Welcome to the Utopia Forums! Register a new account
The current time is Fri Sep 21 12:17:25 2018

Utopia Talk / Politics / Gammas etc.
Seb
Member
Wed Mar 07 08:58:54
jergul:

"The sum of all factors that lessen gamma particle emission to orbital platforms are relevant"
No. E.g. the pressure hull of the sub has some effect, but many orders of magnitude less than ocean, sky or silt. So it can be neglected.

What I've demonstrated is that for the energy range of interests, the silt, sky and any shielding are irrelevant compared to the ocean. Many orders of magnitude less.

"So a casing that gives significant secondary gamma rediation on fusion is the wrong kind."

You've missed the point. So far I'm *only* looking at the prompt gammas, not secondary. I.e. I'm giving you a free ride in ignoring a whole other source of detectable gammas.

The uranium tamper is a much better shield of those prompt gammas than lead.

*If* several cm of lead were physically capable of screening prompt gammas, then the uranium tamper in a thermonuclear bomb would mean that no thermonuclear bomb would ever produce detectable prompt gammas because they'd be absorbed by the uranium tamper before it had a chance to fission and be blown appart.

However, because we do indeed detect the prompt gamma in a nuclear blast, we empirically know that several cm of uranium (several tens of cm of lead) is completely insufficient to screen the prompt gammas produced in MT nuclear explosions.


"I was challenging you to assume a casing suited to the task at hand."


As for replacing the uranium tamper (responsible for over half the yield and in doing so necessarily producing the secondary gammas I've not even considered yet as a detectable signal), if the goal is to choose a material for the radiation case and tamper that effectively reduces gammas then I'd suggest instead of the tamper material, we could replace the plutonium in the primary with lead.

This would very effectively reduce the gammas emitted upon detonation to zero if that (rather than making a big bang) is the task at hand.

"It is not given that increased pressure increases the timeline was my point."

Erm, it is!

"Shockwave dissipation outside a bubble lasts for a long time as force and distance trends towards 0."
Not sure what this means.

"Quoted limitations on TGF detection on the articles I reviewed. We are speaking of detection."

TGFs - I need to check the distribution of energies but I'd guess that many are in the photo-electric effect range which are much more easily screened by the atmosphere, and TGFs are far, far fainter a source than a 20MT device and so just look like cosmic rays to ground detectors. Signal to noise etc.

"You have not even gotten to calculating effective distance to the closest monitoring satelitte yet ;)."

Did i not post the geometric part yet?
I maybe deleted that draft because the scattering part needs to integrate over energies in a way that was too complex for the time available.

But the short version is the number of GPS in 180 degrees is 6-7 for relevant latitudes.

Solid angle geometric effects mean you lose lots of gammas but you still are looking at millions for each detector.

Then you need to look at the water, but I think that's going to need a numerical model.
Sam Adams
Member
Wed Mar 07 11:16:53
Lol
jergul
large member
Wed Mar 07 15:07:39
Seb
And we can just dismiss entropi because it is so much smaller than enthalpy? Its just sloppy.

But water first of course. Starting with an estimate of the angle to observing satelitte.

I do not see how increased pressure increases the timeline then.

The atmosphere and silt + casing matter.

We have not actually established how detonation gamma would compare to TGF. We need to know water effect first.

If you are doing 180 degrees, then you would actually need to model the effects of the continental shelf (and the continent proper for that matter as we are looking at MEO).

The short version is simply estimating the best angle the Russian could time device detonation for.
Sam Adams
Member
Wed Mar 07 17:08:42
"but many orders of magnitude less than ocean, sky or silt. So it can be neglected. "

Funny to see seb using that line since he argues so staunchly that very minor global warming changes can not be neglected.
jergul
large member
Thu Mar 08 01:51:17
Sammy
I think the slam-dunk case he is trying to make is becoming more clearly unlikely.

Even millions of gamma ray particles entering into space are unlikely to pass through a detector (a single particle has something in the region of 1 chance in 1.2x10^16).
Seb
Member
Thu Mar 08 05:12:21
jergul:

Attenuation terms will be a linear combination of terms. Io * exp(a+b).

if a <<< b, then a is negligable.

a few meters of silt the atmosphere is basically negligable compared to 1km of water. That's pretty obvious.

