Thank you Adaobi for specified a fantastic suggestion. All credit to her for the idea, we are motionless workin’ on it!
My brilliant companion Adaobi Adibe sent me a document on making biologic tissues transparent. In “Achieving optical transparency in live animals alongside engaging molecules”, the authors create biologic tissues, akin poultry breast, transparent by soaking them in a resolution of tartrazine. Tartrazine is the extremely brilliant yellowish-orange food dye that is established on Doritos.
We aim to do a additional aspiring replication imminently. But as a heated up, I tried to replicate parts of this protocol by soaking poultry chest in tartrazine solution. I got partial transparency but not to the flat of the paper. The example below is an about 2mm thick raw poultry chest cut from a 5mm piece that was soaked in ~0.19 M tartrazine resolution for 6 hours at area temperature. I took an depiction of this alongside a array light. You can see the grid lines showing through from underneath. The chicken, equal although the texture to the contact is the same, has a ocular norm akin that of gummy bears.
The first document used 0.62 M, 40 °C and a device shaker, but I couldn't discover a way to do this safely, nor did I have the equipment to do so. Hence the altered protocol.
The depiction below is a distinct viewpoint of my sample. The example on the remaining is the identical 2mm thick poultry chest piece as above. And the example on the correct is a authority anywhere I soaked it in h2o (it is slightly yellow since this procedure is excellent messy and I got powder on it).
In the top part of the example we do see several transparency, alongside the 5mm x 5mm grid showing through. The ocular norm of the poultry chest takes on a gummy-bear akin quality. Kinda makes me wanna chew it if it wouldn't provision me Salmonella.
I don't wanna get copyrighted by Doritos so current is a animation of anywhere you can discover tartrazine in the wild.
How does this work?
The current example we have is that the absorbance properties of the tartrazine additionally change the refractive indicator of water. By pushing the refractive indicator of h2o to that of macromolecule and fat solution, we create the entity additional transparent. I’ll explain this in awesome item later.
Tartrazine was from APC Pure. It was >80% dye content, which method several doubt complete what the true molar concentration volition be. I afterward prepared three concentrations of tartrazine at these molar concentrations, one of which was fair a 0 concentration as a control.
Tartrazine: 0 g, Tap water: 20 g, w/w: 0%, Molality (mol/kg water): 0, Tartrazine (mmol): 0, Molarity (approx.): 0 M (control)
Tartrazine: 2 g, Tap water: 18 g, w/w: 10%, Molality (mol/kg water): 0.21, Tartrazine (mmol): 3.7, Molarity (approx.): ~0.19 M
Tartrazine: 6 g, Tap water: 14 g, w/w: 30%, Molality (mol/kg water): 0.80, Tartrazine (mmol): 11.2, Molarity (approx.): ~0.56 M
Tartrazine has M = 534.4 g/mol. Moles use the nominal bulk (the powder is >80% dye, so true values are up to ~15% lower). Molarity assumes a final resolution quantity of ~20 mL.
This is what the tartrazine looked like. It doesn't have an odour, and it seems to be considered food harmless by the FDA, so I didn't annoy wearing gloves for this bit. Here is a image of the tartrazine and the specs.
Product: Tartrazine E102 FG, C.I. 19140
Dye content: >80%
Molecular formula: C16H9N4Na3O9S2
Molecular weight: 534.37 g/mol
λmax (in water): 425 nm
Water-insoluble matter: <0.20%
Subsidiary dyes: <1.00%
Heavy metals (as Pb): <40 mg/kg
To prepared the solutions I used a jeweller's measure to measure out the correct amounts and afterward added tap h2o to the weights listed in the table.
To dissolve the tartrazine, I blended it alongside a pane stirrer at area heat until there were no apparent lumps.¹
Here are the distinct solutions at distinct concentrations, the most concentrated one on the left, the center concentration in the middle, and afterward a h2o authority on the right.
Next, I got poultry chest fillets from Tesco. I cut a 5mm thick slab and afterward cut three pieces out to about 15mm in length.
