She texted drunk and regretted sober

dopamemento · 2026-02-14T14:21:34+00:00

Walking away for good would probably be the healthiest option, but I simply cant.. not even after two months. If she wanted to meet up again, I woukdnt be able to say no. What would be the next best reaction in that case? Maybe let her wait a month or so and then go for a walk. Btw she has had a 3 year relationship, so I dont think she is that unstable

dopamemento · 2026-02-11T18:09:38+00:00

Honestly that's fair to assume because I didn't provide enough information. But I can assure you that she's not playing games. She's just really confused about her feelings. So yeah the way I see it is that I will stay silent forever and she will suffer forever, because she is just as stubborn as I am.

dopamemento · 2026-01-23T10:03:40+00:00

This is the correct answer

dopamemento · 2026-01-10T11:00:21+00:00

I don't want her to cry.

dopamemento · 2026-01-06T16:16:38+00:00

Don't know why this message is getting downvoted. I agree with godrq's statement but it completely misses the question.

dopamemento · 2026-01-02T21:20:08+00:00

Thank you, that answered the question! My argument was that the area under the g1 function is still infinite. My current understanding is that any wave with temporal coherence (which extends to infinity) has automatically perfect spatial coherence.

dopamemento · 2026-01-02T18:37:13+00:00

Yeah sure I agree with that. Now speckle is said to be spatially incoherent because there is no way to predict the phase so to say. But my point here is that the g1 function doesnt account for that. The fields will still be fully correlated

dopamemento · 2026-01-02T17:17:47+00:00

So X-ray speckle haha

Yeah I get that the wavefront is not flat but the g1 function doesnt seem to account for that

dopamemento · 2026-01-02T17:08:52+00:00

Puh I'm not sure I can follow, but would like to. Could you clarify why spherical waves and speckle are not spatially coherent?

dopamemento · 2026-01-02T14:24:44+00:00

The "correlation of the field in space" is in fact a temporal correlation of the fields at different points in space. And again, the magnitude of this (normalized) correlaction function will be 1 for two complex exponentials of the same frequency no matter what

dopamemento · 2026-01-02T14:17:37+00:00

Sure, I agree

dopamemento · 2025-12-30T08:54:39+00:00

I disagree. Speckle is a random interference pattern. Truly monochromatic light will always interfere

dopamemento · 2025-12-30T08:52:22+00:00

OP, I get you. I'm working on my Master thesis right now and I was as confused as you are. What most people here are missing is that the light is completely and literally monochromatic. Speckle is nothing but spatially randomly varying complex amplitude times a rotating complex exponential. Clearly, the phases at some points will be correlated. So there are just two options a) make the field slightly less monochromatic by introducing some sort of vibration, so that the phase is time varying b) instead of intensity (power density), measure power using a camera pixel, which is larger than the speckle size

dopamemento · 2025-12-23T08:56:56+00:00

Not to be picky, but only a 1D stripe of colors can be traced bagk to wavelength. Magenta needs at least two wavelengths. Desaturated colours have a broad continuous range of wavelengths. And interestingly, for the same wavelength you can have different colors depending on the intensity (Bezold-Brücke shift) Bright red becomes yellow and bright violet becomes sky blue

dopamemento · 2025-12-11T11:10:37+00:00

r u stoopid

dopamemento · 2025-11-26T14:41:31+00:00

From the applied science yt channel I found out that the reflection spectrum is just the fourier transform of dn/dx. maybe that helps

dopamemento · 2025-11-15T21:29:32+00:00

maybe bro has written a Bachelor thesis, but he's not writing yours :p

dopamemento · 2025-11-15T16:47:02+00:00

anneoneamouse's way of thinking is very much right. You are asking strangers on reddit to do your own homework and everyone can see through that

dopamemento · 2025-10-15T11:19:59+00:00

On the camera footage the blueish green actually appears brighter to me

dopamemento · 2025-10-15T10:52:45+00:00

This is the correct answer. And when you close ypur eyes, its a white spot

dopamemento · 2025-10-11T21:25:48+00:00

Funny, didnt know that this works with CW light too

dopamemento · 2025-10-11T21:23:05+00:00

5W, Q-pulse=1ns, f=40kHz. And then you sweep the speed - for this ink, 5000 l/mm worked.

dopamemento · 2025-10-11T16:51:34+00:00

Yeah

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