"Starting with an estimate of the angle to observing satelitte."

GPS satleites are in orbits such that a minimum of 4 (6 in US lattitidues) are visxible within a 90 degree arc at any point. So 1km of water minmum, 1.5km maximum. There are 8 at 180, but we only need three. I'll use a 90 degree cone. The difference between 1.5 and 1km of water for an order of magnitude isn't enough (given other approximations) to worry about. At this stage we are just trying to show whether it is a non-starter or plausible - not the preicse sensitivities needed.

"Even millions of gamma ray particles entering into space"
You didn't read correctly.

I said "Solid angle geometric effects mean you lose lots of gammas", it's not millions into space, it's millions in the solid angle that intersects any one detector (I've assume the detector subtends 10cm) I.e. I'd already applied the factor of circa 10^-16 - that should be obvious from the numbers jergul. There are several gammas per fused atom, and 10^16 atoms isn't very much!

It's actually much more than millions as I'm using the peak intensity of the prompt radiation rather than the integration over the entire pulse of prompt radiation.

But other than a basic ballpark figure to show that geometric features alone are insufficient, this is the wrong calculation to do.

It's kind of pointless until we've worked out what the flux at the top of the water looks like.
It's going to look like a diffuse circle with a varying intensity, and then we need to integrate over that circle to work out the flux heading towards any given satelite. In short, it is a bit of a bitch to do fully rigorously.

Roughly, the way you do this calculation is to start with a distirbution of gamma energies and intensity as a point source. Then you need to apply a random walk based on the compton scattering, which modifies your energy distribution as you go. It's, basically, a right royal pain of a calculation that needs a numerical approach to do properly.

But very basically, the efficiency of energy transfer from compton scattering goes as z. H is petty poor with a low cross section for scattering, so mostly it's the O you care about.
The z is too low for 1-10MeV's to have a significant cross section for photo-electric interactions.Potentially, if there are enough scattering events, gammas might end up below 1MeV and then rapidly get screened out by photoelectri effects.

The air and casing can be neglected. Particularly because measurements of prompt radiation measured from nuclear explosions are measurements of what escapes the radiation case and tamper (so that's already taken into account and any realistic lead shield or silt is going to just be a linear factor of that and negligable compared to the water).

The key thing to realise is we are dealing with very, very, very high gamma fluxes, so even very effective screening means you get detectable gammas.

Bottom line, I'm not going to do the calculation, it's too complicated to have time to do so.

But hey, there's also the hydrophones and seismic detectors that would also easily detect and identify a 20MT based on those signatures.


Sam:

We established that the climate system is affected by small chagnes in radiative forcing (otherwise the planet would be an iceball).
jergul
large member
Thu Mar 08 07:32:56
Seb
The depth has to be 2 km.

If you are not going to do the calculation, then we can just use the energy half rate for gamma through a water medium. Which is measured in single digit meters at energy levels we are thinking of.

Meaning of course, that gamma will not reliably reach detection plates in MEO.
Seb
Member
Thu Mar 08 07:42:06
Jergul:

No we can't because - as discussed - that's a calculation for a beam and it treats any scattered photon as no longer part of the beam, even if it intersects the target.

This is analagous to the "atmosphere is saturated for IR absorbtion" fallacy in respect of global warming.

It would be quite trivial for me to show that a single gamma photon will not be absorbed by several meters of water, if that helps.

If that was the case, nuclear reactors wouldn't require nearlly so much shielding (and those are alot less bright sources than megatonne bombs!).




Seb
Member
Thu Mar 08 07:42:58
Important to understand the difference between "does this beam remain coherent through this thickness of material" to "does this point recieve any flux of gammas from this source behind a screen".

jergul
large member
Thu Mar 08 07:56:46
Seb
IF you are not going to do the calculation, THEN we will use what tools are available.

Those tools strongly suggest that gamma will not reliably reach a few detection plates in MEO. Not by a huge margin.
Seb
Member
Thu Mar 08 08:13:13
jergul:

If we do not use the hammer to put the nail in, so this piece of polystyrene will suffice?

