Theory for cause of metastable state - Establishing causality in nature again -

[swansont]

More rapidly means the information from the outside is coming faster in. There are many ways - for instance a higher pressure and temperature in a gas. I could measure a dependency from the pressure. You suggested an electromagnetic field. In space are the times longest, because a single atom is very alone.

 

Information? We're talking about spontaneous decays. Higher temperature and pressure will lead to collisional broadening of the lines, but this is understood as an interaction and not the spontaneous reaction. Limiting ourselves to spontaneous reactions means there are no outside influences.

 

 

What is "Paschen notation"?

I made my research 1966 - Mr. Paschen was obviously later - or?

 

I take it that means no. So what is your notation, then?

 

You may investigate the angular momentum. My theory opens up an entire field of new research.

 

If you haven't already investigated the angular momentum to see if the transitions are disallowed because of the selection rules, then any new proposal is premature and presumptuous. You have to at least investigate that these states should have an allowed transition before you can claim that there is no explanation for them.

 

You have not yet presented your definition of metastable states that other physicist can compare it side by side.

Otherwise we get lost in details.

 

It's not "my" definition. I'm discussing standard physics. Metastable states are those for which there is no allowed first-order transition. (i.e. no intermediate coupling required) which gives rise to longer lifetimes, as compared to those states with allowed transitions.

[Wolfhart Willimczik]

Information? We're talking about spontaneous decays. Higher temperature and pressure will lead to collisional broadening of the lines, but this is understood as an interaction and not the spontaneous reaction. Limiting ourselves to spontaneous reactions means there are no outside influences.

 

 

 

I take it that means no. So what is your notation, then?

 

 

If you haven't already investigated the angular momentum to see if the transitions are disallowed because of the selection rules, then any new proposal is premature and presumptuous. You have to at least investigate that these states should have an allowed transition before you can claim that there is no explanation for them.

 

 

It's not "my" definition. I'm discussing standard physics. Metastable states are those for which there is no allowed first-order transition. (i.e. no intermediate coupling required) which gives rise to longer lifetimes, as compared to those states with allowed transitions.

..and why are there "no allowed first-order transitions" ?

This definition explains not why are there metastable states!

That is the point - everything else you mention is something else.

I have found an explanation why metastable states exist – who else?

You could actually use my definition - test it.

[John Cuthber]

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

so, when you say "I made my research 1966 - Mr. Paschen was obviously later - or?"

It's obviously not true.

So it's clear that you have not studied the subject in any depth

 

"You may investigate the angular momentum. My theory opens up an entire field of new research."

It wouldn't be new research: the effect of angular momentum on the lifetime of excited states is already known.

Again, it is clear that you have not done the research.

 

"..and why are there "no allowed first-order transitions" ?"

usually, the conservation of angular momentum

"This definition explains not why are there metastable states!"

Yes it does- as long as you understand it.

"That is the point - everything else you mention is something else."

is meaningless

"I have found an explanation why metastable states exist – who else?"

For a start, no you have not because (as has been pointed out) it is meaningless.

But in answer to the "who else?" it was a bunch of people roughly a hundred years ago.

"You could actually use my definition - test it."

No we couldn't use it because it isn't clear.

Even if it were, what would we gain?

There's already a known working answer.

[swansont]

..and why are there "no allowed first-order transitions" ?

This definition explains not why are there metastable states!

That is the point - everything else you mention is something else.

I have found an explanation why metastable states exist – who else?

You could actually use my definition - test it.

 

I explained this back in post #25. Photons only carry 1 unit of angular momentum. Atomic states that differ by something other than that are forbidden. As I said before, this is not a mystery, it is introductory QM. You only need a new theory AFTER you have looked and found that QM doesn't explain it.

 

I've raised an objection to your definition: what is the choosable parameter that exists for a metastable state that does not exist for a normal excited state?

[Wolfhart Willimczik]

I explained this back in post #25. Photons only carry 1 unit of angular momentum. Atomic states that differ by something other than that are forbidden. As I said before, this is not a mystery, it is introductory QM. You only need a new theory AFTER you have looked and found that QM doesn't explain it.

 

I did not investigated angular momentum within metastable states.

You are free to establish your own theory, but remember the rules there in “speculations” – you have to prove it.

You have to make experiments – like me – to show what will provide the missing angular momentum to end the metastable state.

Everybody has an objection. This is not a creative move.

 

this is not a mystery, it is introductory QM. You only need a new theory AFTER you have looked and found that QM doesn't explain it.

