Your illustration and presentations made it easier to understand very complex physical concepts. Great job! Could you make a presentation to explain the various types of optical coherent tomography, both time domain and Fourier types. Thank you
5:20 Why does x*k = 2pi implies x = 2pik / |k|^2? Why can the dot product rearranged in that way? Don't we have x*k = |x||k|cos(alpha) where alpha is the angel between x and k. Why is this information missing? EDIT: Okay, I got it, but I guess the explanation is not optimal. It is confusing the say that "x" is the short, because if you move perpedicular to k along the wavefront, any x fulfills U(x,0) = U(0,0). So, there is no shortest non-zero value. But what you do is you look for the shortest x that is parallel to k, i.e., has the form x = 2pi * lambda * k/|k| with lambda > 0. If you plug this into the equation the dot product will reduce to lambda*|k| and you end up with the equation for lambda.
At 11:59 ish where you show the field at z_0 is 1) fourier transform of original field 2) times exp(i*k_z*z_0), 3) then take the inverse fourier transform. I was a bit confused what does "fourier transform of original field times exp(i*k_z*z_0)" mean, semantically. for example, what does "propagate each plane wave to plane z_0" really mean, or what does propagation really mean? it is described as if propagating a field is obvious using the field's FT and exp(i*k_z*z). After searching a bit, I found a potentially better explanation should be if you plug in u(x, y, z) in its iFT expression of u_hat(fx, fy, z) into Helmholtz equation, you get a general solution of u_hat(fx, fy, z)=u_hat(fx, fy, 0)*exp(i*k_z*z), then you take the iFT to get u(x, y, z). Overall, very good presentation, truly allows me to think deeper about the subject that I wasn't paying attention to when taking the course at University.
Very unfortunate that the position-vector r isn't used in this video (4:44). Here you must (unnecessary) spend part of your attention to distinguish between a normal x and a bold x: "Are we in 1D or are we in 3D?". @ 3:30 Fixed phase: dphi/dt = k.dx/dt - w = 0 and thus dx/dt = w/k = phase-velocity v. But notice that here (7:47) phi is defined in a different way! From a didactical point of view a bit disappointing. To some this may come as a surprise, but really the Greek alphabet has 24 characters to choose from. P.S. Any wave has a PHASE (if defined correctly) with a spatial, a temporal and a constant component.
Can you explain the decay? we used the complex representation as a mathematical trick for convenience. it is not trivial that this result is indeed valid (need to show that the same thing happens when analysed in time).
I think what needs to be demonstrated in the end is that it is a valid solution to the wave equation. If you derive the Helmholtz equation from the wave equation by separation of variables, you can then plug in the expression for the evanescent field to verify its validity.
Dear Sander, may I ask you that at 10:16, why you wrote your 2d plane waves as e to the power of 2*pi*(i)*(fx*X+fy*Y)?shouldn't this power be k*x where k=2*pi/lambda, and lamba is c/f, so you are missing a value c which is the speed of light?
Hi Rico, with fx and fy I didn't intend to imply any relation with the oscillation frequency f=c/lambda, my apologies for the confusion. You're correct that a 3D plane wave is written as e^{i*(kx*X+ky*Y+kz*Z)} where kx^2+ky^2+kz^2=k^2, with k=2*pi/lambda. I merely defined fx, fy as the spatial frequencies, which the Fourier transform of U(x,y) is a function of. This is just a mathematical definition that one can use for any 2D function U(x,y), regardless of its physical meaning. By themselves, fx and fy don't indicate that we're dealing with a monochromatic field of wavelength lambda. The physical meaning comes in at 10:50, where fx, fy are related to kx, ky, and kz is related to kx, ky.
Loving your work. The meaning of the mathematical equations has been expressed clearly.
Your illustration and presentations made it easier to understand very complex physical concepts. Great job!
Could you make a presentation to explain the various types of optical coherent tomography, both time domain and Fourier types. Thank you
You are just awesome....thanks for these beautiful lectures... Do make more such videos... Helped me a lot...♥♥♥
Sander, the lecture series are very good!
5:20 Why does x*k = 2pi implies x = 2pik / |k|^2? Why can the dot product rearranged in that way? Don't we have x*k = |x||k|cos(alpha) where alpha is the angel between x and k. Why is this information missing?
EDIT: Okay, I got it, but I guess the explanation is not optimal. It is confusing the say that "x" is the short, because if you move perpedicular to k along the wavefront, any x fulfills U(x,0) = U(0,0). So, there is no shortest non-zero value. But what you do is you look for the shortest x that is parallel to k, i.e., has the form x = 2pi * lambda * k/|k| with lambda > 0. If you plug this into the equation the dot product will reduce to lambda*|k| and you end up with the equation for lambda.
At 11:59 ish where you show the field at z_0 is 1) fourier transform of original field 2) times exp(i*k_z*z_0), 3) then take the inverse fourier transform. I was a bit confused what does "fourier transform of original field times exp(i*k_z*z_0)" mean, semantically. for example, what does "propagate each plane wave to plane z_0" really mean, or what does propagation really mean? it is described as if propagating a field is obvious using the field's FT and exp(i*k_z*z).
After searching a bit, I found a potentially better explanation should be if you plug in u(x, y, z) in its iFT expression of u_hat(fx, fy, z) into Helmholtz equation, you get a general solution of u_hat(fx, fy, z)=u_hat(fx, fy, 0)*exp(i*k_z*z), then you take the iFT to get u(x, y, z).
Overall, very good presentation, truly allows me to think deeper about the subject that I wasn't paying attention to when taking the course at University.
You are the king!
These videos are awesome!!😍tnx
Great vids! Thank you!
Very unfortunate that the position-vector r isn't used in this video (4:44). Here you must (unnecessary) spend part of your attention to distinguish between a normal x and a bold x: "Are we in 1D or are we in 3D?". @ 3:30 Fixed phase: dphi/dt = k.dx/dt - w = 0 and thus dx/dt = w/k = phase-velocity v. But notice that here (7:47) phi is defined in a different way! From a didactical point of view a bit disappointing. To some this may come as a surprise, but really the Greek alphabet has 24 characters to choose from. P.S. Any wave has a PHASE (if defined correctly) with a spatial, a temporal and a constant component.
Can you explain the decay? we used the complex representation as a mathematical trick for
convenience. it is not trivial that this result is indeed valid (need to show that the same thing happens when analysed in time).
I think what needs to be demonstrated in the end is that it is a valid solution to the wave equation. If you derive the Helmholtz equation from the wave equation by separation of variables, you can then plug in the expression for the evanescent field to verify its validity.
superb!
Why is x equal to 2*pi*k/|k|^2?
Dear Sander, may I ask you that at 10:16, why you wrote your 2d plane waves as e to the power of 2*pi*(i)*(fx*X+fy*Y)?shouldn't this power be k*x where k=2*pi/lambda, and lamba is c/f, so you are missing a value c which is the speed of light?
Hi Rico, with fx and fy I didn't intend to imply any relation with the oscillation frequency f=c/lambda, my apologies for the confusion. You're correct that a 3D plane wave is written as e^{i*(kx*X+ky*Y+kz*Z)} where kx^2+ky^2+kz^2=k^2, with k=2*pi/lambda. I merely defined fx, fy as the spatial frequencies, which the Fourier transform of U(x,y) is a function of. This is just a mathematical definition that one can use for any 2D function U(x,y), regardless of its physical meaning. By themselves, fx and fy don't indicate that we're dealing with a monochromatic field of wavelength lambda. The physical meaning comes in at 10:50, where fx, fy are related to kx, ky, and kz is related to kx, ky.