Jump to content

Need vaues for critical conditions of Uranium 235


Recommended Posts

Right, and I'm just trying to do the mathematics to prove that the same events hold true for other frames of reference. And the college I'm at already said they'd be interested in doing a summary/story/article on the paper once its finished, so might as well finish it. I already have the shape stuff figured out if I assume that 20.21g/cm^3 is remotely accurate. Using that number I set up a system of equations and solved for the velocity necessary to create length contraction and mass increase that creates a density increase from 18.075g/cm^2 to the critical 20.21g/cm^3. Now all I need is an equation that can be used in both the rest frame and relativistic frames to show that even though the density has the illusion of being 20.21g/cm^3, the probability of a chain reaction is still close to 0. For that, I need to find an equation for the probability of a chain reaction that is a function of whatever time based stuff causes a chain reaction on its own, assuming that mechanism is correct which Swansont never cleared up, and then anything that might be affected by length contraction which will effect the cross section and probability of slow neutrons striking a nucleus, like the average radius of nuclei and neutrons. I guess I will just have to let the paper not be completely accurate and state that I made assumptions for the sake of seeing the problem through.

MW - you are barking up the wrong tree. Please re read what I wrote above and what Sensei explained in more detail. You get no change in the frame of the bomb! It doesn't matter what we see if the bomb is moving quickly.

 

You must understand that velocity is relative - the same laws apply, in fact the situations are identical. It might seem that I whizz past a stationary you at half the speed of light or you whizz past a stationary me at half the speed of light; but they are identical in physics, in observations, and in reality.

 

You are holding a just about subcritical lump of uranium stationary in space - I whizz past you at .5c. This situation MUST be the same as I am stationary and you whizz past at .5c holding a just critical lump of uranium. The bomb goes off in neither scenario because - as constant motion is relative - both scenarios are actually the same. For the scenarios to be the same we do our sums using transforms, using invariant mass etc; we can thus always change our perspective to that of the frame of reference of the centre of mass of the bomb ie where the bomb is stationary

Link to comment
Share on other sites

Please read my text more thoroughly. I already stated that a bomb that does not detonate in one frame will not detonate in another.

 

As for sensai, I keep trying to tell you that yeah, I agree you still have the same number of atoms as you had at rest, but the coordinate system of an object still rotates, and it rotates in such a way that the effects of time dilation and length contraction are the same as in a gravitational field, so it appears as though mass increases, even though the number of atoms stays the same.

Link to comment
Share on other sites

 

With reference to Swansont, I did actually find the six and four factor formulas, but I guess in my ignorance I still do not see how to use that formula. From what I see, that formula only calculates the neutron multiplication factor. Is that what I need? I don't know, it seems like there's more to probability than a neutron multiplication factor.

 

Yes, you would have to learn the science behind that formula. There is probably no simple answer to what you're asking, no simple plug-and-chug formula for this.

 

Criticality is when the factor = 1. You want prompt supercritical, which happens at a slightly higher value, and you have to modify it for fast neutrons causing fission rather than thermal neutrons.

Link to comment
Share on other sites

Hmm. Maybe instead of focusing on probability, I could focus on conditions for criticality to show that critical conditions still aren't met even though it appears as though it's at a higher density.


Could I use a 4-factor formula an in infinite medium to accomplish that, with all necessary factors affected by time dilation, length contraction and relative mass increase taken into account with that equation?

Link to comment
Share on other sites

Hmm. Maybe instead of focusing on probability, I could focus on conditions for criticality to show that critical conditions still aren't met even though it appears as though it's at a higher density.

Could I use a 4-factor formula an in infinite medium to accomplish that, with all necessary factors affected by time dilation, length contraction and relative mass increase taken into account with that equation?

 

An infinite medium is a poor assumption for this case. Also (again) what relativistic factors? There are no explicitly relativistic terms in the formula. Any effects are already accounted for, e.g. if the cross section has a dependence on the neutron speed, it's already accounted for in the value you use. Not that you'd expect there to be any relativistic correction necessary for a ~1 GeV particle that has a few MeV of kinetic energy.

Link to comment
Share on other sites

Why is an infinite medium a poor assumption in the vacuum of space? Also, did you read the paragraphs where I explained the chunk of Uranium is traveling considerably close to the speed of light? That's where relativity comes in. At the very least, I know for a fact that time dilation would affect the half life and specific activity when measuring it from Earth.


If you can help me with this last hurdle, I could reference you in the paper unless you wish to refrain from that. Although I don't think it is formally permissible to omit a source of information. If I were to reference you as an advisor (or something else??), how would you want it?

Edited by MWresearch
Link to comment
Share on other sites

Why is an infinite medium a poor assumption in the vacuum of space?

 

I thought this was about a critical mass of U-235?

