2}A_{2}]e^{i(k_{2}z-wt)} -\frac{\omega^{2}_{2}}{c^{2}}\epsilon^{(1)}(\omega_{2})A_{2}e^{i(k_{2}z-wt)} =\frac{4\pi \omega^{2}_{n}}{c^2}\bullet2dA^{2}_{1}e^{i(2k_{1}z-\omega_{2}t)} \end{equation*}

(6)   \begin{equation*} -[ \frac{\partial^{2} A_{2}}{\partial z^{2}} + 2ik_{2} \frac{\partial A_{2}}{\partial z}-k^{2}_{2}A_{2} -\frac{\omega^{2}_{2}}{c^{2}}\epsilon^{(1)}(\omega_{2})A_{2}]e^{i(k_{2}z-wt)} =\frac{8\pi \omega^{2}_{2}}{c^2}dA^{2}_{1}e^{i(2k_{1}z-\omega_{2}t)} \end{equation*}

$k^{2}_{2}= \frac{\omega^{2}_{2}}{c^{2}}\epsilon^{(1)}(\omega_{2})$ hence

(7)   \begin{equation*} [ \frac{\partial^{2} A_{2}}{\partial z^{2}} + 2ik_{2} \frac{\partial A_{2}}{\partial z}]e^{i(k_{2}z-wt)} =-\frac{8\pi \omega^{2}_{2}}{c^2}dA^{2}_{1}e^{i(2k_{1}z-\omega_{2}t)} \end{equation*}

replace $ \frac{\partial}{\partial}$ with $\frac{d}{dz}$ because $A_{2}$ is only a function of $z$ and with the slowly-varying-amplitude approximation $|\frac{d^{2}A_{2}}{dz^{2}}|<<|k_{2}\frac{aA_{2}}{dz}|$ hence we get

(8)   \begin{equation*} \frac{dA_{2}}{dz}= \frac{4 \pi id \omega^{2}_{2}}{k_{2}c^{2}}A^{2}_{1}e^{i\bigtriangleup kz} \end{equation*}

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The Finished Structure of 3,3-dimethyl-2-butanol

Now that we’ve analyzed both the ^{1}H-NMR and the ^{13}C-NMR spectra, we can see how the results come together and give the structure of 3,3-dimethyl-2-butanol.  Below we bring back the two figures that show the labeled hydrogens and carbons of the molecule:

If we did not already know the identity of the molecule, we could have determined it using only the four spectra previously shown–the ^{1}H-NMR, the ^{13}C-NMR, the DEPT-90, and the DEPT-135–by following the guidelines laid out in the previous posts.  Peaks on the four spectra work together to give  bits of information about  pieces of the molecule, and those pieces are put back together at the end of the analysis.

To summarize, we began by taking advantage of the magnetic properties of ^{1}H and ^{13}C nuclei to learn about the environment of each nucleus in the molecule.  Radio waves were pulsed at the sample to record small changes in the resonant Larmor frequency of the nuclei.  The change happened because electrons around each nucleus became a current when influenced by the NMR’s external field, and generated their own fields in the opposite direction according to Lenz’s Law.  This new, effective magnetic field resulted in changes in the Larmor frequency of each nucleus.  It is these changes that are displayed on the spectra as chemical shifts.  The chemical shift is measured in ppm of the original Larmor frequency of the nucleus.

Using the Larmor frequency equation below, and the shifted frequencies, we calculated the shielding factor \sigma for each nucleus.  The shielding factor is another measure of the relative electron density around each nucleus.

(1)   \begin{equation*} \nu_{0}=\frac{\gamma B_{0}(1-\sigma)}{2\pi} \end{equation*}

Combining the information from the spectra gives the structure of 3,3-dimethyl-2-butanol, which is shown again below.