I afterward soaked 5mm thick slices of poultry chest in all of the three solutions.
After 6 hours of soaking, these were the results.
First, the poultry in the highest concentration resolution failed, since the tartrazine seemed to crystallise. This is what it looked akin - it got all crusty.
For the center concentration (~0.19 M), the poultry appeared gelatinous. But it wasn't transparent. The authority poultry remained entirely opaque.
I threw these pieces out and afterward tried again alongside the method below.
And afterward I tried again alongside the identical procedure, alongside the aim of soaking it for 12 hours at area temperature. Given the issues alongside the highest concentration (~0.56 M), I lone tried the h2o authority and the center concentration (~0.19 M). But this period I put the poultry in test tubes and covered it, stated that I was gonna depart it out for 12 hours at area temperature. I wanted to mitigate the food safety risk.
This was the 12-hour-soaked piece. Whilst it's difficult to see alongside the camera, there was several minor apparent translucency compared to the control.
To analyze it I put it under several ray above.
I afterward cut about 2mm slices of chicken. This was fairly fiddly to do alongside a small slippery piece. Here is what it looked akin compared to the control.
This method has been used to create fluorescence imaging much easier in live animals.
On my end, doing this fine could add to the set of things that autonomous scientists could depiction at home. We power be capable to investigation inner structures of things additional casually. Making things transparent would provision improved images and would authorize autonomous scientists to appearance at tissue construction additional closely.
Effective study at low disbursal is among the themes of my research. I wrote concerning penning tracing application for Brownian particles using my microscope in my final post. I am additionally thinking concerning trying to equivalent the achievement of CO2 gas sensing using pressure, which would create that category of sensing cheaper. In row alongside this theme, I’m excellent enthusiastic concerning attempting weird and chill techniques to create structures additional easier to probe!
Here’s is additional than you always wanted to cognize concerning optical transparency.
A matter can be opaque for two reasons. First, a matter can assimilate ray and obstacle it. Second, it can scramble ray because of distinct refractive indices.
First, ray rays from the another flank could be absorbed whenever they continue through. If ray is absorbed, the ray can't hit your eyes and you can't see the another side. Light can have multiple wavelengths, and the wavelengths absorbed by the matter are resolute by the material's spectra. To compute the spectra, we can approximate the bodily behavior of molecules to that of an speck about a potential. The Schrödinger equation tells us that this scheme can lone inhabit a discrete set of energies, and it tells us which ones these are. The differences between these energies set the wavelengths that are ultimately absorbed by the material.
In reality, materials are composed of many of distinct things and so we anticipate them to additionally assimilate distinct wavelengths.
Another logic for opacity is small obvious, but additional interesting. The refractive indicator of a matter tells us how quickly ray travels inside the medium. When ray travels between materials alongside distinct refractive indices, it deflects slightly. When a matter is packed alongside sub-materials that have distinct refractive indices, the refraction inside causes ray rays to get scattered since ray rays ping around.
Biological materials merge a mix of distinct sub-materials inner - fibres, cells, proteins, and more. Each of these materials contains a distinct refractive index. When you have a matter alongside one refractive indicator embedded in a matter alongside a distinct refractive index, you get scattering. This scattering causes the ray from the another flank to get scrambled. The diagram below shows this.
We can example biologic materials as a 'base material' of water, alongside chunks of lipids and proteins embedded inside, all of which have distinct refractive indices.
To create item transparent, we need to utilize the two causes of opacity. One method is to try to alter the absorbance spectra of a matter so that it absorbs small wavelengths of light.
Alternatively (and this is the method we'll shield in this post), we could try to alter the refractive indicator of the h2o inner to equivalent the embedded media. The depiction below shows this regulation alongside biologic tissues. If you can alter the refractive indicator of h2o alongside a dye, you can afterward equivalent it to the refractive indicator of proteins and lipids, which is slightly higher.
If you oversee to do this without modifying the absorbance of a material, you have to be capable to create it additional transparent.
Ok, so the equivalent is to alter the refractive index. Here is a map of the calculations we need to do to fig out how to modify the refractive indicator of a material.