The fact I don't have an afternoon to build and run a computer model for this doesn't mean you then get to substitute any old thing and say "hey, we can use this simply because it is not that".

In any case, you have rejected other approaches for approximation on the grounds they fall short of the full calculation.

BTW, in the case of WIGWAM (a tens of kt device at 600m), the hydrophone signals were detected over 2500 miles away.

So it strikes me that even talking about gammas is kinda irrelevant here. We know it would be detected on hydrophones and seismics.
jergul
large member
Thu Mar 08 08:34:58
Seb
Lets first just dispense with the ludicrous idea that a few detector places in MEO would be able to realiably detect gamma.

We could reverse calculate it you know. 0.00001 joules have to reach a plate. How big would would the nuclear device have to be if 2% of energy is gamma, 25% of detonation included 75% loss of gamma every 10 m in water + 75% loss in atmosphere + 1.2x10^16 mulitplier to cover fraction of space detectors cover.

How many 1000nds gigatons would the device have to be for it to register?
jergul
large member
Thu Mar 08 08:39:28
Penta* Sorry.
Sam Adams
Member
Thu Mar 08 09:22:01

"We established that the climate system is affected by small chagnes in radiative forcing (otherwise the planet would be an iceball). "


Ohhh but you just said small effects can be ignored. Your extremist political viewpoints clouds your science judgement. the changes brought about by global warming are usually orders of magnitude less important.
jergul
large member
Thu Mar 08 09:37:22
Please ignore the term "ludicrous". Big numbers are not intuitive, so who the hell knows before they are looked at in detail.
Seb
Member
Thu Mar 08 12:50:17
Jergul:
1. 0.00001 joules correspond to 62 million gammas.

5 would be an unambiguous signal.

2. 4% of the bombs yield is prompt gamma. 12% secondaries.

3. There isn't 75% loss of energy. 50% of gammas from a given source have a scattering event in 10m of water. Scattering is a generally small angle deviation and minute loss of energy. The gamma isn't gone. Your analysis is assuming the gamma source is only detectable if the gamma passes to the detector without interacting with anything in the way.

In snooker terms, you are saying it's impossible to post a ball unless there is a direct line between it and the pocket.

jergul
large member
Thu Mar 08 13:21:47
Seb


1. Redo your calculation on gamma.
2. 2% gamma
3. 50% loss of gamma per 5 m. 75% per 10 m Relative to a trajectory to plates in MEO. Includes scattering and loss of energy.

The problem is easily reduced to simple probability seb.

I merely understand the compound effects of scattering better than you seem to do.
jergul
large member
Thu Mar 08 13:33:05
We are incidentally spending more time and energy arguing the pros- and cons of a 500 billion investment in detection systems than Congress does on any of its 500 billion dollar spending allocations.
jergul
large member
Thu Mar 08 13:41:41
(or to put it another way - if you are arguing low energy gamma, then you would want to reduce the halfing distance in water correspondingly).
Seb
Member
Thu Mar 08 17:13:36
1. 0.00001 joules = 6*10^7 meV
2. 5% yield in prompt radiation, of which 80% gamma
3. No. You are using beam attenuation which treats a scattered photon as not part of the beam. The photon hasn't dumped 1mev into the atom.

Let's zoom in on 3 over the next couple of days. We can use single photon dynamics to demonstrate your error here.
Seb
Member
Thu Mar 08 17:44:05
The system exists Jergul, it's called NDS.

I am using 1-10 meV.

100-500 keV are low energy gamma (photo electric effect dominant)

10MeV and higher are high energy (pair production dominant)
jergul
large member
Thu Mar 08 17:59:22
Seb
If you could specify what value you are attributing gamma, then we could set the half-distance in water more conclusively.

I suspect you are moving into cm territory.

My source said 2% gamma. The difference is a bit indifferent. Similar to say silt and casing, or the atmosphere.

I am just using probability theory. At every halfing threshold, the number of particles on a trajectory that can reach MEO is reduced by 50%.