It is the ability of the NMR spectrometer to detect such small changes in resonant frequency, and convert them into data we can use to piece together molecular structures, that makes NMR spectroscopy so powerful and important in the research world.
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Interpreting a C-13 NMR spectrum

In Vassar’s 300 MHz NMR, the Larmor frequency of an unaltered hydrogen nucleus with shielding factor \sigma=0 is 300 MHz.  ^{13}C nuclei are different than ^{1}H nuclei, and have a different Larmor frequency in the spectrometer’s 7.046 Tmagnetic field.  According to Jacobsen in “NMR Spectroscopy Explained,” the Larmor frequency of unaltered ^{13}C nuclei with \sigma=0 is 75.43 MHz in this magnetic field.  The magnetogyric ratio \gamma, is 672.650*10^{5}\frac{rad}{Ts}  We bring back the equation for Larmor Frequency:

(1)   \begin{equation*} \nu_{0}=\frac{\gamma B_{0}(1-\sigma)}{2\pi} \end{equation*}

You might ask why ^{13}C nuclei are the focus of carbon NMR instead of the much more common and ordinary ^{12}C nuclei.  It comes down to the composition of the nucleus.  Only nuclei with nonzero spin are magnetically active.   Spin is a type of angular momentum intrinsic to subatomic particles.  Particles with nonzero spin can have magnetic moments that can be influenced by a magnetic field.  Any nucleus with an even number of protons and an even number of neutrons will not be magnetically active because its spin is 0 and it has no magnetic moment.  Carbon-12 has 6 protons and 6 neutrons, so it can’t be studied with NMR.  Carbon-13 has 6 protons and 7 neutrons, so we can study it with NMR.  The structure and ^{13}C-NMR spectrum of 3,3-dimethyl-2-butanol is shown below.

At first glance, the spectrum looks almost the same as the H-NMR spectrum.  However, there is little indication of how many carbons a single peak is referring to.  Each peak corresponds to a type of carbon in the molecule.  The triplet peak at 77 ppm represents the deuterated chloroform, CDCl_{3}, used to dissolve the sample.  Excluding that triplet, the molecule has four different kinds of carbon.  To help us find what carbons are represented here, we turn to the other two spectra of interest.  They are called the DEPT-90, and the DEPT 135.  DEPT stands for Distortionless Enhancement by Polarization Transfer.  The technique hits the sample with five successive radio pulses designed to excite either hydrogen or ^{13}C nuclei.  The sequence of pulses is:

  • A hydrogen pulse at 90° from the x-axis
  • A hydrogen pulse at 180° from the x-axis  plus a carbon pulse at 90° from the x-axis.
  • A carbon pulse at 180° from the y-axis plus a hydrogen pulse at either 45°, 90°, or 135° from the y-axis.

The angle of the last hydrogen pulse is the number in the DEPT label.  DEPT-90 ends with a hydrogen pulse at 90° from the y-axis.  The pulse sequence is modeled below for DEPT-90 as an example.  Red waves are hydrogen pulses, green waves are carbon pulses:

DEPT-90 Pulse 1

DEPT-90 Pulse 2

DEPT-90 Pulse 3

These combinations of pulses result in transfers of polarization between the hydrogen and carbon nuclei.  Different combinations pick out carbons with one, two, or three hydrogens attached to them.  The DEPT-90 and DEPT-135 spectra are shown below:

DEPT-90 spectrum

DEPT-135 Spectrum

The chemical shifts of the peaks on these two spectra are the same as they are on the ^{13}C-NMR spectrum.  The sizes of the peaks on the two spectra are different, and this size difference is the key to understanding them.  You may have noticed the peak at 35 ppm on the ^{13}C-NMR spectrum doesn’t appear on either of the DEPT spectra.  This is because DEPT spectra only show carbons with hydrogens attached to them.  Therefore the 35 ppm peak represents the carbon in 3,3-dimethyl-2-butanol with no hydrogens on it.

The three peaks on the DEPT spectra are distinguished by their relative directions and sizes on each spectrum.

  • Peaks that point up on the DEPT-135, point up on the DEPT-90, and are larger on the DEPT-90, represent carbons with one hydrogen attached.
  • Peaks that point down on the DEPT-135 represent carbons with two hydrogens attached.
  • Peaks that point up on the DEPT-135, point up on the DEPT-90, and are smaller on the DEPT-90 represent carbons with three hydrogens attached.
  • Carbons without hydrogens don’t appear on either the DEPT-135 or DEPT-90

These are not the most general rules for interpreting DEPT-90 and DEPT-135 spectra.  Normally, carbons with two and three hydrogens don’t appear on the DEPT-90 spectra.  However, in this experiment the radio pulse used to collect the data was too long.  This causes the DEPT-90 spectrum to contain a small fraction of the DEPT-135 spectrum’s information.