To comprehend anywhere the refractive indicator plays a part in a quantitative model, let's example how ray travels according to Maxwell. Light is an electromagnetic wave, comprised of oscillating electric and magnetic waves. For this essay, let's concentration on the electric site part to get the key intuitions.
Using complex representation, if ray travels downward the z axis, in a vacuum, we can compose downward the equation of ray as the following. Because it's a wave, we can write
\(E(z, t) = E_0 \cos\!\left(\frac{\omega z}{c} - \omega t\right)\)
But for convenience we use complex numbers
\(E(z, t) = \mathrm{Re}\left\{ E_0\, e^{\,i\omega z / c \;-\; i\omega t} \right\}\)
Where E₀ is the amplitude (how powerful the site is), ω is the angular frequence (at a fixed item the site goes circular formerly all T = 2π/ω), c is the speed of light, z is extend alongside the beam and t is time. Crest to crest, the wavelength is λ = 2πc/ω.
In media though, we anticipate the two the speed of ray to alter in the medium, as fine as several absorption. We can rewrite the equation as follows
\(E(z, t) = \mathrm{Re}\left\{ E_0\, e^{\,i(n' + i\kappa)\,\omega z / c \;-\; i\omega t} \right\} = \mathrm{Re}\left\{ E_0\, e^{-\kappa \omega z / c}\, e^{\,i(n' \omega z / c \;-\; \omega t)} \right\}\)
The coefficient in the exponential n′ is a multiplier rearward the spatial term, and represents the genuine refractive index. κ represents how much the motion attenuates, and is meant to signify absorption.
In custom we can collection n′ + iκ as one number, and afterward mention to this as the complex refractive index.
First we need to try to alter the genuine refractive index. If we can addition the refractive indicator of h2o to equivalent the tissue refractive index, we would have a additional transparent material. Water has a refractive indicator of 1.33. And although this figure is variable, macromolecule and fat affluent structures have a refractive indicator of about 1.45. So the equivalent is to modify the refractive indicator of the h2o about it towards 1.45, which would create it additional transparent.
Whatever we do to the water, we additionally need to create certain that we don't create the absorbance asset too high. If κ gets too high, afterward we'd end up engaging all the ray and it won't go transparent.
So how do we change the refractive index? To comprehend refractive indices, we archetypal need to comprehend polarisation. To archetypal order, a matter consists of electrons border to nuclei. When an electric site is applied to a material, the related positions of the electrons to their nuclei change.
from [http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dielec.html#c1]
For a positively charged speck and a negatively charged particle, polarisation is the dimension of that accusation multiplied by the relocation of one from the other. To compute the polarisation, we need to example how far electrons displace from the nucleus under the electric field.
Electric sectors provision a power on charges that are free. To example the consequence of an electric site on a material, we treat the affirmative nucleus as fixed, and the electrons as free. To compute the consequence of the system, we'll use the Lorentz oscillator model.
In this model, the electrons are modelled to be border to the nucleus akin a damped spring, alongside the electric site applying a forcing. We can resolve it exactly how we would resolve a damped fountain problem:
Newton's second law for the electron, alongside the light's site oscillating at ω:
\(m\ddot{x} = -m\omega_0^2\, x \;-\; m\gamma\,\dot{x} \;-\; e\,E_0\, e^{-i\omega t}\)
If we try an Ansatz for the stance of the speck x(t) = x₀ e^(−iωt) (the speck oscillates at the driving frequency), we get that the stance of the speck related to the nucleus is stated by the following expression.
\(x(t) = \frac{-e/m}{\omega_0^2 - \omega^2 - i\gamma\omega}\; E(t)\)
The dipole instant is p = −ex, and alongside N electrons per component quantity the polarisation per component quantity is
\(\mathbf{P} = N p = \frac{N e^2/m}{\omega_0^2 - \omega^2 - i\gamma\omega}\; \mathbf{E}\)
In this setup, ω₀ represents the natural frequence of the border electron, which is the frequence the molecule absorbs at (for tartrazine, the 425 nm peak), γ represents the damping rate, which sets how broad that absorption highest is, and ω is the frequence of the electric field. Now we've established polarisation, we desire to connect it to the refractive indicator itself. To do that, we need to do several gymnastics alongside two identities.