Ergo you have to admit I am the first physicist offering an explanation why metastable states exist. A small step in the right direction.

 

I've raised an objection to your definition: what is the choosable parameter that exists for a metastable state that does not exist for a normal excited state?

I explained that there are in any case another free choosable parameter. Please read my entire article.

Edited April 16, 2013 by Wolfhart Willimczik

[swansont]

I did not investigated angular momentum within metastable states.

You are free to establish your own theory, but remember the rules there in “speculations” – you have to prove it.

 

I'm citing established, mainstream physics. The discussion can be found in most QM texts. That rule applies to the people proposing non-mainstream ideas, for which the evidence is not readily available in easily-accessed literature.

 

You have to make experiments – like me – to show what will provide the missing angular momentum to end the metastable state.

Everybody has an objection. This is not a creative move.

 

Avoiding discussing objections is not a creative move.

 

Ergo you have to admit I am the first physicist offering an explanation why metastable states exist. A small step in the right direction.

 

No, as I've said several times now the explanation for the existence of metastable states is part of standard physics. Literally thousands of physicists have previously offered an explanation for metastable states — every time they teach an intro QM course.

 

That you are not familiar with selection rules does not lend credence to your claim that you are a physicist.

 

I explained that there are in any case another free choosable parameter. Please read my entire article.

 

I'm asking you to explain here what that parameter is. Please stop avoiding the question.

[Wolfhart Willimczik]

I'm citing established, mainstream physics. The discussion can be found in most QM texts. That rule applies to the people proposing non-mainstream ideas, for which the evidence is not readily available in easily-accessed literature.

 

 

Avoiding discussing objections is not a creative move.

 

 

No, as I've said several times now the explanation for the existence of metastable states is part of standard physics. Literally thousands of physicists have previously offered an explanation for metastable states — every time they teach an intro QM course.

 

That you are not familiar with selection rules does not lend credence to your claim that you are a physicist.

 

I'm asking you to explain here what that parameter is. Please stop avoiding the question.

Then please copy at least the right passage - why metastable states exist - what the missing angular momentum provided etc. Until now you have no answer provided.

 

Objections are only valid if at least my article has been read, what is obviously not the case.

 

 

I say again: show it.

If you can't do it I will do it by myself.

 

What have selection rules to do with the topic?

I am the one who generated the first time traces from a single alpha particle. If you thing this could be done only by a non-physicist your opinion would shed a dim light on "real physicists." (I though we dropped personnel attacks?)

 

 

If you would read my article you would find several samples where I pointed out the free parameter:

 

5.1 Water:

Cocking clean water in a clean container in labs lead sometimes to an metastable state, that overheated water will not boil, because any point on the bottom has the same chance to create the first steam bubble. There is no singularity. This is a free choose able parameter for the first bubble. Chemist solve this problem simply by putting a small stone on the bottom. Now exist a singularity - a point on the bottom, what distinguishes itself from all other spots. An explosive boiling is prevented. It seems also the more the normal boiling is delayed the more explosive is the ending of the metastable state. The more extended the delay the higher the difference in the temperature over the normal boiling point of water, the more urgency exists for ending the metastable state and the smaller differences in the parameters may trigger the delayed process. The greater the delay, the more energy is suddenly released, the greater the water explosion, but In the laser technique this effect is desirable.

On the other end very clean water can also exist below the freezing temperature, what is also a metastable state. The free parameter is the location, where the first crystal will occur. Throwing in a small stone defines a certain value of the free parameter and the crystallization begins and goes fast. Normally are there impurities which do the same.

5.1.2. Condensation:

If the humidity in clear air gets over a certain point (dew point) it should condensate as water or fog, but some times nothing happens. A high flying airplane starts the process of condensation with particles in the exhaust and we see the well known stripes in the sky. The free parameter of a metastable state was the exact location of condensation, where the water droplet shell be created. Every particle defines such a point. Every droplet has a particle in it.

etc

Please read my article

Everybody can read the entire article at:

http://science-technology-inventions.weebly.com/

 

I used Groot and Penning. You will not be familiar with this, but you must not know everything.

 

Well,

 

 

Is that Paschen notation for the states?

 

 

Edited April 16, 2013 by Wolfhart Willimczik

[swansont]

Then please copy at least the right passage - why metastable states exist - what the missing angular momentum provided etc. Until now you have no answer provided.