 

 

Also, did you read the paragraphs where I explained the chunk of Uranium is traveling considerably close to the speed of light? That's where relativity comes in. At the very least, I know for a fact that time dilation would affect the half life and specific activity when measuring it from Earth.

Half-lives don;t really matter for criticality calculations. As for transforming the other terms into a moving frame, I don't know of anyone that's bothered to do that.

 

 

If you can help me with this last hurdle, I could reference you in the paper unless you wish to refrain from that. Although I don't think it is formally permissible to omit a source of information. If I were to reference you as an advisor (or something else??), how would you want it?

I've said nothing that's not easily found from multiple sources as basic sources of information, and not something one would reference in a paper.

Link to comment
Share on other sites

It's not about the critical mass of U235, but apparently it's how the neutron multiplication factor must stay the same regardless of the frame of reference. It is essentially what everyone has been saying and assuming I didn't understand, except in addition to just stating the obvious like that the number of atoms is intrinsic, I'm trying to actually show it mathematically in conjunction with the conditions of a nuclear detonation. Even though a chunk of U235 can be accelerated to a speed such that the length contracts and mass appears to increase in such a way that the density appears to be critical, that doesn't necessarily mean that k is greater than or equal to 1. But, the mathematics to actually show that involves both nuclear physics as well as relativity. You said no one bothered transforming such values into other frames, well perfect, that will make the paper even more unique.

Edited by MWresearch
Link to comment
Share on other sites

You are familiar with chemistry I see, then you are aware that some circumstances in chemistry change as a result of relativity, like the fact that the electrons are traveling close to the speed of light around massive nuclei. The principal of this shouldn't be too much different than that.


Things like the specific activity and the rate at which neutrons are given off will be slowed, and I think the atomic radii will contract along the direction of motion which will probably affect some cross section variable.


The same principal is used to explain muon decay in the atmosphere wherein muons reach Earth's surface before decay because of the time dilation lengthening the amount of time before they decay.


If you're not familiar with it which would be completely understandable since industry standards might not warrant a relativistic model, you can google "relativistic chemistry" and the first few links should show what I'm talking about.


You will see than even Planck's constant is affected by relativity.


EDIT: After rereading the thread, I realized "affects Planck's constant" might be misleading. Rather, relativity affects equations involving Planck's constant like relative mass and energy.


At this point though I think I see what Swansont is saying about the probability. Since a chain reaction would never occur when the Neutron Multiplication Factor is less than one, I simply have to show that k (the NMF) is less than one in both frames.


Is that correct Swansont?

Link to comment
Share on other sites

You are familiar with chemistry I see,

But you are not..

 

then you are aware that some circumstances in chemistry change as a result of relativity, like the fact that the electrons are traveling close to the speed of light around massive nuclei.

Electrons don't travel around nucleus with speed close to speed of light..

Link to comment
Share on other sites

I assume you are thinking I am mistakenly using a classical Bhor planetary model. But, even though electrons propagate as waves, that does not mean that oscillations or measurements of changes in position are incapable of yielding near-light velocities in atoms. Thus, we have a corrected form of chemistry called "relativistic chemistry."

Edited by MWresearch
Link to comment
Share on other sites

But you are not..

 

 

Electrons don't travel around nucleus with speed close to speed of light..

It's a tricky point.

Since the electron doesn't have a well defined location it's not easy to define a rate of change of that location.

However MWresearch is quite correct in saying that the electron behaviour within heavy atoms is affected (strongly) by relativity.

It's widely cited as the reason for the colour of gold and the low melting point of mercury.

 

Who cares?

Electrons are not involved in nuclear fission.

 

Fundamentally, if, from your point of view, the uranium is moving at a velocity near the speed of light then the energy it carries as a result of that velocity is bigger than any it will release by fission.

If, on the other hand, you are more or less stationary WRT the uranium, the relativistic effects will be small.

Link to comment
Share on other sites

It's a tricky point.

Since the electron doesn't have a well defined location it's not easy to define a rate of change of that location.

IMHO it's easy by looking to what velocity given energy is accelerating particle.

If I have free electron at rest, and give it f.e. 100 eV, it'll be accelerated to v=5931000 m/s.

 

Electron accelerated to 0.866c has double it's mass!

Half from invariant mass-energy, and half from kinetic energy.

Such electrons are clearly visible in Cloud Chamber, as long thin traces..

They're ionizing medium they're flying by (and losing kinetic energy, slowing down).

If they slowed down in CC to our frame, then how can they *still* travel at speed of light, inside of some nucleus.. ?

 

Now look what energies are needed to ionize elements.

Last electron ionization energy (when element has Z protons, and 1 electron only) E=13.6*Z^2

for Z=1, E=13.6 eV (Hydrogen)

for Z=2, E=13.6*2^2 = 54.4 eV (Helium)

....

for Z=118, E=13.6 eV * 118^2 = ~190 keV (just approximation)

 

This pattern is easy visible in this table:

http://en.wikipedia.org/wiki/Ionization_energies_of_the_elements_(data_page)

 

It's nowhere near 511 keV kinetic energy for v=0.866c, not to mention higher values..