By convention, the polarisation can be written as proportional to the electric field
\(\mathbf{P} = \varepsilon_0\,\chi\,\mathbf{E}\)
where ε₀ is the permittivity of liberated space, χ is the electric susceptibility (a dimensionless figure that says how effortlessly the matter polarises), and E is the electric field. Comparing alongside the Lorentz oscillator outcome above,
\(\chi(\omega) = \frac{N e^2/(\varepsilon_0 m)}{\omega_0^2 - \omega^2 - i\gamma\omega}\)
To connect our polarisation, and electric susceptibility to the refractive index, we use the following trick. The below is an AI-assisted evidence alongside manual checking.
This flat of polarisation and the electric site is encoded in item called the electric relocation site D. And the association between D and the electric site is linear, alongside the coefficient called the related permittivity, εᵣ. P current represents the polarisation per component volume, and is the average induced dipole instant throughout all of the material.
\(\mathbf{D} = \varepsilon_0 \varepsilon_r \mathbf{E} = \varepsilon_0 \mathbf{E} + \mathbf{P}\)
It turns out that the related permittivity is connected to the complex refractive indicator as follows:
\(n_c^2 = (n' + i\kappa)^2 = \varepsilon_r\)
The evidence of the association between refractive indicator and permittivity for a norm electric motion is standard, and is in the footnote below.² And so, to alter the complex refractive index, we need to alter the related permittivity of the material.
To proceed, we use an approximation. The electric relocation site is a function of the polarisation of a material. But whenever we mix two materials together, we can add their polarisation densities.
Before adding tartrazine, we have that h2o obeys
\(\mathbf{D} = \varepsilon_0\,\varepsilon_{r,\mathrm{water}}\,\mathbf{E} = \varepsilon_0\mathbf{E} + \mathbf{P}_{\mathrm{water}}\)
Then whenever we add the dye, we get
\(\mathbf{D} = \varepsilon_0\mathbf{E} + \mathbf{P}_{\mathrm{water}} + \mathbf{P}_{\mathrm{tartrazine}}\)
But we cognize that all polarisation is proportional to the field, alongside its own susceptibility: P(water) = ε₀ χ(water) E and P(tartrazine) = ε₀ χ(dye) E, anywhere χ(dye). Substituting that in, we get
\(\mathbf{D} = \varepsilon_0\left(1 + \chi_w + \chi_d\right)\mathbf{E}\)
Comparing alongside D = ε₀εᵣE, and using that h2o solitary has εᵣ(water) = 1 + χ(water) = n(water)², the related permittivity of the dyed h2o is
\(\varepsilon_r = \varepsilon_{r,\mathrm{water}} + \chi_d = n_w^2 + \chi_d\)
For a dilute dye, χ(dye) is small compared to n(water)², so we can develop the quadrate base and use a Taylor enlargement to get the modified refractive indicator of the h2o alongside the dye.
\(n_c = \sqrt{n_w^2 + \chi_d} = n_w\sqrt{1 + \frac{\chi_d}{n_w^2}} \;\approx\; n_w + \frac{\chi_d}{2 n_w}\)
n_c is complex. If we compose it out as n_c = n′ + iκ and divided the electric susceptibility of χ(dye) that we got into genuine and imaginary parts, the dye shifts the genuine refractive indicator and introduces absorption by
\(\Delta n' = n' - n_w = \frac{\mathrm{Re}\,\chi_d}{2 n_w}, \qquad \kappa = \frac{\mathrm{Im}\,\chi_d}{2 n_w}\)
And putting in the Lorentz form of χ(dye) that we got earlier
\(\Delta n' = \frac{N e^2}{2 n_w \varepsilon_0 m}\,\frac{\omega_0^2 - \omega^2}{(\omega_0^2 - \omega^2)^2 + \gamma^2\omega^2}, \qquad \kappa = \frac{N e^2}{2 n_w \varepsilon_0 m}\,\frac{\gamma\omega}{(\omega_0^2 - \omega^2)^2 + \gamma^2\omega^2}\)
For tartrazine and red light, ω is below the resonance ω₀, so ω₀² − ω² > 0 and Δn′ is positive. This method the dye raises the refractive indicator of the water. Meanwhile κ carries a aspect of γω in the numerator and is lone ample near ω ≈ ω₀,, which method that we don't get absorbance.