 

That's an awkward way of saying it — "missing" angular momentum doesn't "provide" anything. The system has some amount of angular momentum. But the ground state angular momentum is either the same, or very different. Emitting a photon would not get it to the angular momentum state it needs to be in, thus it can't emit a photon.

 

Conservation of angular momentum and selection rules are physics topics that should not need explanation. This is like claiming why a spinning top doesn't fall over is a mystery.

 

 

I say again: show it.

If you can't do it I will do it by myself.

 

What have selection rules to do with the topic?

 

OMG.

 

Because selection rules describe allowed and forbidden transitions, based on angular momentum. 1st year quantum. As I've explained twice already. You will need to be a lot more specific than say "show it". What parts of the explanations aren't clear?

 

 

If you would read my article you would find several samples where I pointed out the free parameter:

 

5.1 Water:

…

 

5.1.2. Condensation:

 

Water, whether superheating, supercooling or condensing, is not an atomic system described by QM.

 

The closest you come in your discussion is radioactive decay, claiming it needs "outside help". Why do some decays need outside help while others don't? I gave an example of two systems with almost identical energy, with one being normal and the other metastable. You gave an example where this is also true. What is this "outside help" and how does it know to "help" only atoms in some states?

 

I used Groot and Penning. You will not be familiar with this, but you must not know everything.

 

Never claimed to.

[Wolfhart Willimczik]

That's an awkward way of saying it — "missing" angular momentum doesn't "provide" anything. The system has some amount of angular momentum. But the ground state angular momentum is either the same, or very different. Emitting a photon would not get it to the angular momentum state it needs to be in, thus it can't emit a photon.

 

Conservation of angular momentum and selection rules are physics topics that should not need explanation. This is like claiming why a spinning top doesn't fall over is a mystery.

 

 

 

OMG.

 

Because selection rules describe allowed and forbidden transitions, based on angular momentum. 1st year quantum. As I've explained twice already. You will need to be a lot more specific than say "show it". What parts of the explanations aren't clear?

 

 

 

Water, whether superheating, supercooling or condensing, is not an atomic system described by QM.

 

The closest you come in your discussion is radioactive decay, claiming it needs "outside help". Why do some decays need outside help while others don't? I gave an example of two systems with almost identical energy, with one being normal and the other metastable. You gave an example where this is also true. What is this "outside help" and how does it know to "help" only atoms in some states?

My peace proposal to you: We should go from easy cases to the most complicated, putting angular momentum on the end. I have to find who made the experiments etc

 

 

What is “OMG”?

Finally you got it. Metastable states are not only in QM!

A physicist is always looking for the law applicable in a wider field.

 

Nevertheless I made also a prediction for radioactive decay. The main suspect for “the outside help” is the neutrino from the sun.

I am proud that latest observations indicate exactly this. Since you are there on the forefront of physics you followed this very closely.

Remember there is only one way to test a new theory: this are experiments and observations!

Objections are everywhere – even against Einstein, in spite of experiments and observations proved what he said.

 

Now (or in 50 years, if the debate ended) we have to admit that he was right to say: “God doesn’t play dice.”

Edited April 16, 2013 by Wolfhart Willimczik

[swansont]

My peace proposal to you: We should go from easy cases to the most complicated, putting angular momentum on the end. I have to find who made the experiments etc

 

No, that sounds like you're just dodging the issue.

 

What is “OMG”?

Oh My God

 

Finally you got it. Metastable states are not only in QM!

A physicist is always looking for the law applicable in a wider field.

 

But not when the law doesn't apply. The lesson of water is that boiling, freezing and condensation are induced, rather than truly being spontaneous transitions.

 

 

Nevertheless I made also a prediction for radioactive decay. The main suspect for “the outside help” is the neutrino from the sun.

I am proud that latest observations indicate exactly this. Since you are there on the forefront of physics you followed this very closely.

Remember there is only one way to test a new theory: this are experiments and observations!

 

That's deceptive to the point that it's wrong. Some decays appear to have an annual variation. If this was because decays required neutrino interaction, then all of them would display this. But we were talking about atomic systems, some of which decay very quickly and others that are metastable, despite both being influenced by the same environment. That's what I want an answer to.

 

 

Objections are everywhere – even against Einstein, in spite of experiments and observations proved what he said.

 

Now (or in 50 years, if the debate ended) we have to admit that he was right to say: “God doesn’t play dice.”

 

Objection aren't some badge of honor. It's a failure of logic to compare your work to Einstein, simply because people objected to both.