Armed with this new information from the DEPT spectra, we can determine which carbons are represented by which peaks on the ^{13}C-NMR spectrum, and calculate their shielding factors.  A table of chemical shift and shielding factor values for each ^{13}C nucleus in 3,3-dimethyl-2-butanol is below.

 

The same trend that applies to hydrogen nuclei applies to ^{13}C nuclei.  The more negative the shielding constant, the lower the electron density around the nucleus, and the higher the effective magnetic field felt by the nucleus.  ^{13}C nuclei also experience higher chemical shifts compared to hydrogen nuclei because they are bonded to more atoms and have more opportunity to be deshielded when those atoms take electron density.

The figures below shows the structure of 3,3-dimethyl-2-butanol with the carbons labeled with their corresponding chemical shifts.  This is the final step in reconstructing the molecule from the NMR data.

 

 

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Brief Description of Liquid Crystal

Liquid Crystal is a state of matter between liquid and crystalline solid in which LC molecules align in an ordered array (Yeh 5). LC molecules are anisotropic to light, meaning that they exhibit properties with different effects when oriented in varied directions (Yeh 5).

These are very rough artistic interpretations of the nematic phase alignment that I made in Mathematica as inspired by Fig. 1.9 on pg. 12 of Liquid Crystal Display Drivers: Techniques and Circuits.

We will focus on the nematic phase (see Figure) of liquid crystal, as when aligned, the optical properties of the nematic phase are helpful in liquid crystal displays. Nematic phases are made up of rod-shaped molecules that are ordered with their long axes (the ones used with LCD are uniaxial) approximately parallel (Cristaldi 12).

References:

Cristaldi, David, Pennisi, Salvatorre and Francesco Pulvirenti. Liquid Crystal Display Drivers: Techniques and Circuits. Springer Science+Business Media B.V, 2009.

Yeh, Pochi and Claire Gu. Optics of Liquid Crystal Displays. Canada: John Wiley & Sons, Inc., 1999.

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Inside the EoM

Inside an Electro optic modulator is a birefringent crystal. One of the more popular crystals to be used in EoMs are lithium niobate (LiNb0_{3}) doped with magnesium oxide (MgO). The magnesium oxide is there to prevent optical damage that can occur with high energy light in the bluish green frequencies. The Index of refraction of this crystal is n_{o} = 2.29  and the linear electro optic coefficient r = 30.9. I could not find any information on R the quadratic electro optic coefficient. Inside the EoM, the crystal has a parallel plate capacitor across it as shown below.

This capacitor allows for a voltage controlled uniform electric field. There is a well-known simple relationship

(1)   \begin{equation*} E=\frac{V}{d} \end{equation*}

Where V is the voltage d is the distance between the plates and E is the electric field strength which points from the positive to negative plate. This gives us a direct relationship between the electric field and the applied voltage. This can be used in determining required voltage to obtain specific types of modulations.

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Amplitude Modulation

Amplitude modulation is very similar to phase modulation. In all previous types of modulation we split the beam up into two different oppositely polarized beams only to use one in the modulation process. In the case of amplitude modulation we still only send one beam through the EoM, but we are still in need of the second beam.  The set up for amplitude modulation can be seen below.

In this set up there is a half wave plate to change the direction of the polarization of the laser beam entering the EoM to be identical to the one not going into the EoM. The EoM is used to modulate the phase of the beam exiting it. There are two polarizing beam splitter cubes that are used as one-way mirrors so the two beams can be joined back together. Being able to change the phase of one of the beams allows you to control how much constructive interference and how much destructive interference occurs when the beams add together as shown in the animation below.

The blue wave is the laser beam that is not going through the EoM, the red wave is the beam that is having its phase modulated from 0 to 2\pi as it passes through the EoM. The yellow wave is resultant amplitude modulation.

As you can see from the animation you are able to control the amplitude of the laser beam any where from 0 to twice the original amplitude.

 

 

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A few observations

Having been stymied by the constraints that mathematica imposes on integrals of absolute value functions, I have forgone my attempt to create a smooth plot and moved on to obtaining discrete values and using them to create my plot.  Plotting the difference in rotation frequency between rotor and stator vs the the total induced emf produced by a single rotation, I have made a surprising discovery.  In my model, it does not seem to matter what I make the rotation rate: the emf induced remains constant.  The plot below gives values 1-10 for $\Delta \omega$ with corresponding values for $\epsilon_{induced}$.