We can certainly tweak it all somewhat (even a particle trajectory change still has more than 25% of being on course to reaching MEO) and include errant particle trajectory recovery (by every new halfing point for errant particles, a certain fraction will course correct to again be relevant).

Its more a probability thing than a single photon thing.
jergul
large member
Thu Mar 08 18:17:22
11-34 cm HVL (half value layer) for water 1-10 meV
jergul
large member
Fri Mar 09 01:05:52
An analogy could be a blind, drunk man tasked with threading a needle in a single jab across a 4 km (2.something mile) wide field.

He takes a step at a time. Each step will either go forward relative to the direction he is facing(50%) or to the left or right (25% each). With each step, there is a chance he will simply give up and lay down to sleep.

What is his chance of threading the needle?
Seb
Member
Fri Mar 09 02:28:50
Jergul:

Jesus jergul.

Energy range for fusion gamma is 1-10MeV. This is stated several times. E.g mar 08 05:12:21.

Your source sounds like it's talking about a bombs.

A 50% difference in gamma flux isn't the casing.

Your probability calc is answering the wrong question - a gamma initially not on a trajectory to intercept a detector can be scattered into a trajectory that does. And those that were initially were and are scattered more than once may also end up on track.

The only way to do the calculation is to integrate over all angles, all impact parameters, all energies (allowing for energy loss over time changing the cross section) until you get the intensity profile at the surface. Hideous. There's probably a trick but I've forgotten it.

That's officially too much work. I'm thinking about what can be done with single photon dynamics to show it's plausible.

Again, don't confuse hlv for beams with hlv for photons (good way to get killed when designing shielding).

In your analogy (yes, a random walk, see post on the 8th referenced above) the distance values you are quoting are based on the idea that as soon as he steps left or right, he lies down to sleep, not how far before he falls asleep. I.e. yes, it is vanishingly unlikely any photons will reach the surface without being scattered. It does not follow that this means no photons will reach the surface having been scattered many times.

The question is how many times, and how much energy has been lost. Compton scattering of gammas in 1-10mev range off low z material is really inefficient energy transfer mechanism, the energy too low for pair production, and too high to do photo electric effect.
jergul
large member
Fri Mar 09 03:00:36
Seb
Scattering will tend to see gamma wander around pretty close to the source until they are degraded.

I have never said impossible, just that it is highly unlikely any gamma will pass through the detector plates required to register.

No, the idea is based on he might give up and go to sleep at every step. Not that there is any certainty that he does.

Probability theory seems a way more elegant tool than what you are considering.
Seb
Member
Fri Mar 09 03:43:57
Jergul:

If they were billiard balls, maybe.

But the physical mechanisms underpinning compton scattering mean small angle scattering is far far more likely than large angle scattering. Analogy is electrons and heavy ions. You need many small angle scattering events until you get the equivalent of randomising the particles direction that you get with solid pcle collidions. That then gives you your effective mfp which is >> than the mfp between compton scattering events. From that you can then do diffusion model. Thinking about it, that's probably the best analytical approach if not doing a numerical model. But unlike the coulomb law, Compton scattering fundamental interaction is going to be some horrible qm. There might be a paramterisation.

RE your model, what I'm saying is the values and hlv you are using correspond to beam attenuation which assumes the man goes to sleep if he scatters (when in fact he is lost from the beam, but not gone). I.e. there is a mismatch between the model you have and the parameters.
jergul
large member
Fri Mar 09 04:03:10
Seb
Small angle scattering is obviously not included in the half-distance as small angles would not be considered scattered (the small angle gamma would still pass through the barrier material).

Essentially, the only scatter that can be considered in a halfing calculation is scatter of 90 degrees or greater.

The man will be overwhelmingly likely to fall asleep for as long as the shielding is thick enough.

The best analytical approach presented so far is probability theory.

You should be able to visualize the spread pattern scattering entails. Gamma overwelmingly likely to wander about close to its initial source until degraded.
Seb
Member
Fri Mar 09 04:31:56
Jergul:

In a word, no.

Small angle scattering matters for working out beam intensity which is useful in many applications.

e.g. in radiotherapy.

