Gradient Force of an optical near field

[Sciencestd]

Source: Principles of Nano-Optics, for Lukas Novotny and Bert Hecht.The equations above represent the electric field in the second medium when a light hit a surface and the condition of TIR (total internal reflection) is satisfied. Actually this is what called Evanescent field. The point is if I want to calculate the gradient force on a particle in this field, F(gradient)=0.5a*gradient|E|^2, where "a" is the Polarizability and "E" is the electric field. How does the gradient force hold in y- direction? If I make the derivative of the field in y direction it will vanish according to the equation above because there is no variable y that exist in the equation in y axis, on another hand how I read in some articles that they say the gradient force is exist in y direction but with no explanation or derivation such as: https://www.ncbi.nlm.nih.gov/pubmed/30417030

 

[hutchphd]

I believe the fields you display from Novotny et al are for uniform illumination on flat surface (for light traveling +x). If the beam is instead of finite extent (in y) then the refraction through beads will provide a bead force towards the region of higher intensity. Good question and the answer is that the beam is likely ~Gaussian profile in y

 

[Sciencestd]

I'm really confused, Gaussian beam that in the sides it decays exponentially inside the second medium..! I didn't get it, so in every point where the light propagate inside the waveguide there is three components x,y and z, one should be the propagation direction, one is the evanescent field decaying (let's say in z) and one should be constant intensity in the space? (by the way if it is const. so what is the length/distance of this intensity) or decaying (has a gradient )?!

 

[Cutter Ketch]

I’m not sure why we would start talking about a finite beam when we haven’t yet answered your question about the expression you have. Your expression is for a plane wave of pseudo infinite extent (doesn’t vary over the size scale of interest) impinging on an interface in the x y plane (normal to z) at an angle theta. The plane of incidence defines s and p in the usual way. x is parallel to the plane of incidence, and y is perpendicular.

As you noted there are no x or y dependences in the components in the bracket. The only spatial dependence is outside the bracket and applies to all components of the field. The spatial dependence in z shows that the field falls off exponentially as you go away from the interface. The x dependence is a wave (see the i in the exponent). There is no dependence on y. This looks like waves on the ocean with parallel wave fronts. As you run time forward they will even propagate.

If you think about waves on the ocean, there is no gradient parallel to the wave front. The reason you get zero gradient force in y is because there is no gradient in the field in the y That isn’t weird or confusing. It is just correct. Your math isn’t lying to you. See for example the animations here:

https://aparajitanathblog.wordpress.com/tag/evanescent-waves/

Now, as hutchphd said, if you are reading something that says there IS a gradient in y it must be because the input field has a gradient in y and they may be talking about a more realistic finite beam rather than an infinite plane wave.

 

[Sciencestd]

Thank you all for your answers :). I totally understand the equations above and I can derive them. My whole confusion was what they considered in the article and also in another articles - the gradient force in y direction. I can partially understand the reason why they considered it from your answers but not totally.

 

[Sciencestd]

Can you please explain how The Gaussian beam cause to the gradient Force.. how one can imagine this?

 

[Cutter Ketch]

Sure. Let’s switch coordinates for a second to a coordinate system that is natural for the incoming beam (rotated by the angle of incidence $\theta_1$ from the coordinates at the interface). Z is the direction of propagation, x is still parallel to the plane of incidence, y is still perpendicular.

In your expression the incoming beam is an infinite plane wave. The electric field amplitudes $E_p$ and $E_s$ do not vary in x and y. However no real beam can actually be infinite and have a constant intensity (and so constant electric field) everywhere. Instead, suppose the incoming light has an intensity profile. For example say it is a Gaussian beam where the intensity is

I = $I_{max} e^{\frac {-(x^2 + y^2)} {2 \sigma ^2 }}$

the intensity is the square of the electric field, so the electric field amplitudes also vary as a Gaussian. Instead of the incoming beam being

$\begin{bmatrix} E_p \\ E_s e^{i\phi}\\ 0 \end{bmatrix}$ ...

You have

$\begin{bmatrix} E_p e^{\frac {-(x^2 + y^2)} {4 \sigma ^2 }} \\ E_s e^{i\phi} e^{\frac {-(x^2 + y^2)} {4 \sigma ^2 }} \\ 0 \end{bmatrix}$ ...

Now the field at the interface has a gradient in y.

 

[Cutter Ketch]

looking at what I wrote I can’t fathom why I didn’t write the exponent after the bracket. I was so focused on figuring out how to LaTeX (not an expert) it didn’t even occur to me.

 

[Sciencestd]

Thank a lot for your answer it really helped me to understand what's going there and I hope I understood it as it should be, otherwise I will ask again :P.
Yeah Don't worry about it, I think it's clear.