Another oddity that comes to our notice is that our expression for $n$ plays exactly the same role in this equation as $\Delta \omega$.  That is to say, increasing the number of wire coils achieves the same objective as increasing the rotation rate of the rotor.

This is very interesting indeed, and quite handy as well.  A problem with some wind turbine designs (fixed-pitch, for example) is that they function suboptimally at lower rotation rates.  This causes a certain amount of inherent inefficiency in the generator (or rather adds to the inefficiency already present).  My model, while far from perfect, suggests that this might be ameliorated by increasing the number of wire loops in the stator.  As this inefficiency is largely a problem in cheaper turbine designs, it is my hope that this technique (being relatively inexpensive) may offer a cost-effective way to improve efficiency.

 

As in the last post, the mathematica code for the above can be found in

https://vspace.vassar.edu/xythoswfs/webview/fileManager.action?entryName=/jamcentire&stk=A6745A2F45FA75A&msgStatus=3%20documents%20have%20been%20successfully%20uploaded%20to%20the%20folder%2C%20jamcentire.

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Interpreting a H-NMR Spectrum

The NMR Spectrometer at Vassar college is graded at 300 MHz.  This means the Larmor frequency of a single, unaltered hydrogen nucleus and its electron with \sigma=0, is 300 MHz.  The magnetogyric ratio \gamma of a hydrogen nucleus is 267.513*10^{6}\frac{rad}{T•s} .  A quick calculation using:

(1)   \begin{equation*} \nu_{0}=\frac{\gamma B_{0}(1-\sigma)}{2\pi} \end{equation*}

shows the magnetic field strength of the NMR is roughly 7.046 T.

Using this information and the ^{1}H-NMR spectrum, we can calculate shielding factors for each type of hydrogen nucleus in 3,3-dimethyl-2-butanol, and determine what each spectrum peak actually means.

Let’s start with the peak farthest to the right on the spectrum.  It is a single peak, called a singlet, that represents nine hydrogens, and is centered at approximately 0.9 ppm on the x-axis.  The ppm scale measures how much the Larmor frequency of hydrogen is changed by the effective magnetic field.  The Larmor frequency of the hydrogens represented by the first peak was increased by 0.9 ppm of the initial 300 MHz frequency.  A calculation of these hydrogens’ shielding factor follows using equation (1):

300 MHz+(\frac{0.9 ppm}{10^{6}}*300 MHz)=\frac{267.513*10^{6}\frac{rad}{T•s}*7.046T*(1-\sigma)}{2\pi}

300000270 Hz=299990611 Hz(1-\sigma)

\sigma=-3.2198*10^{-5}

Shielding factors tend to be small for most hydrogen nuclei.  A table of chemical shift and shielding factor values for each hydrogen nucleus in 3,3-dimethyl-2-butanol is below.

A more negative shielding factor corresponds with a lower electron density around the hydrogen nucleus and with a larger effective magnetic field influencing the nucleus.  The presence of very electronegative atoms, like oxygen, near the hydrogen causes increased chemical shifts like the 3.5 ppm shift in the table.

One peak in the spectrum above is split into two peaks centered around the chemical shift  1.1 ppm.  This splitting occurs because there is another magnetically active hydrogen nucleus nearby in the molecule.  The rule for split peaks is: the number of nearby hydrogens is given by n-1, where n is the number of peaks.  The definition of “nearby” is usually 1 carbon atom over in the molecule from the one the original hydrogen is attached to.

The figure below shows the structure of 3,3-dimethyl-2-butanol with the hydrogens labeled with their corresponding chemical shifts.  This will be the first step in reconstructing the molecule from the NMR data.

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Static Light Scattering-Results

Origin Graph Depicting Scattering Intensity vs. Concentration

Using the theory of Static Light Scattering a graph was made for particle size 105 nm (in diameter) for volumes 1000, 2000, 3000, 4000 microliters and the predicted scattering intensity was found using Equation [3].  The wavelength used was 632.8 nanometers (red in the visible spectrum) so the particle size was within the Rayleigh limit.  The scattering intensity was then plotted as the concentration of the particles was varied.