[Wolfhart Willimczik]

Finally you grasp there are no " spontaneous transitions"

causality is there again.

If you mean with "induced" or "help from the outside" what I said (giving a specific value of a free parameter) - you have something right

 

 

The lesson of water is that boiling, freezing and condensation are induced, rather than truly being spontaneous transitions.

There are different definitions for a metastable state, what is a sign that the right definition is still missing.

1.
Metallurgy: chemically unstable in the absence of certain conditions that would induce stability, but not liable to spontaneous transformation.
2.
Physics, Chemistry . pertaining to a body or system existing at an energy level (metastable state) above that of a more stable state and requiring the addition of a small amount of energy to induce a transition to the more stable state.

a). (of a body or system) having a state of apparent equilibrium although capable of changing to a more stable state
b). (of an atom, molecule, ion, or atomic nucleus) existing in an excited state with a relatively long lifetime

Metallurgy .

“absence of certain conditions” comes close to my definition, whereby I only explain the “certain conditions”.

 

Physics, Chemistry
“…requiring the addition of a small amount of energy to induce a transition to the more stable state.”

Giving an information requires always a small amount of energy. Forgetting the information would mean I send only a small amount of energy to this forum, because letters coming from Florida to you requires a small amount of energy.
Secondly a metastable state has too much energy. Why should an atom in an exited state with too much energy need more energy to get out of it?

1. “apparent equilibrium” seems to be only another word for metastable state.

2. that is right; a metastable state has a relatively long lifetime.

 

ergo:

There are no basic contradictions between the old definitions and my new one.

[swansont]

Finally you grasp there are no " spontaneous transitions"

causality is there again.

If you mean with "induced" or "help from the outside" what I said (giving a specific value of a free parameter) - you have something right

Now how about answering the question of why some (most, actually) states aren't metastable?

Physics, Chemistry

“…requiring the addition of a small amount of energy to induce a transition to the more stable state.”

Giving an information requires always a small amount of energy. Forgetting the information would mean I send only a small amount of energy to this forum, because letters coming from Florida to you requires a small amount of energy.

Secondly a metastable state has too much energy. Why should an atom in an exited state with too much energy need more energy to get out of it?

I have explained this several times already. Energy is not the only conserved quantity in a transition. Angular momentum is, as well.

 

There's not much point to this if you are going to ignore information provided to you.

[Wolfhart Willimczik]

 

1. Now how about answering the question of why some (most, actually) states aren't metastable?

 

I have explained this several times already. Energy is not the only conserved quantity in a transition. Angular momentum is, as well.

 

There's not much point to this if you are going to ignore information provided to you.

1. This is basic physics. You should know it.

 

I find no definition of metastable states with “angular momentum”.

(You should post it, if you have one.)

 

But to satisfy your love in angular momentum there is your answer:

The 2S1/2 state of atomic hydrogen is usually metastable, consequently, the lifetime is increased to an estimated value of 0.15 s.

The necessity for angular momentum and parity conservation forces the second excited state (2S1/2) of atomic hydrogen to decay by simultaneous emission of two photons.

 

Ergo – it changes nothing, except there are 2 photons instead of 1.

Ergo - your objection has nothing to do with the topic of this thread.

q.e.d.

Edited April 16, 2013 by Wolfhart Willimczik

[John Cuthber]

Then please copy at least the right passage - why metastable states exist - what the missing angular momentum provided etc. Until now you have no answer provided.

 

 

This sort of thing

http://www.eng.fsu.edu/~dommelen/quantum/style_a/consem.html

which, whether you like it or not, provides a reason for the existence of the metastable states.

 

Might I make a suggestion?

Your definition is very unclear. Part of that problem may be linguistic.

There are people here whose first language isn't English and who might be better able to understand your suggestion if you wrote it in your first language.

How would you write this

"Any physical process - basically ready to start - will be halted as long as more than one of exact equal and choose able values of a parameter for the process exist."

in your first language?

[Wolfhart Willimczik]

Might I make a suggestion?

Your definition is very unclear. Part of that problem may be linguistic.

There are people here whose first language isn't English and who might be better able to understand your suggestion if you wrote it in your first language.

How would you write this

"Any physical process - basically ready to start - will be halted as long as more than one of exact equal and choose able values of a parameter for the process exist."

in your first language?

Ein physikalisches System verharrt solange in einem metastabilen Zustand solange nicht ein bestimmter Wert eines frei wählbaren Parameters von außen vorgegeben wird.

 

Google.

