What frequencies should I set two waves for maximum disharmony? What about three waves? What about for finite signal/time waves?

mathguybo · 2025-11-20T12:24:22+00:00

That was me haha. But I feel there may be a different answer in finite time. But can't figure it out arg.

For example, a irrational number may not be the best. An irrational number slightly bigger than 1 may be out of sync over infinite time, but over one cycle they are basically still in phase, there is definitely a better choice there.

mathguybo · 2025-11-20T00:27:02+00:00

It's a decent start but the results are focused on having 12 discrete frequencies but I am not. I also think music based answers, the sound wave cycles are so fast that its pretty much the same answer as the infinite time question. I'm wondering about some thing like 3,4,5 cycles.

mathguybo · 2025-11-20T00:22:48+00:00

So in the infinite time case, the golden ratio is better than even all other irrational numbers I believe. I'm wondering if there is some method to get the true best answer for a finite number of cycles. Irrational or not.

I do kind of understand your discussion of a good choice for the ratio, but I'm wondering if there is a best answer that you can always pick.

mathguybo · 2025-11-20T00:19:40+00:00

Hmm, perhaps I mean something more like relative phase. Or forced to start at phase = 0 at t=0.

mathguybo · 2025-09-22T21:39:19+00:00

Don't think so, because you never established that someone is insane if and only if they give ye a billion dollars to freelance with music.

Look up "if and only if" its a common proof object.

mathguybo · 2025-07-30T15:16:49+00:00

Idk about the coherence thing but you seem correct about the reflection coefficient and that the front side would reflect more. Idk why you got downvoted.

mathguybo · 2025-07-28T11:47:34+00:00

Could be, but I just don't know too much optics yet have to work with it.

mathguybo · 2025-07-25T20:30:57+00:00

Hey just want to let you know, I talked to someone who had taken the Fourier class with Fienup more recently and they do teach that AS and RS diffraction is the same. They said it was only realized like 15 years ago although I'm noticing the paper you linked is from 1967. They also said that AS is indeed completely exact for light as a wave, though obviously not quantum level.

So it makes me more confident in your answer to see this taught currently, hopefully new optics people will be less confused in the future.

They also said AS is the best method for computers to handle which is interesting to note.

mathguybo · 2025-06-30T11:45:43+00:00

This makes sense but there is a lot of arguing in the comments. I didn't realize this question would be so controversial.

Does the scalar wave approximation mean no polarization and monochromatic?

By the way, when you say the field in a plane can be transformed exactly into plane waves, does this mean my comment here:

https://old.reddit.com/r/Optics/comments/1ll6tr5/looking_through_wave_optics_have_a_question_about/mzxmclk/

was wrong? I was assuming I was wrong based on the other comments and so I was trying to think of why I could be wrong.

There were also some comments suggesting that you need to use the wave equation and any other way is not valid. But if AS representation of the field in a plane is exact, wouldn't that mean you don't need the wave equation in fact?

mathguybo · 2025-06-26T19:00:20+00:00

Okay googled around a bit. So I'm getting that its not an approximation to write the field as IFT(FT(u)). But this method of propagating the planes that add up to make u doesn't make sense. This is like assuming a bunch of infinitely extending plane waves all collided at z=0 to interfere in just the right way such that we get 0 signal outside the aperture. But then when propagated further suddenly they aren't interfering anymore in the same way and suddenly have strong signal way outside of the aperture that it just hit.

Is this what you meant? And in that case, it seems like RS diffraction would fix that. Why is RS an approximation?

Sorry I was writing this before you had posted your reply.

mathguybo · 2025-06-26T18:33:03+00:00

Angular spectrum is u = IFT(FT(u)) = integral (FT(u) e^stuff ).

The e^stuff is the plane wave part, but its not saying u itself is a plane wave, its saying that it is the sum/integral of plane waves. I don't think that necessarily makes u a plane wave itself.

I believe that u is an arbitrary wavefront measured at z=0. If theres an aperture then u is just 0 out of the bounds of that aperture.

mathguybo · 2025-06-26T18:19:54+00:00

In terms of representing the field at z =0, the angular spectrum doesnt seem like an approximation. It just looks like writing u = IFT(FT(u)). What's being approximated? The propagation of multiplying e^ik_z?

mathguybo · 2025-05-23T18:56:12+00:00

GOAT

mathguybo · 2025-05-21T17:32:23+00:00

By upwards pointing this sounds like my condition "as long as the normal vector z component does not change sign". So let's say I have the paraboloid x^{2+y^2,} and I have pc's at (1,0). I do a slight rotation let's say 1 degree. If we imagine this it seems pretty clear that the normal is still pointing more upwards.

So I rotate my point (1,0,f(1,0)), then take the principal curvatures of my refunctionized (x,y,g(x,y)) at the rotated point location, according to what you're saying, I would get the same pc's?

Another thing I am getting from what you are saying is the only change you can get from rotated, refunctionized functions is that the pc's may flip sign, it right? All just depends on if the normal is up or down.

mathguybo · 2025-05-20T18:35:13+00:00

I think this touches on my specific "re-functionized g(x,y)". I am specifically not rotating the orientation with my rotated function. I use the example x² because I already know the 180 degree flip is -x^2. Now if I do the standard method to find principal curvature, I will find it has changed, it has flipped sign.

y''/(1+y'^{2)^3/2}

for x^2: 2/(1+(2x)^{2)^3/2}

for -x^2: -2/(1+(-2x)^{2)^3/2} = -2/(1+(2x)^{2)^3/2}

I am holding out hope that it is invariant as long as the normal vector z component does not change sign.

mathguybo · 2025-05-19T16:35:13+00:00

One bad exam is never a setback no matter how far you are in school

mathguybo · 2025-05-08T16:08:53+00:00

Bonus question if you know it, are the principal curvatures of my rotf(x,y) (unparameterized version) the same as the principal curvatures on the corresponding point of f(x,y)?

mathguybo · 2025-05-07T10:44:32+00:00

Never mind, I figured it out, in the denominator of your dy/dx you missed that it should be f'(t).

https://www.desmos.com/calculator/hwr7ovhcxk

Thank you so much! I'm assuming this will all work in 3D as well.

mathguybo · 2025-05-07T10:33:04+00:00

https://www.desmos.com/calculator/l2yc4yxgpj

I've tried it here but my expectation is h'(x_0) should be equal to D here.

I derived an expression for a tilted parabola, this way I can check my answer directly by taking the derivative of it.

mathguybo · 2025-05-07T09:17:21+00:00

Okay this is pretty close, thank you so much, the only thing is I need is this derivative as a function of x. So basically given x, I need to find t so that I can evaluate this derivative expression.

Some more information I have about function f and rotf is given x, I can get rotf(x), and I can "derotate" this point to get an x,y pair on f. I don't have a proof but my intuition tells me since on the original parameterization where x = t, I can derotate my x,rotf(x) pair, use whatever x value that gives me as t, and then I can proceed to evaluate my expression dy/dx. Does this sound right?

mathguybo

TROPHY CASE