Clinically relevant doses are achieved by intersections of well collimated beams from e.g. cobalt sources. This is achieved by putting the source in a very thick sided tube with a small hole at the end.

Filters of shielding material placed in the tube are used to calibrate the intensity in each beam.

In those situations small angle scattering events take photons out of the beam then either into much thicker shielding of the tube, or sufficiently far away (geometry) from the intersection point that they don't matter clinically and so can be neglected because the linear combination of scattered photons intensity is well below the dose that would be dangerous.

What matters there is the intensity of the beam, hlf for beam intensity is different to the 50% probability depth of penetration for a single gamma or even effective mfp of a gamma for direction to be randomised.

There are a number of such applications - indeed most applications involve creating beams this way - and most literature on hlv for a beam is based on those applications.


If you look at wherever you are getting your data from, there's probably a caveat about how you shouldn't use it for shielding as it leads to gross underestimates.

For shielding, there are some parameterized corrections called build up factors, but they break down quite badly in the kinds of regime we are looking at which are many orders of magnitude higher in intensity than the experimental derived semi-emperical parameters for build up factors.

Consider why spent reactor rods sit under lead bricks and at least ten metres of water (and why radiation alarms go off if the water gets below 10m) before transferring out off the core containment in most nuclear reactors.

Spent nuclear fuel is nowhere near as active a source of gammas and neutrons compared to an exploding megaton nuclear bomb, yet is capable of delivering doses through lead bricks and tens of metres of water.

Surely you can see there's an inconsistency in your numbers here - if most of the gamma from a nuke was screened out from a nuclear blast by a few cm lead as a few meters of water, then you wouldn't need anything like that much screening in a power plant.

Seb
Member
Fri Mar 09 04:35:55
Re wandering around, sounds like you've made a conceptual boo boo.

If the average direction of the particle is continually randomised it still doesn't mean displacement from source over time is 0.

Random walk is the physical mechanism for diffusion. Cf. Einstein.

So actually what you get it a diffusion of gammas from the point source until such time as they lose all their energy.



jergul
large member
Fri Mar 09 05:00:43
Seb
In a word yes.

Boy, that was fun.

You seem to have made a conceptual boo boo by not quite grasping what half-distance means. It is the distance at which half of gamma will penetrate a given material.

Its fine that you feel civilian nuclear power plants are overdimensioned and have fail safe redundancies. Your point?

Most gamma would remain relatively close to its source. Most and close being relative terms. 500 kilotons of gamma energy is still a hell of a lot even when halved every 30 odd centimeters.

I think you made a comprehension reading boo boo. I never said displacement from source is 0.

"You should be able to visualize the spread pattern scattering entails. Gamma overwelmingly likely to wander about close to its initial source until degraded."

Seb
Member
Fri Mar 09 08:01:48
jergul:

Yeah, except I have several paragraphs explaining why no, whereas you have not addressed that.

"It is the distance at which half of gamma will penetrate a given material"

No, it is the distance after which an inicndent beams intensity will be halved. I.e. it doesn't look at intensity of radiation off-beam.

The values you are using are consistent with HLVs for water conforming to that definition, but sure, show me the source of your data, and I'll show you the difference.

"I never said displacement from source is 0."

That's why I said "think" - "Gamma overwelmingly likely to wander about close to its initial source until degraded." isn't a precise statement and, but if I were to pick it would make it sound like "the gammas will bounce back and forth around the source because their direction is randomised" rather than "Gammas will diffuse out a certain distance from the source, depending on how quickly they lose energy in compton scattering events, before being fully absorbed".

The question is "how far will they diffuse".

And the HLV numbers you quote seem to be the wrong parameter for this given they look to correspond to the distance at which half the photons in the beam has undergone scattering, rather than the distance at which half the photons have lost all their energy.



Seb
Member
Fri Mar 09 08:02:33
(most of the energy lost from the beam in the former is still being carried by gamma photons, just gamma photons that are not part of the beam anymore).
jergul
large member
Fri Mar 09 10:20:35
Seb
I have one word explaining it: "Half-distance".

It corresponds to scattering, not energy loss. Obviously.