A physical system in a metastable state remains as long as not a particular value of an arbitrary parameter is determined from outside.

 

oops

Edited April 16, 2013 by Wolfhart Willimczik

[swansont]

1. This is basic physics. You should know it.

 

You aren't discussing basic physics, you are proposing a new theory. How am I supposed to know aspects of your theory that you haven't yet explained?

 

I find no definition of metastable states with “angular momentum”.

(You should post it, if you have one.)

 

http://encyclopedia2.thefreedictionary.com/Metastable+State+of+Quantum+Systems

http://en.wikipedia.org/wiki/Metastability#Atomic_and_molecular_physics

http://www.answers.com/topic/metastable-state-1

 

Selection rules are based on angular momentum.

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

 

 

But to satisfy your love in angular momentum there is your answer:

The 2S1/2 state of atomic hydrogen is usually metastable, consequently, the lifetime is increased to an estimated value of 0.15 s.

The necessity for angular momentum and parity conservation forces the second excited state (2S1/2) of atomic hydrogen to decay by simultaneous emission of two photons.

 

Ergo – it changes nothing, except there are 2 photons instead of 1.

Ergo - your objection has nothing to do with the topic of this thread.

q.e.d.

 

Emitting two photons at once takes longer, on average. It's a second-order coupling. That's why it's metastable. I don't see how this isn't applicable to the subject. The 2P state couples directly to the ground state and happens in around 1.5 ns. There is no external effect that causes the 2S state to require more time.

[Wolfhart Willimczik]

 

You aren't discussing basic physics, you are proposing a new theory. How am I supposed to know aspects of your theory that you haven't yet explained?

 

 

http://encyclopedia2.thefreedictionary.com/Metastable+State+of+Quantum+Systems

http://en.wikipedia.org/wiki/Metastability#Atomic_and_molecular_physics

http://www.answers.com/topic/metastable-state-1

 

Selection rules are based on angular momentum.

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

 

 

 

Emitting two photons at once takes longer, on average. It's a second-order coupling. That's why it's metastable. I don't see how this isn't applicable to the subject. The 2P state couples directly to the ground state and happens in around 1.5 ns. There is no external effect that causes the 2S state to require more time.

That's right - you can't know what I have not yet explained.

 

Unfortunately, you have not understood (or you are not willing to) what I have explained - which is the problem with you.

 

Stay on the simpler cases which I described.

 

There is no external effect that causes the 2S state to require more time, but only to end it. (It shows again that you have nothing understood)

 

You have to start there:

A physical system in a metastable state remains as long as not a particular value of an arbitrary parameter is determined from outside.

[swansont]

That's right - you can't know what I have not yet explained.

 

Unfortunately, you have not understood (or you are not willing to) what I have explained - which is the problem with you.

 

Stay on the simpler cases which I described.

 

There is no external effect that causes the 2S state to require more time, but only to end it. (It shows again that you have nothing understood)

 

I'll rephrase: if there is an external effect that requires ending the excitation, why does it happen so much faster for the 2P state than the 2S state? Or any other normal state with nearby metastable states?

 

What is the effect?

 

[Wolfhart Willimczik]

 

I'll rephrase: if there is an external effect that requires ending the excitation, why does it happen so much faster for the 2P state than the 2S state? Or any other normal state with nearby metastable states?

 

What is the effect?

 

It has something to do with the environment the system is in, not only with the specific state.

Start with putting your system in space without any outside influence and see what happens. I expect nothing for a long time.

(In space are the longest times observed.)

Then do something to your metastable system (anything - even shooting with a pistol) and you will see by what things you end the metastable system.

Tell the result.

The Russians bring you to the space station.

For testing radioactive decay by neutrinos you must leave the sun system until there are no more neutrinos.

Good flight

[Przemyslaw.Gruchala]

For testing radioactive decay by neutrinos you must leave the sun system until there are no more neutrinos.

 

In cosmic space there are neutrinos from the all stars.

 

But of course less than close to Sun.

 

If there is 65 billion neutrinos per cm^2 on Earth from Sun,

we should detect 260 billion neutrinos per cm^2 75 mln km from Sun (twice closer than Earth),

and 16.25 billion per cm^2 300 mln km from Sun (double Earth orbit radius).

and so on the further from the Sun.

 

Experiment confirming or rejecting idea should be launching rocket with radioactive isotopes and decay rate counter in the cosmos, in direction reverse to Sun. And send back decay rate result to us.

 

Radioactive Polonium was already used as power generator in cosmic devices.

