Polarisation through a chiral nematic crystal

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SUMMARY

This discussion focuses on the behavior of light as it passes through a chiral nematic crystal, specifically analyzing the electric field expression, $$\vec{E}(z,t)$$, under varying fast and slow directions. The participants explore the implications of splitting the crystal into slices to derive the electric field at a specific height, $$z=w$$, and consider the use of Bloch waves and Jones calculus for a more comprehensive understanding. The conversation highlights the complexity of internal reflections and the periodic nature of the crystal structure, defined by the pitch $$2\pi/\mu$$.

PREREQUISITES
  • Birefringent materials and their properties
  • Understanding of electric field representations in optics
  • Familiarity with Bloch waves in solid-state physics
  • Basic knowledge of Jones calculus for light transmission analysis
NEXT STEPS
  • Study the derivation of electric field expressions in birefringent materials
  • Learn about Bloch waves and their application in optical systems
  • Explore Jones calculus and its use in analyzing light transmission through layered media
  • Investigate the relationship between pitch and optical properties in chiral nematic crystals
USEFUL FOR

Optical physicists, materials scientists, and engineers working with liquid crystal displays or studying the properties of birefringent materials.

etotheipi
If light passes into a birefringent material with constant fast and slow directions, ##\hat{x}## and ##\hat{y}##, that are oriented the same way at any point in the crystal, then the electric field is$$\vec{E}(z,t) = E_0\hat{x}\cos{(\theta_0)}e^{i\omega(t - \frac{n_x}{c}z)} + E_0\hat{y}\cos{(\theta_0)}e^{i\omega(t - \frac{n_y}{c}z)}$$if the electric field is initially oriented at ##\theta_0## to the ##\hat{x}## axis before it enters the crystal. For a chiral nematic crystal, the fast and slow directions are functions of position, i.e. ##\hat{x} = \cos{(\mu z)} \hat{x}_{0} + \sin{(\mu z)} \hat{y}_0## and ##\hat{y} = -\sin{(\mu z)} \hat{x}_{0} + \cos{(\mu z)} \hat{y}_0##, where ##\mu## is just a constant. I want to find an expression for ##\vec{E}(z=w)##, where ##w## is the length of the crystal.

I'm not really sure what the best way to go about this is. I thought to try and split the crystal into ##N## slices of width ##w/N##, with lower faces at heights ##z_n = nw/N, n \in [0,N-1]##, and ##\hat{x}_n = \cos{(\mu z_n)} \hat{x}_{0} + \sin{(\mu z_n)} \hat{y}_0## and ##\hat{y}_n = -\sin{(\mu z_n)} \hat{x}_{0} + \cos{(\mu z_n)} \hat{y}_0##. Hence, the electric field at height ##z_1## would be$$\begin{align*}

\vec{E}(z = z_1,t) &= E_0\hat{x}_0\cos{(\theta_0)}e^{i\omega(t - \frac{n_x w}{cN})} + E_0\hat{y}_0\cos{(\theta_0)}e^{i\omega(t - \frac{n_y w}{cN})} \\

&= E_0(\cos{\mu z_1} \hat{x}_1 - \sin{\mu z_1} \hat{y}_1)\cos{(\theta_0)}e^{i\omega(t - \frac{n_x w}{cN})} + E_0(\sin{\mu z_1} \hat{x}_1 + \cos{\mu z_1} \hat{y}_1)\cos{(\theta_0)}e^{i\omega(t - \frac{n_y w}{cN})} \\

&= \left[ E_0 \cos{\mu z_1} \cos{\theta_0} e^{i\omega(t - \frac{n_x w}{cN})} + E_0 \sin{\mu z_1}\cos{\theta_0} e^{i\omega(t - \frac{n_y w}{cN})} \right] \hat{x}_1 +

\left[ E_0 \cos{\mu z_1} \cos{\theta_0} e^{i\omega(t - \frac{n_y w}{cN})} - E_0 \sin{\mu z_1}\cos{\theta_0} e^{i\omega(t - \frac{n_x w}{cN})} \right] \hat{y}_1
\end{align*}
$$and then between ##z_1## and ##z_2##, these two components will propagate at two different speeds along the ##\hat{x}_1## and ##\hat{y}_1## directions, so we need to write another equation for this layer, and so on. I really don't know if I'm going to get anything useful out of this, so I wondered if someone could let me know if there is a better approach? Thanks
 
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Perhaps I still have sleepies in my eyes but can't you simplify this expression greatly by pulling all the time dependence out front and writing the the ##cos(\mu z)## as ##etothei\mu z+ etotheminusi\mu z ## and similar?
 
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etotheipi said:
If light passes into a birefringent material with constant fast and slow directions, ##\hat{x}## and ##\hat{y}##, that are oriented the same way at any point in the crystal, then the electric field is$$\vec{E}(z,t) = E_0\hat{x}\cos{(\theta_0)}e^{i\omega(t - \frac{n_x}{c}z)} + E_0\hat{y}\cos{(\theta_0)}e^{i\omega(t - \frac{n_y}{c}z)}$$if the electric field is initially oriented at ##\theta_0## to the ##\hat{x}## axis before it enters the crystal. For a chiral nematic crystal, the fast and slow directions are functions of position, i.e. ##\hat{x} = \cos{(\mu z)} \hat{x}_{0} + \sin{(\mu z)} \hat{y}_0## and ##\hat{y} = -\sin{(\mu z)} \hat{x}_{0} + \cos{(\mu z)} \hat{y}_0##, where ##\mu## is just a constant. I want to find an expression for ##\vec{E}(z=w)##, where ##w## is the length of the crystal.