"You should be able to visualize the spread pattern scattering entails. Gamma overwelmingly likely to wander about close to its initial source until degraded."

"Scattering will tend to see gamma wander around pretty close to the source until they are degraded."

"He takes a step at a time. Each step will either go forward relative to the direction he is facing(50%) or to the left or right (25% each). With each step, there is a chance he will simply give up and lay down to sleep."

The question "how far will they diffuse" is answered by probability theory and the half-distance of the material in question.

The distance is measured in the 10s of meters in water (depending on how many fractional grams of TNT equivalent you want to go down to) Which leads us to understand why water at NPPs are called coolant, not shielding. To describe its primary purpose - sequestering energy from radiation.

It also helps us understand why the ban on nuclear tests stipulate hydrophones, not gamma detectors to detect nuclear explosions under water, or why gps satelittes are equipped with xray detectors, not gamma ray detectors, or why NDS is clear that it detects atmospheric and space detonations, not subterranian, or subsea detonations.

So no, a gamma ray detector in MEO will not be detecting any gamma from a nuclear detonation under water.

Shall we move on to hydrophones and seismic detection?
jergul
large member
Fri Mar 09 10:21:29
under 2km of water*
Seb
Member
Fri Mar 09 17:02:51
Jergul:

I'm glad you now accept the network exists. Even if you are going to continue to pretend the NDS packages on NAVSTAR don't detect gammas. Hard x-ray detection is done with scintillators which also detects gammas - a bit of instrumentation knowledge is useful here. Note, as x-rays lose energy to the atmosphere more quickly than 1mev gammas, that tends to suggest that a scintillator that can reliably pick up x-rays would be more sensitive as fewer x-rays would get to the detector than gammas would.


I suggest we talk hydrophones as they are the first thing I suggested. We are not going to make progress on gammas until I have an hour at a desk free of other concerns, so let's park that - and the issue is "can the US detect a megaton nuclear explosion 1-2 km underwater" (yes, in a number of ways) and does it have infrastructure to do so (yes, and demonstrably so).
Cthulhu
Tentacle Rapist
Fri Mar 09 18:19:50
I bet your mom's ass is big enough to block all emissions!
jergul
large member
Sat Mar 10 03:32:11
Seb
You are using way to many strawmen. It makes you seem desperate to maintain smartest man in room syndrome at any cost.

Any source on what xray imaging sensor the gps satelittes are equipped with.

The design specification would be something that can detect atmospheric and space explosions reliably.

The question is if the US can detect and analyse a nuclear detonation in a manner that gives nuclear response decision makers robust data within a 48 hour window.
jergul
large member
Sat Mar 10 03:36:55
Also, your challenge is more analytical, then for lack of time to sit at your desk.

Radioactive isotopes will move considerably further though mechanical forces than by wavelength through water.
jergul
large member
Sat Mar 10 03:40:23
You can rephrase the dataflow question to "Do US decision makers consider every nuclear test, conventional explosions that resemble nuclear tests, and natural events that resemble nuclear tests as potential nuclear attacks?"
Nimatzo
iChihuaha
Sat Mar 10 04:01:39
”You are using way to many strawmen. It makes you seem desperate to maintain smartest man in room syndrome at any cost.”

Here is a guy, who is not scared of throwing grenades in the glass fort.
Seb
Member
Sat Mar 10 06:03:09
Jergul:

Any reason to think they would be substantially different from the ones on VELA which held the same mission and detected GRBs?

Any reason to think they are not scintillators, being the best instrumentation for measuring hard x-rays and which necessarily detect gammas?

Amy reason you have ignored the specific links I've posted that state NDS packages do have gammas?

You are hanging a lot on a source that's simply conflated gammas and x-rays.

Deep underwater explosions will produce radio isotopes but probably not intense enough sources to be seen from space.

I covered radio isotopes in the last thread.
Seb
Member
Sat Mar 10 06:04:00
1-2km deep, radio isotopes are going to be very distributed by the time they are upwelled to the surface.
jergul
large member
Sat Mar 10 06:29:04
Seb
Feel free to repost that NDS packages can detect gammas. That would be a good indication that they in fact do that.