But wiki is just mentioning about devices with distances from Earth just max. ~400,000 km.

Far less to notice significant difference (not to mention that Moon half time is closer to Sun than Earth).

 

 

Polonium-based sources of alpha particles were produced in the former Soviet Union.[51] Such sources were applied for measuring the thickness of industrial coatings via attenuation of alpha radiation.[52] Because of intense alpha radiation, a one-gram sample of 210Po will spontaneously heat up to above 500 °C (932 °F) generating about 140 watts of energy. Therefore, 210Po is used as an atomic heat source to power radioisotope thermoelectric generators via thermoelectric materials.[3][14][53][54] For instance, 210Po heat sources were used in the Lunokhod 1 (1970) and Lunokhod 2 (1973) Moon rovers to keep their internal components warm during the lunar nights, as well as the Kosmos 84 and 90 satellites (1965).[51][55]

The alpha particles emitted by polonium can be converted to neutrons

using beryllium oxide, at a rate of 93 neutrons per million alpha

particles.[53] Thus Po-BeO mixtures or alloys are used as a neutron source, for example in a neutron trigger or initiator for nuclear weapons[14][56]

and for inspections of oil wells. About 1500 of such sources with an

individual activity of 1850 Ci have been used annually in the Soviet

Union.[57]

 

Does anybody know whether polonium was used by NASA in devices that are farther than Moon?

Edited April 17, 2013 by Przemyslaw.Gruchala

[Wolfhart Willimczik]

In cosmic space there are neutrinos from the all stars.

 

But of course less than close to Sun.

 

If there is 65 billion neutrinos per cm^2 on Earth from Sun,

we should detect 130 billion neutrinos per cm^2 75 mln km from Sun (twice closer than Earth),

and 32.5 billion per cm^2 300 mln km from Sun (double Earth radius).

and so on the further from the Sun.

 

Experiment confirming or rejecting idea should be launching rocket with radioactive isotopes and decay rate counter in the cosmos, in direction reverse to Sun. And send back decay rate result to us.

 

Radioactive Polonium was already used as power generator in cosmic devices.

But wiki is just mentioning about devices with distances from Earth just max. ~400,000 km.

Far less to notice significant difference (not to mention that Moon half time is closer to Sun than Earth).

 

You are right we can expect neutrinos everywhere - therefore everywhere radioactive decay, - if my theory is right.

(I was a little sloppy, because I wished the man behind Sean Connery as far away as possible...)

We can only use changes in the density of the neutrino flow.

On earth we had to build a strong neutrino source - (not only for a short time like a bomb),

what would be a challenge.

[Przemyslaw.Gruchala]

On earth we had to build a strong neutrino source

 

Reverse.

Find material capable to intercept neutrinos.

And use it, together with aluminium, lead and other cosmic rays stopping materials as walls of our houses and buildings.

Each particle from the space which is hitting our cells and DNA is destroying them.

[Wolfhart Willimczik]

Reverse.

Find material capable to intercept neutrinos.

And use it, together with aluminium, lead and other cosmic rays stopping materials as walls of our houses and buildings.

Each particle from the space which is hitting our cells and DNA is destroying them.

Unfortunately there is no material on earth to stop neutrinos. They are going through the entire earth without stopping (hitting a nucleus). The earth is for them like glass for us. Otherwise it would be easy to make measurements at day and night - ergo behind the earth. In the night would be the decay lower. Changing the distance to the sun seems to be the best thing what we could do. A little bit is the earth itself doing it for us, but the best thing would be a probe to the edge of the solar system as we just had.

[John Cuthber]

The experiment has been done. Lots of isotope powered generators have been launched into space, some of them towards and more of them away from the sun.

The effect of neutron on decay rates which this new idea predicts would have been noticed.

Not least, because space craft designers are very careful to monitor power available.

 

Since no such effect was found we can rule out this idea.

 

If someone's idea does not agree with reality, it is not because reality has made a mistake.

[Wolfhart Willimczik]

The experiment has been done. Lots of isotope powered generators have been launched into space, some of them towards and more of them away from the sun.

The effect of neutron on decay rates which this new idea predicts would have been noticed.

Not least, because space craft designers are very careful to monitor power available.

 

Since no such effect was found we can rule out this idea.

 

If someone's idea does not agree with reality, it is not because reality has made a mistake.

then this case is closed for you and you can leave this threat.

Thanks for your contributions.

Edited April 17, 2013 by Wolfhart Willimczik