References
Ou Z, Duh YS, Rommelfanger NJ, Keck CHC, Jiang S, Brinson K Jr, Zhao S, Schmidt EL, Wu X, Yang F, Cai B, Cui H, Qi W, Wu S, Tantry A, Roth R, Ding J, Chen X, Kaltschmidt JA, Brongersma ML, Hong G. Achieving optical transparency in live animals alongside engaging molecules. Science. 2024 Sep 6;385(6713):eadm6869. doi: 10.1126/science.adm6869. Epub 2024 Sep 6. PMID: 39236186; PMCID: PMC11931656.
² Here are Maxwell's equations for electric and magnetic sectors in a material. This is generated by Claude.
\(\nabla \cdot \mathbf{D} = \rho_f, \qquad \nabla \cdot \mathbf{B} = 0, \qquad \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}, \qquad \nabla \times \mathbf{H} = \mathbf{J}_f + \frac{\partial \mathbf{D}}{\partial t}\)
\(\mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P} = \varepsilon_0 \varepsilon_r \mathbf{E}, \qquad \mathbf{B} = \mu_0 \mathbf{H} \quad \text{(non-magnetic)}\)
The inquiry is afterward - how do any of these connect to the refractive index?
No liberated charges or currents (ρ = 0, J = 0), non-magnetic. Faraday's law says that the curl of an electric site is the negative of the magnetic field's period derivative. Take the curl of Faraday's law, and afterward substitute in
\(\nabla \times (\nabla \times \mathbf{E}) = -\frac{\partial}{\partial t}\,\nabla \times \mathbf{B}\)
Ampère–Maxwell alongside J = 0:
\(\nabla \times \mathbf{H} = \frac{\partial \mathbf{D}}{\partial t}\)
Substitute B = μ₀H and D = ε₀E + P = ε₀εᵣE:
\(\nabla \times \mathbf{B} = \mu_0 \frac{\partial \mathbf{D}}{\partial t} = \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t} + \mu_0 \frac{\partial \mathbf{P}}{\partial t} = \mu_0 \varepsilon_0 \varepsilon_r \frac{\partial \mathbf{E}}{\partial t}\)
This is anywhere εᵣ enters: the polarisation current ∂P/∂t adds to the vacuum relocation current. Then (taking εᵣ changeless in time, ie operating at a sole frequency)
\(\nabla \times (\nabla \times \mathbf{E}) = -\frac{\partial}{\partial t}\,\nabla \times \mathbf{B} = -\mu_0 \varepsilon_0 \varepsilon_r \frac{\partial^2 \mathbf{E}}{\partial t^2}\)
\(\nabla \times (\nabla \times \mathbf{E}) = \nabla(\nabla \cdot \mathbf{E}) - \nabla^2 \mathbf{E} = -\nabla^2 \mathbf{E}\)
Then if we equate the two sides, we get a aircraft motion equation
\(\Rightarrow \quad \nabla^2 \mathbf{E} = \frac{\varepsilon_r}{c^2}\frac{\partial^2 \mathbf{E}}{\partial t^2}, \qquad c^2 = \frac{1}{\mu_0 \varepsilon_0}\)
Plane motion E = E₀ e^(i(kz − ωt))
\(k^2 = \varepsilon_r \frac{\omega^2}{c^2}\)
Comparing alongside k = (n′ + iκ) ω/c from the attenuating motion written earlier,
\(\varepsilon_r(\omega) = n_c(\omega)^2 = (n' + i\kappa)^2 = n'^2 - \kappa^2 + 2 i n' \kappa\)
