I tutor people and just came across a particle in a ring problem. I was pretty rusty, but this video was the perfect refresher. It was really well-made, and thank you so much for sharing it!
You're welcome! Particle in a ring is a common next homework problem after particle in a box because it uses the same principles, but different boundary conditions
If you wanted to think of the wavefunction as a two-dimensional function, like Ψ(r,θ), then a 2-dimensional *spatial* integral would require ∫∫ r dr dθ. And a line integral around the circumference of a circle would be ∫ r dθ. But our wavefunction only has angular dependence: Ψ(θ). And the integral we're performing is not spatial.
Thanks for your video. Your viewers might enjoy seeing this visual aid. The sheet of spring steel buckled to a Gaussian curve represents a two dimensional field with the ends bounded. Seeing the mechanical effect takes some of the mystery of what the math is showing. Your viewers might be interested in watching the test video of the model. See the load verse deflection graph. th-cam.com/video/wrBsqiE0vG4/w-d-xo.htmlsi=waT8lY2iX-wJdjO3
That's a great visual analogy, thanks! I think the parallel between the two cases arises because in QM, the curvature (2nd derivative) corresponds to kinetic energy, while in mechanics, the curvature of your sheet of steel corresponds to its strain energy. Minimizing the curvature in order to minimize the (kinetic or strain) energy gives rise to the same sinusoidal curve in each case. Very cool! Is the initial shape of your sheet a Gaussian, not sinusoidal?? And is the behavior in the last few seconds because the sheet slips out of the clamp on the right side?
Thanks for checking it out and the supportive word for my amateur science studies! Some math purest say it's not a sinusoidal and other say it's not a Gaussian either. To me it looks to be a single wavelength sinusoidal. When I see Gaussian curves in textbooks, they are in a portrait view. Personally, I think if the X axis goes toward infinity in two directions and the number of samples is finite, then they should show a relative flat shallow curve in a landscape view. I used to try to maximize the area under the curve when I was making truss core panels using the process. I learn that the process did not work if I buckled the height greater than one half the overall length. I trimmed the end of the five-minute video down because it was fairly long and somewhat boring to see it return to the original curve and shape. I think what you notice was due to the thickness of the material and internal stresses. Hope you find my visual antilog worth checking out firsthand in your lab. I think you might find some interesting insights with this model. Might make a young science student a good science fair project. Also hope you get to check out the sawtooth load vs deflection graph. PS: doing Physical Chemistry backward on a window is a awesome feat in-it-self!!! @@PhysicalChemistry
If you're using e^imφ as your wavefunction, that is right. But if you're using cos φ, as in this video, then A² = 1/π. You can confirm this is true by performing the normalization integral.
Please don't stop, you are doing great job . It helps me alot
Thanks, that's good to hear
I tutor people and just came across a particle in a ring problem. I was pretty rusty, but this video was the perfect refresher. It was really well-made, and thank you so much for sharing it!
You're welcome! Particle in a ring is a common next homework problem after particle in a box because it uses the same principles, but different boundary conditions
@@PhysicalChemistry It's definitely good to be able to think outside of the box!
Great explanation👍
Thx! Why the solution is just COS? What about the SIN?
Thanks ! Why in the integral for the normalization the line element is not r*dtheta ?
If you wanted to think of the wavefunction as a two-dimensional function, like Ψ(r,θ), then a 2-dimensional *spatial* integral would require ∫∫ r dr dθ. And a line integral around the circumference of a circle would be ∫ r dθ. But our wavefunction only has angular dependence: Ψ(θ). And the integral we're performing is not spatial.
Thanks for your video. Your viewers might enjoy seeing this visual aid. The sheet of spring steel buckled to a Gaussian curve represents a two dimensional field with the ends bounded.
Seeing the mechanical effect takes some of the mystery of what the math is showing.
Your viewers might be interested in watching the test video of the model.
See the load verse deflection graph.
th-cam.com/video/wrBsqiE0vG4/w-d-xo.htmlsi=waT8lY2iX-wJdjO3
That's a great visual analogy, thanks!
I think the parallel between the two cases arises because in QM, the curvature (2nd derivative) corresponds to kinetic energy, while in mechanics, the curvature of your sheet of steel corresponds to its strain energy. Minimizing the curvature in order to minimize the (kinetic or strain) energy gives rise to the same sinusoidal curve in each case. Very cool!
Is the initial shape of your sheet a Gaussian, not sinusoidal??
And is the behavior in the last few seconds because the sheet slips out of the clamp on the right side?
Hope my response to your question went through. Thanks again for your video and supportive words.
Thanks for checking it out and the supportive word for my amateur science studies!
Some math purest say it's not a sinusoidal and other say it's not a Gaussian either. To me it looks to be a single wavelength sinusoidal.
When I see Gaussian curves in textbooks, they are in a portrait view. Personally, I think if the X axis goes toward infinity in two directions and the number of samples is finite, then they should show a relative flat shallow curve in a landscape view.
I used to try to maximize the area under the curve when I was making truss core panels using the process. I learn that the process did not work if I buckled the height greater than one half the overall length.
I trimmed the end of the five-minute video down because it was fairly long and somewhat boring to see it return to the original curve and shape. I think what you notice was due to the thickness of the material and internal stresses.
Hope you find my visual antilog worth checking out firsthand in your lab. I think you might find some interesting insights with this model. Might make a young science student a good science fair project.
Also hope you get to check out the sawtooth load vs deflection graph.
PS: doing Physical Chemistry backward on a window is a awesome feat in-it-self!!!
@@PhysicalChemistry
A^2 is 1/2pi professor
If you're using e^imφ as your wavefunction, that is right. But if you're using cos φ, as in this video, then A² = 1/π. You can confirm this is true by performing the normalization integral.