I'm not really sure what the best way to go about this is. I thought to try and split the crystal into ##N## slices of width ##w/N##, with lower faces at heights ##z_n = nw/N, n \in [0,N-1]##, and ##\hat{x}_n = \cos{(\mu z_n)} \hat{x}_{0} + \sin{(\mu z_n)} \hat{y}_0## and ##\hat{y}_n = -\sin{(\mu z_n)} \hat{x}_{0} + \cos{(\mu z_n)} \hat{y}_0##. Hence, the electric field at height ##z_1## would be$$\begin{align*}

\vec{E}(z = z_1,t) &= E_0\hat{x}_0\cos{(\theta_0)}e^{i\omega(t - \frac{n_x w}{cN})} + E_0\hat{y}_0\cos{(\theta_0)}e^{i\omega(t - \frac{n_y w}{cN})} \\

&= E_0(\cos{\mu z_1} \hat{x}_1 - \sin{\mu z_1} \hat{y}_1)\cos{(\theta_0)}e^{i\omega(t - \frac{n_x w}{cN})} + E_0(\sin{\mu z_1} \hat{x}_1 + \cos{\mu z_1} \hat{y}_1)\cos{(\theta_0)}e^{i\omega(t - \frac{n_y w}{cN})} \\

&= \left[ E_0 \cos{\mu z_1} \cos{\theta_0} e^{i\omega(t - \frac{n_x w}{cN})} + E_0 \sin{\mu z_1}\cos{\theta_0} e^{i\omega(t - \frac{n_y w}{cN})} \right] \hat{x}_1 +

\left[ E_0 \cos{\mu z_1} \cos{\theta_0} e^{i\omega(t - \frac{n_y w}{cN})} - E_0 \sin{\mu z_1}\cos{\theta_0} e^{i\omega(t - \frac{n_x w}{cN})} \right] \hat{y}_1
\end{align*}
$$and then between ##z_1## and ##z_2##, these two components will propagate at two different speeds along the ##\hat{x}_1## and ##\hat{y}_1## directions, so we need to write another equation for this layer, and so on. I really don't know if I'm going to get anything useful out of this, so I wondered if someone could let me know if there is a better approach? Thanks
Pardon my slightly "flip" answer earlier but I could not resist and it had some germ of truth. Of course the layer approach might work but you need to keep track of all the internal reflections backwards and forwards at the boundaries. I think it very difficult.
I note that you are imposing a periodic ##length=2\pi/\mu## condition with the chirality and I think something more like Bloch waves may be more reasonable.
Did any particular problem generate this line of investigation, or were you just thinking again (it will lead to no good)?
 
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hutchphd said:
Pardon my slightly "flip" answer earlier but I could not resist and it had some germ of truth. Of course the layer approach might work but you need to keep track of all the internal reflections backwards and forwards at the boundaries. I think it very difficult.
I note that you are imposing a periodic ##length=2\pi/\mu## condition with the chirality and I think something more like Bloch waves may be more reasonable.
Did any particular problem generate this line of investigation, or were you just thinking again (it will lead to no good)?

lol, no worries! It made me smile. Anyway, yeah, my ##2\pi / \mu## is the 'pitch' of the crystal. I wonder, how would one go about using Bloch waves in this context, I've only come across them in the context of electron wave-functions in lattices?

It's not particularly important that I get the solution, I was more just playing around with it and trying to understand how a liquid crystal display works a bit more quantitatively :wink:
 
OK so I took a look at wikipedia and now realize I don't understand what the pitch has to do with it ( or if it does.) You do understand the fundamental rotation angle of the plane of polarization because of "normal" birefringence? (it goes like ##(k_{ordinary}-k_{extraordinary})z##. After that I am unsure...
 
This is roughly the type of arrangement I'm thinking of, my ##z## axis lies horizontally:

1604156894726.png


The director, and thus the fast and slow axes, rotate with increasing ##z## through the crystal, so the distance ##\Delta z## for one complete rotation in the orientation of the lattice planes is ##2\pi / \mu##, where I defined ##\mu## to be the equivalent of my wavenumber, i.e. ##2\pi / \mu## is the distance after which the structure repeats itself, one pitch.
 
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etotheipi said:
<snip> I wondered if someone could let me know if there is a better approach? Thanks

I think (but I'm not sure) this problem has been worked out already using the Jones calculus.

https://www.osapublishing.org/josaa/abstract.cfm?uri=josaa-10-5-966

Abstract: We derive an extended Jones matrix method to treat the transmission of light through birefringent networks, where the incident angle of light and the optical axis of the birefringent media are arbitrary. As an example, we employ the method to analyze the leakage problem of a twisted nematic liquid-crystal display and to suggest its possible solutions. A generalization of the method covers all dielectric media, including uniaxial and biaxial crystals and also gyrotropic materials that exhibit optical rotation and Faraday rotation.
 
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That's awesome! I'm a little tired to work through the document this evening, but I'll try and do that tomorrow morning. When I skimmed it just now, it looks like the model of the crystal they're using is not too different, i.e. they're also considering stacked layers:
"LCD medium divided into N layers. Each layer can be considered as a uniaxial medium. The orientation of the c axis may change from layer to layer (e.g., a TNLCD). The ordinary and extraordinary indices of refraction, ##n_0## and ##n_e## respectively, are constants for all layer"
I have no experience with the Jones calculus, so maybe I'll need to read up on that as well before I can attempt the paper. Thanks for letting me know if this!
 
When you figure it out you could post your digested version (if it is digestible) of the physics! I too will have a peek...but we shall see how far that goes.
 
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Sure, yeah, I'll post status updates. I should have all of my 'actual' work done by mid-Thursday so I'm pretty sure I can throw something together by the end of the week :smile:
 

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