 

Fundamentally, if, from your point of view, the uranium is moving at a velocity near the speed of light then the energy it carries as a result of that velocity is bigger than any it will release by fission.

If Uranium traveling near speed of light will decay, products it will emit will be f.e. blue shifted and observer at rest will notice even more powerful particles than he would expect from Uranium at rest.

Otherwise energy conservation would be violated..

 

Another thing:

when something is traveling fast, photons are blue (red) shifted (depending on direction),

so shouldn't the same be with MBR?

If we have Deuterium traveling so fast that MBR (not to mention visible light from stars) is blue shifted to E=2.2+ MeV then it's enough energy for photo disintegration of nucleus, and splitting it to proton and neutron, yes? no?

(to blue shift 2.33 eV green photon to 2.3 MeV there is needed 1 mln blue shift... quite extreme situation, and even more for MBR->2.2+ MeV)

Link to comment
Share on other sites

As I said, the effects are well known and well documented.

http://en.wikipedia.org/wiki/Relativistic_quantum_chemistry

 

The electron has a finite probability of being arbitrarily close to the nucleus and, in that circumstance it's "velocity" is high enough for relativistic effects to be significant.

I really don't see why you are arguing about this well established physics.

 

Incidentally, you don't seem to have yet provided an answer to the questions asked here

http://www.scienceforums.net/topic/87510-sulfuric-acid-in-ppm/?p=850748

Link to comment
Share on other sites

I still can't figure out if the mechanism is set up right though, john. So if I have critical mass in say, a bare sphere of 17cm diameter with 52kg of assuming pure U235 (but without knowledge of the actual grade used), the delayed neutrons alone in that sphere will start a chain reaction themselves? Because if that's true, all I need to do is show that the effective neutron multiplication factor stays less than one in both frames, but if I need some initiator like a neutron gun or plutonium pit, that complicates things a lot more than is necessary to show the point of the research.

Link to comment
Share on other sites

If I'm interpreting that information correctly, a critical mass piece of U235 in a critical spherical volume will explode on its own but it will take a little time to ramp up since the number of fission events follows an exponential growth? But, wouldn't the neutrons released by the first fission increase the number of fission events more than just delayed neutrons? Or does that equation consider that delayed neutrons initiate the reaction and then the fast and slow neutrons resulting from critical events start the take over and sustain the chain reaction?

Edited by MWresearch
Link to comment
Share on other sites

 

At this point though I think I see what Swansont is saying about the probability. Since a chain reaction would never occur when the Neutron Multiplication Factor is less than one, I simply have to show that k (the NMF) is less than one in both frames.

Is that correct Swansont?

 

A stronger statement is that k is the same in all frames. Since k is normalized to a "generation", or cycle, of the process, time dilation should not (directly) affect it.

If I'm interpreting that information correctly, a critical mass piece of U235 in a critical spherical volume will explode on its own but it will take a little time to ramp up since the number of fission events follows an exponential growth? But, wouldn't the neutrons released by the first fission increase the number of fission events more than just delayed neutrons? Or does that equation consider that delayed neutrons initiate the reaction and then the fast and slow neutrons resulting from critical events start the take over and sustain the chain reaction?

 

My understanding is that a critical mass will get hot and force itself apart, rather than go "boom" . You have the possibility that thermal expansion alone would make it subcritical.

Link to comment
Share on other sites

I also think k is the "same" but I still think it's worth showing mathematically. Off the top of my head, I would guess that the gamma transformations cancel out with some being in the numerator and some being in the denominator.

 

As for the temperature increase, is there anything I could introduce that would stop the temperature from increasing that wouldn't affect the dimensions and density of Uranium that much? Otherwise, I'm neglecting a lot of things. If I neglect the density expansion due to the energy released in fission events, is it scientific still? Just how much neglecting can I get away with?

Link to comment
Share on other sites

I also think k is the "same" but I still think it's worth showing mathematically. Off the top of my head, I would guess that the gamma transformations cancel out with some being in the numerator and some being in the denominator.

 

 

k is not time dependent, it is cycle dependent, and everyone agrees that a cycle has occurred (fission —> neutrons —> next fission)

Link to comment
Share on other sites

But a cycle takes time, so on an individual scale, it must be affected by time dilation even though the final result after all calculations should show it isn't. Not everyone can agree that a cycle occurs at the "same time" as I'm sure you know and they also can't agree that a cycle takes the same amount of time to happen. But that's beside the point, I already know it's not suppose to change between frames, I'm just trying to show it mathematically in a particular circumstance. Can you address the point about the temperature?

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.