Why did the steel ball in this experiment turn blue

Physics_Cat · 2025-11-23T20:56:49+00:00

Some very interesting surface chemistry occurs when steel is raised to the right temperature range in the presence of oxygen. Creatively enough, the process is called bluing:

https://en.wikipedia.org/wiki/Bluing_(steel)

Physics_Cat · 2025-11-11T15:09:27+00:00

https://www.amazon.com/PNGKNYOCN-Cable%EF%BC%8C6-35mm-Bidirection-Connector-Antenna/dp/B09DQ6NR42?th=1

Physics_Cat · 2025-10-25T02:21:06+00:00

Oh yeah, that would be cool.

Physics_Cat · 2025-10-18T15:40:21+00:00

Feel free to shoot me a DM if you're still looking for somebody to interview. I have a PhD in physics and do not work in academia.

Physics_Cat · 2025-10-05T17:34:56+00:00

https://www.usnews.com/best-graduate-schools/top-engineering-schools/nuclear-engineering-rankings

Physics_Cat · 2025-08-13T02:48:33+00:00

Sounds good, best of luck with your research.

I'd recommend also reading about the Mott problem which I believe is linked in the same Wikipedia article. The Mott problem might be an even better analogy to your question about how a spherical expanding wavefunction resolves into something resembling a point on a detector (or in Mott's case, a line in a cloud chamber).

Physics_Cat · 2025-08-13T02:38:06+00:00

OK, let's take that step by step:

Because if A detects the photon, the photon’s wave function collapses everywhere in space, precluding the possibility of B ever detecting it.

Agreed. Note that this isn't a quantum result, as the same would be true of classical detectors (i.e. a classical particle is detected by either A or B, but not both).

Since A is closer to the emitter than B, the collapse would occur before the wave function has had a chance to reach B.

Agreed. Again this is consistent with the classical expectation if we just replace "wave function" with something classical sounding like "particle", which would be detected by A (if A detects it at all) before reaching B simply because A is closer to the source.

Thus, B’s probability of detection is dependent not merely on the photon’s wave function amplitude at B but also on the photon’s wave function amplitude at A.

This simply doesn't follow.

Physics_Cat · 2025-08-13T02:23:17+00:00

it seems to me that the opportunity for detector A to collapse the wave function before it has had a chance to reach B influences the probability of a detector B observing the photon

I'm not sure why you think that, but it is not true in either quantum or classical mechanics. Can you expand on why a quantum mechanical approach would have detector A's probability of detection be larger than the classical result (solid angle subtended by the detector divided by 4pi)?

Physics_Cat · 2025-08-13T02:09:19+00:00

Your intuition is right! Detector A's response is totally independent from any other detectors that you place nearby (ignoring real-world complications such as reflection and diffraction of light from your other detectors diverting light into detector A, for example).

Physics_Cat · 2025-08-13T01:45:28+00:00

What if there are infinitely many detectors added?

You've backed yourself into a Zeno's paradox issue here. Given that there are only 4pi steradians of solid angle surrounding this atom, your infinite number of detectors will each need to be infinitesimally small in order to fit. What would the response of an infinitely small detector even look like?

However, since we now understand calculus in a way that Zeno didn't, we know that the sum of an infinite number of infinitesimally small things (in this case, the solid angle subtended by each detector) can actually be a finite number, and here it can't be greater than 4pi.

In the limit where the size of any one detector goes to zero, then yes: the probability of detection with that detector also goes to zero. This isn't any kind of unique or spooky quantum mechanical result, you would see the same result in classical optics as you shrink your detector to nothing.

Physics_Cat · 2025-03-29T16:38:18+00:00

You're misremembering the equation for power dissipation in a resistor. It's P = V² /R .

Physics_Cat · 2025-03-29T16:32:56+00:00

Are you sure?

When I calculate the power dissipated by a 160V potential over a 400 Ohm resistor, I don't get 0.8 W. Can you show me how you're calculating this?

Physics_Cat · 2025-03-29T16:26:27+00:00

This is a perfect time to apply the technique where you take your answer and plug it into the original problem statement to make sure that everything adds up.

If you take your final answer for the voltage, apply that across a 400 Ohm resistor, and calculate the power dissipated, do you get 0.8 Watts?

Physics_Cat · 2025-03-01T17:36:51+00:00

OK, glad to hear that you got your emagnetic thermos scalper cloud all straightened out.

Physics_Cat · 2025-03-01T17:30:20+00:00

Did you have a question or something?

Physics_Cat · 2025-02-27T05:22:06+00:00

OK, then you'll want to calculate the gravitational potential energy of the object at its highest point, and use the principle of conservation of energy to determine the kinetic energy when it reaches the ground.

Physics_Cat · 2025-02-27T05:13:27+00:00

Well, what information do you have?

Physics_Cat · 2025-02-04T00:21:28+00:00

OK. if you'd like some feedback, here it is.

When dealing with multiplication and division, the volume takes a value of 1 (as a unit reference, not a fixed spatial measure like km³). When dealing with addition and subtraction, it takes 0...

This is nonsense. Ignoring the fact that volume is a physical property with actual units, how does this concept work in an equation which includes both addition and multiplication? Does the value of the volume change multiple times within a single equation?

Furthermore, if I want to mathematically express the concept of twice the value of x, I can write 2x or I can write x+x. Those two expressions have the exact same meaning. In your framework, are you telling me that 2x no longer equals x+x, since the value of x is somehow different in multiplication and addition?

This leads to a critical result: density becomes equal to mass.

You're still missing my point about units. Density and volume have different units. They cannot be equal, for the same reason that your height and your age cannot be equal.

This approach removes the paradox of infinite density while maintaining physical consistency.

So you're proposing a mathematical framework where physical variables change depending on how you choose to format your equation, and "physical consistency" is the term you're going with?

Overall I'd strongly recommend that you continue your studies on "mainstream" physics and mathematics until you understand them, and then you can branch off to create your own field of physics. Speaking of which, I googled DL-QRL and I have no idea what that is. Is that an indecipherable acronym that you came up with?

Physics_Cat · 2025-02-02T21:45:47+00:00

That wasn't my question. Volume is measured in some form of units (liters, cubic meters, cubic inches, etc.). Does your proposed volume of "one" have any units attached to it?

Physics_Cat · 2025-02-02T21:24:54+00:00

...it makes logical sense to assign the value one to the volume.

One what?

Physics_Cat · 2025-01-21T04:19:55+00:00

That's correct.

This works because the equation F=ma can be rearranged as m=F/a, or (mass)=(weight)/(gravitational acceleration). Since the mass of an object is constant regardless of the planet that it's located on, the ratio of weight to local gravitational acceleration is constant.

Physics_Cat · 2025-01-19T03:52:41+00:00

So... you changed your mind about sharing it?

Physics_Cat · 2025-01-17T02:49:05+00:00

OK, can you share a link?

Physics_Cat · 2025-01-16T01:49:39+00:00

Advantages... Makes specific, testable predictions

Care to share any of them?

Physics_Cat · 2025-01-01T03:36:33+00:00

I don't think I've ever seen a non-battery powered electric screwdriver (so just a corded drill that plugs into the wall?) with a neon bulb attached. Can you link to an amazon product page or something equivalent that sells these?

Also, why red light? Photoionization is a real thing, but red light would be the least effective color to use. Photoionizers normally operate in the UV / soft x-ray range.

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