1/y-intercept was supposed to yield the molecular mass of a particle in Daltons.                    (1 Dalton=1amu).  A linear fit was applied to the graph in order to find the y-intercept.  As shown on the graph the y-intercept was found to be  $2.3*{10^{-12}}$

which yields a molecular mass of approximately  $4.5*{10^{11}} $ Daltons

In order to calculate the actual molecular mass of the microspheres the 2SPI website was used.  Here’s the link: http://www.2spi.com/catalog/standards/microspheres.shtml

The approximate concentration listed on the website of the 0.105 micrometer spheres was ${10^{3}}$ particles/mL

The bead density was approximately $1.05$ grams/mL

which meant that each particle was approximately $1.05*{10^{-16}}$ kg

In order to convert between kilograms and Daltons it is important to note that $1.022*{10^{-27}}$ equals approximately 1 Dalton.

This yielded a molecular mass of $1.1*{10^{11}}$ Daltons

While the calculated molecular mass and the actual molecular mass are on the same scale (both ${10^{11}}$) the answer is still off by a factor of about 3 to 4.

This error may stem from the many approximations used when calculating the actual molecular mass.  The number of particles/mL was estimated on the SPI website.  Also the bead density was approximate.  Since these polystyrene microspheres are mass produced it cannot be guaranteed that every particle is the exact same size and a perfect spherical shape.  There will also be slight variations in the particle densities across different containers.  However the particles need to meet certain standards, so the error is limited, but could possibly account for the small differences between the actual and calculated molecular mass.

In addition, since the slope was found to be $1.057*{10^{-9}}$ (which is greater than zero) it would appear, based on Static Light Scattering Theory, that the particles will stay in a stable solution.  This information was confirmed on the 2SPI MSDS data and safety sheet for these microspheres (the sheet claims the solution formed with these sphere is stable).

Just for reference it seems like the method I used for calculating the molecular mass based on Static Light Scattering Theory is not the easiest, quickest, nor most accurate method.  Equipment can be purchased that measures the intensity of the samples at different concentrations, compares it to a known standard, and outputs a Debye Plot which calculates the molecular mass and the gradient of the graph.  However, it can be important to test theories in a more controlled method where every step of the process is known in order to check the accuracy of the theory and learn exactly how the equipment works.

 

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Light Scattering-Results

For my project I also wanted to look at the Rayleigh predicted scattering intensity for a particle size larger than the Rayleigh limit and compare that to the Rayleigh-Debye predicted scattering intensity.

Polar plot depicting the Rayleigh Relative Scattering Intensity for scattering angles from 0 to 360 degrees for Particle Sizes larger than the Rayleigh limit

The wavelength used was 400 nm or violet light in the visible spectrum since this is the shortest wavelength in the visible spectrum and as a result has the smallest Rayleigh limit.  The particle sizes used were 50, 60, 70, 80, 90, 100 nanometers in diameter.

Polar Plot depicting the Rayleigh-Debye Relative Scattering Intensity for scattering angles from 0 to 360 degrees.

The same particle sizes and wavelengths were used as in the Rayleigh scattering graph above.

As you can seen in the graphs above the Rayleigh predicted scattering intensity is the same for forward and back scattering.  While the Rayleigh-Debye predicted scattering intensity occurs mainly in the forward direction.  Larger particles (outside the Rayleigh limit) should scatter more light in the forward direction, as in the Rayleigh-Debye graph, which shows that the form factor introduced in Equation [5] is necessary to predict the scattering intensity of larger particles.

Here is a link to show the differences between Rayleigh Scattering and Mie Scattering as particles become larger in size. http://www.mwit.ac.th/~Physicslab/hbase/atmos/imgatm/mie.gif

The similarities between the image of Mie Scattering depicted in this link and the Rayleigh-Debye scattering shown in the graph above suggest that the Rayleigh-Debye equation is both an accurate and simpler method for calculating the scattering for particles larger than the Rayleigh limit.

Below is a link to all the Mathematica files used to create the preliminary results, these results, and find the concentration and scattering intensity used in the testing of the Static Light Scattering Theory.

https://vspace.vassar.edu/xythoswfs/webview/fileManager?stk=60D409B61FE706E&entryName=%2Freeells%2FScattering&msgStatus=

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