I was mentioning isotopes as an example of something that would spread much further than gamma through water.

It would be more interesting a consideration.
Seb
Member
Sat Mar 10 10:31:33
Jergul:

I think we already covered it though - you didn't pick up on it to dispute, focusing instead on the acausal idea that subsequent landslide could somehow obscure the preceding nuclear detonation.



Seb
Member
Sat Mar 10 11:16:51
http://boo...AhWkJZoKHTknD-kQ6AEwBHoECAAQAQ


P296

jergul
large member
Sat Mar 10 15:19:49
Sigh, you could have found something a bit more conclusive. Here is the device:

"Los Alamos National Laboratory (LANL) Combined X-ray and Dosimeter (CXD)"

Data has been released this year by EO for weather stuff. Saturday night and all (wife beckons) but thanks to Trump we can probably see exactly what the CXD does.

What is a nuclear blast signature (it follows the same principles for both hydro-accustic and seismic monitoring)?

Or more specifically, how much time does it take a blast to sign its name on seismic and hydro-accustic monitors?
Seb
Member
Sun Mar 11 12:11:17
Jergul:

So your argument is based on the name of the device rather than a statement of its capability and a consideration of the physics of the instrumentation?



jergul
large member
Sun Mar 11 12:15:29
Seb
Uhm, no. Now that we know the name, we can look at the specifics. Just for kicks as my argument was demonstrated long ago (gamma is not making it to MEO).
Seb
Member
Sun Mar 11 12:20:07
But using that as a search term (I'd fpund stuff on the older bdx) you can get a publication that goes into the energy range for the high end sensor.

It's a scintillator, and it covers 500-1500 kev (low energy gammas) and 1.5kev upward.

So yeah, they can call it an x-ray sensor but we are talking about photons in the 500mev above. We can argue whether that's a gamma ray or an xray, but we are happily into the 1-10 1mev range from nuclear destinations commonly referred to as gammas.
jergul
large member
Mon Mar 12 04:23:16
Seb
The Executive Order has to have involved the release of full instrumentation information. The data released is otherwise not valid.

Yes, it is a scintillator (a very broad term). The purpose was to establish what wavelength of particles it scintilated. That depends on the device specs.

Of academic interest as the details are no longer pertinent to the point and even older designs do cover the gamma spectrum (I fully read and understood an older declassified specification)

Seb
Member
Mon Mar 12 04:43:15
Well I'm glad you're satisified there are indeed gamma ray sensors designed for detecting nuclear explosions on gas.

So, absent me finding a few spare hours to do a painful calculation, let's discuss hydrophones and seismic.
Seb
Member
Mon Mar 12 09:01:10
Oh wow. So I've been fascinated that for some reason people are describing a 50 MeV photon as an x-ray.

Back in the day - 2000ish - my lecturers were always a bit vague about the formal differentiation. Generally we'd call them xray if < 500kev and gamma if above, though weirdly any photon in nuclear and particle was always a gamma even if it's in the lower kev.

Apparently the formal definition is it's a gamma if it's coming from a nucleus or a cosmic source, and an x-ray if it's coming from electron dynamics. (N.b. We now think most gammas from cosmic sources are actually from electron dynamics).

That's the most fucked up definition in physics - which is possibly why I'd never come across it. It's frankly embarrassing.


Seb
Member
Mon Mar 12 09:03:33
N.b. This means the reason these are described as xrays is because technically, in terms of energy, most is the 100kev and higher radiation they'd be detecting would be brhemstrahlung which apparently is in the x-ray box.

jergul
large member
Mon Mar 12 09:13:19
Seb
Yah, I noticed the fuck up definitionwise while researching. Better to just use the eV range to avoid the issue entirely.
Seb
Member
Mon Mar 12 10:58:52
https://cdn.miniphysics.com/wp-content/uploads/2011/07/electromagneticspectrum.jpg

This is more in line with the kind of thing I think about.

KreeL
Special Member
Wed Mar 14 20:56:49
Jergul wins.
show deleted posts

Your Name:
Your Password:
Your Message:
Bookmark and Share