The h Bar

7:53 AM

Can someone tell me the name of this particular potential(if it has one)?

10:05 AM

@ManasDogra I think the second condition should be $\sigma\lt r\lt\lambda\sigma$.

10:55 AM

still a weird potential, since the middle term doesn't go to $\infty$ but to $\epsilon$ at $r\to \sigma$?

John Rennie

11:41 AM

@ACuriousMind It's a hard core with a soft attractive potential outside the core. This sort of potential is widely used in colloid science as it's cheaper to compute than a 12-6 potential and still reasonably representative of real life. I don't know of any official name for it though.

ah, so the middle term isn't supposed to interpolate between $\infty$ and $0$, the $\infty$ really is a sudden hard barrier

John Rennie

Yes

Computationally convenient unless you need to compute ∇U at r = σ :-)

2:15 PM

@JohnRennie floating points have infinities

the $df$ is gonna be steep certainly

2:38 PM

@Loong What's the hot object in that photo? The cable? Or just the general vicinity?

The object is not in the picture. The black cable is connected to a probe. And, fortunately, the probe is under water.

Ah.

The source was a bundle of old dummy rods left in the spent fuel pool.

Oops.

Are dummy rods moderator rods? Or some kind of place-holder rod that's supposed to not interact (much) with the neutrons?

Dummy rods are used to replace individual fuel rods in a fuel assembly. For example, if a fuel rod gets damaged.

Dummy rods are just stainless steel or zircaloy. They don't contain any nuclear fuel.

2:50 PM

Ok. But they're supposed to not have much effect on the neutron flux, right?

Yes, correct

More important is that they have the same effect on the coolant flow as a normal fuel rod.

I vaguely remembered that zirconium has a small cross-section for thermal neutrons. :)

There should be the same flow rate of coolant through all the fuel assemblies in the reactor. So you should not leave an open hole somewhere.

Yeah, having the same effect on the coolant flow is what I was referring to with "place-holder".

okay, yes, correct

And, yes, Zr has a small cross-section for thermal neutrons. Also a quite high melting point.

Unfortunately, its thermal conductivity is not so good.

3:01 PM

And I guess the stuff you alloy with the Zr to improve its conductivity tends to have a larger neutron cross-section...

Yes. And there are always impurities in real materials.

A little while ago, I found a nicely illustrated page about Pu-238 for RTGs. But I couldn't find much solid info on who's currently making Pu-238.

Usual Zr contains Hf. The secret of nuclear-grade Zr is how to remove the Hf.

Good question. According to this recent article, "By 2024, the DOE plans to produce 1.5 kilograms (3.3 pounds) of Plutonium-238 dioxide per year, enough to fuel a full RTG every 3-4 years". — PM 2Ring Nov 18 at 4:42

Oh. Hafnium has an "interesting" nuclear isomer, IIRC.

I don't know about Pu-238 production. But when we got our Cf-252 for Olkiluoto-3, it was from Oak Ridge.

3:09 PM

From en.wikipedia.org/wiki/Hafnium The high energy content of 178m2Hf was the concern of a DARPA-funded program in the US. This program eventually concluded that using the above-mentioned 178m2Hf nuclear isomer of hafnium to construct high-yield weapons with X-ray triggering mechanisms—an application of induced gamma emission—was infeasible because of its expense. See hafnium controversy.

^ Has a half-life of 31 years.

Multiple national laboratories are used to generate, refine, and assemble Plutonium-238-fueled RTGs.

planetary.org/space-images/…

Stupid .webp image, won't one-box.

Such an RTG is the reason why the Pu in the southern hemisphere has a different nuclide vector than the Pu in the northern hemisphere.

The tricky thing about making Pu-238 from Np-237 is minimising the Pu-239 that's inevitably produced. ;)

@Loong What's a nuclide vector?

3:24 PM

the ratio of the (activities or masses) of the individual nuclides

Northern hemisphere Fallout has about 0.02 % Pu-238 by mass. In the southern hemisphere, it's 0.09 %.

Which doesn't sound like much, but in terms of activity, it gets a bit more interesting because of the relatively short half-life.

Are you talking about the fallout created during the era of above-ground testing?

Yes. It's still there since Pu doesn't travel so much in the ground.

I know that some of the messiest Pu fallout in the world is the result of the British test in the South Australian desert, at Maralinga.

Maralinga

Maralinga, in the remote western areas of South Australia, was the site, measuring about 3,300 square kilometres (1,300 sq mi) in area, of British nuclear tests in the mid-1950s. In January 1985 native title was granted to the Maralinga Tjarutja, a southern Pitjantjatjara Aboriginal Australian people, over some land, but around the same time, the McClelland Royal Commission identified significant residual nuclear contamination at some sites. Under an agreement between the governments of the United Kingdom and Australia, efforts were made to clean up the site before the Maralinga people resettled...

Well, the worst test made by a western country was probably Castle Bravo.

3:49 PM

Good call. Admittedly, it wasn't supposed to be that powerful.

From en.wikipedia.org/wiki/Castle_Bravo The yield of 15 megatons was triple that of the 5 Mt predicted by its designers. The cause of the higher yield was an error made by designers of the device at Los Alamos National Laboratory. They considered only the lithium-6 isotope in the lithium-deuteride secondary to be reactive; the lithium-7 isotope, accounting for 60% of the lithium content, was assumed to be inert

5:52 PM

Projection operator is defined by P^2=P, so it should have eigenvalues 0,1.
Identity matrix(2x2) satisfies the above equation but it has eigenvalues 1,1. not 0.
What's going on?

Is the identity matrix a projection matrix or not?

It is a projection operator on the same subspace as the original space, yes

So when I am plugging P^2=P in the eigenvalue equation, the eigenvalues I obtain are all the possible eigenvalues of P, but not all of them necessarily have to be eigenvalue for a particular P right?

$0$ is always an eigenvalue

Since for any matrix, $M 0 = 0$

For the identity matrix 0 is not an eigenvalue

Well, an eigenvector I should say

5:59 PM

@Slereah uh...that's not how eigenvalue is defined

Ah well

@ManasDogra yes

I was thinking P^2=P is quadratic so all projection operators must be 2 Dimensional but then I realized P^2=P is not the characteristic equation always so Cayley Hamilton doesn't apply...this is how I came to this question.

B. Brekke

6:38 PM

For Feynman diagrams we sometimes write interaction vertices as lines and sometimes as dots. What determines which should be used? I am considering an example which confuses me

Using a dot or a line seems to determine which type of diagrams are possible

I don't know what you mean by "writing interaction vertices as lines"

the lines are propagators, not vertices

note that vertex is just a sophisticated name for "dot" :P

B. Brekke

Ah, that is true, so for some interaction mediated by a boson, we often write a line for the boson propagator. And if there is only a contact interaction we only use a dot

So now I am considering this non-linear sigma model, and I do not understand why I have these lines

yeah, for a force carrier boson the typical diagram is two charged particles interacting via a virtual boson, but that's not one interaction drawn as a line, it's two interaction vertices (particle-particle-boson) connected by a boson propagator